## MATH1111 Quizzes

The Partial Derivative Quiz
Web resources available Questions

This quiz tests the work covered in the lecture on partial derivatives and corresponds to Section 14.1 of the textbook Calculus: Single and Multivariable (Hughes-Hallett, Gleason, McCallum et al.).
There is a discussion on partial derivatives at http://www.bio.brandeis.edu/biomath/populate/surface.html

There are more web quizzes at Wiley, select Section 1. This quiz has 15 questions.

Suppose $f\left(x,y\right)=2{x}^{2}y$. Which of the following gives the best estimate of the quotient ${f}_{x}\left(2,1\right)$ and ${f}_{y}\left(2,1\right)$ where $h=1\phantom{\rule{0.3em}{0ex}}?$ Exactly one option must be correct)
 a) ${f}_{x}\left(2,1\right)=8$ and ${f}_{y}\left(2,1\right)=8\phantom{\rule{0.3em}{0ex}}.$ b) ${f}_{x}\left(2,1\right)=2$ and ${f}_{y}\left(2,1\right)=1\phantom{\rule{0.3em}{0ex}}.$ c) ${f}_{x}\left(2,1\right)=1$ and ${f}_{y}\left(2,1\right)=4\phantom{\rule{0.3em}{0ex}}.$ d) ${f}_{x}\left(2,1\right)=10$ and ${f}_{y}\left(2,1\right)=8\phantom{\rule{0.3em}{0ex}}.$

Choice (a) is incorrect
Try again, these are the exact values of the partial derivatives.
Choice (b) is incorrect
Choice (c) is incorrect
Try again, you may have forgotten to square your $x$ values.
Choice (d) is correct!
$\underset{h\to 0}{lim}\frac{2×{\left(2+h\right)}^{2}×1-2×{2}^{2}×1}{h}=\frac{18-8}{1}=10$ when $h=1$
and $\underset{h\to 0}{lim}\frac{2×{2}^{2}×\left(1+h\right)-2×{2}^{2}×1}{h}=\frac{16-8}{1}=8$ when $h=1\phantom{\rule{0.3em}{0ex}}.$
Refer to the table on page 686 of Calculus: Single and Multivariable (Hughes-Hallett, Gleason, McCallum et al.) giving the temperature, $T\phantom{\rule{0.3em}{0ex}},$ in ${o}^{}C\phantom{\rule{0.3em}{0ex}}$ of a plate, as a function of its distance from the bottom corner of the plate.

Which of the following statements are the most accurate? (Zero or more options can be correct)
 a) ${T}_{x}\left(3,2\right)\approx 4{5}^{o}C∕m$ when $h=1$ b) ${T}_{x}\left(3,2\right)\approx 2{5}^{o}C∕m$ when $h=1$ c) ${T}_{y}\left(3,2\right)\approx -1{5}^{o}C∕m$ when $h=1$ d) ${T}_{y}\left(3,2\right)\approx -1{0}^{o}C∕m$ when $h=1$

There is at least one mistake.
For example, choice (a) should be True.
$T\left(3,2\right)=145$ and $T\left(4,2\right)=190$ so ${T}_{x}\left(3,2\right)\approx \frac{190-145}{1}=4{5}^{o}C∕m$ when $h=1\phantom{\rule{0.3em}{0ex}}.$
There is at least one mistake.
For example, choice (b) should be False.
You may have been looking at $T\left(2,3\right)\phantom{\rule{0.3em}{0ex}}.$ This is also a good approximation for ${T}_{x}\left(2,2\right)\phantom{\rule{0.3em}{0ex}}.$
There is at least one mistake.
For example, choice (c) should be False.
This is a good approximation for ${T}_{y}\left(3,1\right)\phantom{\rule{0.3em}{0ex}}.$
There is at least one mistake.
For example, choice (d) should be True.
$T\left(3,2\right)=145$ and $T\left(3,3\right)=135$ so ${T}_{y}\left(3,2\right)\approx \frac{135-145}{1}=-1{0}^{o}C∕m$ when $h=1\phantom{\rule{0.3em}{0ex}}.$
Correct!
1. True $T\left(3,2\right)=145$ and $T\left(4,2\right)=190$ so ${T}_{x}\left(3,2\right)\approx \frac{190-145}{1}=4{5}^{o}C∕m$ when $h=1\phantom{\rule{0.3em}{0ex}}.$
2. False You may have been looking at $T\left(2,3\right)\phantom{\rule{0.3em}{0ex}}.$ This is also a good approximation for ${T}_{x}\left(2,2\right)\phantom{\rule{0.3em}{0ex}}.$
3. False This is a good approximation for ${T}_{y}\left(3,1\right)\phantom{\rule{0.3em}{0ex}}.$
4. True $T\left(3,2\right)=145$ and $T\left(3,3\right)=135$ so ${T}_{y}\left(3,2\right)\approx \frac{135-145}{1}=-1{0}^{o}C∕m$ when $h=1\phantom{\rule{0.3em}{0ex}}.$
Use the difference quotient with $△x=0.2$ and $△y=0.1$ to estimate ${f}_{x}\left(1,2\right)$ and ${f}_{y}\left(1,2\right)$ where $f\left(x,y\right)={e}^{-2x}cosy\phantom{\rule{0.3em}{0ex}}.$
Which of the following are the correct estimations? Exactly one option must be correct)
 a) ${f}_{x}\left(1,2\right)\approx \frac{{e}^{-2.4}cos2-{e}^{-2}cos2}{0.2}$ and ${f}_{y}\left(1,2\right)\approx \frac{{e}^{-2}cos2.2-{e}^{-2}cos2}{0.2}$ b) ${f}_{x}\left(1,2\right)\approx \frac{{e}^{-2.4}cos2-{e}^{-2}cos2}{0.2}$ and ${f}_{y}\left(1,2\right)\approx \frac{{e}^{-2}cos2.1-{e}^{-2}cos2}{0.1}$ c) ${f}_{x}\left(1,2\right)\approx \frac{{e}^{-2.2}cos2-{e}^{-2}cos2}{0.2}$ and ${f}_{y}\left(1,2\right)\approx \frac{{e}^{-2}cos2.2-{e}^{-2}cos2}{0.2}$ d) ${f}_{x}\left(1,2\right)\approx \frac{{e}^{-2.2}cos2-{e}^{-2}cos2}{0.2}$ and ${f}_{y}\left(1,2\right)\approx \frac{{e}^{-2}cos2.1-{e}^{-2}cos2}{0.1}$

Choice (a) is incorrect
Try again, you have correctly estimated ${f}_{x}\left(2,1\right)$ but you have not used the correct value for $△y\phantom{\rule{0.3em}{0ex}}.$
Choice (b) is correct!
You have carefully substitute into the formulae
${f}_{x}\left(1,2\right)\approx \frac{{e}^{-2×\left(1+△x\right)}cos2-{e}^{-2×1}cos2}{△x}$ and
${f}_{y}\left(1,2\right)\approx \frac{{e}^{-2×1}cos\left(2+△y\right)-{e}^{-2×1}cos2}{△y}\phantom{\rule{0.3em}{0ex}}.$
Choice (c) is incorrect
Try again, neither answer is correct. Carefully substitute into the formulae
$\frac{{e}^{-2×\left(1+△x\right)}cos2-{e}^{-2×1}cos2}{△x}$ and
$\frac{{e}^{-2×1}cos\left(2+△y\right)-{e}^{-2×1}cos2}{△y}\phantom{\rule{0.3em}{0ex}}.$
Choice (d) is incorrect
Try again, you have correctly estimated ${f}_{y}\left(2,1\right)$ but you have not correctly used the value for $△x\phantom{\rule{0.3em}{0ex}}.$
Consider the contour diagram below for $f\left(x,y\right)\phantom{\rule{0.3em}{0ex}}.$

Which of the following are the most accurate estimates for $f\left(2,4\right)\phantom{\rule{0.3em}{0ex}},\phantom{\rule{1em}{0ex}}{f}_{x}\left(2,4\right)$ and ${f}_{y}\left(2,4\right)\phantom{\rule{0.3em}{0ex}}?$ Exactly one option must be correct)
 a) $f\left(2,4\right)=6\phantom{\rule{0.3em}{0ex}}\phantom{\rule{1em}{0ex}}{f}_{x}\left(2,4\right)=-3$ and ${f}_{y}\left(2,4\right)=\frac{3}{2}$ b) $f\left(2,4\right)=-3\phantom{\rule{0.3em}{0ex}}\phantom{\rule{1em}{0ex}}{f}_{x}\left(2,4\right)=\frac{3}{2}$ and ${f}_{y}\left(2,4\right)=-3$ c) $f\left(2,4\right)=6\phantom{\rule{0.3em}{0ex}}\phantom{\rule{1em}{0ex}}{f}_{x}\left(2,4\right)=\frac{3}{2}$ and ${f}_{y}\left(2,4\right)=-3$ d) $f\left(2,4\right)=-3\phantom{\rule{0.3em}{0ex}}\phantom{\rule{1em}{0ex}}{f}_{x}\left(2,4\right)=-3$ and ${f}_{y}\left(2,4\right)=\frac{3}{2}$

Choice (a) is correct!
The point $\left(2,4\right)$ is on the +6 contour. As $x$ increases $f\left(x,y\right)$ decreases so ${f}_{x}\left(x,y\right)$ is negative. The contours are evenly spread and as $x$ increases by 1 $f\left(x,y\right)$ decreases by 3, so ${f}_{x}\left(2,4\right)=-3\phantom{\rule{0.3em}{0ex}}.$
Similarly ${f}_{y}\left(x,y\right)$ is positive as we see $f\left(x,y\right)$ increases as $y$ is increases. The contours are evenly spread and as $y$ increases by 2 $f\left(x,y\right)$ increases by 3, so ${f}_{x}\left(2,4\right)=\frac{3}{2}\phantom{\rule{0.3em}{0ex}}.$
Choice (b) is incorrect
Try again, you seem to have confused your $x$ and $y$ values.
Choice (c) is incorrect
Try again, you have the correct value for $f\left(2,4\right)\phantom{\rule{0.3em}{0ex}}.$
Choice (d) is incorrect
Try again, you do no have the correct value for $f\left(2,4\right)\phantom{\rule{0.3em}{0ex}}.$