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Quiz 12: Directional derivatives and the gradient vector

Question

Question 1

Let

f (x, y) = e^{x^{2}} cos y

. What is

\nabla f (x, y) ?

a)	$e^{x^{2}} i + cos y j$	b)	$e^{x^{2}} cos y i - e^{x^{2}} sin y j$
c)	$2 x e^{x^{2}} cos y - e^{x^{2}} sin y$	d)	$e^{x^{2}} + cos y$
e)	$2 x e^{x^{2}} cos y i - e^{x^{2}} sin y j$

Not correct. Choice (a) is false.

Not correct. Choice (b) is false.

Not correct. Choice (c) is false.

Not correct. Choice (d) is false.

Your answer is correct.

Question 2

Let

f (x, y) = \frac{1}{x + y^{2}}

. Find the gradient vector

\nabla f (1, 1)

at the point (1,1).

a)	$- \frac{1}{4} i - \frac{1}{2} j$	b)	$- i - \frac{1}{2} j$
c)	$- \frac{1}{2} i - j$	d)	$\frac{1}{4} i + \frac{1}{2} j$
e)	$i + \frac{1}{2} j$

Your answer is correct.

Not correct. Choice (b) is false.

Not correct. Choice (c) is false.

Not correct. Choice (d) is false.

Not correct. Choice (e) is false.

Question 3

The directional derivative of

f (x, y) = x^{2} y^{3} + 2 x^{4} y

at the point (1,-2) in the direction

3 i - 4 j

a)	$\frac{1}{4} i + \frac{1}{2} j$	b)	$- 96 i - 56 j$
c)	-152	d)	-30.4
e)	$- 32 i + 14 j$

Not correct. Choice (a) is false.

Not correct. Choice (b) is false.

Not correct. Choice (c) is false.

Your answer is correct.

D_{u} f (1, - 2) = \nabla f (1, - 2) \cdot \hat{u}

. Here

\hat{u} = \frac{3}{5} i - \frac{4}{5} j

and

\nabla f (1, - 2) = - 32 i + 14 j

. The directional derivative is always a scalar as it is the dot product of the two vectors.

Not correct. Choice (e) is false.

Question 4

Find the direction where the directional derivative is greatest for the function

f (x, y) = 3 x^{2} y^{2} - x^{4} - y^{4}

at the point (1,2).

a)	$\frac{1}{\sqrt{2}} (- i + j)$	b)	$\frac{1}{\sqrt{2}} (i - j)$
c)	$\frac{1}{\sqrt{2}} (i + j)$	d)	$\frac{1}{\sqrt{5}} (2 i + j)$
e)	$- \frac{1}{\sqrt{5}} (i - j)$

Not correct. Choice (a) is false.

Your answer is correct.

The gradient vector gives the direction where the directional derivative is steepest.

\nabla f (1, - 2) = 20 i - 20 j

, so any positive scalar multiple of this vector would provide an answer to this question. One such vector is the unit vector in this direction,

\frac{1}{\sqrt{2}} (i - j)

Not correct. Choice (c) is false.

Not correct. Choice (d) is false.

Not correct. Choice (e) is false.

Question 5

Find the maximum directional derivative of the function

f (x, y) = x ln y + x^{2} y^{2}

at the point (-1,1).

a)	$- 2 i + j$	b)	$\frac{1}{\sqrt{5}} (- 2 i + j)$
c)	$1$	d)	$\sqrt{5}$
e)	$\frac{1}{\sqrt{5}}$

Not correct. Choice (a) is false.

Not correct. Choice (b) is false.

Not correct. Choice (c) is false.

Your answer is correct.

The maximum directional derivative is equal to the magnitude of the gradient vector. Here

| \nabla f | = | - 2 i + j | = | \sqrt{5} | .

Not correct. Choice (e) is false.

Question 6

Let the temperature at the point

(x, y)

in a flat plate be given by the function

T (x, y) = 3 x^{2} + 2 x y .

A tub of margarine is placed at (3,-6) in what direction should it be moved to cool most quickly?

a)	$6 i + 6 j$	b)	$i + j$
c)	$- i - j$	d)	$6 i - 12 j$
e)	(3,-6) is already the coolest point on the plate.

Not correct. Choice (a) is false.

Not correct. Choice (b) is false.

Your answer is correct.

\nabla T (3, - 6) = 6 i + 6 j

so the direction of most rapid increase in

T

i + j

. For the most rapid decrease the tub of margarine must be moved in the opposite direction,

- i - j

Not correct. Choice (d) is false.

Not correct. Choice (e) is false.

Question 7

Find a vector normal to the curve

x^{2} y + ln y - 2 x = 0

at the point (2,1).

a)	$2 i + 5 j$	b)	$5 i - 2 j$
c)	$- \frac{2}{5} i + j$	d)	$2 i + j$
e)	None of the above

Your answer is correct.

If a planar curve in the

x y

plane is defined implicitly by

f (x, y) = c

then the vector

\nabla f

is normal to the curve. Here

\nabla f = (2 x y - 2) i + (x^{2} + \frac{1}{y}) j

Not correct. Choice (b) is false.

Not correct. Choice (c) is false.

Not correct. Choice (d) is false.

Not correct. Choice (e) is false.

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