Exam 2 Review Sheet

asked 2014-07-16 07:09:35 -0600

Anonymous
740 ●2 ●27 ●49

Number 4, part c, of the exam 2 review sheet asks: Let $f(x,y)=xy^3$ (c) From the point $(2,1)$, is there any direction $u$ so that $D_uf(2,1)=10$?

So far I have calculated the gradient to be $$\bigtriangledown f(2,1)=<8,6>$$ The idea I had was to create a dot product of $\bigtriangledown f(2,1)$ and a general $u$ vector, and set it equal to 10. The trouble I'm having is figuring out how to solve the two unknowns because I only have one equation.

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answered 2014-07-16 14:20:34 -0600

Justin
1009 ●2 ●20 ●38

I agree with Tiffany regarding $\nabla f(2, 1) = \langle 1, 6 \rangle$. This is actually the only piece of information that you need to solve part (c). Since $\nabla f$ points in the direction that $f$ is changing the fastest, there will be no direction vector where $f$ changes faster than $|| \nabla f ||$.

$$|| \ \nabla f \ || = \sqrt{1^2 + 6^2} = \sqrt{37} $$

So is $\sqrt{37}$ greater than or less than $10$? We can square both sides of the equation to find out.

$$\sqrt{37} \stackrel{?}{\geq} 10 $$ $$37 \stackrel{?}{\geq} 100 $$

This is clearly false and therefore there is no direction such that $D_{\ \vec u}\ f(2, 1) = 10$.

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answered 2014-07-16 12:10:31 -0600

Tiffany
656 ●3 ●18 ●34

I agree with you on setting $$ D_\vec u f = 10$$ then solve backwards to find the $\vec u. $ or you could keep plugging a bunch of different numbers in to try to find a dot product vector that would equal 10, but I feel like it would be easier to work the problem backwards to find it.

I'm also getting a different answer for my gradient. I'm not very good a derivatives though so it could be why. But the partial derivative with respect to x should be $ y^3$ right? Since the derivative of x would be 1 then you leave the second part alone. Then since y is constant the first left alone times the second part would be 0. Which would make the partial derivative with respect to y be $3xy^2$ which plugging in $ \langle 2,1\rangle$ would be $\nabla f =\langle 1,6 \rangle $ if I'm thinking about that wrong could someone explain where my thought process went bad?

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answered 2014-07-17 06:58:19 -0600

Anonymous
740 ●2 ●27 ●49

Okay, I see what I did wrong; I plugged the x value into y for some reason on the x partial derivative.

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Exam 2 Review Sheet

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