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Partial Derivatives

2.6 Linear Approximations and Error

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

1Stage 1Conceptual

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

Q1Stage 1

Let x0x_0 and y0y_0 be constants and let mm and nn be integers. If m<0m<0 assume that x0≠0x_0\ne 0, and if n<0n<0 assume that y0≠0y_0\ne 0. Define P(x,y)=xmynP(x,y) = x^m y^n.

  1. Find the linear approximation to P(x0+Δx,y0+Δy)P(x_0+\De x,y_0+\De y).

  2. Denote by

    P%=100∣P(x0+Δx,y0+Δy)−P(x0,y0)P(x0,y0)∣x%=100∣Δxx0∣y%=100∣Δyy0∣\begin{equation*} P_\% = 100\left|\frac{P(x_0+\De x,y_0+\De y)-P(x_0,y_0)}{P(x_0,y_0)}\right| \qquad x_\% = 100\left|\frac{\De x}{x_0}\right| \qquad y_\% = 100\left|\frac{\De y}{y_0}\right| \end{equation*}

    the percentage errors in PP, xx and yy respectively. Use the linear approximation to find an (approximate) upper bound on P%P_\% in terms of mm, nn, x%x_\% and y%y_\%.

Hint

Review Example 2.6.11 in the CLP-3 text. Be careful when taking absolute values.

Answer

(a) P(x0+Δx,y0+Δy)≈P(x0,y0)+mx0m−1y0n Δx+nx0my0n−1 ΔyP(x_0+\De x,y_0+\De y) \approx P(x_0,y_0) + mx_0^{m-1}y_0^n\,\De x + nx_0^m y_0^{n-1}\,\De y

(b) P%≤∣m∣ x%+∣n∣ y%P_\% \le |m|\,x_\% + |n|\,y_\%

Full solution

(a) The first order partial derivatives of P(x,y)P(x,y) at x=x0x=x_0 and y=y0y=y_0 are

Px(x0,y0)=mx0m−1y0nPy(x0,y0)=nx0my0n−1\begin{equation*} P_x(x_0,y_0) = m x_0^{m-1} y_0^n\qquad P_y(x_0,y_0) = n x_0^m y_0^{n-1} \end{equation*}

So, by (2.6.1) in the CLP-3 text, the linear approximation is

P(x0+Δx,y0+Δy)≈P(x0,y0)+Px(x0,y0) Δx+Py(x0,y0) Δy≈P(x0,y0)+mx0m−1y0n Δx+nx0my0n−1 Δy\begin{align*} P(x_0+\De x,y_0+\De y) &\approx P(x_0,y_0) + P_x(x_0,y_0)\,\De x + P_y(x_0,y_0)\,\De y\\ &\approx P(x_0,y_0) + mx_0^{m-1}y_0^n\,\De x + nx_0^m y_0^{n-1}\,\De y \end{align*}

(b) By part (a)

P(x0+Δx,y0+Δy)−P(x0,y0)P(x0,y0)≈mx0m−1y0n Δx+nx0my0n−1 Δyx0my0n=mΔxx0+nΔyy0\begin{align*} \frac{P(x_0+\De x,y_0+\De y)-P(x_0,y_0)}{P(x_0,y_0)} &\approx \frac{mx_0^{m-1}y_0^n\,\De x + nx_0^m y_0^{n-1}\,\De y}{x_0^my_0^n} =m\frac{\De x}{x_0} + n\frac{\De y}{y_0} \end{align*}

Hence

P%≈100∣mΔxx0+nΔyy0∣≤∣m∣100∣Δxx0∣+∣n∣100∣Δyy0∣≤∣m∣x%+∣n∣y%\begin{align*} P_\% &\approx 100\left|m\frac{\De x}{x_0} + n\frac{\De y}{y_0}\right| \\ &\le |m| 100\left|\frac{\De x}{x_0}\right| + |n| 100\left|\frac{\De y}{y_0}\right| \\ &\le |m| x_\% + |n| y_\% \end{align*}

Warning. The answer m x%+n y%m\,x_\% +n\,y_\%, without absolute values on mm and nn, can be seriously wrong. As an example, suppose that m=1m=1, n=−1n=-1, x0=y0=1x_0=y_0=1, Δx=0.05\De x = 0.05 and Δy=−0.05\De y=-0.05. Then

P%≈100∣mΔxx0+nΔyy0∣=100∣(1)0.051+(−1)−0.051∣=10%\begin{align*} P_\% &\approx 100\left|m\frac{\De x}{x_0} + n\frac{\De y}{y_0}\right| \\ &= 100\left|(1)\frac{0.05}{1} + (-1)\frac{-0.05}{1}\right| \\ & = 10\% \end{align*}

while

m x%+n y%=m 100∣Δxx0∣+n 100∣Δyy0∣=(1)100∣0.051∣+(−1)100∣−0.051∣=0\begin{align*} m\,x_\% +n\,y_\% &= m\, 100\left|\frac{\De x}{x_0}\right| + n\, 100\left|\frac{\De y}{y_0}\right| \\ &= (1) 100\left|\frac{0.05}{1}\right| + (-1) 100\left|\frac{-0.05}{1}\right| \\ &= 0 \end{align*}

The point is that mm and nn being of opposite sign does not guarantee that there is a cancelation between the two terms of mΔxx0+nΔyy0m\frac{\De x}{x_0} + n\frac{\De y}{y_0}, because Δxx0\frac{\De x}{x_0} and Δyy0\frac{\De y}{y_0} can also be of opposite sign.

Q2Stage 1

Consider the following work.

We compute, approximately, the yy-coordinate of the point whose polar coordinates are r=0.9r=0.9 and θ=2∘\theta=2^\circ. In general, the yy-coordinate of the point whose polar coordinates are rr and θ\theta is Y(r,θ)=rsin⁡θY(r,\theta) = r\sin\theta. The partial derivatives

Yr(r,θ)=sin⁡θYθ(r,θ)=rcos⁡θ\begin{equation*} Y_r(r,\theta)=\sin\theta\qquad Y_\theta(r,\theta) = r\cos\theta \end{equation*}

So the linear approximation to Y(r0+Δr,θ0+Δθ)Y(r_0+\De r,\theta_0+\De\theta) with r0=1r_0=1 and θ0=0\theta_0=0 is

Y(1+Δr,0+Δθ)≈Y(1,0)+Yr(1,0) Δr+Yθ(1,0) Δθ=0 + (0) Δr + (1)Δθ\begin{align*} Y(1+\De r,0+\De\theta) &\approx Y(1,0) + Y_r(1,0)\,\De r + Y_\theta(1,0)\,\De\theta \\ &= 0\ +\ (0)\,\De r\ +\ (1)\De\theta \end{align*}

Applying this with Δr=−0.1\De r=-0.1 and Δθ=2\De\theta=2 gives the (approximate) yy-coordinate

Y(0.9,2)=Y(1−0.1 , 0+2)≈0 + (0) (−0.1) + (1)(2)=2\begin{align*} Y(0.9,2) = Y(1-0.1\,,\, 0+2)\approx 0\ +\ (0)\,(-0.1)\ +\ (1)(2) =2 \end{align*}

This conclusion is ridiculous. We're saying that the yy-coordinate is more than twice the distance from the point to the origin. What was the mistake?

Hint

Units!

Answer

We used that ddθsin⁡θ=cos⁡θ\diff{}{\theta}\sin\theta = \cos\theta. That is true only if θ\theta is given in radians, not degrees.

Full solution

We used that ddθsin⁡θ=cos⁡θ\diff{}{\theta}\sin\theta = \cos\theta. That is true only if θ\theta is given in radians, not degrees. (See Lemma 2.8.3 and Warning 3.4.23 in the CLP-1 text.) So we have to convert 2∘2^\circ to radians, which is 2×π180=π902\times\frac{\pi}{180}=\frac{\pi}{90}. The correct computation is

Y(0.9,π90)=Y(1−0.1 , 0+π90)≈0 + (0) (−0.1) + (1)(π90)=π90≈0.035\begin{align*} Y\big(0.9,\tfrac{\pi}{90}\big) = Y\big(1-0.1\,,\, 0+\tfrac{\pi}{90}\big) \approx 0\ +\ (0)\,(-0.1)\ +\ (1)\big(\tfrac{\pi}{90}\big) =\tfrac{\pi}{90} \approx 0.035 \end{align*}

Just out of general interest, 0.9sin⁡π90=0.03140.9\sin\tfrac{\pi}{90} =0.0314 to four decimal places.

2Stage 2Procedural

Practising the skill itself, until applying it is automatic.

Q3Stage 2

Find an approximate value for f(x,y)=sin⁡(πxy+ln⁡y)f(x,y)=\sin(\pi xy+\ln y) at (0.01,1.05)(0.01,1.05) without using a calculator or computer.

Answer

0.01 π+0.05≈0.08140.01\,\pi + 0.05 \approx 0.0814

Full solution

Apply the linear approximation f(0.01,1.05)≈f(0,1)+fx(0,1)(0.01)+fy(0,1)(0.05)f(0.01,1.05)\approx f(0,1)+f_x(0,1)(0.01)+f_y(0,1)(0.05), with

f(x,y)=sin⁡(πxy+ln⁡y)f(0,1)=sin⁡0=0fx(x,y)=πycos⁡(πxy+ln⁡y)fx(0,1)=πcos⁡0=πfy(x,y)=(πx+1y)cos⁡(πxy+ln⁡y)fy(0,1)=cos⁡0=1\begin{align*} f(x,y)&=\sin(\pi xy+\ln y) & f(0,1)&=\sin 0=0\\ f_x(x,y)&=\pi y\cos(\pi xy+\ln y) & f_x(0,1)&=\pi\cos 0=\pi\\ f_y(x,y)&=\left(\pi x+\frac{1}{y}\right)\cos(\pi xy+\ln y) & f_y(0,1)&=\cos 0=1 \end{align*}

This gives

f(0.01,1.05)≈f(0,1)+fx(0,1)(0.01)+fy(0,1)(0.05)=0+π(0.01)+1(0.05)=0.01 π+0.05≈0.0814\begin{align*} f(0.01,1.05)&\approx f(0,1)+f_x(0,1)(0.01)+f_y(0,1)(0.05) =0+\pi(0.01)+1(0.05) \\ &=0.01\,\pi + 0.05 \approx 0.0814 \end{align*}
Q4Stage 2Past exam · M200 2009A

Let f(x,y)=x2yx4+2y2\displaystyle f(x,y)=\frac{x^2y}{x^4+2y^2}. Find an approximate value for f(−0.9 , 1.1)f (-0.9\,,\, 1.1) without using a calculator or computer.

Answer

0.30.3

Full solution

We are going to need the first order derivatives of f(x,y)f(x,y) at (x,y)=(−1,1)(x,y)=(-1,1). So we find them first.

fx(x,y)=2xyx4+2y2−x2y(4x3)(x4+2y2)2fx(−1,1)=−23+432=−29fy(x,y)=x2x4+2y2−x2y(4y)(x4+2y2)2fy(−1,1)=13−432=−19\begin{alignat*}{3} f_x(x,y)&=\frac{2xy}{x^4+2y^2}-\frac{x^2y(4x^3)}{{(x^4+2y^2)}^2}\qquad & f_x(-1,1)&=-\frac{2}{3} +\frac{4}{3^2}=-\frac{2}{9} \\ f_y(x,y)&=\frac{x^2}{x^4+2y^2}-\frac{x^2y(4y)}{{(x^4+2y^2)}^2}\qquad & f_y(-1,1)&=\frac{1}{3} -\frac{4}{3^2}=-\frac{1}{9} \end{alignat*}

The linear approximation to f(x,y)f(x,y) about (−1,1)(-1,1) is

f(x,y)≈=f(−1,1)+fx(−1,1) (x+1)+fy(−1,1) (y−1)=13−29 (x+1)−19 (y−1)\begin{align*} f(x,y)\approx=f(-1,1) + f_x(-1,1)\,(x+1) + f_y(-1,1)\,(y-1) =\frac{1}{3} -\frac{2}{9}\,(x+1) -\frac{1}{9}\,(y-1) \end{align*}

In particular

f(−0.9,1.1)≈13−29 (0.1)−19 (0.1)=2790=0.3\begin{align*} f(-0.9,1.1)\approx \frac{1}{3} -\frac{2}{9}\,(0.1) -\frac{1}{9}\,(0.1) =\frac{27}{90}=0.3 \end{align*}
Q5Stage 2

Four numbers, each at least zero and each at most 50, are rounded to the first decimal place and then multiplied together. Estimate the maximum possible error in the computed product.

Hint

Let the four numbers be x1x_1, x2x_2, x3x_3 and x4x_4. Let the four rounded numbers be x1+ε1x_1+\veps_1, x2+ε2x_2+\veps_2, x3+ε3x_3+\veps_3 and x4+ε4x_4+\veps_4. If P(x1,x2,x3,x4)=x1x2x3x4P(x_1,x_2,x_3,x_4)=x_1x_2x_3x_4, then the error in the product introduced by rounding is ∣P(x1+ε1,x2+ε2,x3+ε3,x4+ε4)−P(x1,x2,x3,x4)∣\big|P(x_1+\veps_1,x_2+\veps_2,x_3+\veps_3,x_4+\veps_4) -P(x_1,x_2,x_3,x_4)\big|.

Answer

2500025000

Full solution

Let the four numbers be x1x_1, x2x_2, x3x_3 and x4x_4. Let the four rounded numbers be x1+ε1x_1+\veps_1, x2+ε2x_2+\veps_2, x3+ε3x_3+\veps_3 and x4+ε4x_4+\veps_4. Then 0≤x1,x2,x3,x4≤500\le x_1,x_2,x_3,x_4\le 50 and ∣ε1∣,∣ε2∣,∣ε3∣,∣ε4∣≤0.05|\veps_1|,|\veps_2|,|\veps_3|,|\veps_4|\le 0.05. If P(x1,x2,x3,x4)=x1x2x3x4P(x_1,x_2,x_3,x_4)=x_1x_2x_3x_4, then the error in the product introduced by rounding is, using the four variable variant of the linear approximation (2.6.2) of the CLP-3 text,

∣P(x1+ε1,x2+ε2,x3+ε3,x4+ε4)−P(x1,x2,x3,x4)∣≈∣∂P∂x1(x1,x2,x3,x4)ε1+∂P∂x2(x1,x2,x3,x4)ε2+∂P∂x3(x1,x2,x3,x4)ε3+∂P∂x4(x1,x2,x3,x4)ε4∣=∣x2x3x4ε1+x1x3x4ε2+x1x2x4ε3+x1x2x3ε4∣≤4×50×50×50×0.05=25000\begin{align*} &\big|P(x_1+\veps_1,x_2+\veps_2,x_3+\veps_3,x_4+\veps_4)-P(x_1,x_2,x_3,x_4)\big| \\ &\approx\big|\pdiff{P}{x_1}(x_1,x_2,x_3,x_4)\veps_1 +\pdiff{P}{x_2}(x_1,x_2,x_3,x_4)\veps_2 +\pdiff{P}{x_3}(x_1,x_2,x_3,x_4)\veps_3 +\pdiff{P}{x_4}(x_1,x_2,x_3,x_4)\veps_4\big|\\ &=\big|x_2x_3x_4\veps_1 +x_1x_3x_4\veps_2 +x_1x_2x_4\veps_3 +x_1x_2x_3\veps_4\big|\\ &\le 4\times 50\times 50\times 50\times 0.05 =25000 \end{align*}
Q6Stage 2Past exam · M200 2005D

One side of a right triangle is measured to be 33 with a maximum possible error of ±0.1\pm 0.1, and the other side is measured to be 44 with a maximum possible error of ±0.2\pm 0.2. Use the linear approximation to estimate the maximum possible error in calculating the length of the hypotenuse of the right triangle.

Hint

Use Pythagoras to express the length of the hypotenuse in terms of the lengths of the other two sides.

Answer

0.220.22

Full solution

Denote by xx and yy the lengths of sides with x=3±0.1x=3\pm 0.1 and y=4±0.2y=4\pm 0.2. Then the length of the hypotenuse is f(x,y)=x2+y2f(x,y)=\sqrt{x^2+y^2}. Note that

f(x,y)=x2+y2f(3,4)=5fx(x,y)=xx2+y2fx(3,4)=35fy(x,y)=yx2+y2fy(3,4)=45\begin{alignat*}{5} f(x,y)&=\sqrt{x^2+y^2} & f(3,4)&=5 \\ f_x(x,y)&=\frac{x}{\sqrt{x^2+y^2}}\qquad & f_x(3,4)&=\frac{3}{5} \\ f_y(x,y)&=\frac{y}{\sqrt{x^2+y^2}} & f_y(3,4)&=\frac{4}{5} \end{alignat*}

By the linear approximation

f(x,y)≈f(3,4)+fx(3,4) (x−3)+fy(3,4) (y−4)=5+35 (x−3)+45 (y−4)\begin{align*} f(x,y)\approx f(3,4) + f_x(3,4)\,(x-3) + f_y(3,4)\,(y-4) =5 +\frac{3}{5}\,(x-3) +\frac{4}{5}\,(y-4) \end{align*}

So the approximate maximum error in calculating the length of the hypotenuse is

35 (0.1)+45 (0.2)=1.15=0.22\begin{align*} \frac{3}{5}\,(0.1) +\frac{4}{5}\,(0.2) =\frac{1.1}{5} = 0.22 \end{align*}
Q7Stage 2Past exam · M200 2006A

If two resistors of resistance R1R_1 and R2R_2 are wired in parallel, then the resulting resistance R satisfies the equation 1R=1R1+1R2\frac{1}{R} =\frac{1}{R_1}+\frac{1}{R_2}. Use the linear approximation to estimate the change in RR if R1R_1 decreases from 22 to 1.91.9 ohms and R2R_2 increases from 8 to 8.1 ohms.

Answer

−0.06-0.06

Full solution

The function R(R1,R2)R(R_1,R_2) is defined implictly by

1R(R1,R2)=1R1+1R2\begin{equation*} \frac{1}{R(R_1,R_2)} =\frac{1}{R_1}+\frac{1}{R_2} \tag{$*$}\end{equation*}

In particular

1R(2,8)=12+18=58  ⟹  R(2,8)=85\begin{equation*} \frac{1}{R(2,8)} = \frac{1}{2}+\frac{1}{8} = \frac{5}{8} \implies R(2,8) = \frac{8}{5} \end{equation*}

We wish to use the linear approximation

R(R1,R2)≈R(2,8)+∂R∂R1(2,8) (R1−2)+∂R∂R2(2,8) (R2−8)\begin{equation*} R(R_1,R_2) \approx R(2,8) + \pdiff{R}{R_1}(2,8)\,(R_1-2) + \pdiff{R}{R_2}(2,8)\,(R_2-8) \end{equation*}

To do so, we need the partial derivatives ∂R∂R1(2,8)\pdiff{R}{R_1}(2,8) and ∂R∂R2(2,8)\pdiff{R}{R_2}(2,8). To find them, we differentiate (∗)(*) with respect to R1R_1 and R2R_2:

−1R(R1,R2)2∂R∂R1(R1,R2)=−1R12−1R(R1,R2)2∂R∂R2(R1,R2)=−1R22\begin{align*} -\frac{1}{R(R_1,R_2)^2}\pdiff{R}{R_1}(R_1,R_2) &= -\frac{1}{R_1^2} \\ -\frac{1}{R(R_1,R_2)^2}\pdiff{R}{R_2}(R_1,R_2) &= -\frac{1}{R_2^2} \\ \end{align*}

Setting R1=2R_1=2 and R2=8R_2=8 gives

−1(8/5)2∂R∂R1(2,8)=−14  ⟹  ∂R∂R1(2,8)=1625−1(8/5)2∂R∂R2(2,8)=−164  ⟹  ∂R∂R2(2,8)=125\begin{alignat*}{5} -\frac{1}{(8/5)^2}\pdiff{R}{R_1}(2,8) &= -\frac{1}{4} &\quad\implies\quad & &\pdiff{R}{R_1}(2,8) &=\frac{16}{25} \\ -\frac{1}{(8/5)^2}\pdiff{R}{R_2}(2,8) &= -\frac{1}{64} &\quad\implies\quad & &\pdiff{R}{R_2}(2,8) &=\frac{1}{25} \end{alignat*}

So the specified change in RR is

R(1.9,8.1)−R(2,8)≈1625(−0.1)+125(0.1)=−15250=−0.06\begin{align*} R(1.9,8.1)-R(2,8) \approx \frac{16}{25} (-0.1) + \frac{1}{25}(0.1) =-\frac{15}{250} = -0.06 \end{align*}
Q8Stage 2

The total resistance RR of three resistors, R1R_1, R2R_2, R3R_3, connected in parallel is determined by

1R=1R1+1R2+1R3\begin{equation*} \frac{1}{R}=\frac{1}{R_1}+\frac{1}{R_2}+\frac{1}{R_3} \end{equation*}

If the resistances, measured in Ohms, are R1=25ΩR_1=25\Om, R2=40ΩR_2=40\Om and R3=50ΩR_3=50\Om, with a possible error of 0.5% in each case, estimate the maximum error in the calculated value of RR.

Hint

Review the relationship between absolute error and percentage error given in Definition 2.6.6 in the CLP-3 text.

Answer

117≈0.059\frac{1}{17}\approx 0.059

Full solution

First, we compute the values of the partial derivatives of R(R1,R2,R3)R(R_1,R_2,R_3) at the measured values of R1R_1, R2R_2, R3R_3. Applying ∂∂Ri\pdiff{}{R_i}, with i=1,2,3i=1,2,3 to both sides of the defining equation

1R(R1,R2,R3)=1R1+1R2+1R3\begin{equation*} \frac{1}{R(R_1,R_2,R_3)}=\frac{1}{R_1}+\frac{1}{R_2}+\frac{1}{R_3} \end{equation*}

for R(R1,R2,R3)R(R_1,R_2,R_3) gives

−1R(R1,R2,R3)2 ∂R∂R1(R1,R2,R3)=−1R12−1R(R1,R2,R3)2 ∂R∂R2(R1,R2,R3)=−1R22−1R(R1,R2,R3)2 ∂R∂R3(R1,R2,R3)=−1R32\begin{align*} -\frac{1}{R(R_1,R_2,R_3)^2}\ \pdiff{R}{R_1}(R_1,R_2,R_3) &=-\frac{1}{R_1^2}\\ -\frac{1}{R(R_1,R_2,R_3)^2}\ \pdiff{R}{R_2}(R_1,R_2,R_3) &=-\frac{1}{R_2^2}\\ -\frac{1}{R(R_1,R_2,R_3)^2}\ \pdiff{R}{R_3}(R_1,R_2,R_3) &=-\frac{1}{R_3^2} \end{align*}

When R1=25ΩR_1=25\Om, R2=40ΩR_2=40\Om and R3=50ΩR_3=50\Om

1R(25,40,50)=125+140+150=8+5+4200  ⟹  R(25,40,50)=20017=11.765\begin{equation*} \frac{1}{R(25,40,50)}=\frac{1}{25}+\frac{1}{40}+\frac{1}{50} =\frac{8+5+4}{200}\implies R(25,40,50)=\frac{200}{17}=11.765 \end{equation*}

Substituting in these values of R1R_1, R2R_2, R3R_3 and RR,

∂R∂R1(25,40,50)=R(25,40,50)2252=64172=0.221∂R∂R2(25,40,50)=R(25,40,50)2402=25172=0.0865∂R∂R3(25,40,50)=R(25,40,50)2502=16172=0.0554\begin{alignat*}{5} \pdiff{R}{R_1}(25,40,50) &=\frac{R(25,40,50)^2}{25^2}&=\frac{64}{17^2}&=0.221\\ \pdiff{R}{R_2}(25,40,50) &=\frac{R(25,40,50)^2}{40^2}&=\frac{25}{17^2}&=0.0865\\ \pdiff{R}{R_3}(25,40,50) &=\frac{R(25,40,50)^2}{50^2}&=\frac{16}{17^2}&=0.0554 \end{alignat*}

If the absolute errors in measuring R1R_1, R2R_2 and R3R_3 are denoted ε1\veps_1, ε2\veps_2 and ε3\veps_3, respectively, then, , using the linear approximation (2.6.2) of the CLP-3 text, the corresponding error EE in RR is

E=R(25+ε1,40+ε2,50+ε3)−R(25,40,50)≈∂R∂R1(25,40,50)ε1+∂R∂R2(25,40,50)ε2+∂R∂R3(25,40,50)ε3\begin{align*} E &= R(25+\veps_1,40+\veps_2,50+\veps_3)-R(25,40,50) \\ &\approx \pdiff{R}{R_1}(25,40,50)\veps_1 +\pdiff{R}{R_2}(25,40,50)\veps_2 +\pdiff{R}{R_3}(25,40,50)\veps_3 \end{align*}

and obeys

∣E∣≤64172∣ε1∣+25172∣ε2∣+16172∣ε3∣or∣E∣≤0.221∣ε1∣+0.0865∣ε2∣+0.0554∣ε3∣\begin{align*} |E|&\le \frac{64}{17^2}|\veps_1|+\frac{25}{17^2}|\veps_2| +\frac{16}{17^2}|\veps_3| \\ \text{or}\quad|E|&\le 0.221|\veps_1|+0.0865|\veps_2|+0.0554|\veps_3| \end{align*}

We are told that the percentage error in each measurement is no more that 0.5%0.5\%. So

∣ε1∣≤0.510025=18=0.125∣ε2∣≤0.510040=15=0.2∣ε3∣≤0.510050=14=0.25\begin{equation*} |\veps_1|\le \frac{0.5}{100}25=\frac{1}{8}=0.125\qquad |\veps_2|\le \frac{0.5}{100}40=\frac{1}{5}=0.2\qquad |\veps_3|\le \frac{0.5}{100}50=\frac{1}{4}=0.25 \end{equation*}

so that

∣E∣≤8172+5172+4172=117or∣E∣≤0.221×0.125+0.0865×0.2+0.0554×0.25=0.059\begin{align*} |E|&\le \frac{8}{17^2}+\frac{5}{17^2}+\frac{4}{17^2} =\frac{1}{17}\\ \text{or}\quad|E|&\le 0.221\times0.125+0.0865\times0.2+0.0554\times0.25 =0.059 \end{align*}
Q9Stage 2

The specific gravity SS of an object is given by S=AA−WS=\frac{A}{A-W} where AA is the weight of the object in air and WW is the weight of the object in water. If A=20±.01A=20\pm .01 and W=12±.02W=12\pm.02 find the approximate percentage error in calculating SS from the given measurements.

Hint

Be very careful about signs. There is a trap hidden in this question. As an example of the trap, suppose you know that ∣ε1∣≤0.2|\veps_1|\le 0.2 and ∣ε2∣≤0.1|\veps_2|\le 0.1. It does not follow from this that ∣ε1−ε2∣≤0.2−0.1=0.1\big|\veps_1-\veps_2|\le 0.2 -0.1 =0.1. The reason is that it is possible to have ε1=0.2\veps_1=0.2 and ε2=−0.1\veps_2=-0.1 and then ε1−ε2=0.3\veps_1-\veps_2=0.3. The correct way to bound ∣ε1−ε2∣\big|\veps_1-\veps_2| is

∣ε1−ε2∣≤∣ε1∣+∣ε2∣≤0.2+0.1≤0.3\begin{equation*} \big|\veps_1-\veps_2| \le |\veps_1|+ |\veps_2|\le 0.2+0.1\le 0.3 \end{equation*}
Answer

1340%=0.325%\frac{13}{40}\%=0.325\%

Full solution

By the linear approximation

ΔS≈∂S∂A(20,12) ΔA+∂S∂W(20,12) ΔW\begin{align*} \De S\approx \pdiff{S}{A}(20,12)\,\De A +\pdiff{S}{W}(20,12)\,\De W \end{align*}

with S(A,W)=AA−W=1+WA−WS(A,W)=\frac{A}{A-W} =1+\frac{W}{A-W}. So

S(A,W)=AA−WS(20,12)=208=52SA(A,W)=−W(A−W)2SA(20,12)=−1282=−316SW(A,W)=A(A−W)2SW(20,12)=2082=516\begin{align*} S(A,W)&=\frac{A}{A-W} & S(20,12)&=\frac{20}{8}=\frac{5}{2}\\ S_A(A,W)&=-\frac{W}{(A-W)^2} & S_A(20,12)&=-\frac{12}{8^2}=-\frac{3}{16}\\ S_W(A,W)&=\frac{A}{(A-W)^2} & S_W(20,12)&=\frac{20}{8^2}=\frac{5}{16} \end{align*}

For any given ΔA\De A and ΔW\De W, the percentage error is

∣100ΔSS∣=∣10025(−316ΔA+516ΔW)∣\begin{equation*} \left|100\frac{\De S}{S}\right| =\left|100\frac{2}{5}\Big(-\frac{3}{16}\De A+\frac{5}{16}\De W\Big)\right| \end{equation*}

We are told that ∣ΔA∣≤0.01|\De A|\le 0.01 and ∣ΔW∣≤0.02|\De W|\le 0.02. To maximize ∣10025(−316ΔA+516ΔW)∣\big|100\frac{2}{5}\big(-\frac{3}{16}\De A+\frac{5}{16}\De W\big)\big| take ΔA=−0.01\De A=- 0.01 and ΔW=+0.02\De W=+ 0.02. So the maximum percentage error is

10025[−316(−0.01)+516(0.02)]=25×1316=1340=0.325%\begin{equation*} 100\frac{2}{5}\left[-\frac{3}{16}(- 0.01)+\frac{5}{16}(0.02)\right] = \frac{2}{5}\times \frac{13}{16} =\frac{13}{40}=0.325\% \end{equation*}
Q10Stage 2Past exam · M200 2008D

The pressure in a solid is given by

P(s,r)=sr(4s2−r2−2)\begin{equation*} P(s,r) = sr(4s^2 - r^2 - 2) \end{equation*}

where ss is the specific heat and rr is the density. We expect to measure (s,r)(s,r) to be approximately (2,2)(2,2) and would like to have the most accurate value for PP. There are two different ways to measure ss and rr. Method 11 has an error in ss of ±0.01\pm 0.01 and an error in rr of ±0.1\pm 0.1, while method 2 has an error of ±0.02\pm 0.02 for both ss and rr.

Should we use method 1 or method 2? Explain your reasoning carefully.

Answer

Method 1 is better.

Full solution

The linear approximation to P(s,r)P(s,r) at (2,2)(2,2) is

P(s,r)≈P(2,2)+Ps(2,2) (s−2)+Pr(2,2) (r−2)\begin{align*} P(s,r)\approx P(2,2) +P_s(2,2)\,(s-2) + P_r(2,2)\,(r-2) \end{align*}

As

P(2,2)=(2)(2)[4(2)2−(2)2−2]=40(which we don’t actually need) Ps(2,2)=[12s2r−r3−2r]s=r=2=84Pr(2,2)=[4s3−3sr2−2s]s=r=2=4\begin{align*} P(2,2) &= (2)(2)\big[4(2)^2-(2)^2-2\big] = 40 \qquad\text{(which we don't actually need) } \\ P_s(2,2) &= \Big[12s^2r-r^3-2r\Big]_{s=r=2} = 84 \\ P_r(2,2) &= \Big[4s^3-3sr^2-2s\Big]_{s=r=2} = 4 \end{align*}

the linear approximation is

P(s,r)≈40+84 (s−2)+4 (r−2)\begin{align*} P(s,r)\approx 40 +84\,(s-2) + 4\,(r-2) \end{align*}

Under method 1, the maximum error in PP will have magnitude at most (approximately)

84(0.01)+4(0.1)=1.24\begin{align*} 84(0.01) + 4(0.1) = 1.24 \end{align*}

Under method 2, the maximum error in PP will have magnitude at most (approximately)

84(0.02)+4(0.02)=1.76\begin{align*} 84(0.02) + 4(0.02) = 1.76 \end{align*}

Method 1 is better.

Q11Stage 2

A rectangular beam that is supported at its two ends and is subjected to a uniform load sags by an amount

S=Cpℓ4wh3\begin{equation*} S=C\frac{p\ell^4}{w h^3} \end{equation*}

where p=loadp={\rm load}, ℓ=length\ell={\rm length}, h=heighth={\rm height}, w=widthw={\rm width} and CC is a constant. Suppose p≈100p\approx 100, ℓ≈4\ell\approx 4, w≈.1w\approx .1 and h≈.2h\approx.2. Will the sag of the beam be more sensitive to changes in the height of the beam or to changes in the width of the beam.

Hint

Determine, approximately, the change in sag when the height changes by a small amount ε\veps and also when the width changes by a small amount ε\veps. Which is bigger?

Answer

The sag will be more sensitive to changes in height.

Full solution

Using the four variable variant of the linear approximation (2.6.2) of the CLP-3 text,

ΔS≈∂S∂pΔp+∂S∂ℓΔℓ+∂S∂wΔw+∂S∂hΔh=Cpℓ4wh3[Δpp+4Δℓℓ−Δww−3Δhh]\begin{equation*} \De S\approx \pdiff{S}{p}\De p +\pdiff{S}{\ell}\De \ell +\pdiff{S}{w}\De w +\pdiff{S}{h}\De h =C\frac{p\ell^4}{w h^3}\left[\frac{\De p}{p}+4\frac{\De\ell}{\ell} -\frac{\De w}{w}-3\frac{\De h}{h} \right] \end{equation*}

When w≈0.1w\approx0.1 and h≈0.2h\approx0.2,

Δww≈10Δw3Δhh≈15Δh\begin{equation*} \frac{\De w}{w}\approx 10\De w\qquad 3\frac{\De h}{h}\approx 15\De h \end{equation*}

So a change in height by Δh=ε\De h=\veps produces a change in sag of about ΔS=15ε\De S = 15\veps times −Cpℓ4wh3-C\frac{p\ell^4}{w h^3}, while a change Δw\De w in width by the same ε\veps produces a change in sag of about ΔS=10ε\De S=10\veps times the same −Cpℓ4wh3-C\frac{p\ell^4}{w h^3}. The sag is more sensitive to Δh\De h.

Q12Stage 2Past exam · M200 2010D

Let z=f(x,y)=2yx2+y2z=f(x,y)=\frac{2y}{x^2+y^2}. Find an approximate value for f(−0.8,2.1)f(-0.8,2.1).

Answer

0.840.84

Full solution

The first order partial derivatives of ff are

fx(x,y)=−4xy(x2+y2)2fx(−1,2)=825fy(x,y)=2x2+y2−4y2(x2+y2)2fy(−1,2)=25−1625=−625\begin{alignat*}{3} f_x(x,y) & = -\frac{4xy}{{(x^2+y^2)}^2}\quad & f_x(-1,2) & = \frac{8}{25} \\ f_y(x,y) & = \frac{2}{x^2+y^2}-\frac{4y^2}{{(x^2+y^2)}^2}\quad & f_y(-1,2) & = \frac{2}{5}-\frac{16}{25} =-\frac{6}{25} \\ \end{alignat*}

The linear approximation of f(x,y)f(x,y) about (−1,2)(-1,2) is

f(x,y)≈f(−1,2)+fx(−1,2) (x+1)+fy(−1,2) (y−2)=45+825 (x+1)−625 (y−2)\begin{align*} f(x,y)&\approx f(-1,2) + f_x(-1,2)\,(x+1) + f_y(-1,2)\,(y-2) \\ &=\frac{4}{5} +\frac{8}{25}\,(x+1) - \frac{6}{25}\,(y-2) \end{align*}

In particular, for x=−0.8x=-0.8 and y=2.1y=2.1,

f(−0.8,2.1)≈45+825 (0.2)−625 (0.1)=0.84\begin{align*} f(-0.8,2.1)&\approx \frac{4}{5} +\frac{8}{25}\,(0.2) - \frac{6}{25}\,(0.1) \\ &=0.84 \end{align*}
Q13Stage 2Past exam · M200 2012D

Suppose that a function z=f(x,y)z = f (x, y) is implicitly defined by an equation:

xyz+x+y2+z3=0\begin{equation*} xyz + x + y^2 + z^3 = 0 \end{equation*}

  1. Find ∂z∂x\pdiff{z}{x}.

  2. If f(−1,1)<0f(-1, 1) < 0, find the linear approximation of the function z=f(x,y)z = f (x, y) at (−1,1)(-1, 1).

  3. If f(−1,1)<0f(-1, 1) < 0, use the linear approximation in (b) to approximate f(−1.02,0.97)f(-1.02, 0.97).

Answer

(a) fx(x,y)=−y f(x,y)+13f(x,y)2+xyf_x(x,y) = -\frac{y\,f(x,y) +1}{3f(x,y)^2 +xy} (b) f(x,y)≈−1−32(y−1)f(x,y) \approx -1 -\frac{3}{2} (y-1) (c) −0.955-0.955

Full solution

(a) The function f(x,y)f(x,y) obeys

xy f(x,y)+x+y2+f(x,y)3=0\begin{equation*} xy\,f(x,y) + x + y^2 + f(x,y)^3 =0 \tag{$*$}\end{equation*}

for all xx and yy (sufficiently close to (−1,1)(-1,1)). Differentiating (∗)(*) with respect to xx gives

y f(x,y)+xy fx(x,y)+1+3f(x,y)2fx(x,y)=0  ⟹  fx(x,y)=−y f(x,y)+13f(x,y)2+xy\begin{align*} y\,f(x,y) +xy\,f_x(x,y) + 1 + 3f(x,y)^2 f_x(x,y) = 0 \implies f_x(x,y) = -\frac{y\,f(x,y) +1}{3f(x,y)^2 +xy} \end{align*}

Without knowing f(x,y)f(x,y) explicitly, there's not much that we can do with this.

(b) f(−1,1)f(-1,1) obeys

(−1)(1) f(−1,1)+(−1)+(1)2+f(−1,1)3=0  ⟺  f(−1,1)3−f(−1,1)=0\begin{equation*} (-1)(1)\,f(-1,1) + (-1) + (1)^2 + f(-1,1)^3 =0 \iff f(-1,1)^3 -f(-1,1) =0 \end{equation*}

Since f(−1,1)<0f(-1, 1) < 0 we may divide this equation by f(−1,1)<0f(-1, 1) < 0, giving f(−1,1)2−1=0f(-1,1)^2 - 1=0. Since f(−1,1)<0f(-1, 1) < 0, we must have f(−1,1)=−1f(-1, 1)=-1. By part (a)

fx(−1,1)=−(1) f(−1,1)+13f(−1,1)2+(−1)(1)=0\begin{align*} f_x(-1, 1) = -\frac{(1)\,f(-1, 1) +1}{3f(-1, 1)^2 +(-1)(1)} = 0 \end{align*}

To get the linear approximation, we still need fy(−1,1)f_y(-1,1). Differentiating (∗)(*) with respect to yy gives

x f(x,y)+xy fy(x,y)+2y+3f(x,y)2fy(x,y)=0\begin{equation*} x\,f(x,y) + xy\,f_y(x,y) + 2y + 3f(x,y)^2 f_y(x,y) =0 \end{equation*}

Then setting x=−1x=-1, y=1y=1 and f(−1,1)=−1f(-1,1)=-1 gives

(−1) (−1)+(−1)(1) fy(−1,1)+2(1)+3(−1)2fy(−1,1)=0  ⟹  fy(−1,1)=−32\begin{equation*} (-1)\,(-1) + (-1)(1)\,f_y(-1,1) + 2(1) + 3(-1)^2 f_y(-1,1) =0 \implies f_y(-1,1) =-\frac{3}{2} \end{equation*}

So the linear approximation is

f(x,y)≈f(−1,1)+fx(−1,1) (x+1)+fy(−1,1) (y−1)=−1−32(y−1)\begin{equation*} f(x,y) \approx f(-1,1) + f_x(-1,1)\,(x+1) + f_y(-1,1)\,(y-1) = -1 -\frac{3}{2} (y-1) \end{equation*}

(c) By part (b),

f(−1.02,0.97)≈−1−32(0.97−1)=−0.955\begin{align*} f(-1.02,0.97) \approx -1 -\frac{3}{2} (0.97-1) =-0.955 \end{align*}
Q14Stage 2Past exam · M200 2014D

Let z=f(x,y)z = f(x,y) be given implicitly by

ez+yz=x+y.\begin{equation*} e^z + yz = x + y. \end{equation*}

  1. Find the differential dz\dee{z}.

  2. Use linear approximation at the point (1,0)(1,0) to approximate f(0.99,0.01)f(0.99,0.01).

Answer

(a) The differential at x=ax=a, y=by=b is dxef(a,b)+b+1−f(a,b)ef(a,b)+b dy\frac{\dee{x}}{e^{f(a,b)}+b} + \frac{1-f(a,b)}{e^{f(a,b)}+b} \,\dee{y}

(b) f(0.99 , 0.01)≈0f\big(0.99\,,\,0.01\big) \approx 0

Full solution

By definition, the differential at x=ax=a, y=by=b is

fx(a,b) dx+fy(a,b) dy\begin{equation*} f_x(a,b)\,\dee{x} + f_y(a,b)\,\dee{y} \end{equation*}

so we have to determine the partial derivatives fx(a,b)f_x(a,b) and fy(a,b)f_y(a,b). We are told that

ef(x,y)+y f(x,y)=x+y\begin{equation*} e^{f(x,y)} + y\,f(x,y) = x + y \end{equation*}

for all xx and yy. Differentiating this equation with respect to xx and with respect to yy gives, by the chain rule,

ef(x,y)fx(x,y)+y fx(x,y)=1ef(x,y)fy(x,y)+f(x,y)+y fy(x,y)=1\begin{align*} e^{f(x,y)}f_x(x,y) + y\,f_x(x,y) = 1 \\ e^{f(x,y)}f_y(x,y) + f(x,y) +y\,f_y(x,y) = 1 \end{align*}

Solving the first equation for fxf_x and the second for fyf_y gives

fx(x,y)=1ef(x,y)+yfy(x,y)=1−f(x,y)ef(x,y)+y\begin{align*} f_x(x,y) &= \frac{1}{e^{f(x,y)}+y} \\ f_y(x,y) &= \frac{1-f(x,y)}{e^{f(x,y)}+y} \end{align*}

So the differential at x=ax=a, y=by=b is

dxef(a,b)+b+1−f(a,b)ef(a,b)+b dy\begin{equation*} \frac{\dee{x}}{e^{f(a,b)}+b} + \frac{1-f(a,b)}{e^{f(a,b)}+b} \,\dee{y} \end{equation*}

Since we can't solve explicitly for f(a,b)f(a,b) for general aa and bb. There's not much more that we can do with this.

(b) In particular, when a=1a=1 and b=0b=0, we have

ef(1,0)+0 f(1,0)=1+0  ⟹  ef(1,0)=1  ⟹  f(1,0)=0\begin{equation*} e^{f(1,0)} + 0\,f(1,0) = 1 + 0 \implies e^{f(1,0)} = 1 \implies f(1,0)=0 \end{equation*}

and the linear approximation simpifies to

f(1+dx , dy)≈f(1,0)+dxef(1,0)+0+1−f(1,0)ef(1,0)+0 dy=dx+dy\begin{equation*} f\big(1+\dee{x}\,,\,\dee{y}\big) \approx f(1,0) + \frac{\dee{x}}{e^{f(1,0)}+0} + \frac{1-f(1,0)}{e^{f(1,0)}+0} \,\dee{y} = \dee{x} +\dee{y} \end{equation*}

Choosing dx=−0.01\dee{x} = -0.01 and dy=0.01\dee{y} = 0.01, we have

f(0.99 , 0.01)≈−0.01+0.01=0\begin{equation*} f\big(0.99\,,\,0.01\big) \approx -0.01 + 0.01 =0 \end{equation*}
Q15Stage 2Past exam · M200 2004A

Two sides and the enclosed angle of a triangle are measured to be 3±.13\pm.1m, 4±.14\pm.1m and 90±1∘90\pm 1^\circ respectively. The length of the third side is then computed using the cosine law C2=A2+B2−2ABcos⁡θC^2=A^2+B^2-2AB\cos\theta. What is the approximate maximum error in the computed value of CC?

Hint

1∘=π1801^\circ = \frac{\pi}{180} radians

Answer

π75+0.14≈0.182\frac{\pi}{75}+0.14\approx 0.182

Full solution

Let C(A,B,θ)=A2+B2−2ABcos⁡θC(A,B,\theta)=\sqrt{A^2+B^2-2AB\cos\theta}. Then C(3,4,π2)=5C\big(3,4,\frac{\pi}{2}\big)=5. Differentiating C2=A2+B2−2ABcos⁡θC^2=A^2+B^2-2AB\cos\theta gives

2C∂C∂A(A,B,θ)=2A−2Bcos⁡θ  ⟹  10∂C∂A(3,4,π2)=62C∂C∂B(A,B,θ)=2B−2Acos⁡θ  ⟹  10∂C∂B(3,4,π2)=82C∂C∂θ(A,B,θ)=2ABsin⁡θ  ⟹  10∂C∂θ(3,4,π2)=24\begin{alignat*}{5} 2C\frac{\partial C}{\partial A}(A,B,\theta)&=2A-2B\cos\theta &\quad&\implies\quad & 10\frac{\partial C}{\partial A}\big(3,4,\tfrac{\pi}{2}\big)&=6\cr 2C\frac{\partial C}{\partial B}(A,B,\theta)&=2B-2A\cos\theta & &\implies & 10\frac{\partial C}{\partial B}\big(3,4,\tfrac{\pi}{2}\big)&=8\cr 2C\frac{\partial C}{\partial \theta}(A,B,\theta)&=2AB\sin\theta & &\implies & 10\frac{\partial C}{\partial \theta}\big(3,4,\tfrac{\pi}{2}\big)&=24\cr \end{alignat*}

Hence the approximate maximum error in the computed value of CC is

∣ΔC∣≈∣∂C∂A(3,4,π2)ΔA+∂C∂B(3,4,π2)ΔB+∂C∂θ(3,4,π2)Δθ∣≤(0.6)(0.1)+(0.8)(0.1)+(2.4)π180=π75+0.14≤0.182\begin{align*} |\De C|&\approx \left| \frac{\partial C}{\partial A}\big(3,4,\tfrac{\pi}{2}\big)\De A +\frac{\partial C}{\partial B}\big(3,4,\tfrac{\pi}{2}\big)\De B +\frac{\partial C}{\partial \theta}\big(3,4,\tfrac{\pi}{2}\big)\De\theta\right| \\ &\le(0.6)(0.1)+(0.8)(0.1)+(2.4)\frac{\pi}{180}\cr &=\frac{\pi}{75}+0.14\le 0.182 \end{align*}
Q16Stage 2Past exam · M200 2003D

Use differentials to find a reasonable approximation to the value of f(x,y)=xyx2+y2f(x,y)=xy\sqrt{x^2+y^2} at x=3.02x=3.02, y=3.96y=3.96. Note that 3.02≈33.02\approx 3 and 3.96≈43.96\approx 4.

Answer

59.56059.560

Full solution

Substituting (x0,y0)=(3,4)(x_0,y_0)=(3,4) and (x,y)=(3.02,3.96)(x,y)=(3.02,3.96) into

f(x,y)≈f(x0,y0)+fx(x0,y0)(x−x0)+fy(x0,y0)(y−y0)\begin{equation*} f(x,y)\approx f(x_0,y_0)+f_x(x_0,y_0)(x-x_0)+f_y(x_0,y_0)(y-y_0) \end{equation*}

gives

f(3.02,3.96)≈f(3,4)+0.02fx(3,4)−0.04fy(3,4)=60+0.02(20+365)−0.04(15+485)=59.560\begin{align*} f(3.02,3.96)&\approx f(3,4)+0.02f_x(3,4)-0.04f_y(3,4)\\ &=60+0.02\left(20+\frac{36}{5}\right)-0.04\left(15+\frac{48}{5}\right)\\ &=59.560 \end{align*}

since

fx(x,y)=yx2+y2+x2yx2+y2fy(x,y)=xx2+y2+xy2x2+y2\begin{equation*} f_x(x,y)=y\sqrt{x^2+y^2}+\frac{x^2y}{\sqrt{x^2+y^2}}\qquad f_y(x,y)=x\sqrt{x^2+y^2}+\frac{xy^2}{\sqrt{x^2+y^2}} \end{equation*}
Q17Stage 2Past exam · M200 2000D

Use differentials to estimate the volume of metal in a closed metal can with diameter 8cm and height 12cm if the metal is 0.04cm thick.

Answer

π×128×0.04=5.12π≈16.1\pi\times 128\times 0.04= 5.12\pi\approx 16.1cc

Full solution

The volume of a cylinder of diameter dd and height hh is V(d,h)=π(d2)2hV(d,h)=\pi\big(\frac{d}{2}\big)^2h. The wording of the question is a bit ambiguous in that it does not specify if the given dimensions are inside dimensions or outside dimensions. Assume that they are outside dimensions. Then the volume of the can, including the metal, is V(8,12)V(8,12) and the volume of the interior, excluding the metal, is

V(8−2×0.04 , 12−2×0.04)≈V(8,12)+Vd(8,12)(−2×0.04)+Vh(8,12)(−2×0.04)=V(8,12)+12π×8×12×(−2×0.04)+π(82)2 ⁣ ⁣(−2×0.04)=V(8,12)−π×128×0.04\begin{align*} V(8-2\times0.04\,,\,12-2\times 0.04) &\approx V(8,12) +V_d(8,12)(-2\times 0.04) +V_h(8,12)(-2\times 0.04)\\ &=V(8,12) +\frac{1}{2}\pi\times 8\times 12\times(-2\times 0.04) +\pi\left(\frac{8}{2}\right)^2\!\!(-2\times 0.04)\\ &=V(8,12) -\pi\times 128\times 0.04 \end{align*}

So the volume of metal is approximately π×128×0.04=5.12π≈16.1\pi\times 128\times 0.04= 5.12\pi\approx 16.1cc. (To this level of approximation, it doesn't matter whether the dimensions are inside or outside dimensions.)

Q18Stage 2Past exam · M200 2000A

Let zz be a function of xx, yy such that

z3−z+2xy−y2=0,z(2,4)=1.\begin{equation*} z^3 - z + 2xy - y^2 = 0,\qquad z(2, 4) = 1. \end{equation*}

  1. Find the linear approximation to zz at the point (2,4)(2, 4).

  2. Use your answer in (a) to estimate the value of zz at (2.02,3.96)(2.02, 3.96).

Answer

(a) z(x,y)≈1−4x+2yz(x,y)\approx 1-4x+2y (b) 0.840.84

Full solution

(a) The function z(x,y)z(x,y) obeys

z(x,y)3−z(x,y)+2xy−y2=0\begin{equation*} z(x,y)^3-z(x,y)+2xy-y^2=0 \end{equation*}

for all (x,y)(x,y) near (2,4)(2,4). Differentiating with respect to xx and yy

3z(x,y)2∂z∂x(x,y)−∂z∂x(x,y)+2y=03z(x,y)2∂z∂y(x,y)−∂z∂y(x,y)+2x−2y=0\begin{align*} 3z(x,y)^2\pdiff{z}{x}(x,y) -\pdiff{z}{x}(x,y)+2y&=0 \\ 3z(x,y)^2\pdiff{z}{y}(x,y) -\pdiff{z}{y}(x,y)+2x-2y&=0 \end{align*}

Substituting in x=2x=2, y=4y=4 and z(2,4)=1z(2,4)=1 gives

3∂z∂x(2,4)−∂z∂x(2,4)+8=0  ⟺  ∂z∂x(2,4)=−43∂z∂y(2,4)−∂z∂y(2,4)+4−8=0  ⟺  ∂z∂y(2,4)=2\begin{alignat*}{5} 3\pdiff{z}{x}(2,4) -\pdiff{z}{x}(2,4)+8&=0 & &\iff & \pdiff{z}{x}(2,4) & = -4 \\ 3\pdiff{z}{y}(2,4) -\pdiff{z}{y}(2,4)+4-8&=0 & &\iff & \pdiff{z}{y}(2,4) & = 2 \end{alignat*}

The linear approximation is

z(x,y)≈z(2,4)+zx(2,4)(x−2)+zy(2,4)(y−4)=1−4(x−2)+2(y−4)=1−4x+2y\begin{align*} z(x,y)&\approx z(2,4)+z_x(2,4)(x-2)+z_y(2,4)(y-4)= 1-4(x-2)+2(y-4) \\ &=1-4x+2y \end{align*}

(b) Substituting in x=2.02x=2.02 and y=3.96y=3.96 gives

z(2.02,3.96)≈1−4×0.02+2×(−0.04)=0.84\begin{align*} z(2.02,3.96)\approx 1-4\times0.02+2\times(-0.04) =0.84 \end{align*}

3Stage 3Application

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

Q19Stage 3Past exam · M200 2006D

Consider the surface given by:

z3−xyz2−4x=0.\begin{equation*} z^3 - xyz^2 - 4x = 0. \end{equation*}

  1. Find expressions for ∂z∂x\pdiff{z}{x}, ∂z∂y\pdiff{z}{y} as functions of xx, yy, zz.

  2. Evaluate ∂z∂x\pdiff{z}{x}, ∂z∂y\pdiff{z}{y} at (1,1,2)(1, 1, 2).

  3. Measurements are made with errors, so that x=1±0.03x = 1 \pm 0.03 and y=1±0.02y = 1 \pm 0.02. Find the corresponding maximum error in measuring zz.

  4. A particle moves over the surface along the path whose projection in the xyxy–plane is given in terms of the angle θ\theta as

    x(θ)=1+cos⁡θ, y(θ)=sin⁡θ\begin{equation*} x(\theta) = 1 + \cos\theta,\ y(\theta) = \sin\theta \end{equation*}

    from the point A:x=2, y=0A : x = 2,\ y = 0 to the point B:x=1, y=1B : x = 1,\ y = 1. Find dzdθ\diff{z}{\theta} at points AA and BB.

Answer

(a) ∂z∂x=4+yz23z2−2xyz\pdiff{z}{x} = \frac{4+yz^2}{3z^2-2xyz}, ∂z∂x=xz23z2−2xyz\pdiff{z}{x} = \frac{xz^2}{3z^2-2xyz} (b) ∂z∂x(1,1)=1\pdiff{z}{x}(1,1) = 1, ∂z∂y(1,1)=12\pdiff{z}{y}(1,1) = \frac{1}{2} (c) ±0.04\pm 0.04 (d) At AA, dzdθ=23\diff{z}{\theta} =\frac{2}{3}. At BB, dzdθ=−1\diff{z}{\theta} = -1.

Full solution

(a) We are told that

z(x,y)3−xy z(x,y)2−4x=0\begin{equation*} z(x,y)^3 - xy\,z(x,y)^2 - 4x = 0 \end{equation*}

for all (x,y)(x,y) (sufficiently near (1,1)(1,1)). Differentiating this equation with respect to xx gives

3z(x,y)2 ∂z∂x(x,y)−y z(x,y)2−2xy z(x,y)∂z∂x(x,y)−4=0  ⟹  ∂z∂x=4+yz23z2−2xyz\begin{align*} & 3z(x,y)^2\,\pdiff{z}{x}(x,y) -y\, z(x,y)^2 -2xy\,z(x,y)\pdiff{z}{x}(x,y) - 4=0 \\ \implies &\pdiff{z}{x} = \frac{4+yz^2}{3z^2-2xyz} \end{align*}

and differentiating with respect to yy gives

3z(x,y)2 ∂z∂y(x,y)−x z(x,y)2−2xy z(x,y)∂z∂y(x,y)=0  ⟹  ∂z∂y=xz23z2−2xyz\begin{align*} & 3z(x,y)^2\,\pdiff{z}{y}(x,y) -x\, z(x,y)^2 -2xy\,z(x,y)\pdiff{z}{y}(x,y) =0 \\ \implies &\pdiff{z}{y} = \frac{xz^2}{3z^2-2xyz} \end{align*}

(b) When (x,y,z)=(1,1,2)(x,y,z)=(1,1,2),

∂z∂x(1,1)=4+(1)(2)23(2)2−2(1)(1)(2)=1∂z∂y(1,1)=(1)(2)23(2)2−2(1)(1)(2)=12\begin{align*} \pdiff{z}{x}(1,1) = \frac{4+(1)(2)^2}{3(2)^2-2(1)(1)(2)} = 1\qquad \pdiff{z}{y}(1,1) = \frac{(1)(2)^2}{3(2)^2-2(1)(1)(2)} = \frac{1}{2} \end{align*}

(c) Under the linear approximation at (1,1)(1,1)

z(x,y)≈z(1,1)+zx(1,1) (x−1)+zy (1,1) (y−1)=2+(x−1)+12(y−1)\begin{equation*} z(x,y) \approx z(1,1) + z_x(1,1)\,(x-1) + z_y\,(1,1)\,(y-1) = 2 + (x-1) +\frac{1}{2}(y-1) \end{equation*}

So errors of ±0.03\pm 0.03 in xx and ±0.02\pm 0.02 in yy leads of errors of about

±[0.03+12(0.02)]=±0.04\begin{equation*} \pm\left[0.03 + \frac{1}{2}(0.02)\right] =\pm 0.04 \end{equation*}

in zz.

(d) By the chain rule

ddθz(x(θ),y(θ))=zx(x(θ),y(θ)) x′(θ)+zy(x(θ),y(θ)) y′(θ)=−zx(1+cos⁡θ,sin⁡θ) sin⁡θ+zy(1+cos⁡θ,sin⁡θ) cos⁡θ\begin{align*} \diff{}{\theta} z\big(x(\theta),y(\theta)\big) &= z_x\big(x(\theta),y(\theta)\big)\,x'(\theta) +z_y\big(x(\theta),y(\theta)\big)\,y'(\theta) \\ &=-z_x\big(1 + \cos\theta,\sin\theta\big)\,\sin\theta +z_y\big(1 + \cos\theta,\sin\theta\big)\,\cos\theta \end{align*}

At AA, x=2x=2, y=0y=0, z=2z=2 (since z3−(2)(0)z2−4(2)=0z^3-(2)(0)z^2-4(2)=0) and θ=0\theta=0, so that

∂z∂x(2,0)=4+(0)(2)23(2)2−2(2)(0)(2)=13∂z∂y(2,0)=(2)(2)23(2)2−2(2)(0)(2)=23\begin{align*} \pdiff{z}{x}(2,0) = \frac{4+(0)(2)^2}{3(2)^2-2(2)(0)(2)} = \frac{1}{3}\qquad \pdiff{z}{y}(2,0) = \frac{(2)(2)^2}{3(2)^2-2(2)(0)(2)} = \frac{2}{3} \end{align*}

and

dzdθ=−13sin⁡(0)+23cos⁡(0)=23\begin{align*} \diff{z}{\theta} &=-\frac{1}{3}\sin(0) +\frac{2}{3}\cos(0) =\frac{2}{3} \end{align*}

At BB, x=1x=1, y=1y=1, z=2z=2 and θ=π2\theta=\frac{\pi}{2}, so that, by part (b),

∂z∂x(1,1)=1∂z∂y(1,1)=12\begin{align*} \pdiff{z}{x}(1,1) = 1\qquad \pdiff{z}{y}(1,1) = \frac{1}{2} \end{align*}

and

dzdθ=−sin⁡π2+12cos⁡π2=−1\begin{align*} \diff{z}{\theta} &=-\sin\frac{\pi}{2} +\frac{1}{2}\cos\frac{\pi}{2} = -1 \end{align*}
Q20Stage 3Past exam · M200 2007A

Consider the function ff that maps each point (x,y)(x, y) in R2\bbbr^2 to ye−xye^{-x}.

  1. Suppose that x=1x = 1 and y=ey = e, but errors of size 0.10.1 are made in measuring each of xx and yy. Estimate the maximum error that this could cause in f(x,y)f(x,y).

  2. The graph of the function ff sits in R3\bbbr^3 , and the point (1,e,1)(1, e, 1) lies on that graph. Find a nonzero vector that is perpendicular to that graph at that point.

Answer

(a) 1+e−110\frac{1+e^{-1}}{10} (b) any nonzero constant times <−1 , e−1 , −1>\llt -1\,,\,e^{-1}\,,\,-1\rgt

Full solution

We are going to need the first order partial derivatives of f(x,y)=ye−xf(x,y)=ye^{-x} at (x,y)=(1,e)(x,y)=(1,e). Here they are.

fx(x,y)=−ye−xfx(1,e)=−e e−1=−1fy(x,y)=e−xfy(1,e)=e−1\begin{align*} f_x(x,y)&= -ye^{-x} & f_x(1,e)&=-e\,e^{-1}=-1 \\ f_y(x,y)&=e^{-x} & f_y(1,e)&= e^{-1} \end{align*}

(a) The linear approximation to f(x,y)f(x,y) at (x,y)=(1,e)(x,y)=(1,e) is

f(x,y)≈f(1,e)+fx(1,e) (x−1)+fy(1,e) (y−e)=1−(x−1)+e−1(y−e)\begin{align*} f(x,y) \approx f(1,e) + f_x(1,e)\,(x-1) +f_y(1,e)\,(y-e) = 1 -(x-1) +e^{-1}(y-e) \end{align*}

The maximum error is then approximately

−1(−0.1)+e−1(0.1)=1+e−110\begin{align*} -1(-0.1) +e^{-1}(0.1) =\frac{1+e^{-1}}{10} \end{align*}

(b) The equation of the graph is g(x,y,z)=f(x,y)−z=0g(x,y,z) = f(x,y) -z =0. Any vector that is a nonzero constant times

∇g(1,e,1)=<fx(1,e) , fy(1,e) , −1>=<−1 , e−1 , −1>\begin{align*} \vnabla g(1,e,1) =\llt f_x(1,e)\,,\,f_y(1,e)\,,\,-1\rgt =\llt -1\,,\,e^{-1}\,,\,-1\rgt \end{align*}

is perpendicular to g=0g=0 at (1,e,1)(1,e,1).

Q21Stage 3Past exam · M200 2009D

A surface is defined implicitly by z4−xy2z2+y=0z^4 - xy^2 z^2 + y = 0.

  1. Compute ∂z∂x\pdiff{z}{x}, ∂z∂y\pdiff{z}{y} in terms of xx, yy, zz.

  2. Evaluate ∂z∂x\pdiff{z}{x} and ∂z∂y\pdiff{z}{y} at (x,y,z)=(2,−1/2,1)(x, y, z) = (2, -1/2, 1).

  3. If xx decreases from 22 to 1.941.94, and yy increases from −0.5-0.5 to −0.4-0.4, find the approximate change in zz from 11.

  4. Find the equation of the tangent plane to the surface at the point (2,−1/2,1)(2, -1/2, 1).

Answer

(a) ∂z∂x=y2z24z3−2xy2z\pdiff{z}{x} = \frac{y^2 z^2}{4z^3-2xy^2z}, ∂z∂y=2xy z2−14z3−2xy2z\pdiff{z}{y} = \frac{2xy\, z^2-1}{4z^3-2xy^2 z}

(b) ∂z∂x(2,−1/2)=112\pdiff{z}{x}(2,-1/2) =\frac{1}{12}, ∂z∂y(2,−1/2)=−1\pdiff{z}{y}(2,-1/2) =-1

(c) f(1.94,−0.4)−1≈−0.105f(1.94,-0.4) - 1 \approx -0.105

(d) x12−y−z=−13\frac{x}{12} -y -z = -\frac{1}{3}

Full solution

(a) We are told that for all x,yx,y (with (x,y,z)(x,y,z) near (2,−1/2,1)(2,-1/2,1)), the function z(x,y)z(x,y) obeys

z(x,y)4−xy2z(x,y)2+y=0\begin{align*} z(x,y)^4 -xy^2 z(x,y)^2 +y=0 \tag{$*$}\end{align*}

Differentiating (∗)(*) with respect to xx gives

4z(x,y)3∂z∂x(x,y)−y2z(x,y)2−2xy2z(x,y)∂z∂x(x,y)=0  ⟹  ∂z∂x(x,y)=y2z(x,y)24z(x,y)3−2xy2z(x,y)\begin{align*} &4z(x,y)^3\pdiff{z}{x}(x,y) - y^2z(x,y)^2 -2xy^2z(x,y)\pdiff{z}{x}(x,y) =0 \\ &\implies \pdiff{z}{x}(x,y) = \frac{y^2 z(x,y)^2}{4z(x,y)^3-2xy^2z(x,y)} \end{align*}

Similarly, differentiating this equation with respect to yy gives

4z(x,y)3∂z∂y(x,y)−2xy z(x,y)2−2xy2z(x,y)∂z∂y(x,y)+1=0  ⟹  ∂z∂y(x,y)=2xy z(x,y)2−14z(x,y)3−2xy2z(x,y)\begin{align*} &4z(x,y)^3\pdiff{z}{y}(x,y) - 2xy\,z(x,y)^2 -2xy^2z(x,y)\pdiff{z}{y}(x,y) +1=0 \\ &\implies \pdiff{z}{y}(x,y) = \frac{2xy\, z(x,y)^2-1}{4z(x,y)^3-2xy^2z(x,y)} \end{align*}

(b) Substituting (x,y,z)=(2,−1/2,1)(x, y, z) = (2, -1/2, 1) into the results of part (a) gives

∂z∂x(2,−1/2)=1/44−1=112∂z∂y(2,−1/2)=−2−14−1=−1\begin{align*} \pdiff{z}{x}(2,-1/2) &= \frac{1/4}{4-1} =\frac{1}{12}\\ \pdiff{z}{y}(2,-1/2) &= \frac{-2-1}{4-1}=-1 \end{align*}

(c) Under the linear approximation about (2,−1/2)(2,-1/2),

f(x,y)≈f(2,−1/2)+fx(2,−1/2) (x−2)+fy(2,−1/2) (y+1/2)=1+112(x−2)−(y+0.5)\begin{align*} f(x,y) &\approx f(2,-1/2) + f_x(2,-1/2)\,(x-2) + f_y(2,-1/2)\,(y+1/2) \\ &= 1 +\frac{1}{12}(x-2) - (y+0.5) \end{align*}

In particular

f(1.94,−0.4)≈1−0.0612−0.1\begin{align*} f(1.94,-0.4) \approx 1 -\frac{0.06}{12}-0.1 \end{align*}

so that

f(1.94,−0.4)−1≈−0.105\begin{align*} f(1.94,-0.4) - 1 \approx -0.105 \end{align*}

(d) The tangent plane is

z=f(2,−1/2)+fx(2,−1/2) (x−2)+fy(2,−1/2) (y+1/2)=1+112(x−2)−(y+0.5)\begin{align*} z&=f(2,-1/2) + f_x(2,-1/2)\,(x-2) + f_y(2,-1/2)\,(y+1/2) \\ &= 1 +\frac{1}{12}(x-2) - (y+0.5) \end{align*}

or

x12−y−z=−13\begin{align*} \frac{x}{12} -y -z = -\frac{1}{3} \end{align*}
Q22Stage 3Past exam · M200 2010A

A surface z=f(x,y)z = f (x, y) has derivatives ∂f∂x=3\pdiff{f}{x}=3 and ∂f∂y=−2\pdiff{f}{y}=-2 at (x,y,z)=(1,3,1)(x, y, z) = (1, 3, 1).

  1. If xx increases from 11 to 1.21.2, and yy decreases from 33 to 2.62.6, find the change in zz using a linear approximation.

  2. Find the equation of the tangent plane to the surface at the point (1,3,1)(1, 3, 1).

Answer

(a) 1.41.4 (b) 3x−2y−z=−43x-2y-z = -4

Full solution

(a) The linear approximation to f(x,y)f(x,y) at (1,3)(1,3) is

f(x,y)≈f(1,3)+fx(1,3) (x−1)+fy(1,3) (y−3)=1+3(x−1)−2(y−3)\begin{align*} f(x,y) \approx f(1,3) + f_x(1,3)\,(x-1) +f_y(1,3)\,(y-3) = 1 + 3(x-1) -2(y-3) \end{align*}

So the change is zz is approximately

3(1.2−1)−2(2.6−3)=1.4\begin{align*} 3(1.2-1) -2(2.6-3) = 1.4 \end{align*}

(b) The equation of the tangent plane is

z=f(1,3)+fx(1,3) (x−1)+fy(1,3) (y−3)=1+3(x−1)−2(y−3)\begin{align*} z = f(1,3) + f_x(1,3)\,(x-1) +f_y(1,3)\,(y-3) = 1 + 3(x-1) -2(y-3) \end{align*}

or

3x−2y−z=−4\begin{equation*} 3x-2y-z = -4 \end{equation*}
Q23Stage 3Past exam · M200 2011A

According to van der Waal's equation, a gas satisfies the equation

(pV2+16)(V−1)=TV2,\begin{equation*} (pV^2 + 16)(V - 1) = T V^2 , \end{equation*}

where pp, VV and TT denote pressure, volume and temperature respectively. Suppose the gas is now at pressure 11, volume 22 and temperature 55. Find the approximate change in its volume if pp is increased by 0.20.2 and TT is increased by 0.30.3.

Answer

0.10.1

Full solution

Think of the volume as being the function V(p,T)V(p,T) of pressure and temperature that is determined implicitly (at least for p≈1p\approx 1, T≈5T\approx 5 and V≈2V\approx 2) by the equation

(pV(p,T)2+16)(V(p,T)−1)=TV(p,T)2\begin{equation*} \big(pV(p,T)^2 + 16\big)\big(V(p,T) - 1\big) = T V(p,T)^2 \tag{$*$}\end{equation*}

To determine the approximate change in VV, we will use the linear approximation to V(p,T)V(p,T) at p=1p=1, T=5T=5. So we will need the partial derivatives Vp(1,5)V_p(1,5) and VT(1,5)V_T(1,5). As the equation (∗)(*) is valid for all pp near 11 and TT near 5, we may differentiate (∗)(*) with respect to pp, giving

(V2+2pVVp)(V−1)+(pV2+16)Vp=2TVVp\begin{align*} &\big(V^2 +2 pV V_p\big)\big(V - 1\big) + \big(pV ^2 + 16\big)V_p = 2T V V_p \end{align*}

and we may also differentiate (∗)(*) with respect to TT, giving

(2pVVT)(V−1)+(pV2+16)VT=V2+2TVVT\begin{align*} &\big(2 pV V_T\big)\big(V - 1\big) + \big(pV ^2 + 16\big)V_T = V^2 + 2T V V_T \end{align*}

In particular, when p=1p=1, V=2V=2, T=5T=5,

(4+4Vp(1,5))(2−1)+(4+16)Vp(1,5)=20Vp(1,5)  ⟹  Vp(1,5)=−14VT(1,5)(2−1)+(4+16)VT(1,5)=4+20VT(1,5)  ⟹  VT(1,5)=1\begin{alignat*}{3} \big(4 +4 V_p(1,5)\big)\big(2 - 1\big) + \big(4 + 16\big)V_p(1,5) &= 20 V_p(1,5) &\quad\implies V_p(1,5)&=-1 \\ 4 V_T(1,5)\big(2 - 1\big) + \big(4 + 16\big)V_T(1,5) &= 4 + 20 V_T(1,5) &\quad\implies V_T(1,5)&=1 \end{alignat*}

so that the change in VV is

V(1.2 , 5.3)−V(1,5)≈Vp(1,5) (0.2)+VT(1,5) (0.3)=−0.2+0.3=0.1\begin{align*} V(1.2\,,\,5.3)-V(1,5) \approx V_p(1,5)\,(0.2) + V_T(1,5)\,(0.3) = -0.2+0.3 = 0.1 \end{align*}
Q24Stage 3Past exam · M200 2011D

Consider the function f(x,y)=e−x2+4y2f(x, y) = e^{-x^2 +4y^2}.

  1. Find the equation of the tangent plane to the graph z=f(x,y)z = f (x,y) at the point where (x,y)=(2,1)(x, y) = (2, 1).

  2. Find the tangent plane approximation to the value of f(1.99,1.01)f(1.99, 1.01) using the tangent plane from part (a).

Answer

(a) z=1−4x+8yz=1-4x+8y (b) 1.121.12

Full solution

Since

fx(2,1)=−2xe−x2+4y2∣(x,y)=(2,1)=−4fy(2,1)=−8ye−x2+4y2∣(x,y)=(2,1)=8\begin{align*} f_x(2,1) & = -2x e^{-x^2 +4y^2}\Big|_{(x,y)=(2,1)} = -4 \\ f_y(2,1) & = \phantom{-}8y e^{-x^2 +4y^2}\Big|_{(x,y)=(2,1)} = 8 \end{align*}

The tangent plane to z=f(x,y)z=f(x,y) at (2,1)(2,1) is

z=f(2,1)+fx(2,1) (x−2)+fy(2,1) (y−1)=1−4(x−2)+8(y−1)=1−4x+8y\begin{align*} z &= f(2,1) +f_x(2,1)\,(x-2) +f_y(2,1)\,(y-1) = 1 -4 (x-2) +8(y-1) \\ &= 1-4x +8y \end{align*}

and the tangent plane approximation to the value of f(1.99,1.01)f(1.99, 1.01) is

f(1.99,1.01)≈1−4(1.99−2)+8(1.01−1)=1.12\begin{align*} f(1.99, 1.01) \approx 1 -4 (1.99-2) +8(1.01-1) = 1.12 \end{align*}
Q25Stage 3Past exam · M200 2012a

Let z=f(x,y)=ln⁡(4x2+y2)z = f (x, y) = \ln(4x^2 + y^2).

  1. Use a linear approximation of the function z=f(x,y)z = f (x, y) at (0,1)(0, 1) to estimate f(0.1,1.2)f(0.1, 1.2).

  2. Find a point P(a,b,c)P (a, b, c) on the graph of z=f(x,y)z = f(x, y) such that the tangent plane to the graph of z=f(x,y)z = f (x, y) at the point PP is parallel to the plane 2x+2y−z=32x + 2y - z = 3.

Answer

(a) f(0.1,1.2)≈0.4f(0.1,1.2) \approx 0.4 (b) (15 , 45 , ln⁡45)\left(\frac{1}{5}\,,\,\frac{4}{5}\,,\,\ln\frac{4}{5}\right)

Full solution

(a) The linear approximation to f(x,y)f(x,y) at (a,b)(a,b) is

f(x,y)≈f(a,b)+fx(a,b) (x−a)+fy(a,b) (y−b)=ln⁡(4a2+b2)+8a4a2+b2 (x−a)+2b4a2+b2 (y−b)\begin{align*} f(x,y) &\approx f(a,b) + f_x(a,b)\,(x-a) + f_y(a,b)\,(y-b) \\ &= \ln(4a^2 + b^2) + \frac{8a}{4a^2+b^2}\, (x-a) + \frac{2b}{4a^2+b^2}\, (y-b) \end{align*}

In particular, for a=0a=0 and b=1b=1,

f(x,y)≈2 (y−1)\begin{align*} f(x,y) &\approx 2\, (y-1) \end{align*}

and, for x=0.1x=0.1 and y=1.2y=1.2,

f(0.1,1.2)≈0.4\begin{align*} f(0.1,1.2) &\approx 0.4 \end{align*}

(b) The point (a,b,c)(a,b,c) is on the surface z=f(x,y)z=f(x,y) if and only if

c=f(a,b)=ln⁡(4a2+b2)\begin{equation*} c = f(a,b) = \ln(4a^2 + b^2) \end{equation*}

Note that this forces 4a2+b24a^2 + b^2 to be nonzero. The tangent plane to the surface z=f(x,y)z = f (x, y) at the point (a,b,c)(a,b,c) is parallel to the plane 2x+2y−z=32x + 2y - z = 3 if and only if <2 , 2 , −1>\llt 2\,,\,2\,,\,-1\rgt is a normal vector for the tangent plane. That is, there is a nonzero number tt such that

<2 , 2 , −1>=t<fx(a,b) , fy(a,b) , −1>=t<8a4a2+b2 , 2b4a2+b2 , −1>\begin{align*} \llt 2\,,\,2\,,\,-1\rgt =t\llt f_x(a,b)\,,\,f_y(a,b)\,,\ -1\rgt =t\llt \frac{8a}{4a^2+b^2} \,,\, \frac{2b}{4a^2+b^2} \,,\,-1\rgt \end{align*}

For the zz–coordinates to be equal, tt must be 11. Then, for the xx– and yy–coordinates to be equal, we need

8a4a2+b2=22b4a2+b2=2\begin{align*} \frac{8a}{4a^2+b^2} &= 2 \\ \frac{2b}{4a^2+b^2} &= 2 \end{align*}

Note that these equations force both aa and bb to be nonzero. Dividing these equations gives 8a2b=1\frac{8a}{2b}=1 and hence b=4ab=4a. Substituting b=4ab=4a into either of the two equations gives

8a20a2=2  ⟹  a=15\begin{align*} \frac{8a}{20a^2}=2 \implies a=\frac{1}{5} \end{align*}

So a=15a=\frac{1}{5}, b=45b=\frac{4}{5} and

c=ln⁡(452+4252)=ln⁡45\begin{align*} c=\ln\left(\frac{4}{5^2}+\frac{4^2}{5^2}\right) =\ln\frac{4}{5} \end{align*}
Q26Stage 3Past exam · M200 2013D

  1. Find the equation of the tangent plane to the surface x2z3+ysin⁡(πx)=−y2x^2 z^3 + y \sin(\pi x) = -y^2 at the point P=(1,1,−1)P = (1,1,-1).

  2. Let zz be defined implicitly by x2z3+ysin⁡(πx)=−y2x^2 z^3 + y \sin(\pi x) = -y^2. Find ∂z∂x\pdiff{z}{x} at the point P=(1,1,−1)P = (1,1,-1).

  3. Let zz be the same implicit function as in part (ii), defined by the equation x2z3+ysin⁡(πx)=−y2x^2 z^3 + y \sin(\pi x) = -y^2. Let x=0.97x = 0.97, and y=1y = 1. Find the approximate value of zz.

Answer

(a) −(2+π)x+2y+3z=−π−3-(2+\pi)x +2y +3z = -\pi-3

(b) ∂z∂x(1,1)=π+23\pdiff{z}{x}(1,1) = \frac{\pi+2}{3}

(c) z(0.97,1)≈−π+102100z(0.97,1) \approx -\frac{\pi+102}{100}

Full solution

(a) The surface has equation G(x,y,z)=x2z3+ysin⁡(πx)+y2=0G(x,y,z) = x^2 z^3 + y \sin(\pi x) + y^2 =0. So a normal vector to the surface at (1,1−1)(1,1-1) is

∇G(1,1,−1)=[(2xz3+πycos⁡(πx))ı^+(sin⁡(πx)+2y)ȷ^+3z2x2 k^](x,y,z)=(1,1,−1)=(−2−π)ı^+2 ȷ^+3 k^\begin{align*} \vnabla G(1,1,-1) &= \big[ \big(2x z^3 +\pi y\cos(\pi x)\big)\hi +\big(\sin(\pi x) +2y\big)\hj +3z^2 x^2\,\hk\big]_{(x,y,z)=(1,1,-1)} \\ &= \big(-2-\pi\big)\hi +2\,\hj +3\,\hk \end{align*}

So the equation of the tangent plane is

(−2−π)(x−1)+2(y−1)+3(z+1)=0or−(2+π)x+2y+3z=−π−3\begin{align*} \big(-2-\pi\big)(x-1) + 2(y-1) +3 (z+1) =0\quad\text{or}\quad -(2+\pi)x +2y +3z = -\pi-3 \end{align*}

(b) The functions z(x,y)z(x,y) obeys

x2z(x,y)3+ysin⁡(πx)+y2=0\begin{align*} x^2 z(x,y)^3 +y\sin(\pi x) +y^2 = 0 \end{align*}

for all xx and yy. Differentiating this equation with respect to xx gives

2xz(x,y)3+3x2z(x,y)2∂z∂x(x,y)+πycos⁡(πx)=0\begin{align*} 2x z(x,y)^3 + 3 x^2 z(x,y)^2 \pdiff{z}{x}(x,y) + \pi y \cos(\pi x) =0 \end{align*}

Evaluating at (1,1,−1)(1,1,-1) gives

−2+3∂z∂x(1,1)−π=0  ⟹  ∂z∂x(1,1)=π+23\begin{align*} -2 + 3 \pdiff{z}{x}(1,1) -\pi =0 \implies \pdiff{z}{x}(1,1) = \frac{\pi+2}{3} \end{align*}

(c) Using the linear approximation about (x,y)=(1,1)(x,y)=(1,1),

z(x,1)≈z(1,1)+∂z∂x(1,1) (x−1)\begin{align*} z(x,1) \approx z(1,1) +\pdiff{z}{x}(1,1)\ (x-1) \end{align*}

gives

z(0.97,1)≈−1+π+23 (−0.03)=−1−π+2100=−π+102100\begin{align*} z(0.97,1) \approx -1 +\frac{\pi+2}{3}\ (-0.03) = -1 -\frac{\pi+2}{100} = -\frac{\pi+102}{100} \end{align*}
Q27Stage 3Past exam · M200 2001A

The surface x4+y4+z4+xyz=17x^4+y^4+z^4+xyz=17 passes through (0,1,2)(0,1,2), and near this point the surface determines xx as a function, x=F(y,z)x=F(y,z), of yy and zz.

  1. Find FyF_y and FzF_z at (x,y,z)=(0,1,2)(x,y,z)=(0,1,2).

  2. Use the tangent plane approximation (also known as linear, first order or differential approximation) to find the approximate value of xx (near 00) such that (x,1.01,1.98)(x,1.01, 1.98) lies on the surface.

Answer

(a) Fy(1,2)=−2F_y(1,2)=-2, Fz(1,2)=−16F_z(1,2)=-16 (b) 0.30.3

Full solution

(a) The function F(y,z)F(y,z) obeys F(y,z)4+y4+z4+F(y,z)yz=17F(y,z)^4+y^4+z^4+F(y,z)yz=17 for all yy and zz near y=1y=1, z=2z=2. Applying the derivatives ∂∂y\pdiff{}{y} and ∂∂z\pdiff{}{z} to this equation gives

4F(y,z)3Fy(y,z)+4y3+Fy(y,z)yz+F(y,z)z=04F(y,z)3Fz(y,z)+4z3+Fz(y,z)yz+F(y,z)y=0\begin{align*} 4F(y,z)^3F_y(y,z)+4y^3+F_y(y,z)yz +F(y,z)z&=0 \\ 4F(y,z)^3F_z(y,z)+4z^3+F_z(y,z)yz +F(y,z)y&=0 \end{align*}

Substiututing F(1,2)=0F(1,2)=0, y=1y=1 and z=2z=2 gives

4+2Fy(1,2)=0   ⟹   Fy(1,2)=−232+2Fz(1,2)=0   ⟹   Fz(1,2)=−16\begin{alignat*}{3} 4+2F_y(1,2)&=0 & &\ \implies\ & &F_y(1,2)=-2 \\ 32+2F_z(1,2)&=0 & &\ \implies\ & & F_z(1,2)=-16 \end{alignat*}

(b) Using the tangent plane to x=F(y,z)x=F(y,z) at y=1y=1 and z=2z=2, which is

x≈F(1,2)+Fy(1,2) (y−1)+Fz(1,2) (z−2)\begin{equation*} x \approx F(1,2) +F_y(1,2)\,(y-1) +F_z(1,2)\,(z-2) \end{equation*}

with y=1.01y=1.01 and z=1.98z=1.98 gives

x=F(1.01,1.98)≈F(1,2)+Fy(1,2)(1.01−1)+Fz(1,2)(1.98−2)=0−2(.01)−16(−0.02)=0.3\begin{align*} x=F(1.01, 1.98) &\approx F(1,2)+F_y(1,2)(1.01-1)+F_z(1,2)(1.98-2) \\ &=0-2(.01)-16(-0.02) =0.3 \end{align*}

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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.