- #1

sss1

- 50

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- Homework Statement
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- Relevant Equations
- NA

I just wanted to know if my solution to part (b) is correct. Here's what I did:

I took the partial derivative with respect to x and y, which gave me
respectively.

Then I computed the partial derivatives at (-3,4) which gave me 3/125 for partial derivative wrt x and -4/125 for partial derivative wrt y

Then since directional derivative requires a direction, I just chose an arbitrary one, uhat=(a,b)

since u is a unit vector that means sqrt(a^2+b^2)=1, or a^2+b^2=1.

I then solved for a, which is the plus minus of sqrt(1-b^2).

I just chose to use the positive answer here instead.

So the directional derivative is 3sqrt(1-b^2)/125-4b/125

To maximise this I took the derivative, which is -3b/125sqrt(1-b^2)-4/125

I set it to 0 and solved for b, which gave me -4/5.

So there is a local maximum for the directional derivative at b=-4/5 (I evaluated the second derivative and it was positive).

So that means on both sides of b=-4/5 the directional derivative decreases.

Subsituting b=-0.8 into my formula for a, a=sqrt(1-b^2), gives me a=0.6

So the directional derivative should be a maximum in the direction given by the unit vector (0.6, -0.8, 0) with magnitude 0.6(3/125)-0.8(-4/125) which is 0.04?

Although I didn't try the negative answer for a, a=-sqrt(1-b^2), I believe that this will yield a smaller answer because if a is negative, then a negative number times a positive number (3/125) will decrease the answer overall? Is my logic correct?

I took the partial derivative with respect to x and y, which gave me

Then I computed the partial derivatives at (-3,4) which gave me 3/125 for partial derivative wrt x and -4/125 for partial derivative wrt y

Then since directional derivative requires a direction, I just chose an arbitrary one, uhat=(a,b)

since u is a unit vector that means sqrt(a^2+b^2)=1, or a^2+b^2=1.

I then solved for a, which is the plus minus of sqrt(1-b^2).

I just chose to use the positive answer here instead.

So the directional derivative is 3sqrt(1-b^2)/125-4b/125

To maximise this I took the derivative, which is -3b/125sqrt(1-b^2)-4/125

I set it to 0 and solved for b, which gave me -4/5.

So there is a local maximum for the directional derivative at b=-4/5 (I evaluated the second derivative and it was positive).

So that means on both sides of b=-4/5 the directional derivative decreases.

Subsituting b=-0.8 into my formula for a, a=sqrt(1-b^2), gives me a=0.6

So the directional derivative should be a maximum in the direction given by the unit vector (0.6, -0.8, 0) with magnitude 0.6(3/125)-0.8(-4/125) which is 0.04?

Although I didn't try the negative answer for a, a=-sqrt(1-b^2), I believe that this will yield a smaller answer because if a is negative, then a negative number times a positive number (3/125) will decrease the answer overall? Is my logic correct?

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