Learn4Less Logo

Navigation

Partial Derivatives

2.7 Directional Derivatives and the Gradient

34 problems · hints, answers and solutions shown beside each one

1Stage 1Conceptual

Do you understand the idea? Usually little or no calculation.

Q1Stage 1Past exam · M200 2008A

Find the directional derivative of f(x,y,z)=exyzf(x,y,z) = e^{xyz} in the <0,1,1>\llt 0,1,1\rgt direction at the point (0,1,1)(0,1,1).

Answer

00

Full solution

The partial derivatives, at a general point (x,y,z)(x,y,z) and also at the point of interest (0,1,1)(0,1,1), are

fx(x,y,z)=yzexyzfx(0,1,1)=1fy(x,y,z)=xzexyzfy(0,1,1)=0fz(x,y,z)=xyexyzfz(0,1,1)=0\begin{alignat*}{3} f_x(x,y,z)&=yz e^{xyz}\qquad & f_x(0,1,1)&= 1 \\ f_y(x,y,z)&=xz e^{xyz}\qquad & f_y(0,1,1)&= 0 \\ f_z(x,y,z)&=xy e^{xyz}\qquad & f_z(0,1,1)&= 0 \end{alignat*}

So ∇f(0,1,1)=<1,0,0>\vnabla f(0,1,1) =\llt 1,0,0\rgt and the specified directional derivative is

D<0,1,1>2f(0,1,1)=<1,0,0>⋅<0,1,1>2=0\begin{align*} D_{\frac{\llt 0,1,1\rgt}{\sqrt{2}}}f(0,1,1) =\llt 1,0,0\rgt\cdot\frac{\llt 0,1,1\rgt}{\sqrt{2}} =0 \end{align*}
Q2Stage 1Past exam · M200 2008A

Find ∇(y2+sin⁡(xy))\vnabla\big(y^2 + \sin(xy)\big).

Answer

ycos⁡(xy) ı^+[2y+xcos⁡(xy)] ȷ^y\cos(xy)\,\hi + [2y + x\cos(xy)]\,\hj

Full solution

In two dimensions, write g(x,y)=y2+sin⁡(xy)g(x,y) = y^2+\sin(xy). Then

∇g=<gx , gy>=<ycos⁡(xy) , 2y+xcos⁡(xy)>\begin{align*} \vnabla g =\llt g_x\,,\,g_y\rgt =\llt y\cos(xy)\,,\, 2y + x\cos(xy) \rgt \end{align*}

In three dimensions, write g(x,y,z)=y2+sin⁡(xy)g(x,y,z) = y^2+\sin(xy). Then

∇g=<gx , gy , gz>=<ycos⁡(xy) , 2y+xcos⁡(xy) , 0>\begin{align*} \vnabla g =\llt g_x\,,\,g_y\,,\,g_z\rgt =\llt y\cos(xy)\,,\, 2y + x\cos(xy)\,,\, 0 \rgt \end{align*}

2Stage 2Procedural

Practising the skill itself, until applying it is automatic.

Q3Stage 2

Find the rate of change of the given function at the given point in the given direction.

  1. f(x,y)=3x−4yf(x,y)=3x-4y at the point (0,2)(0,2) in the direction −2ı^-2\hi.

  2. f(x,y,z)=x−1+y−1+z−1f(x,y,z)=x^{-1}+y^{-1}+z^{-1} at (2,−3,4)(2,-3,4) in the direction ı^+ȷ^+k^\hi+\hj+\hk.

Answer

(a) −3-3 (b) −611443≈−0.2446-\frac{61}{144\sqrt{3}}\approx-0.2446

Full solution

(a) The gradient of ff is ∇f(x,y)=<3,−4>\vnabla f(x,y)=\llt 3,-4\rgt. So the specified rate of change is

<3,−4>⋅<−2,0>∣<−2,0>∣=−3\begin{equation*} \llt 3,-4\rgt\cdot\frac{\llt -2,0\rgt}{|\llt -2,0\rgt|}=-3 \end{equation*}

(b) The gradient of ff is ∇f(x,y,z)=<−x−2,−y−2,−z−2>\vnabla f(x,y,z)=\llt -x^{-2},-y^{-2},-z^{-2}\rgt. In particular, the gradient of ff at the point (2,−3,4)(2,-3,4) is ∇f(2,−3,4)=<−14,−19,−116>\vnabla f(2,-3,4)=\llt -\frac{1}{4},-\frac{1}{9},-\frac{1}{16}\rgt. So the specified rate of change is

<−14,−19,−116>⋅<1,1,1>3=−611443≈−0.2446\begin{equation*} \llt -\frac{1}{4},-\frac{1}{9},-\frac{1}{16}\rgt\cdot \frac{\llt 1,1,1\rgt}{\sqrt{3}}= -\frac{61}{144\sqrt{3}}\approx-0.2446 \end{equation*}
Q4Stage 2

In what directions at the point (2,0)(2,0) does the function f(x,y)=xyf(x,y)=xy have the specified rates of change?

  1. −1-1

  2. −2-2

  3. −3-3

Hint

The rate of change in the direction that makes angle θ\theta with respect to the xx-axis, that is, in the direction <cos⁡θ,sin⁡θ>\llt \cos\theta,\sin\theta\rgt is <cos⁡θ,sin⁡θ>⋅∇f(2,0)\llt \cos\theta,\sin\theta\rgt \cdot\vnabla f(2,0).

Answer

(a) <±32,−12>\llt\pm\frac{\sqrt{3}}{2} ,-\frac{1}{2} \rgt (b) <0,−1>\llt 0 ,-1 \rgt (c) No direction works!

Full solution

The gradient of f(x,y)f(x,y) is ∇f(x,y)=<y,x>\vnabla f(x,y)=\llt y,x\rgt. In particular, the gradient of ff at the point (2,0)(2,0) is ∇f(2,0)=<0,2>\vnabla f(2,0)=\llt 0,2 \rgt. So the rate of change in the direction that makes angle θ\theta with respect to the xx-axis, that is, in the direction <cos⁡θ,sin⁡θ>\llt \cos\theta,\sin\theta\rgt is

<cos⁡θ,sin⁡θ>⋅∇f(2,0)=<cos⁡θ,sin⁡θ>⋅<0,2>=2sin⁡θ\begin{equation*} \llt \cos\theta,\sin\theta\rgt \cdot\vnabla f(2,0) =\llt \cos\theta,\sin\theta\rgt \cdot\llt 0, 2\rgt =2\sin\theta \end{equation*}

(a) To get a rate −1-1, we need

sin⁡θ=−12  ⟹  θ=−30∘, −150∘\begin{equation*} \sin\theta=-\frac{1}{2} \implies \theta=-30^\circ,\ -150^\circ \end{equation*}

So the desired directions are

<cos⁡θ,sin⁡θ>=<±32,−12>\begin{equation*} \llt \cos\theta,\sin\theta\rgt =\llt\pm\frac{\sqrt{3}}{2} ,-\frac{1}{2} \rgt \end{equation*}

(b) To get a rate −2-2, we need

sin⁡θ=−1  ⟹  θ=−90∘\begin{equation*} \sin\theta=-1 \implies \theta=-90^\circ \end{equation*}

So the desired direction is

<cos⁡θ,sin⁡θ>=<0,−1>\begin{equation*} \llt \cos\theta,\sin\theta\rgt =\llt 0 ,-1 \rgt \end{equation*}

(c) To get a rate −3-3, we need

sin⁡θ=−32\begin{equation*} \sin\theta=-\frac{3}{2} \end{equation*}

No θ\theta obeys this, since −1≤sin⁡θ≤1-1\le\sin\theta\le 1 for all θ\theta. So no direction works!

Q5Stage 2

Find ∇f(a,b)\vnabla f(a,b) given the directional derivatives

D(ı^+ȷ^)/2f(a,b)=32D(3ı^−4ȷ^)/5f(a,b)=5\begin{equation*} D_{(\hi+\hj)/\sqrt{2}}f(a,b)=3\sqrt{2}\qquad D_{(3\hi-4\hj)/5}f(a,b)=5 \end{equation*}
Hint

Denote ∇f(a,b)=<α,β>\vnabla f(a,b)=\llt \al,\be\rgt.

Answer

∇f(a,b)=<7,−1>\vnabla f(a,b)=\llt 7,-1\rgt

Full solution

Denote ∇f(a,b)=<α,β>\vnabla f(a,b)=\llt \al,\be\rgt. We are told that

<α,β>⋅<12,12>=32orα+β=6<α,β>⋅<35,−45>=5or3α−4β=25\begin{alignat*}{5} \llt\al,\be\rgt\cdot\llt\frac{1}{\sqrt{2}},\frac{1}{\sqrt{2}}\rgt&=3\sqrt{2} & &\quad\text{or}\quad & \al+\be&=6\\ \llt\al,\be\rgt\cdot\llt\frac{3}{5},-\frac{4}{5}\rgt&=5 & &\quad\text{or}\quad & 3\al-4\be&=25 \end{alignat*}

Adding 4 times the first equation to the second equation gives 7α=497\al=49. Substituting α=7\al=7 into the first equation gives β=−1\be=-1. So ∇f(a,b)=<7,−1>\vnabla f(a,b)=\llt 7,-1\rgt.

Q6Stage 2Past exam · M200 2005D

You are standing at a location where the surface of the earth is smooth. The slope in the southern direction is 44 and the slope in the south–eastern direction is 2\sqrt{2}. Find the slope in the eastern direction.

Hint

Use a coordinate system with the positive yy–axis pointing north, with the positive xx–axis pointing east and with our current location being x=y=0x=y=0. Denote by z(x,y)z(x,y) the elevation of the earth's surface at (x,y)(x,y). Express the various slopes in terms of ∇z(0,0)\vnabla z(0,0).

Answer

−2-2

Full solution

Use a coordinate system with the positive yy–axis pointing north, with the positive xx–axis pointing east and with our current location being x=y=0x=y=0. Denote by z(x,y)z(x,y) the elevation of the earth's surface at (x,y)(x,y). We are told that

∇z(0,0)⋅(−ȷ^)=4∇z(0,0)⋅(ı^−ȷ^2)=2\begin{align*} \vnabla z(0,0)\cdot(-\hj) &= 4 \\ \vnabla z(0,0)\cdot\left(\frac{\hi-\hj}{\sqrt{2}}\right) &= \sqrt{2} \end{align*}

The first equation implies that zy(0,0)=−4z_y(0,0)=-4 and the second equation implies that

zx(0,0)−zy(0,0)2=2  ⟹  zx(0,0)=zy(0,0)+2=−2\begin{align*} \frac{z_x(0,0)-z_y(0,0)}{\sqrt{2}}=\sqrt{2} \implies z_x(0,0)=z_y(0,0)+2 = -2 \end{align*}

So the slope in the eastern direction is

∇z(0,0)⋅ı^=zx(0,0)=−2\begin{align*} \vnabla z(0,0)\cdot\hi = z_x(0,0) = -2 \end{align*}
Q7Stage 2Past exam · M200 2006A

Assume that the directional derivative of w=f(x,y,z)w = f(x,y,z) at a point PP is a maximum in the direction of the vector 2ı^−ȷ^+k^2\hi - \hj + \hk, and the value of the directional derivative in that direction is 363\sqrt{6}.

  1. Find the gradient vector of w=f(x,y,z)w = f(x,y,z) at PP.

  2. Find the directional derivative of w=f(x,y,z)w = f(x,y,z) at PP in the direction of the vector ı^+ȷ^\hi + \hj

Answer

(a) 6ı^−3ȷ^+3k^6\hi - 3\hj + 3\hk (b) 32\frac{3}{\sqrt{2}}

Full solution

(a) Use ∇f(P)\vnabla f(P) to denote the gradient vector of ff at PP. We are told that

  • directional derivative of ff at PP is a maximum in the direction 2ı^−ȷ^+k^2\hi - \hj + \hk, which implies that ∇f(P)\vnabla f(P) is parallel to 2ı^−ȷ^+k^2\hi - \hj + \hk, and

  • the magnitude of the directional derivative in that direction is 363\sqrt{6}, which implies that ∣∇f(P)∣=36|\vnabla f(P)|=3\sqrt{6}.

So

∇f(P)=362ı^−ȷ^+k^∣2ı^−ȷ^+k^∣=6ı^−3ȷ^+3k^\begin{align*} \vnabla f(P) = 3\sqrt{6} \frac{2\hi - \hj + \hk}{|2\hi - \hj + \hk|} = 6\hi - 3\hj + 3\hk \end{align*}

(b) The directional derivative of ff at PP in the direction ı^+ȷ^\hi+\hj is

∇f(P)⋅ı^+ȷ^∣ı^+ȷ^∣=12(6ı^−3ȷ^+3k^)⋅(ı^+ȷ^)=32\begin{equation*} \vnabla f(P) \cdot\frac{\hi+\hj}{|\hi+\hj|} =\frac{1}{\sqrt{2}}\big(6\hi - 3\hj + 3\hk\big)\cdot\big(\hi+\hj\big) =\frac{3}{\sqrt{2}} \end{equation*}
Q8Stage 2Past exam · M200 2006D

A hiker is walking on a mountain with height above the z=0z = 0 plane given by

z=f(x,y)=6−xy2\begin{equation*} z = f(x,y) = 6 - xy^2 \end{equation*}

The positive xx–axis points east and the positive yy–axis points north, and the hiker starts from the point P(2,1,4)P(2, 1, 4).

  1. In what direction should the hiker proceed from PP to ascend along the steepest path? What is the slope of the path?

  2. Walking north from PP, will the hiker start to ascend or descend? What is the slope?

  3. In what direction should the hiker walk from PP to remain at the same height?

Answer

(a) The path of steepest ascent is in the direction −117<1 , 4>-\frac{1}{\sqrt{17}}\llt 1\,,\, 4\rgt, which is a little west of south. The slope is ∣∇f(2,1)∣=∣<−1 , −4>∣=17|\vnabla f(2,1)| = |\llt -1\,,\, -4\rgt| = \sqrt{17}.

(b) So the hiker descends with slope 44.

(c) ±117<4,−1>\pm\frac{1}{\sqrt{17}}\llt 4, -1\rgt

Full solution

(a) The gradient of ff at (x,y)=(2,1)(x,y)=(2,1) is

∇f(2,1)=<−y2 , −2xy>∣(x,y)=(2,1)=<−1 , −4>\begin{align*} \vnabla f(2,1) = \llt -y^2\,,\,-2xy \rgt\Big|_{(x,y)=(2,1)} = \llt -1\,,\, -4\rgt \end{align*}

So the path of steepest ascent is in the direction −117<1 , 4>-\frac{1}{\sqrt{17}}\llt 1\,,\, 4\rgt, which is a little west of south. The slope is

∣∇f(2,1)∣=∣<−1 , −4>∣=17\begin{equation*} |\vnabla f(2,1)| = |\llt -1\,,\, -4\rgt| = \sqrt{17} \end{equation*}

(b) The directional derivative in the north direction is

D<0,1>f(2,1)=∇f(2,1)⋅<0,1>=<−1 , −4>⋅<0,1>=−4\begin{equation*} D_{\llt 0,1\rgt}f(2,1) = \vnabla f(2,1)\cdot\llt 0,1\rgt = \llt -1\,,\, -4\rgt \cdot \llt 0,1\rgt = -4 \end{equation*}

So the hiker descends with slope ∣−4∣=4|-4|=4.

(c) To contour, i.e. remain at the same height, the hiker should walk in a direction perpendicular to ∇f(2,1)=<−1 , −4>\vnabla f(2,1)= \llt -1\,,\, -4\rgt. Two unit vectors perpendicular to <−1 , −4>\llt -1\,,\, -4\rgt are ±117<4,−1>\pm\frac{1}{\sqrt{17}}\llt 4, -1\rgt.

Q9Stage 2

Two hikers are climbing a (small) mountain whose height is z=1000−2x2−3y2.z=1000-2x^2-3y^2. They start at (1,1,995)(1,1,995) and follow the path of steepest ascent. Their (x,y)(x,y) coordinates obey y=axby=ax^b for some constants a,ba, b. Determine aa and bb.

Hint

In order for y=axby=ax^b to give the (x,y)(x,y) coordinates of the path of steepest ascent, the tangent vector to y=axby=ax^b must be parallel to the height gradient ∇h(x,y)\vnabla h(x,y) at all points on y=axby=ax^b. Also, don't forget that (1,1)(1,1) must be on y=axby=ax^b.

Answer

a=1a=1, b=32b=\frac{3}{2}

Full solution

The gradient of h(x,y)=1000−2x2−3y2h(x,y)=1000-2x^2-3y^2 is ∇h(x,y)=(−4x,−6y)\vnabla h(x,y)=(-4x,-6y). This gradient (which points in the direction of steepest ascent) must be parallel to the tangent to y=axby=ax^b at all points on y=axby=ax^b. A tangent to y=axby=ax^b is <1,dydx>=<1,abxb−1>\llt 1,\diff{y}{x}\rgt =\llt 1, abx^{b-1}\rgt.

<−4x,−6y>∥<1,abxb−1>  ⟹  abxb−11=−6y−4x  ⟹  32y=abxb\begin{equation*} \llt -4x,-6y\rgt \parallel \llt 1, abx^{b-1}\rgt \implies \frac{abx^{b-1}}{1}=\frac{-6y}{-4x} \implies \frac{3}{2}y=abx^b \end{equation*}

This is true at all points on y=axby=ax^b if and only if b=32b=\frac{3}{2}. As (1,1)(1,1) must also be on y=axby=ax^b, we need 1=a1b1=a1^b, which forces a=1a=1, b=32b=\frac{3}{2}. Here is a contour map showing the hiking trail.

Figure from prob_s2.7, line 398

Figure from prob_s2.7, line 398

Q10Stage 2Past exam · M200 2007A

A mosquito is at the location (3,2,1)(3, 2, 1) in R3\bbbr^3. She knows that the temperature TT near there is given by T=2x2+y2−z2T = 2x^2 + y^2 - z^2.

  1. She wishes to stay at the same temperature, but must fly in some initial direction. Find a direction in which the initial rate of change of the temperature is 00.

  2. If you and another student both get correct answers in part (a), must the directions you give be the same? Why or why not?

  3. What initial direction or directions would suit the mosquito if she wanted to cool down as fast as possible?

Answer

(a) Any nonzero <a , b , c>\llt a\,,\,b\,,\,c\rgt that obeys 12a+4b−2c=012a+4b-2c=0 is an allowed direction. Four allowed unit vectors are ±<0 , 1 , 2>5\pm\frac{\llt 0\,,\,1\,,\,2\rgt}{\sqrt{5}} and ±<1 , −3 , 0>10\pm\frac{\llt 1\,,\,-3\,,\,0\rgt}{\sqrt{10}}.

(b) No they need not be the same. Four different explicit directions were given in part (a).

(c) −<6 , 2 , −1>41-\frac{\llt 6\,,\,2\,,\,-1\rgt}{\sqrt{41}}

Full solution

(a) The temperature gradient at (3,2,1)(3,2,1) is

∇T(3,2,1)=<4x , 2y , −2z>∣(x,y,z)=(3,2,1)=<12 , 4 , −2>\begin{align*} \vnabla T(3,2,1) = \llt 4x\,,\,2y\,,\,-2z\rgt\Big|_{(x,y,z)=(3,2,1)} = \llt 12\,,\,4\,,\,-2\rgt \end{align*}

She wishes to fly in a direction that is perpendicular to ∇T(3,2,1)\vnabla T(3,2,1). That is, she wishes to fly in a direction <a , b , c>\llt a\,,\,b\,,\,c\rgt that obeys

0=<12 , 4 , −2>⋅<a , b , c>=12a+4b−2c\begin{equation*} 0 = \llt 12\,,\,4\,,\,-2\rgt\cdot \llt a\,,\,b\,,\,c\rgt = 12a+4b-2c \end{equation*}

Any nonzero <a , b , c>\llt a\,,\,b\,,\,c\rgt that obeys 12a+4b−2c=012a+4b-2c=0 is an allowed direction. Four allowed unit vectors are ±<0 , 1 , 2>5\pm\frac{\llt 0\,,\,1\,,\,2\rgt}{\sqrt{5}} and ±<1 , −3 , 0>10\pm\frac{\llt 1\,,\,-3\,,\,0\rgt}{\sqrt{10}}.

(b) No they need not be the same. Four different explicit directions were given in part (a).

(c) To cool down as quickly as possible, she should move in the direction opposite to the temperature gradient. A unit vector in that direction is −<6 , 2 , −1>41-\frac{\llt 6\,,\,2\,,\,-1\rgt}{\sqrt{41}}.

Q11Stage 2Past exam · M200 2008D

The air temperature T(x,y,z)T(x,y,z) at a location (x,y,z)(x,y,z) is given by:

T(x,y,z)=1+x2+yz.\begin{equation*} T(x,y,z) = 1 + x^2 + yz. \end{equation*}

  1. A bird passes through (2,1,3)(2,1,3) travelling towards (4,3,4)(4,3,4) with speed 22. At what rate does the air temperature it experiences change at this instant?

  2. If instead the bird maintains constant altitude (z=3z = 3) as it passes through (2,1,3)(2,1,3) while also keeping at a fixed air temperature, T=8T = 8, what are its two possible directions of travel?

Answer

(a) 1010 (b) ±15<3 ,−4 ,0>\pm \frac{1}{5}\llt 3\,,-4\,,0\rgt

Full solution

The temperature gradient at (2,1,3)(2,1,3) is

∇T(2,1,3)=<2x , z , y>∣(x,y,z)=(2,1,3)=<4 , 3 , 1>\begin{align*} \vnabla T(2,1,3) = \llt 2x\,,\,z\,,\,y \rgt\Big|_{(x,y,z)=(2,1,3)} = \llt 4\,,\,3\,,\,1 \rgt \end{align*}

(a) The bird is flying in the direction <4−2 , 3−1 , 4−3>=<2 , 2 , 1>\llt 4-2\,,\,3-1\,,\,4-3 \rgt =\llt 2\,,\,2\,,\,1\rgt at speed 22 and so has velocity v=2<2 , 2 , 1>∣<2 , 2 , 1>∣=23<2 , 2 , 1>\vv=2\frac{\llt 2\,,\,2\,,\,1\rgt}{|\llt 2\,,\,2\,,\,1\rgt|} =\frac{2}{3} \llt 2\,,\,2\,,\,1\rgt. The rate of change of air temperature experienced by the bird at that instant is

∇T(2,1,3)⋅v=23<4 , 3 , 1>⋅<2 , 2 , 1>=10\begin{align*} \vnabla T(2,1,3) \cdot\vv =\frac{2}{3} \llt 4\,,\,3\,,\,1 \rgt \cdot \llt 2\,,\,2\,,\,1\rgt = 10 \end{align*}

(b) To maintain constant altitude (while not being stationary), the bird's direction of travel has to be of the form <a , b , 0>\llt a\,,\,b\,,\,0\rgt, for some constants aa and bb, not both zero. To keep the air temperature fixed, its direction of travel has to be perpendicular to ∇T(2,1,3)=<4 , 3 , 1>\vnabla T(2,1,3)= \llt 4\,,\,3\,,\,1 \rgt. So aa and bb have to obey

0=<a , b , 0>⋅<4 , 3 , 1>=4a+3b  ⟺  b=−43a\begin{align*} 0 = \llt a\,,\,b\,,\,0\rgt \cdot \llt 4\,,\,3\,,\,1 \rgt = 4a+3b \iff b=-\frac{4}{3}a \end{align*}

and the direction of travel has to be a nonzero constant times <3 ,−4 ,0>\llt 3\,,-4\,,0\rgt. The two such unit vectors are ±15<3 ,−4 ,0>\pm \frac{1}{5}\llt 3\,,-4\,,0\rgt.

Q12Stage 2Past exam · M200 2009A

Let f(x,y)=2x2+3xy+y2f(x,y) = 2x^2 + 3xy + y^2 be a function of xx and yy.

  1. Find the maximum rate of change of f(x,y)f(x,y) at the point P(1,−43)P\left(1, -\frac{4}{3}\right).

  2. Find the directions in which the directional derivative of f(x,y)f(x,y) at the point P(1,−43)P\left(1, -\frac{4}{3}\right) has the value 15\frac{1}{5}.

Answer

(a) 13\frac{1}{3} (b) <±45 , 35>\llt \pm\frac{4}{5}\,,\,\frac{3}{5}\rgt

Full solution

We are going to need, in both parts of this question, the gradient of f(x,y)f(x,y) at (x,y)=(1,−43)(x,y)=\left(1, -\frac{4}{3}\right). So we find it first.

fx(x,y)=4x+3yfx(1,−4/3)=0fy(x,y)=3x+2yfy(1,−4/3)=13\begin{alignat*}{3} f_x(x,y)&=4x+3y\qquad & f_x(1,-4/3)&=0 \\ f_y(x,y)&=3x+2y\qquad & f_y(1,-4/3)&=\frac{1}{3} \end{alignat*}

so ∇f(1,−43)=<0,13>\vnabla f\left(1, -\frac{4}{3}\right) =\llt 0,\frac{1}{3}\rgt.

(a) The maximum rate of change of ff at PP is

∣∇f(1,−43)∣=∣<0,13>∣=13\begin{align*} \left|\vnabla f\left(1, -\tfrac{4}{3}\right)\right| =\left|\llt 0,\tfrac{1}{3}\rgt\right| =\tfrac{1}{3} \end{align*}

(b) If <a,b>\llt a,b\rgt is a unit vector, the directional derivative of ff at PP in the direction <a,b>\llt a,b\rgt is

D<a,b>f(1,−43)=∇f(1,−43)⋅<a,b>=<0,13>⋅<a,b>=b3\begin{align*} D_{\llt a,b\rgt}f\left(1, -\tfrac{4}{3}\right) =\vnabla f\left(1, -\tfrac{4}{3}\right)\cdot \llt a,b\rgt =\llt 0,\tfrac{1}{3}\rgt\cdot \llt a,b\rgt =\tfrac{b}{3} \end{align*}

So we need b3=15\frac{b}{3}=\frac{1}{5} and hence b=35b=\frac{3}{5}. For <a,b>\llt a,b\rgt to be a unit vector, we also need

a2+b2=1  ⟺  a2=1−b2=1−3252=1625  ⟺  a=±45\begin{align*} a^2+b^2=1 \iff a^2=1-b^2=1-\frac{3^2}{5^2}=\frac{16}{25} \iff a=\pm\frac{4}{5} \end{align*}

So the allowed directions are <±45 , 35>\llt \pm\frac{4}{5}\,,\,\frac{3}{5}\rgt.

Q13Stage 2Past exam · M200 2010A

The temperature T(x,y)T(x, y) at a point of the xyxy–plane is given by

T(x,y)=yex2\begin{equation*} T(x, y) = ye^{x^2} \end{equation*}

A bug travels from left to right along the curve y=x2y = x^2 at a speed of 0.010.01m/sec. The bug monitors T(x,y)T(x, y) continuously. What is the rate of change of TT as the bug passes through the point (1,1)(1, 1)?

Answer

0.04e5\frac{0.04 e}{\sqrt{5}}

Full solution

The slope of y=x2y=x^2 at (1,1)(1,1) is ddxx2∣x=1=2\diff{}{x}x^2\Big|_{x=1}=2. So a unit vector in the bug's direction of motion is <1,2>5\frac{\llt 1,2\rgt}{\sqrt{5}} and the bug's velocity vector is v=0.01<1,2>5\vv=0.01\frac{\llt 1,2\rgt}{\sqrt{5}}.

The temperature gradient at (1,1)(1,1) is

∇T(1,1)=<2xyex2 , ex2>∣(x.y)=(1,1)=<2e , e>\begin{align*} \vnabla T(1,1) = \llt 2xy e^{x^2}\,,\,e^{x^2}\rgt\Big|_{(x.y)=(1,1)} =\llt 2e\,,\,e\rgt \end{align*}

and the rate of change of TT (per unit time) that the bug feels as it passes through the point (1,1)(1, 1) is

∇T(1,1)⋅v=0.015<2e , e>⋅<1,2>=0.04e5\begin{align*} \vnabla T(1,1)\cdot \vv =\frac{0.01}{\sqrt{5}} \llt 2e\,,\,e\rgt\cdot \llt 1,2\rgt =\frac{0.04 e}{\sqrt{5}} \end{align*}
Q14Stage 2Past exam · M200 2010D

Suppose the function T=F(x,y,z)=3+xy−y2+z2−xT=F(x,y,z)=3+xy-y^2+z^2-x describes the temperature at a point (x,y,z)(x,y,z) in space, with F(3,2,1)=3F(3,2,1)=3.

  1. Find the directional derivative of TT at (3,2,1)(3, 2, 1), in the direction of the point (0,1,2)(0,1,2).

  2. At the point (3,2,1)(3, 2, 1), in what direction does the temperature decrease most rapidly?

  3. Moving along the curve given by x=3etx=3e^t, y=2cos⁡ty = 2 \cos t, z=1+tz= \sqrt{1 + t}, find dTdt\diff{T}{t}, the rate of change of temperature with respect to tt, at t=0t = 0.

  4. Suppose ı^+5ȷ^+ak^\hi+5\hj+a\hk is a vector that is tangent to the temperature level surface T(x,y,z)=3T(x, y, z) = 3 at (3,2,1)(3, 2, 1). What is aa?

Answer

(a) 00 (b) 16<−1 , 1 , −2>\frac{1}{\sqrt{6}}\llt -1\,,\,1\,,\,-2 \rgt (c) 44 (d) a=2a=2

Full solution

(a) We are to find the directional derivative in the direction

<0−3 , 1−2 , 2−1>=<−3 , −1 , 1>\begin{equation*} \llt 0-3\,,\,1-2\,,\,2-1\rgt = \llt -3\,,\,-1\,,\,1\rgt \end{equation*}

As the gradient of FF is

∇F(x,y,z)=<y−1 , x−2y , 2z>\begin{align*} \vnabla F(x,y,z) = \llt y-1\,,\,x-2y\,,\,2z \rgt \end{align*}

the directional derivative is

D<−3 , −1 , 1>∣<−3 , −1 , 1>∣F(3,2,1)=∇F(3,2,1)⋅<−3 , −1 , 1>∣<−3 , −1 , 1>∣=<2−1 , 3−2(2) , 2(1)>⋅<−3 , −1 , 1>∣<−3 , −1 , 1>∣=<1 , −1 , 2>⋅<−3 , −1 , 1>11=0\begin{align*} D_{\frac{\llt -3\,,\,-1\,,\,1\rgt}{|\llt -3\,,\,-1\,,\,1\rgt|}}F(3,2,1) &=\vnabla F(3,2,1)\cdot \frac{\llt -3\,,\,-1\,,\,1\rgt}{|\llt -3\,,\,-1\,,\,1\rgt|} \\ &= \llt 2-1\,,\,3-2(2)\,,\,2(1) \rgt \cdot \frac{\llt -3\,,\,-1\,,\,1\rgt}{|\llt -3\,,\,-1\,,\,1\rgt|} =\llt 1\,,\,-1\,,\,2\rgt \cdot \frac{\llt -3\,,\,-1\,,\,1\rgt}{\sqrt{11}}\\ &=0 \end{align*}

(b) The temperature decreases most rapidly in the direction opposite the gradient. A unit vector in that direction is

−∇F(3,2,1)∣∇F(3,2,1)∣=−<1 , −1 , 2>∣<1 , −1 , 2>∣=16<−1 , 1 , −2>\begin{align*} -\frac{\vnabla F(3,2,1)}{|\vnabla F(3,2,1)|} = -\frac{\llt 1\,,\,-1\,,\,2 \rgt}{|\llt 1\,,\,-1\,,\,2 \rgt|} = \frac{1}{\sqrt{6}}\llt -1\,,\,1\,,\,-2 \rgt \end{align*}

(c) The velocity vector at time 00 is

v=<x′(0) , y′(0) , z′(0)>=<3et , −2sin⁡t , 121+t>∣t=0=<3 , 0 , 12>\begin{align*} \vv=\llt x'(0)\,,\,y'(0)\,,\,z'(0)\rgt =\llt 3e^t\,,\,-2\sin t\,,\,\frac{1}{2\sqrt{1+t}}\rgt\Big|_{t=0} =\llt 3\,,\,0\,,\,\frac{1}{2}\rgt \end{align*}

So the rate of change of temperature with respect to tt at t=0t=0 is

∇F(3,2,1)⋅v=<1 , −1 , 2>⋅<3 , 0 , 12>=4\begin{align*} \vnabla F(3,2,1)\cdot \vv =\llt 1\,,\,-1\,,\,2 \rgt\cdot \llt 3\,,\,0\,,\,\frac{1}{2}\rgt =4 \end{align*}

(d) For ı^+5ȷ^+ak^\hi+5\hj+a\hk to be tangent to the level surface F(x,y,z)=3F(x, y, z) = 3 at (3,2,1)(3, 2, 1), ı^+5ȷ^+ak^\hi+5\hj+a\hk must be perpendicular to ∇F(3,2,1)\vnabla F(3,2,1). So

0=<1 , 5 , a>⋅<1 , −1 , 2>=−4+2a\begin{align*} 0=\llt 1\,,\,5\,,\,a \rgt \cdot \llt 1\,,\,-1\,,\,2 \rgt = -4+2a \end{align*}

So a=2a=2.

Q15Stage 2Past exam · M200 2011A

Let

f(x,y,z)=(2x+y)e−(x2+y2+z2)g(x,y,z)=xz+y2+yz+z2\begin{align*} f(x, y, z) &= (2x + y)e^{-(x^2 +y^2 +z^2)} \\ g(x, y, z) &= xz + y^2 + yz + z^2 \end{align*}

  1. Find the gradients of ff and gg at (0,1,−1)(0,1,-1).

  2. A bird at (0,1,−1)(0,1,-1) flies at speed 66 in the direction in which f(x,y,z)f(x, y, z) increases most rapidly. As it passes through (0,1,−1)(0,1,-1), how quickly does g(x,y,z)g(x, y, z) appear (to the bird) to be changing?

  3. A bat at (0,1,−1)(0,1,-1) flies in the direction in which f(x,y,z)f (x, y, z) and g(x,y,z)g(x, y, z) do not change, but zz increases. Find a vector in this direction.

Answer

(a) ∇f(0,1,−1)=e−2<2,−1,2>\vnabla f(0,1,-1) =e^{-2}\llt 2,-1,2\rgt, ∇g(0,1,−1)=<−1,1,−1>\vnabla g(0,1,-1) =\llt -1,1,-1\rgt (b) 1010

(c) Any vector which is a (strictly) positive constant times <−1 , 0 , 1>\llt -1 \,,\, 0 \,,\, 1 \rgt is fine.

Full solution

(a) The first order partial derivatives of ff and gg are

∂f∂x(x,y,z)=2e−(x2+y2+z2)−2x(2x+y)e−(x2+y2+z2)  ⟹  ∂f∂x(0,1,−1)=2e−2∂f∂y(x,y,z)=e−(x2+y2+z2)−2y(2x+y)e−(x2+y2+z2)  ⟹  ∂f∂y(0,1,−1)=−e−2∂f∂z(x,y,z)=−2z(2x+y)e−(x2+y2+z2)  ⟹  ∂f∂z(0,1,−1)=2e−2∂g∂x(x,y,z)=z  ⟹  ∂g∂x(0,1,−1)=−1∂g∂y(x,y,z)=2y+z  ⟹  ∂g∂y(0,1,−1)=1∂g∂z(x,y,z)=x+y+2z  ⟹  ∂g∂z(0,1,−1)=−1\begin{alignat*}{3} \pdiff{f}{x}(x,y,z)&= 2e^{-(x^2 +y^2 +z^2)} -2x(2x + y)e^{-(x^2 +y^2 +z^2)} &\quad\implies \pdiff{f}{x}(0,1,-1)&=2e^{-2} \\ \pdiff{f}{y}(x,y,z)&= e^{-(x^2 +y^2 +z^2)} -2y(2x + y)e^{-(x^2 +y^2 +z^2)} &\quad\implies \pdiff{f}{y}(0,1,-1)&=-e^{-2} \\ \pdiff{f}{z}(x,y,z)&= -2z(2x + y)e^{-(x^2 +y^2 +z^2)} &\quad\implies \pdiff{f}{z}(0,1,-1)&=2e^{-2} \\ \pdiff{g}{x}(x,y,z)&= z &\quad\implies \pdiff{g}{x}(0,1,-1)&=-1 \\ \pdiff{g}{y}(x,y,z)&= 2y+z &\quad\implies \pdiff{g}{y}(0,1,-1)&= 1 \\ \pdiff{g}{z}(x,y,z)&= x+y+2z &\quad\implies \pdiff{g}{z}(0,1,-1)&=-1 \end{alignat*}

so that gradients are

∇f(0,1,−1)=e−2<2,−1,2>∇g(0,1,−1)=<−1,1,−1>\begin{equation*} \vnabla f(0,1,-1) =e^{-2}\llt 2,-1,2\rgt\qquad \vnabla g(0,1,-1) =\llt -1,1,-1\rgt \end{equation*}

(b) The bird's velocity is the vector of length 66 in the direction of ∇f(0,1,−1)\vnabla f(0,1,-1), which is

v=6<2,−1,2>∣<2,−1,2>∣=<4,−2,4>\begin{equation*} \vv= 6\frac{\llt 2,-1,2\rgt}{|\llt 2,-1,2\rgt|} =\llt 4,-2,4\rgt \end{equation*}

The rate of change of gg (per unit time) seen by the bird is

∇g(0,1,−1)⋅v=<−1,1,−1>⋅<4,−2,4>=−10\begin{align*} \vnabla g(0,1,-1) \cdot \vv =\llt -1,1,-1\rgt \cdot \llt 4,-2,4\rgt = -10 \end{align*}

(c) The direction of flight for the bat has to be perpendicular to both ∇f(0,1,−1)=e−2<2,−1,2>\vnabla f(0,1,-1) =e^{-2}\llt 2,-1,2\rgt and ∇g(0,1,−1)=<−1,1,−1>\vnabla g(0,1,-1) =\llt -1,1,-1\rgt. Any vector which is a non zero constant times

<2,−1,2>×<−1,1,−1>=det⁡[ı^ȷ^k^2−12−11−1]=<−1 , 0 , 1>\begin{align*} \llt 2,-1,2\rgt \times \llt -1,1,-1\rgt =\det\left[\begin{matrix} \hi & \hj & \hk \\ 2 & -1 & 2 \\ -1 & 1 & -1 \end{matrix}\right] =\llt -1 \,,\, 0 \,,\, 1 \rgt \end{align*}

is perpendicular to both ∇f(0,1,−1)\vnabla f(0,1,-1) and ∇g(0,1,−1)\vnabla g(0,1,-1). In addition, the direction of flight for the bat must have a positive zz–component. So any vector which is a (strictly) positive constant times <−1 , 0 , 1>\llt -1 \,,\, 0 \,,\, 1 \rgt is fine.

Q16Stage 2Past exam · M200 2011D

A bee is flying along the curve of intersection of the surfaces 3z+x2+y2=23z + x^2 + y^2 = 2 and z=x2−y2z = x^2 - y^2 in the direction for which zz is increasing. At time t=2t = 2, the bee passes through the point (1,1,0)(1, 1, 0) at speed 66.

  1. Find the velocity (vector) of the bee at time t=2t = 2.

  2. The temperature TT at position (x,y,z)(x, y, z) at time tt is given by T=xy−3x+2yt+zT = xy - 3x+2yt+z. Find the rate of change of temperature experienced by the bee at time t=2t = 2.

Answer

(a) v=<−2 , −4 , 4>\vv = \llt -2 \,,\, -4 \,,\, 4 \rgt (b) −10-10

Full solution

(a) Let's use v\vv to denote the bee's velocity vector at time t=2t=2.

  • The bee's direction of motion is tangent to the curve. That tangent is perpendicular to both the normal vector to 3z+x2+y2=23z + x^2 + y^2 = 2 at (1,1,0)(1,1,0), which is

    <2x , 2y , 3>∣(x,y,z)=(1,1,0)=<2 , 2 , 3>\begin{equation*} \llt 2x\,,\,2y\,,\,3\rgt\Big|_{(x,y,z)=(1,1,0)} = \llt 2\,,\,2\,,\,3\rgt \end{equation*}

    and the normal vector to z=x2−y2z = x^2 - y^2 at (1,1,0)(1,1,0), which is

    <2x , −2y , −1>∣(x,y,z)=(1,1,0)=<2 , −2 , −1>\begin{equation*} \llt 2x\,,\,-2y\,,\,-1\rgt\Big|_{(x,y,z)=(1,1,0)} = \llt 2\,,\,-2\,,\,-1\rgt \end{equation*}

    So v\vv has to be some constant times

    <2 , 2 , 3>×<2 , −2 , −1>=det⁡[ı^ȷ^k^2232−2−1]=<4 , 8 , −8>\begin{align*} \llt 2\,,\,2\,,\,3\rgt \times \llt 2\,,\,-2\,,\,-1\rgt =\det\left[\begin{matrix} \hi & \hj & \hk \\ 2 & 2 & 3 \\ 2 & -2 & -1 \end{matrix}\right] =\llt 4 \,,\, 8 \,,\, -8 \rgt \end{align*}

    or, equivalently, some constant times <1 , 2 , −2>\llt 1 \,,\, 2 \,,\, -2 \rgt.

  • Since the zz–component of v\vv has to be positive, v\vv has to be a positive constant times <−1 , −2 , 2>\llt -1 \,,\, -2 \,,\, 2 \rgt.

  • Since the speed has to be 66, v\vv has to have length 66.

As ∣<−1 , −2 , 2>∣=3|\llt -1 \,,\, -2 \,,\, 2 \rgt|=3

v=2<−1 , −2 , 2>=<−2 , −4 , 4>\begin{equation*} \vv = 2\llt -1 \,,\, -2 \,,\, 2 \rgt =\llt -2 \,,\, -4 \,,\, 4 \rgt \end{equation*}

(b) Solution 1:
Suppose that the bee is at (x(t),y(t),z(t))\big(x(t),y(t),z(t)\big) at time tt. Then the temperature that the bee feels at time tt is

T(x(t),y(t),z(t),t)=x(t)y(t)−3x(t)+2y(t)t+z(t)\begin{align*} T\big(x(t),y(t),z(t),t\big) = x(t) y(t) -3x(t) +2y(t) t +z(t) \end{align*}

Then the rate of change of temperature (per unit time) felt by the bee at time t=2t=2 is

ddtT(x(t),y(t),z(t),t)∣t=2=x′(2)y(2)+x(2)y′(2)−3x′(2)+2y′(2)2+2y(2)+z′(2)\begin{align*} \diff{}{t}T\big(x(t),y(t),z(t),t\big)\Big|_{t=2} &=x'(2)y(2) + x(2)y'(2) -3x'(2) +2y'(2)2+2y(2) +z'(2) \\ \end{align*}

Recalling that, at time t=2t=2, the bee is at (1,1,0)(1,1,0) and has velocity <−2 , −4 , 4>\llt -2 \,,\, -4 \,,\, 4 \rgt

ddtT(x(t),y(t),z(t),t)∣t=2=(−2)(1)+(1)(−4)−3(−2)+2(−4)2+2(1)+4=−10\begin{align*} \diff{}{t}T\big(x(t),y(t),z(t),t\big)\Big|_{t=2} &=(-2)(1) + (1)(-4) -3(-2) +2(-4)2+2(1) +4 \\ &=-10 \end{align*}

(b) Solution 2:
Suppose that the bee is at (x(t),y(t),z(t))\big(x(t),y(t),z(t)\big) at time tt. Then the temperature that the bee feels at time tt is

T(x(t),y(t),z(t),t)\begin{align*} T\big(x(t),y(t),z(t),t\big) \end{align*}

By the chain rule, the rate of change of temperature (per unit time) felt by the bee at time t=2t=2 is

ddtT(x(t),y(t),z(t),t)∣t=2=[∂T∂x(x(t),y(t),z(t),t) x′(t)+∂T∂y(x(t),y(t),z(t),t) y′(t)+∂T∂z(x(t),y(t),z(t),t) z′(t)+∂T∂t(x(t),y(t),z(t),t)]t=2\begin{align*} \diff{}{t}T\big(x(t),y(t),z(t),t\big)\Big|_{t=2} &=\bigg[\pdiff{T}{x}\big(x(t),y(t),z(t),t\big)\,x'(t) +\pdiff{T}{y}\big(x(t),y(t),z(t),t\big)\,y'(t) \\&\hskip0.5in +\pdiff{T}{z}\big(x(t),y(t),z(t),t\big)\,z'(t) +\pdiff{T}{t}\big(x(t),y(t),z(t),t\big)\bigg]_{t=2} \end{align*}

Recalling that T=xy−3x+2yt+zT = xy - 3x+2yt+z, we have

ddtT(x(t),y(t),z(t),t)∣t=2=[y(2)−3]x′(2)+[x(2)+2×2]y′(2)+z′(2)+2y(2)\begin{align*} \diff{}{t}T\big(x(t),y(t),z(t),t\big)\Big|_{t=2} &=[y(2)-3]x'(2) + [x(2)+2\times 2]y'(2) +z'(2) +2y(2) \end{align*}

Also recalling that, at time t=2t=2, the bee is at (1,1,0)(1,1,0) and has velocity <−2 , −4 , 4>\llt -2 \,,\, -4 \,,\, 4 \rgt

ddtT(x(t),y(t),z(t),t)∣t=2=[−2](−2)+[5](−4)+4+2=−10\begin{align*} \diff{}{t}T\big(x(t),y(t),z(t),t\big)\Big|_{t=2} &=[-2](-2) + [5](-4) +4 +2 \\ &=-10 \end{align*}
Q17Stage 2Past exam · M200 2012a

The temperature at a point (x,y,z)(x, y, z) is given by T(x,y,z)=5e−2x2−y2−3z2T(x, y, z) = 5e^{-2x^2-y^2-3z^2}, where TT is measured in centigrade and xx, yy, zz in meters.

  1. Find the rate of change of temperature at the point P(1,2,−1)P(1, 2, -1) in the direction toward the point (1,1,0)(1, 1, 0).

  2. In which direction does the temperature decrease most rapidly?

  3. Find the maximum rate of decrease at PP.

Answer

(a) 25 2 e−925\,\sqrt{2}\,e^{-9} (b) <2,2,−3>17\frac{\llt 2,2,-3\rgt}{\sqrt{17}} (c) −1017e−9-10\sqrt{17} e^{-9}

Full solution

(a) We are to find the rate of change of T(x,y,z)T(x,y,z) at (1,2,−1)(1, 2, -1) in the direction <1,1,0>−<1,2,−1>=<0,−1,1>\llt 1, 1, 0\rgt - \llt 1, 2, -1\rgt = \llt 0,-1,1 \rgt. That rate of change (per unit distance) is the directional derivative

D<0,−1,1>2T(1,2,−1)=∇T(1,2,−1)⋅<0,−1,1>2\begin{align*} D_{\frac{\llt 0,-1,1 \rgt}{\sqrt{2}}} T(1,2,-1) =\vnabla T(1,2,-1)\cdot \frac{\llt 0,-1,1 \rgt}{\sqrt{2}} \end{align*}

As

∂T∂x(x,y,z)=−20x e−2x2−y2−3z2∂T∂x(1,2,−1)=−20 e−9∂T∂y(x,y,z)=−10y e−2x2−y2−3z2∂T∂y(1,2,−1)=−20 e−9∂T∂z(x,y,z)=−30z e−2x2−y2−3z2∂T∂z(1,2,−1)=30 e−9\begin{align*} \pdiff{T}{x}(x,y,z)&= -20 x\,e^{-2x^2-y^2-3z^2} & \pdiff{T}{x}(1,2,-1)&= -20\, e^{-9} \\ \pdiff{T}{y}(x,y,z)&= -10 y\,e^{-2x^2-y^2-3z^2} & \pdiff{T}{y}(1,2,-1)&= -20\, e^{-9} \\ \pdiff{T}{z}(x,y,z)&= -30 z\,e^{-2x^2-y^2-3z^2} & \pdiff{T}{z}(1,2,-1)&= 30\, e^{-9} \end{align*}

the directional derivative

D<0,−1,1>2T(1,2,−1)=e−9<−20,−20,30>⋅<0,−1,1>2=502e−9=25 2 e−9\begin{align*} D_{\frac{\llt 0,-1,1 \rgt}{\sqrt{2}}} T(1,2,-1) =e^{-9}\llt -20,-20,30\rgt\cdot \frac{\llt 0,-1,1 \rgt}{\sqrt{2}} =\frac{50}{\sqrt{2}}e^{-9} =25\,\sqrt{2}\,e^{-9} \end{align*}

(b) The direction of maximum rate of decrease is −∇T(1,2,−1)-\vnabla T(1,2,-1). A unit vector in that direction is <2,2,−3>17\frac{\llt 2,2,-3\rgt}{\sqrt{17}}.

(c) The maximum rate of decrease at PP is −∣∇T(1,2,−1)∣=−10e−9∣<−2,−2,3>∣=−1017e−9-|\vnabla T(1,2,-1)|=-10 e^{-9} |\llt -2,-2,3\rgt| = -10\sqrt{17} e^{-9}.

Q18Stage 2Past exam · M200 2012D

The directional derivative of a function w=f(x,y,z)w = f(x, y, z) at a point PP in the direction of the vector ı^\hi is 2, in the direction of the vector ı^+ȷ^\hi+\hj is −2-\sqrt{2}, and in the direction of the vector ı^+ȷ^+k^\hi+\hj+\hk is −53-\frac{5}{\sqrt{3}}. Find the direction in which the function w=f(x,y,z)w = f(x, y, z) has the maximum rate of change at the point PP . What is this maximum rate of change?

Answer

The unit vector in the direction of maximum rate of change is <2 , −4 , −3>29\frac{\llt 2\,,\,-4\,,\,-3\rgt}{\sqrt{29}}. The maximum rate of change is 29\sqrt{29}.

Full solution

Denote by <a , b , c>\llt a\,,\,b\,,\,c\rgt the gradient of the function ff at PP. We are told

<a , b , c>⋅<1 , 0 , 0>=2<a , b , c>⋅12<1 , 1 , 0>=−2<a , b , c>⋅13<1 , 1 , 1>=−53\begin{align*} \llt a\,,\,b\,,\,c\rgt \cdot \llt 1\,,\,0\,,\,0\rgt &= 2 \\ \llt a\,,\,b\,,\,c\rgt \cdot \frac{1}{\sqrt{2}}\llt 1\,,\,1\,,\,0\rgt &= -\sqrt{2} \\ \llt a\,,\,b\,,\,c\rgt \cdot \frac{1}{\sqrt{3}}\llt 1\,,\,1\,,\,1\rgt &= -\frac{5}{\sqrt{3}} \end{align*}

Simplifying

a=2a+b=−2a+b+c=−5\begin{align*} a&=2 \\ a+b&=-2 \\ a+b+c&=-5 \end{align*}

From these equations we read off, in order, a=2a=2, b=−4b=-4 and c=−3c=-3. The function ff has maximum rate of change at PP in the direction if the gradient of ff. The unit vector in that direction is

<2 , −4 , −3>∣<2 , −4 , −3>∣=<2 , −4 , −3>29\begin{equation*} \frac{\llt 2\,,\,-4\,,\,-3\rgt}{|\llt 2\,,\,-4\,,\,-3\rgt|} =\frac{\llt 2\,,\,-4\,,\,-3\rgt}{\sqrt{29}} \end{equation*}

The maximum rate of change is the magnitude of the gradient, which is 29\sqrt{29}.

Q19Stage 2Past exam · M200 2013D

Suppose it is known that the direction of the fastest increase of the function f(x,y)f(x,y) at the origin is given by the vector <1,2>\llt 1, 2\rgt. Find a unit vector uu that is tangent to the level curve of f(x,y)f(x,y) that passes through the origin.

Answer

±15<2,−1>\pm\frac{1}{\sqrt{5}}\llt 2, -1\rgt

Full solution

We are told that the direction of fastest increase for the function f(x,y)f(x,y) at the origin is given by the vector <1,2>\llt 1, 2\rgt. This implies that ∇f(0,0)\vnabla f(0,0) is parallel to <1,2>\llt 1, 2\rgt. This in turn implies that <1,2>\llt 1, 2\rgt is normal to the level curve of f(x,y)f(x,y) that passes through the origin. So <2,−1>\llt 2, -1\rgt, being perpendicular to <1,2>\llt 1, 2\rgt, is tangent to the level curve of f(x,y)f(x,y) that passes through the origin. The unit vectors that are parallel to <2,−1>\llt 2, -1\rgt are ±15<2,−1>\pm\frac{1}{\sqrt{5}}\llt 2, -1\rgt.

Q20Stage 2Past exam · M200 2013D

The shape of a hill is given by z=1000−0.02x2−0.01y2z = 1000 - 0.02x^2 - 0.01y^2. Assume that the xx–axis is pointing East, and the yy–axis is pointing North, and all distances are in metres.

  1. What is the direction of the steepest ascent at the point (0,100,900)(0, 100, 900)? (The answer should be in terms of directions of the compass).

  2. What is the slope of the hill at the point (0,100,900)(0, 100, 900) in the direction from (a)?

  3. If you ride a bicycle on this hill in the direction of the steepest descent at 55 m/s, what is the rate of change of your altitude (with respect to time) as you pass through the point (0, 100, 900)?

Answer

(a) South (b) 22 (c) −25-2\sqrt{5}

Full solution

Write h(x,y)=1000−0.02 x2−0.01y2h(x,y) = 1000 -0.02\,x^2-0.01 y^2 so that the hill is z=h(x,y)z=h(x,y).

(a) The direction of steepest ascent at (0,100,900)(0,100,900) is the direction of maximum rate of increase of h(x,y)h(x,y) at (0,100)(0,100) which is ∇h(0,100)=<0 , −0.01(2)(100)>=<0 , −2>\vnabla h(0,100) = \llt 0\,,\, -0.01(2)(100)\rgt = \llt 0\,,\,-2\rgt. In compass directions that is South.

(b) The slope of the hill there is

∇h(0,100)⋅<0,−1>=−∂h∂y(0,100)=2\begin{align*} \vnabla h(0,100)\cdot\llt 0,-1\rgt =-\pdiff{h}{y}(0,100) = 2 \end{align*}

(c) Denote by (x(t),y(t),z(t))\big(x(t),y(t),z(t)\big) your position at time tt and suppose that you are at (0,100,900)(0,100,900) at time t=0t=0. Then we know

  • z(t)=1000−0.02 x(t)2−0.01y(t)2z(t) = 1000 -0.02\,x(t)^2-0.01 y(t)^2, so that z′(t)=−0.04 x(t)x′(t)−0.02y(t)y′(t)z'(t) = -0.04\,x(t)x'(t)-0.02 y(t)y'(t), since you are on the hill and

  • x′(0)=0x'(0)=0 and y′(0)>0y'(0)>0 since you are going in the direction of steepest descent and

  • x′(0)2+y′(0)2+z′(0)2=25x'(0)^2+y'(0)^2+z'(0)^2=25 since you are moving at speed 55.

Since x(0)x(0) and y(0)=100y(0)=100, we have z′(0)=−0.02(100)y′(0)=−2y′(0)z'(0)= -0.02(100)y'(0) = -2y'(0). So

25=x′(0)2+y′(0)2+z′(0)2=5 y′(0)2  ⟹  y′(0)=5  ⟹  <x′(0) , y′(0) , z′(0)>=<0 , 5 , −25>\begin{align*} 25 = x'(0)^2+y'(0)^2+z'(0)^2 =5\ y'(0)^2 &\implies y'(0) = \sqrt{5} \\ &\implies \llt x'(0)\,,\,y'(0)\,,\,z'(0) \rgt =\llt 0\,,\, \sqrt{5}\,,\, -2\sqrt{5}\rgt \end{align*}

and your rate of change of altitude is

ddth(x(t) , y(t))∣t=0=∇h(0,100)⋅<x′(0) , y′(0)>=<0 , −2>⋅<0 , 5>=−25\begin{align*} \diff{}{t} h\big(x(t)\,,\,y(t)\big)\Big|_{t=0} &=\vnabla h(0,100)\cdot \llt x'(0)\,,\,y'(0) \rgt =\llt 0\,,\,-2\rgt \cdot \llt 0\,,\, \sqrt{5}\rgt =-2\sqrt{5} \end{align*}
Q21Stage 2Past exam · M200 2014A

Let the pressure PP and temperature TT at a point (x,y,z)(x, y, z) be

P(x,y,z)=x2+2y21+z2,T(x,y,z)=5+xy−z2\begin{equation*} P(x,y,z) = \frac{x^2+2y^2}{1+z^2},\qquad T(x,y,z) = 5 + xy - z^2 \end{equation*}

  1. If the position of an airplane at time tt is

    (x(t),y(t),z(t))=(2t,t2−1,cos⁡t)\begin{equation*} (x(t), y(t), z(t)) = (2t, t^2 - 1, \cos t) \end{equation*}

    find ddt(PT)2\diff{}{t} (PT)^2 at time t=0t = 0 as observed from the airplane.

  2. In which direction should a bird at the point (0,−1,1)(0,-1,1) fly if it wants to keep both PP and TT constant. (Give one possible direction vector. It does not need to be a unit vector.)

  3. An ant crawls on the surface z3+zx+y2=2z^3 + zx + y^2 = 2. When the ant is at the point (0,−1,1)(0,-1,1), in which direction should it go for maximum increase of the temperature T=5+xy−z2T = 5 + xy - z^2? Your answer should be a vector <a,b,c>\llt a, b, c\rgt, not necessarily of unit length. (Note that the ant cannot crawl in the direction of the gradient because that leads off the surface. The direction vector <a,b,c>\llt a, b, c\rgt has to be on the tangent plane to the surface.)

Answer

(a) −16-16 (b) C<4 , 1 , −2>C \llt 4\,,\, 1 \,,\, -2\rgt for any nonzero constant CC

(c) Any positive non zero multiple of −<1 , 2 , 1>-\llt 1\,,\, 2\,,\,1\rgt will do.

Full solution

Reading through the question as a whole we see that we will need

  • for part (a), the gradient of PTPT at (2t,t2−1,cos⁡t)∣t=0=(0,−1,1)(2t, t^2 - 1, \cos t)\Big|_{t=0} =(0,-1,1)

  • for part (b), the gradients of both PP and TT at (0,−1,1)(0,-1,1) and

  • for part (c), the gradient of TT at (0,−1,1)(0,-1,1) and the gradient of S=z3+xz+y2S=z^3+xz+y^2 at (0,−1,1)(0,-1,1) (to get the normal vector to the surface at that point).

So, by way of preparation, let's compute all of these gradients.

∇P(x,y,z)=2x1+z2ı^+4y1+z2ȷ^−(x2+2y2)2z(1+z2)2k^∇P(0,−1,1)=−2 ȷ^−k^∇T(x,y,z)=y ı^+x ȷ^−2z k^∇T(0,−1,1)=−ı^−2k^∇S(x,y,z)=z ı^+2y ȷ^+(x+3z2) k^∇S(0,−1,1)=ı^−2ȷ^+3 k^\begin{align*} \vnabla P(x,y,z) &= \frac{2x}{1+z^2}\hi + \frac{4y}{1+z^2}\hj -\frac{(x^2+2y^2)2z}{{(1+z^2)}^2}\hk & \vnabla P(0,-1,1) &= -2\,\hj - \hk \\ \vnabla T(x,y,z) &= y\,\hi + x\,\hj -2z\,\hk & \vnabla T(0,-1,1) &= -\hi - 2\hk \\ \vnabla S(x,y,z) &= z\,\hi + 2y\,\hj +(x+3z^2)\,\hk & \vnabla S(0,-1,1) &= \hi - 2\hj +3\,\hk \end{align*}

To get the gradient of PTPT we use the product rule

∇(PT)(x,y,z)=T(x,y,z) ∇P(x,y,z)+P(x,y,z) ∇T(x,y,z)\begin{equation*} \vnabla(PT)(x,y,z) =T(x,y,z)\,\vnabla P(x,y,z) +P(x,y,z)\,\vnabla T(x,y,z) \end{equation*}

so that

∇(PT)(0,−1,1)=T(0,−1,1) ∇P(0,−1,1)+P(0,−1,1) ∇T(0,−1,1)=(5+0−1)(−2 ȷ^−k^)+0+21+1(−ı^−2k^)=−ı^−8 ȷ^−6 k^\begin{align*} \vnabla(PT)(0,-1,1) &=T(0,-1,1)\,\vnabla P(0,-1,1) +P(0,-1,1)\,\vnabla T(0,-1,1) \\ &=(5+0-1) \big(-2\,\hj - \hk\big) + \frac{0+2}{1+1}\big(-\hi - 2\hk\big) \\ &= -\hi -8\,\hj -6\,\hk \end{align*}

(a) Since ddt(PT)2=2(PT)ddt(PT)\diff{}{t}(PT)^2 = 2(PT)\diff{}{t}(PT), and the velocity vector of the plane at time 00 is

ddt<2t,t2−1,cos⁡t>∣t=0=<2,2t,−sin⁡t>∣t=0=<2,0,0>\begin{equation*} \diff{}{t}\llt 2t, t^2 - 1, \cos t\rgt\Big|_{t=0} =\llt 2, 2t, -\sin t\rgt\Big|_{t=0} =\llt 2,0,0\rgt \end{equation*}

we have

ddt(PT)2∣t=0=2 P(0,−1,1) T(0,−1,1) ∇(PT)(0,−1,1)⋅<2,0,0>=2 0+21+1 (5+0−1)<−1,−8,−6>⋅<2,0,0>=−16\begin{align*} \diff{}{t}(PT)^2\Big|_{t=0} &= 2\,P(0,-1,1)\,T(0,-1,1)\ \vnabla(PT)(0,-1,1)\cdot \llt 2,0,0\rgt \\ &= 2\ \frac{0+2}{1+1}\ (5+0-1)\llt -1,-8,-6\rgt \cdot \llt 2,0,0\rgt \\ &= -16 \end{align*}

(b) The direction should be perpendicular to ∇P(0,−1,1)\vnabla P(0,-1,1) (to keep PP constant) and should also be perpendicular to ∇T(0,−1,1)\vnabla T(0,-1,1) (to keep TT constant). So any nonzero constant times

±∇P(0,−1,1)×∇T(0,−1,1)=±<0 , −2 , −1>×<−1 , 0 , −2>=±det⁡[ı^ȷ^k^021102]=±<4 , 1 , −2>\begin{align*} \pm \vnabla P(0,-1,1) \times \vnabla T(0,-1,1) &=\pm \llt 0 \,,\, -2 \,,\, -1\rgt \times \llt -1\,,\,0 \,,\,- 2\rgt =\pm \det\left[\begin{matrix} \hi & \hj & \hk \\ 0 & 2 & 1 \\ 1 & 0 & 2 \end{matrix}\right] \\ &=\pm \llt 4\,,\, 1 \,,\, -2\rgt \end{align*}

are allowed directions.

(c) We want the direction to be as close as possible to ∇T(0,−1,1)=<−1 , 0 , −2>\vnabla T(0,-1,1) =\llt -1 \,,\, 0 \,,\, -2\rgt while still being tangent to the surface, i.e. being perpendicular to the normal vector ∇S(0,−1,1)=<1 , −2 , 3>\vnabla S(0,-1,1)=\llt 1 \,,\, -2 \,,\, 3\rgt. We can get that optimal direction by subtracting from ∇T(0,−1,1)\vnabla T(0,-1,1) the projection of ∇T(0,−1,1)\vnabla T(0,-1,1) onto the normal vector.

Figure from prob_s2.7, line 1182

Figure from prob_s2.7, line 1182

The projection of ∇T(0,−1,1)\vnabla T(0,-1,1) onto the normal vector ∇S(0,−1,1)\vnabla S(0,-1,1) is

proj∇S(0,−1,1)∇T(0,−1,1)=∇T(0,−1,1)⋅∇S(0,−1,1)∣∇S(0,−1,1)∣2∇S(0,−1,1)=<−1 , 0 , −2>⋅<1 , −2 , 3>∣<1 , −2 , 3>∣2<1 , −2 , 3>=−714<1 , −2 , 3>\begin{align*} \text{proj}_{\vnabla S(0,-1,1)}\vnabla T(0,-1,1) &=\frac{\vnabla T(0,-1,1)\cdot \vnabla S(0,-1,1)} {|\vnabla S(0,-1,1)|^2}\vnabla S(0,-1,1) \\ &=\frac{\llt -1\,,\,0 \,,\,- 2\rgt\cdot \llt 1\,,\, -2 \,,\,3\rgt} {|\llt 1\,,\, -2 \,,\,3\rgt|^2}\llt 1\,,\, -2 \,,\,3\rgt \\ &=\frac{-7}{14}\llt 1\,,\, -2 \,,\,3\rgt \end{align*}

So the optimal direction is

d=∇T(0,−1,1)−proj∇S(0,−1,1)∇T(0,−1,1)=<−1 , 0 , −2>−−714<1 , −2 , 3>=<−12 , −1 , −12>\begin{align*} \vd&=\vnabla T(0,-1,1) - \text{proj}_{\vnabla S(0,-1,1)}\vnabla T(0,-1,1) \\ &=\llt -1\,,\,0 \,,\,- 2\rgt -\frac{-7}{14}\llt 1\,,\, -2 \,,\,3\rgt \\ &= \llt -\frac{1}{2}\,,\,-1 \,,\,- \frac{1}{2}\rgt \end{align*}

So any positive non zero multiple of −<1 , 2 , 1>-\llt 1\,,\, 2\,,\,1\rgt will do. Note, as a check, that −<1 , 2 , 1>-\llt 1\,,\, 2\,,\,1\rgt has dot product zero, i.e. is perpendicular to, ∇S(0,−1,1)=<1 , −2 , 3>\vnabla S(0,-1,1)=\llt 1 \,,\, -2 \,,\, 3\rgt.

Q22Stage 2Past exam · M200 2014D

Suppose that f(x,y,z)f(x,y,z) is a function of three variables and let u=16<1,1,2>\vu = \frac{1}{\sqrt{6}} \llt 1, 1, 2\rgt and v=13<1,−1,−1>\vv = \frac{1}{\sqrt{3}} \llt 1, -1, -1\rgt and w=13<1,1,1>\vw = \frac{1}{\sqrt{3}} \llt 1, 1, 1\rgt. Suppose that at a point (a,b,c)(a,b,c),

Duf=0Dvf=0Dwf=4\begin{align*} D_\vu f&=0 \\ D_\vv f&=0 \\ D_\vw f&=4 \end{align*}

Find ∇f\vnabla f at (a,b,c)(a,b,c).

Answer

∇f(a,b,c)=3<2,6,−4>\vnabla f(a,b,c) =\sqrt{3} \llt 2,6,-4\rgt

Full solution

Write ∇f(a,b,c)=<F,G,H>\vnabla f(a,b,c) =\llt F,G,H\rgt. We are told that

Duf=16<1,1,2>⋅<F,G,H>=0Dvf=13<1,−1,−1>⋅<F,G,H>=0Dwf=13<1,1,1>⋅<F,G,H>=4\begin{align*} D_\vu f&=\frac{1}{\sqrt{6}} \llt 1, 1, 2\rgt \cdot \llt F,G,H\rgt = 0 \\ D_\vv f&=\frac{1}{\sqrt{3}} \llt 1, -1, -1\rgt \cdot \llt F,G,H\rgt = 0 \\ D_\vw f&=\frac{1}{\sqrt{3}} \llt 1, 1, 1\rgt \cdot \llt F,G,H\rgt = 4 \end{align*}

so that

F+G+2H=0F−G−H=0F+G+H=43\begin{align*} F + G + 2H &= 0 \tag{E1}\\ F - G - H &= 0 \tag{E2}\\ F + G + H &= 4\sqrt{3} \tag{E3} \end{align*}

Adding (E2) and (E3) gives 2F=432F=4\sqrt{3} or F=23F=2\sqrt{3}. Substituting F=23F=2\sqrt{3} into (E1) and (E2) gives

G+2H=−23−G−H=−23\begin{align*} G + 2H &= -2\sqrt{3} \tag{E1}\\ - G - H &= -2\sqrt{3} \tag{E2} \end{align*}

Adding (E1) and (E2) gives H=−43H=-4\sqrt{3} and substituting H=−43H=-4\sqrt{3} back into (E2) gives G=63G=6\sqrt{3}. All together

∇f(a,b,c)=3<2,6,−4>\begin{equation*} \vnabla f(a,b,c) =\sqrt{3} \llt 2,6,-4\rgt \end{equation*}
Q23Stage 2Past exam · M200 2003D

The elevation of a hill is given by the equation f(x,y)=x2y2e−x−yf(x,y)=x^2y^2e^{-x-y}. An ant sits at the point (1,1,e−2)(1,1,e^{-2}).

  1. Find the unit vector u=<u1,u2>\vu=\llt u_1,u_2\rgt that maximizes

    lim⁡t→0f((1,1)+tu)−f(1,1)t\begin{equation*} \lim_{t\rightarrow 0}\frac{f\big((1,1)+t\vu\big)-f(1,1)}{t} \end{equation*}

  2. Find a vector v=<v1,v2,v3>\vv=\llt v_1,v_2,v_3\rgt pointing in the direction of the path that the ant could take in order to stay on the same elevation level e−2e^{-2}.

  3. Find a vector v=<v1,v2,v3>\vv=\llt v_1,v_2,v_3\rgt pointing in the direction of the path that the ant should take in order to maximize its instantaneous rate of level increase.

Answer

(a) 12<1,1>\frac{1}{\sqrt{2}}\llt 1,1\rgt (b) v=c<1,−1,0>\vv=c\llt 1,-1,0\rgt for any nonzero constant cc

(c) v=12<1,1,2e−2>\vv=\frac{1}{\sqrt{2}}\llt 1,1,2e^{-2}\rgt. Any positive multiple of this vector is also a correct answer.

Full solution

(a) The expression lim⁡t→0f((1,1)+tu)−f(1,1)t\lim_{t\rightarrow 0}\frac{f((1,1)+t\vu)-f(1,1)}{t} is the directional derivative of ff at (1,1)(1,1) in the direction u\vu, which is Duf(1,1)=∇f(1,1)⋅uD_\vu f(1,1)=\vnabla f(1,1)\cdot\vu. This is mazimized when u\vu is parallel to ∇f(1,1)\vnabla f(1,1). Since

fx(x,y)=2xy2e−x−y−x2y2e−x−yfy(x,y)=2x2ye−x−y−x2y2e−x−y\begin{equation*} f_x(x,y)=2xy^2e^{-x-y}-x^2y^2e^{-x-y}\qquad f_y(x,y)=2x^2ye^{-x-y}-x^2y^2e^{-x-y} \end{equation*}

we have

∇f(1,1)=e−2<1,1>\begin{equation*} \vnabla f(1,1)=e^{-2}\llt 1,1\rgt \end{equation*}

so that the desired unit vector u\vu is 12<1,1>\frac{1}{\sqrt{2}}\llt 1,1\rgt.

(b) In order to remain at elevation e−2e^{-2}, the ant must move so that Duf(1,1)=0D_\vu f(1,1)=0. This is the case if u⊥∇f(1,1)\vu\perp\vnabla f(1,1). For example, we can take u=<1,−1>\vu=\llt 1,-1\rgt. When the ant moves in this direction, while remaining on the surface of the hill, its vertical component of velocity is zero. So v=c<1,−1,0>\vv=c\llt 1,-1,0\rgt for any nonzero constant cc.

(c) In order to maximize its instantaneous rate of level increase, the ant must choose the xx and yy coordinates of its velocity vector in the same direction as ∇f(1,1)\vnabla f(1,1). Namely u=c<1,1>\vu=c\llt 1,1\rgt for any c>0c>0. To make u\vu a unit vector, we choose c=12c=\frac{1}{\sqrt{2}}. The corresponding value of the zz coordinate of its velocity vector is the rate of change of ff per unit horizontal distance travelled, which is the directional derivative

Duf(1,1)=∇f(1,1)⋅u=e−2<1,1>⋅<c,c>=2ce−2\begin{equation*} D_\vu f(1,1)=\vnabla f(1,1)\cdot\vu=e^{-2}\llt 1,1\rgt \cdot\llt c,c\rgt =2ce^{-2} \end{equation*}

So v=12<1,1,2e−2>\vv=\frac{1}{\sqrt{2}}\llt 1,1,2e^{-2}\rgt. Any positive multiple of this vector is also a correct answer.

Q24Stage 2Past exam · M200 2001D

Let the temperature in a region of space be given by T(x,y,z)=3x2+12y2+2z2T(x,y,z)=3x^2+\half y^2+2z^2 degrees.

  1. A sparrow is flying along the curve r(s)=(13s3,2s,s2)\vr(s)=\big(\frac{1}{3}s^3,2s,s^2\big) at a constant speed of 3ms−13{\rm ms}^{-1}. What is the velocity of the sparrow when s=1s=1?

  2. At what rate does the sparrow feel the temperature is changing at the point A(13,2,1)A\big(\frac{1}{3},2,1\big) for which s=1s=1.

  3. At the point A(13,2,1)A\big(\frac{1}{3},2,1\big) in what direction will the temperature be decreasing at maximum rate?

  4. An eagle crosses the path of the sparrow at A(13,2,1)A\big(\frac{1}{3},2,1\big), is moving at right angles to the path of the sparrow, and is also moving in a direction in which the temperature remains constant. In what directions could the eagle be flying as it passes through the point AA?

Answer

(a) <1,2,2>\llt 1,2,2\rgt (b) 14∘/s14^\circ/{\rm s} (c) any positive constant times −<2,2,4>-\llt 2,2,4\rgt

(d) any positive constant times ±<2,0,−1>\pm\llt 2,0,-1\rgt

Full solution

(a) The direction of motion at s=1s=1 is given by the tangent vector

r′(s)=<s2,2,2s>∣s=1=<1,2,2>\begin{equation*} \vr'(s)=\llt s^2,2,2s\rgt\big|_{s=1}=\llt 1,2,2\rgt \end{equation*}

Since the length of the velocity vector must be 33,

velocity=v=3<1,2,2>∣<1,2,2>∣=<1,2,2>\begin{equation*} \text{velocity}=\vv=3\frac{\llt 1,2,2\rgt}{|\llt 1,2,2\rgt|}=\llt 1,2,2\rgt \end{equation*}

(b) The rate of change of temperature per unit distance felt by the sparrow at s=1s=1 is ∇T(13,2,1)⋅v∣v∣\vnabla T\big(\frac{1}{3},2,1\big)\cdot\frac{\vv}{|\vv|}. The rate of change of temperature per unit time felt by the sparrow at s=1s=1 is

∇T(13,2,1)⋅v∣v∣ ∣v∣=∇T(13,2,1)⋅v=v⋅<6x,y,4z>∣(13,2,1)=<1,2,2>⋅<2,2,4>=14∘/s\begin{align*} \vnabla T\left(\frac{1}{3},2,1\right)\cdot\frac{\vv}{|\vv|}\ |\vv| &=\vnabla T\left(\frac{1}{3},2,1\right)\cdot\vv =\vv\cdot\llt 6x,y,4z\rgt\Big|_{({1\over3},2,1\big)} \\ &=\llt 1,2,2\rgt\cdot\llt 2,2,4\rgt=14^\circ/{\rm s} \end{align*}

(c) The temperature decreases at maximum rate in the direction opposite the temperature gradient, which is (any positive constant times) −<2,2,4>-\llt 2,2,4\rgt.

(d) The eagle is moving at right angles to the direction of motion of the sparrow, which is <1,2,2>\llt 1,2,2\rgt. As the eagle is also moving in a direction for which the temperature remains constant, it must be moving perpendicularly to the temperature gradient, <2,2,4>\llt 2,2,4\rgt. So the direction of the eagle must be (a posiitve constant times) one of

±<1,2,2>×<2,2,4>=±det⁡[ı^ȷ^k^122224]=±<4,0,−2>\begin{align*} \pm\llt 1,2,2\rgt \times\llt 2,2,4\rgt =\pm\det\left[\begin{matrix} \hi & \hj & \hk \\ 1 & 2 & 2 \\ 2 & 2 & 4 \end{matrix}\right] =\pm\llt 4,0,-2\rgt \end{align*}

or equivalently, any positive constant times ±<2,0,−1>\pm\llt 2,0,-1\rgt.

Q25Stage 2Past exam · M200 2001A

Assume that the temperature TT at a point (x,y,z)(x,y,z) near a flame at the origin is given by

T(x,y,z)=2001+x2+y2+z2\begin{equation*} T(x,y,z)=\frac{200}{1+x^2+y^2+z^2} \end{equation*}

where the coordinates are given in meters and the temperature is in degrees Celsius. Suppose that at some moment in time, a moth is at the point (3,4,0)(3,4,0) and is flying at a constant speed of 1m/s1 {\rm m/s} in the direction of maximum increase of temperature.

  1. Find the velocity vector v\vv of the moth at this moment.

  2. What rate of change of temperature does the moth feel at that moment?

Answer

(a) v=−<35,45,0>\vv=-\llt \frac{3}{5},\frac{4}{5},0\rgt (b) 500169≈2.96∘/s\frac{500}{169}\approx 2.96^\circ/{\rm s}

Full solution

(a) The moth is moving the direction of the temperature gradient at (3,4,0)(3,4,0), which is

∇T(3,4,0)=−2002xı^+2yȷ^+2zk^(1+x2+y2+z2)2∣(3,4,0)=−4003ı^+4ȷ^262\begin{equation*} \vnabla T(3,4,0) =-200\frac{2x\hi+2y\hj+2z\hk}{{(1+x^2+y^2+z^2)}^2}\bigg|_{(3,4,0)} =-400\frac{3\hi+4\hj}{26^2} \end{equation*}

Since the speed of the moth is 1m/s1 {\rm m/s} its velocity vector is a vector of length one in direction −400262<3,4,0>-\frac{400}{26^2}\llt 3,4,0\rgt and hence is v=−<3,4,0>∣<3,4,0>∣=−<35,45,0>\vv=-\frac{\llt 3,4,0\rgt}{|\llt 3,4,0\rgt|} =-\llt\frac{3}{5},\frac{4}{5},0\rgt.

(b) The rate of change of temperature (per unit time) the moth feels at that time is

∇T(3,4,0)⋅v=400262<3,4,0>⋅<35,45,0>=400×25262×5=500169≈2.96∘/s\begin{equation*} \vnabla T(3,4,0)\cdot\vv =\frac{400}{26^2}\llt 3,4,0\rgt\cdot \llt\frac{3}{5},\frac{4}{5},0\rgt =\frac{400\times25}{26^2\times 5} =\frac{500}{169}\approx 2.96^\circ/{\rm s} \end{equation*}
Q26Stage 2Past exam · M200 2000D

We say that uu is inversely proportional to vv if there is a constant kk so that u=k/vu=k/v. Suppose that the temperature TT in a metal ball is inversely proportional to the distance from the centre of the ball, which we take to be the origin. The temperature at the point (1,2,2)(1,2,2) is 120∘120^\circ.

  1. Find the constant of proportionality.

  2. Find the rate of change of TT at (1,2,2)(1,2,2) in the direction towards the point (2,1,3)(2,1,3).

  3. Show that at most points in the ball, the direction of greatest increase is towards the origin.

Answer

(a) 360360 (b) −4033≈−7.70-\frac{40}{3\sqrt{3}}\approx-7.70 (c) ∇T(x,y,z)=−360xı^+yȷ^+zk^(x2+y2+z2)3/2\vnabla T(x,y,z)=-360\frac{x\hi+y\hj+z\hk}{{(x^2+y^2+z^2)}^{3/2}}

Full solution

(a) We are told that T(x,y,z)=k∣<x,y,z>∣=kx2+y2+z2T(x,y,z)=\frac{k}{|\llt x,y,z\rgt|} =\frac{k}{\sqrt{x^2+y^2+z^2}} for some constant kk and that

120=T(1,2,2)=k∣<1,2,2>∣  ⟹  k=120×1+22+22=360\begin{equation*} 120=T(1,2,2)=\frac{k}{|\llt 1,2,2\rgt|} \implies k=120\times\sqrt{1+2^2+2^2} = 360 \end{equation*}

(b) The (unit) direction from (1,2,2)(1,2,2) to (2,1,3)(2,1,3) is d=<2,1,3>−<1,2,2>∣<2,1,3>−<1,2,2>∣=<1,−1,1>∣<1,−1,1>∣=13<1,−1,1>\vd=\frac{\llt 2,1,3\rgt-\llt 1,2,2\rgt}{|\llt 2,1,3\rgt-\llt 1,2,2\rgt|} =\frac{\llt 1,-1,1\rgt}{|\llt 1,-1,1\rgt|}=\frac{1}{\sqrt{3}}\llt 1,-1,1\rgt. The desired rate of change of temperature is

DdT(1,2,2)=∇T(1,2,2)⋅d=−360xı^+yȷ^+zk^(x2+y2+z2)3/2∣<1,2,2>⋅d=−360<1,2,2>27⋅<1,−1,1>3=−4033≈−7.70\begin{align*} D_{\vd} T(1,2,2) &=\vnabla T(1,2,2)\cdot\vd =-360\frac{x\hi+y\hj+z\hk}{{(x^2+y^2+z^2)}^{3/2}}\Big|_{\llt 1,2,2\rgt}\cdot\vd \\ &=-360\frac{\llt 1,2,2\rgt}{27}\cdot\frac{\llt 1,-1,1\rgt}{\sqrt{3}} =-\frac{40}{3\sqrt{3}}\approx-7.70 \end{align*}

degrees per unit distance.

(c) At (x,y,z)(x,y,z), the direction of greatest increase is in the direction of the temperature gradient at (x,y,z)(x,y,z), which is ∇T(x,y,z)=−360xı^+yȷ^+zk^(x2+y2+z2)3/2\vnabla T(x,y,z) =-360\frac{x\hi+y\hj+z\hk}{{(x^2+y^2+z^2)}^{3/2}} and which points opposite to the radius vector. That is, it points towards the origin. This argument only fails at (x,y,z)=(0,0,0)(x,y,z)=(0,0,0), where the gradient, and indeed T(x,y,z)T(x,y,z), is not defined.

Q27Stage 2Past exam · M200 2000A

The depth of a lake in the xyxy-plane is equal to f(x,y)=32−x2−4x−4y2f(x, y) = 32-x^2-4x-4y^2 meters.

  1. Sketch the shoreline of the lake in the xyxy-plane.

Your calculus instructor is in the water at the point (−1,1)(-1, 1). Find a unit vector which indicates in which direction he should swim in order to:

  1. stay at a constant depth?

  2. increase his depth as rapidly as possible (i.e. be most likely to drown)?

Answer

(a)

Figure from prob_s2.7, line 1629

Figure from prob_s2.7, line 1629

(b) ±117<4,−1>\pm\frac{1}{\sqrt{17}}\llt 4,-1\rgt (c) −117<1,4>-\frac{1}{\sqrt{17}}\llt 1,4\rgt

Full solution

(a) The shoreline is f(x,y)=0f(x,y)=0 or x2+4x+4y2=32x^2+4x+4y^2=32 or (x+2)2+4y2=36(x+2)^2+4y^2=36, which is an ellipse centred on (−2,0)(-2,0) with semiaxes 66 in the xx-direction and 33 in the yy-direction.

Figure from prob_s2.7, line 1629

Figure from prob_s2.7, line 1629

(b,c) The gradient of ff at (−1,1)(-1,1) is

∇f(−1,1)=[(−2x−4) ı^−8y ȷ^](−1,1)=−2 ı^−8 ȷ^\begin{equation*} \vnabla f(-1,1)=\big[(-2x-4)\,\hi-8y\,\hj\big]_{(-1,1)}=-2\,\hi-8\,\hj \end{equation*}

To remain at constant depth, he should swim perpendicular to the depth gradient. So he should swim in direction ±117<4,−1>\pm\frac{1}{\sqrt{17}}\llt 4,-1\rgt. To increase his depth as rapidly as possible, he should swim in the direction of the depth gradient, which is −117<1,4>-\frac{1}{\sqrt{17}}\llt 1,4\rgt.

3Stage 3Application

Further than practice: several ideas at once, or an unfamiliar situation.

Q28Stage 3

The temperature T(x,y)T(x,y) at points of the xyxy-plane is given by T(x,y)=x2−2y2T(x,y)=x^2-2y^2.

  1. Draw a contour diagram for TT showing some isotherms (curves of constant temperature).

  2. In what direction should an ant at position (2,−1)(2,-1) move if it wishes to cool off as quickly as possible?

  3. If the ant moves in that direction at speed vv at what rate does its temperature decrease?

  4. What would the rate of decrease of temperature of the ant be if it moved from (2,−1)(2,-1) at speed vv in direction <−1,−2>\llt -1,-2\rgt?

  5. Along what curve through (2,−1)(2,-1) should the ant move to continue experiencing maximum rate of cooling?

Hint

Review §2.7 in the CLP-3 text.

(e) Suppose that the ant moves along the curve y=y(x)y=y(x). For the ant to always experience maximum rate of cooling (or maximum rate of heating), the tangent to this curve must be parallel to ∇T(x,y)\vnabla T(x,y) at every point of the curve. This gives a separable differential equation for the function y(x)y(x). Also, don't forget that (2,−1)(2,-1) must be on the curve.

Answer

(a) Here is a sketch which show the isotherms T=0, 1, −1T=0,\ 1,\ -1 as well as the branch of the T=2T=2 isotherm that contains the ant's location (2,−1)(2,-1).

Figure from prob_s2.7, line 1694

Figure from prob_s2.7, line 1694

(b) <−1,−1>/2\llt -1,-1\rgt/\sqrt{2} (c) 42 v4\sqrt{2}\,v (d) 125 v\frac{12}{\sqrt{5}}\,v (e) y=−4x2y=-\frac{4}{x^2}

Full solution

(a) The curve on which the temperature is T0T_0 is x2−2y2=T0x^2-2y^2=T_0. If T0=0T_0=0, this is the pair of straight lines y=±x2y=\pm\frac{x}{\sqrt{2}}. If T0>0T_0>0, it is a hyperbola on which x2=2y2+T0≥T0x^2=2y^2+T_0\ge T_0. If T0<0T_0<0, it is a hyperbola on which 2y2=x2−T0≥∣T0∣2y^2=x^2-T_0\ge |T_0|. Here is a sketch which show the isotherms T=0, 1, −1T=0,\ 1,\ -1 as well as the branch of the T=2T=2 isotherm that contains the ant's location (2,−1)(2,-1).

Figure from prob_s2.7, line 1694

Figure from prob_s2.7, line 1694

Note that the temperature gradient is ∇T(x,y)=<2x,−4y>\vnabla T(x,y)=\llt 2x,-4y\rgt. In particular, the temperature gradient at (2,−1)(2,-1) is ∇T(2,−1)=<4,4>\vnabla T(2,-1)=\llt 4,4\rgt.

(b) To achieve maximum rate of cooling, the ant should move in the direction opposite the temperature gradient at (2,−1)(2,-1). So the direction of maximum rate of cooling is

−<4,4>42=<−1,−1>2\begin{equation*} -\frac{\llt 4,4\rgt}{4\sqrt{2}} =\frac{\llt -1,-1\rgt}{\sqrt{2}} \end{equation*}

(c) If the ant moves in the direction of part (b), its rate of cooling per unit distance is ∣∇T(2,−1)∣=∣<4,4>∣=42|\vnabla T(2,-1)|=|\llt 4,4\rgt| = 4\sqrt{2}. It the ant is moving at speed vv, its rate of cooling per unit time is 42 v4\sqrt{2}\,v.

(d) If the ant moves from (2,−1)(2,-1) in direction <−1,−2>\llt -1,-2\rgt its temperature increases at the rate

D<−1,−2>5T(2,−1)=<4,4>⋅<−1,−2>5=−125\begin{equation*} D_{\frac{\llt -1,-2\rgt}{\sqrt{5}}} T(2,-1) = \llt 4,4\rgt\cdot \frac{\llt -1,-2\rgt}{\sqrt{5}} = -\frac{12}{\sqrt{5}} \end{equation*}

per unit distance. So, if the ant is moving at speed vv, its rate of decrease of temperature per unit time is 125 v\frac{12}{\sqrt{5}}\,v

(e) Suppose that the ant moves along the curve y=y(x)y=y(x). For the ant to always experience maximum rate of cooling (or maximum rate of heating), the tangent to this curve must be parallel to ∇T(x,y)\vnabla T(x,y) at every point of the curve. A tangent to the curve at (x,y)(x,y) is <1,dydx(x)>\llt 1,\diff{y}{x}(x)\rgt. This is parallel to ∇T(x,y)=<2x,−4y>\vnabla T(x,y)=\llt 2x,-4y\rgt when

dydx1=−4y2x  ⟹  dyy=−2dxx  ⟹  ln⁡y=−2ln⁡x+C  ⟹  y=C′x−2\begin{align*} \frac{\diff{y}{x}}{1}=\frac{-4y}{2x} \implies \frac{dy}{y}=-2\frac{dx}{x} \implies \ln y=-2\ln x+C \implies y=C'x^{-2} \end{align*}

To pass through (2,−1)(2,-1), we need C′=−4C'=-4, so y=−4x2y=-\frac{4}{x^2}.

Q29Stage 3Past exam · M200 2008A

Consider the function f(x,y,z)=x2+cos⁡(yz)f(x,y,z) = x^2 + \cos(yz).

  1. Give the direction in which ff is increasing the fastest at the point (1,0,π/2)(1, 0, \pi/2).

  2. Give an equation for the plane TT tangent to the surface

    S={ (x,y,z) ∣ f(x,y,z)=1 }\begin{equation*} S = \Set{(x,y,z)}{f(x,y,z) = 1} \end{equation*}

    at the point (1,0,π/2)(1, 0, \pi/2).

  3. Find the distance between TT and the point (0,1,0)(0, 1, 0).

  4. Find the angle between the plane TT and the plane

    P={ (x,y,z) ∣ x+z=0 }.\begin{equation*} P = \Set{(x,y,z)}{x + z = 0}. \end{equation*}
Answer

(a) ı^\hi (b) x=1x=1 (c) 11 (d) π4\frac{\pi}{4}

Full solution

The first order partial derivatives of ff, both at a general point (x,y,z)(x,y,z) and at the point (1,0,π/2)(1, 0, \pi/2), are

fx(x,y,z)=2xfx(1,0,π/2)=2fy(x,y,z)=−zsin⁡(yz)fy(1,0,π/2)=0fz(x,y,z)=−ysin⁡(yz)fz(1,0,π/2)=0\begin{alignat*}{3} f_x(x,y,z)&= 2x\qquad & f_x(1, 0, \pi/2)&= 2 \\ f_y(x,y,z)&= -z\sin(yz)\qquad & f_y(1, 0, \pi/2)&= 0 \\ f_z(x,y,z)&= -y\sin(yz)\qquad & f_z(1, 0, \pi/2)&= 0 \end{alignat*}

(a) The rate of increase of ff is largest in the direction of ∇f(1,0,π/2)=<2,0,0>\vnabla f(1, 0, \pi/2)=\llt 2,0,0\rgt. A unit vector in that direction is ı^\hi.

(b) The gradient vector ∇f(1,0,π/2)=<2,0,0>\vnabla f(1, 0, \pi/2)=\llt 2,0,0\rgt is a normal vector to the surface f=1f=1 at (1,0,π/2)(1, 0, \pi/2). So the specified tangent plane is

<2,0,0>⋅<x−1 , y−0 , z−π/2>=0orx=1\begin{align*} \llt 2,0,0\rgt \cdot \llt x-1\,,\,y-0\,,\,z-\pi/2\rgt=0\qquad\text{or}\qquad x=1 \end{align*}

(c) The vector from the point (0,1,0)(0,1,0) to the point (1,1,0)(1,1,0), on TT, is <1,0,0>\llt 1,0,0\rgt, which is perpendicular to TT. So (1,1,0)(1,1,0) is the point on TT nearest (0,1,0)(0,1,0) and the distance from (0,1,0)(0,1,0) to TT is ∣<1,0,0>∣=1|\llt 1,0,0\rgt|=1.

(d) The vector <1,0,1>\llt 1,0,1\rgt is perpendicular to the plane x+z=0x+z=0. So the angle between the planes TT and x+z=0x+z=0 is the same as the angle θ\theta between the vectors <1,0,0>\llt 1,0,0\rgt and <1,0,1>\llt 1,0,1\rgt, which obeys

∣<1,0,0>∣ ∣<1,0,1>∣ cos⁡θ=∣<1,0,0>⋅<1,0,1>∣=1  ⟹  cos⁡θ=12  ⟹  θ=π4\begin{align*} & |\llt 1,0,0\rgt| \ |\llt 1,0,1\rgt|\ \cos\theta =|\llt 1,0,0\rgt\cdot\llt 1,0,1\rgt| =1 \\ &\implies \cos\theta=\frac{1}{\sqrt{2}} \implies \theta=\frac{\pi}{4} \end{align*}
Q30Stage 3Past exam · M200 2015D

A function T(x,y,z)T(x,y,z) at P=(2,1,1)P = (2,1,1) is known to have T(P)=5T(P) = 5, Tx(P)=1T_x (P) = 1, Ty(P)=2T_y(P) = 2, and Tz(P)=3T_z(P) = 3.

  1. A bee starts flying at PP and flies along the unit vector pointing towards the point Q=(3,2,2)Q = (3,2,2). What is the rate of change of T(x,y,z)T(x,y,z) in this direction?

  2. Use the linear approximation of TT at the point PP to approximate T(1.9,1,1.2)T(1.9,1,1.2).

  3. Let S(x,y,z)=x+zS(x,y,z) = x + z. A bee starts flying at PP; along which unit vector direction should the bee fly so that the rate of change of T(x,y,z)T(x,y,z) and of S(x,y,z)S(x,y,z) are both zero in this direction?

Answer

(a) 232\sqrt{3} (b) 5.55.5 (c) ±<1,1,−1>3\pm\frac{\llt 1,1,-1\rgt}{\sqrt{3}}

Full solution

(a) We are being asked for the directional derivative of TT in the direction of the unit vector from P=(2,1,1)P=(2,1,1) to Q=(3,2,2)Q=(3,2,2), which is <1,1,1>3\frac{\llt 1,1,1\rgt}{\sqrt{3}}. That directional derivative is

∇T(P)⋅<1,1,1>3=<1,2,3>⋅<1,1,1>3=23\begin{align*} \vnabla T(P)\cdot \frac{\llt 1,1,1\rgt}{\sqrt{3}} =\llt 1,2,3\rgt \cdot \frac{\llt 1,1,1\rgt}{\sqrt{3}} =2\sqrt{3} \end{align*}

(b) The linear approximation to TT at PP is

T(2+Δx , 1+Δy , 1+Δz)≈T(P)+Tx(P) Δx+Ty(P) Δy+Tz(P) Δz=5+Δx+2 Δy+3 Δz\begin{align*} T(2+\De x\,,\,1+\De y\,,\,1+\De z) &\approx T(P) + T_x(P)\,\De x + T_y(P)\,\De y + T_z(P)\,\De z \\ &= 5 +\De x +2\,\De y +3\,\De z \end{align*}

Applying this with Δx=−0.1\De x = -0.1, Δy=0\De y = 0, Δz=0.2\De z=0.2 gives

T(1.9 , 1 , 1.2)≈5+(−0.1)+2 (0)+3 (0.2)=5.5\begin{align*} T(1.9\,,\,1\,,\,1.2) &\approx 5 +(-0.1) +2\,(0) +3\,(0.2) =5.5 \end{align*}

(c) For the rate of change of TT to be zero, the direction of motion must be perpendicular to ∇T(P)=<1,2,3>\vnabla T(P) = \llt 1,2,3\rgt. For the rate of change of SS to also be zero, the direction of motion must also be perpendicular to ∇S(P)=<1,0,1>\vnabla S(P) = \llt 1,0,1\rgt. The vector

<1,2,3>×<1,0,1>=det⁡[ı^ȷ^k^123101]=<2,2,−2>\begin{align*} \llt 1,2,3\rgt \times \llt 1,0,1\rgt &=\det\left[\begin{matrix} \hi & \hj & \hk\\ 1 & 2 & 3 \\ 1 & 0 & 1\end{matrix}\right] =\llt 2,2,-2\rgt \end{align*}

is perpendicular to both ∇T(P)\vnabla T(P) and ∇S(P)\vnabla S(P). So the desired unit vectors are ±<1,1,−1>3\pm\frac{\llt 1,1,-1\rgt}{\sqrt{3}}.

Q31Stage 3Past exam · M200 2016D

Consider the functions F(x,y,z)=z3+xy2+xzF(x,y,z) = z^3 +xy^2 +xz and G(x,y,z)=3x−y+4zG(x,y,z)=3x-y+4z. You are standing at the point P(0,1,2)P(0,1,2).

  1. You jump from PP to Q(0.1 , 0.9 , 1.8)Q(0.1\,,\,0.9\,,\,1.8). Use the linear approximation to determine approximately the amount by which FF changes.

  2. You jump from PP in the direction along which GG increases most rapidly. Will FF increase or decrease?

  3. You jump from PP in a direction <a , b , c>\llt a\,,\,b\,,\,c\rgt along which the rates of change of FF and GG are both zero. Give an example of such a direction (need not be a unit vector).

Answer

(a) −2.1-2.1 (b) FF increases. (c) Any nonzero constant times <4 , 8 , −1>\llt 4 \,,\, 8 \,,\, -1 \rgt.

Full solution

We are going to need the gradients of both FF and GG at (0,1,2)(0,1,2). So we compute

∂F∂x(x,y,z)=y2+z∂F∂y(x,y,z)=2xy∂F∂z(x,y,z)=3z2+x∂G∂x(x,y,z)=3∂G∂y(x,y,z)=−1∂G∂z(x,y,z)=4\begin{align*} \pdiff{F}{x}(x,y,z)&=y^2+z & \pdiff{F}{y}(x,y,z)&=2xy & \pdiff{F}{z}(x,y,z)&=3z^2+x \\ \pdiff{G}{x}(x,y,z)&=3 & \pdiff{G}{y}(x,y,z)&=-1 & \pdiff{G}{z}(x,y,z)&=4 \end{align*}

and then

∇F(0,1,2)=<3,0,12>∇G(0,1,2)=<3,−1,4>\begin{equation*} \vnabla F(0,1,2) = \llt 3,0,12 \rgt\qquad \vnabla G(0,1,2) = \llt 3,-1,4 \rgt \end{equation*}

(a) The linear approximation to FF at (0,1,2)(0,1,2) is

F(x,y,z)≈F(0,1,2)+Fx(0,1,2) x+Fy(0,1,2) (y−1)+Fz(0,1,2) (z−2)=8+3x+12(z−2)\begin{align*} F(x,y,z) &\approx F(0,1,2) + F_x(0,1,2)\,x + F_y(0,1,2)\,(y-1) + F_z(0,1,2)\,(z-2) \\ &=8+ 3 x + 12 (z-2) \end{align*}

In particular

F(0.1 , 0.9 , 1.8)−F(0,1,2)≈3(0.1)+12(−0.2)=−2.1\begin{align*} F(0.1\,,\,0.9\,,\,1.8) - F(0,1,2) &\approx 3(0.1) + 12(-0.2) =-2.1 \end{align*}

(b) The direction along which GG increases most rapidly at PP is ∇G(0,1,2)=<3,−1,4>\vnabla G(0,1,2) = \llt 3,-1,4 \rgt. The directional derivative of FF in that direction is

D<3,−1,4>26F(0,1,2)=∇F(0,1,2)⋅<3,−1,4>26=<3,0,12>⋅<3,−1,4>26>0\begin{align*} D_{\frac{\llt 3,-1,4\rgt}{\sqrt{26}}}F(0,1,2) = \vnabla F(0,1,2) \cdot \frac{\llt 3,-1,4\rgt}{\sqrt{26}} = \llt 3,0,12 \rgt\cdot \frac{\llt 3,-1,4\rgt}{\sqrt{26}} > 0 \end{align*}

So FF increases.

(c) For the rate of change of FF to be zero, <a , b , c>\llt a\,,\,b\,,\,c\rgt must be perpendicular to ∇F(0,1,2)=<3,0,12>\vnabla F(0,1,2) = \llt 3,0,12 \rgt.

For the rate of change of GG to be zero, <a , b , c>\llt a\,,\,b\,,\,c\rgt must be perpendicular to ∇G(0,1,2)=<3,−1,4>\vnabla G(0,1,2) = \llt 3,-1,4 \rgt.

So any nonzero constant times

det⁡[ı^ȷ^k^30123−14]=<12 , 24 , −3>=3<4 , 8 , −1>\begin{align*} \det\left[\begin{matrix} \hi & \hj & \hk \\ 3 & 0 & 12 \\ 3 & -1 & 4 \end{matrix}\right] =\llt 12 \,,\, 24 \,,\, -3 \rgt =3 \llt 4 \,,\, 8 \,,\, -1 \rgt \end{align*}

is an allowed direction.

Q32Stage 3Past exam · M200 2004A

A meteor strikes the ground in the heartland of Canada. Using satellite photographs, a model

z=f(x,y)=−100x2+2x+4y2+11\begin{equation*} z=f(x,y)=-\frac{100}{x^2+2x+4y^2+11} \end{equation*}

of the resulting crater is made and a plan is drawn up to convert the site into a tourist attraction. A car park is to be built at (4,5)(4,5) and a hiking trail is to be made. The trail is to start at the car park and take the steepest route to the bottom of the crater.

  1. Sketch a map of the proposed site clearly marking the car park, a few level curves for the function ff and the trail.

  2. In which direction does the trail leave the car park?

Answer

(a)

Figure from prob_s2.7, line 2050

Figure from prob_s2.7, line 2050

(b) −<1,4>-\llt 1,4\rgt

Full solution

(a) Since

z=−100x2+2x+4y2+11=−100(x+1)2+4y2+10\begin{equation*} z=-\frac{100}{x^2+2x+4y^2+11}=-\frac{100}{(x+1)^2+4y^2+10} \end{equation*}

the bottom of the crater is at x=−1x=-1, y=0y=0 (where the denominator is a minimum) and the contours (level curves) are ellipses having equations (x+1)2+4y2=C(x+1)^2+4y^2=C. In the sketch below, the filled dot represents the bottom of the crater and the open dot represents the car park. The contours sketched are (from inside out) z=−7.5,−5,−2.5,−1z=-7.5, -5, -2.5, -1. Note that the trail crosses the contour lines at right angles.

Figure from prob_s2.7, line 2050

Figure from prob_s2.7, line 2050

(b) The trail is to be parallel to

∇z=100(x2+2x+4y2+11)2(2x+2,8y)\vnabla z = \frac{100}{{(x^2+2x+4y^2+11)}^2}(2x+2,8y)

At the car park ∇z(4,5)∥<10,40>∥<1,4>\vnabla z(4,5)\parallel \llt 10, 40\rgt\parallel \llt 1,4\rgt. To move towards the bottom of the crater, we should leave in the direction
−<1,4>-\llt 1,4\rgt.

Q33Stage 3Past exam · M200 2002D

You are standing at a lone palm tree in the middle of the Exponential Desert. The height of the sand dunes around you is given in meters by

h(x,y)=100e−(x2+2y2)h(x,y)=100 e^{-(x^2+2y^2)}

where xx represents the number of meters east of the palm tree (west if xx is negative) and yy represents the number of meters north of the palm tree (south if yy is negative).

  1. Suppose that you walk 33 meters east and 2 meters north. At your new location, (3,2)(3,2), in what direction is the sand dune sloping most steeply downward?

  2. If you walk north from the location described in part (a), what is the instantaneous rate of change of height of the sand dune?

  3. If you are standing at (3,2)(3,2) in what direction should you walk to ensure that you remain at the same height?

  4. Find the equation of the curve through (3,2)(3,2) that you should move along in order that you are always pointing in a steepest descent direction at each point of this curve.

Answer

(a) any positive multiple of <3,4>\llt 3,4\rgt (b) −800e−17-800 e^{-17} (c) ±(45,−35)\pm\big(\frac{4}{5},-\frac{3}{5}\big) (d) y=29x2y=\frac{2}{9}x^2

Full solution

We have

∇h(x,y)=−200e−(x2+2y2)<x,2y> and, in particular, ∇h(3,2)=−200e−17<3,4>\begin{equation*} \vnabla h(x,y)= -200 e^{-(x^2+2y^2)}\llt x,2y\rgt\text{ and, in particular, } \vnabla h(3,2)= -200 e^{-17}\llt 3,4\rgt \end{equation*}

(a) At (3,2)(3,2) the dune slopes downward the most steeply in the direction opposite ∇h(3,2)\vnabla h(3,2), which is (any positive multiple of) <3,4>\llt 3,4\rgt.

(b) The rate is Dȷ^h(3,2)=∇h(3,2)⋅ȷ^=−800e−17D_{\hj} h(3,2)=\vnabla h(3,2)\cdot\hj=-800 e^{-17}.

(c) To remain at the same height, you should walk perpendicular to ∇h(3,2)\vnabla h(3,2). So you should walk in one of the directions ±(45,−35)\pm\big(\frac{4}{5},-\frac{3}{5}\big).

(d) Suppose that you are walking along a steepest descent curve. Then the direction from (x,y)(x,y) to (x+dx,y+dy)(x+\dee{x}, y+\dee{y}), with (dx,dy)(\dee{x},\dee{y}) infinitesmal, must be opposite to ∇h(x,y)=−200e−(x2+2y2)(x,2y)\vnabla h(x,y)= -200 e^{-(x^2+2y^2)}(x,2y). Thus (dx,dy)(\dee{x},\dee{y}) must be parallel to (x,2y)(x,2y) so that the slope

dydx=2yx  ⟹  dyy=2dxx  ⟹  ln⁡y=2ln⁡x+C\begin{align*} \diff{y}{x}=\frac{2y}{x} \implies \frac{\dee{y}}{y}=2\frac{\dee{x}}{x} \implies \ln y=2\ln x+C \end{align*}

We must choose CC to obey ln⁡2=2ln⁡3+C\ln 2=2\ln 3+C in order to pass through the point (3,2)(3,2). Thus C=ln⁡29C=\ln\frac{2}{9} and the curve is ln⁡y=2ln⁡x+ln⁡29\ln y=2\ln x+\ln\frac{2}{9} or y=29x2y=\frac{2}{9}x^2.

Q34Stage 3Past exam · M200 2002A

Let f(x,y)f(x,y) be a differentiable function with f(1,2)=7f(1,2)=7. Let

u=35 ı^+45 ȷ^,v=35 ı^−45 ȷ^\begin{equation*} \vu=\frac{3}{5}\,\hi+\frac{4}{5}\,\hj,\qquad \vv=\frac{3}{5}\,\hi-\frac{4}{5}\,\hj \end{equation*}

be unit vectors. Suppose it is known that the directional derivatives Duf(1,2)D_\vu f(1,2) and Dvf(1,2)D_\vv f(1,2) are equal to 1010 and 22 respectively.

  1. Show that the gradient vector ∇f\vnabla f at (1,2)(1,2) is 10ı^+5ȷ^10\hi+5\hj.

  2. Determine the rate of change of ff at (1,2)(1,2) in the direction of the vector ı^+2ȷ^\hi+2\hj.

  3. Using the tangent plane approximation, estimate the value of f(1.01,2.05)f(1.01,2.05).

Answer

(a) See the solution. (b) 45≈8.9444\sqrt{5}\approx 8.944 (c) 7.357.35

Full solution

(a) Denote ∇f(1,2)=<a,b>\vnabla f(1,2)= \llt a,b\rgt. We are told that

Duf(1,2)=u⋅(a,b)=35a+45b=10Dvf(1,2)=v⋅(a,b)=35a−45b=2\begin{alignat*}{3} D_\vu f(1,2)&=\vu\cdot(a,b)&&=\frac{3}{5}a+\frac{4}{5}b&&=10 \\ D_\vv f(1,2)&=\vv\cdot(a,b)&&=\frac{3}{5}a-\frac{4}{5}b&&=2 \end{alignat*}

Adding these two equations gives 65a=12\frac{6}{5}a=12, which forces a=10a=10, and subtracting the two equations gives 85b=8\frac{8}{5}b=8, which forces b=5b=5, as desired.

(b) The rate of change of ff at (1,2)(1,2) in the direction of the vector ı^+2ȷ^\hi+2\hj is

ı^+2ȷ^∣ı^+2ȷ^∣⋅∇f(1,2)=15<1,2>⋅<10,5>=45≈8.944\begin{align*} \frac{\hi+2\hj}{|\hi+2\hj|}\cdot \vnabla f(1,2) =\frac{1}{\sqrt{5}}\llt 1,2\rgt\cdot\llt 10,5\rgt =4\sqrt{5}\approx 8.944 \end{align*}

(c) Applying (2.6.1 in the CLP3 text, which is

f(x0+Δx , y0+Δy)≈f(x0 , y0)+∂f∂x(x0 , y0) Δx+∂f∂y(x0 , y0) Δy\begin{align*} f\big(x_0+\De x\,,\,y_0+\De y\big) &\approx f\big(x_0\,,\,y_0\big) + \pdiff{f}{x}\big(x_0\,,\,y_0\big)\,\De x + \pdiff{f}{y}\big(x_0\,,\,y_0\big)\,\De y \end{align*}

with x0=1x_0=1, Δx=0.01\De x=0.01, y0=2y_0=2, and Δy=0.05\De y=0.05, gives

f(1.01,2.05)≈f(1,2)+fx(1,2)×(1.01−1)+fy(1,2)×(2.05−2)=7+10×0.01+5×0.05=7.35\begin{align*} f(1.01,2.05) &\approx f(1,2)+f_x(1,2)\times(1.01-1)+f_y(1,2)\times(2.05-2) \\ &=7+10\times0.01+5\times0.05 \\ &=7.35 \end{align*}

My list

nothing marked yet

Loading…

Open the whole list →

Your tutor can open this list with you. It follows your account, so it is there on whichever device you study on.

From the UBC Math 200 open textbook project, © Joel Feldman, Andrew Rechnitzer, Elyse Yeager and others. Licensed CC BY-NC-SA 4.0; this HTML adaptation is released under the same licence. Full attribution.