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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
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

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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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

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
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
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
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
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
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
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

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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.