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The Partial Derivative Quiz

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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.

Question 1

Suppose $f(x,y) = 2x2y$ . Which of the following gives the best estimate of the quotient $fx(2,1)$ and $fy(2,1)$ where $h = 1?$

a)	$fx(2,1) = 8$ and $fy(2,1) = 8.$	b)	$fx(2,1) = 2$ and $fy(2,1) = 1.$
c)	$fx(2,1) = 1$ and $fy(2,1) = 4.$	d)	$fx(2,1) = 10$ and $fy(2,1) = 8.$

Not correct. Choice (a) is false.

Try again, these are the exact values of the partial derivatives.

Not correct. Choice (b) is false.

Try again, check your multiplication.

Not correct. Choice (c) is false.

Try again, you may have forgotten to square your $x$ values.

Your answer is correct.

$2×-(2+-h)2 ×-1--2-×-22-×-1 18--8- lhim→0 h = 1 = 10$ when $h = 1$
and $2 2 lim 2×-2-×-(1+-h)--2-×-2-×-1= 16--8-= 8 h→0 h 1$ when $h = 1.$

Question 2

Refer to the table on page 686 of Calculus: Single and Multivariable (Hughes-Hallett, Gleason, McCallum et al.) giving the temperature, $T,$ in $oC$ 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?

a)		$Tx(3,2) ≈ 45oC∕m$ when $h = 1$		b)		$Tx(3,2) ≈ 25oC ∕m$ when $h = 1$
c)		$Ty(3,2) ≈ - 15oC∕m$ when $h = 1$		d)		$Ty(3,2) ≈ - 10oC∕m$ when $h = 1$

There is at least one mistake.
For example, choice (a) should be true.

$T (3,2) = 145$ and $T(4,2) = 190$ so $190---145 o Tx(3,2) ≈ 1 = 45 C∕m$ when $h = 1.$

There is at least one mistake.
For example, choice (b) should be false.

You may have been looking at $T (2,3).$ This is also a good approximation for $Tx (2,2).$

There is at least one mistake.
For example, choice (c) should be false.

This is a good approximation for $Ty(3,1).$

There is at least one mistake.
For example, choice (d) should be true.

$T (3,2) = 145$ and $T(3,3) = 135$ so $Ty(3,2) ≈ 135---145-= - 10oC ∕m 1$ when $h = 1.$

Your answers are correct

True. $T (3,2) = 145$ and $T(4,2) = 190$ so $190---145 o Tx(3,2) ≈ 1 = 45 C∕m$ when $h = 1.$
False. You may have been looking at $T (2,3).$ This is also a good approximation for $Tx (2,2).$
False. This is a good approximation for $Ty(3,1).$
True. $T (3,2) = 145$ and $T(3,3) = 135$ so $Ty(3,2) ≈ 135---145-= - 10oC ∕m 1$ when $h = 1.$

Question 3

Use the difference quotient with $△x = 0.2$ and $△y = 0.1$ to estimate $fx(1,2)$ and $fy(1,2)$ where $f(x,y) = e-2xcosy.$
Which of the following are the correct estimations?

a)		$e-2.4cos2- e-2cos2 fx(1,2) ≈--------0.2--------$ and $e-2cos2.2 - e-2cos2 fy(1,2) ≈ --------0.2--------$
b)		$e-2.4-cos2---e-2cos2 fx(1,2) ≈ 0.2$ and $e-2cos2.1--e-2cos2 fy(1,2) ≈ 0.1$
c)		$e-2.2cos2 - e-2 cos 2 fx(1,2) ≈ -------0.2--------$ and $e-2cos2.2- e-2cos2 fy(1,2) ≈--------0.2--------$
d)		$f (1,2) ≈ e-2.2cos2--e-2cos2 x 0.2$ and $e-2cos2.1--e-2cos2 fy(1,2) ≈ 0.1$

Not correct. Choice (a) is false.

Try again, you have correctly estimated $fx(2,1)$ but you have not used the correct value for $△y.$

Your answer is correct.

You have carefully substitute into the formulae
$fx(1,2) ≈ e-2×(1+△x)cos2--e-2×1cos2 △x$ and
$e-2×1cos(2+ △y )- e-2×1 cos2 fy(1,2) ≈------------△y-------------.$

Not correct. Choice (c) is false.

Try again, neither answer is correct. Carefully substitute into the formulae
$e--2×-(1+△x)cos2--e-2×1-cos2 △x$ and
$e--2×1-cos(2+-△y-)--e-2×1cos2. △y$

Not correct. Choice (d) is false.

Try again, you have correctly estimated $fy(2,1)$ but you have not correctly used the value for $△x.$

Question 4

Consider the contour diagram below for $f(x,y).$

Which of the following are the most accurate estimates for $f(2,4), fx(2,4)$ and $fy(2,4)?$

a)	$f (2,4) = 6 fx(2,4) = - 3$ and $fy(2,4) = 32$	b)	$f(2,4) = - 3 fx(2,4) = 32$ and $fy(2,4) = - 3$
c)	$f(2,4) = 6 fx(2,4) = 32$ and $fy(2,4) = - 3$	d)	$f(2,4) = - 3 fx(2,4) = - 3$ and $fy(2,4) = 32$

Your answer is correct.

The point $(2,4)$ is on the +6 contour. As $x$ increases $f(x,y)$ decreases so $fx(x,y)$ is negative. The contours are evenly spread and as $x$ increases by 1 $f(x,y)$ decreases by 3, so $fx(2,4) = - 3.$
Similarly $fy(x,y)$ is positive as we see $f(x,y)$ increases as $y$ is increases. The contours are evenly spread and as $y$ increases by 2 $f(x,y)$ increases by 3, so $fx(2,4) = 32.$

Not correct. Choice (b) is false.