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Find a directional derivative

  • Thread starter betamu
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  • #1
12
3

Homework Statement


[/B]
Find the directional derivative of the function at the given point in the direction of the vector v.

$$g(s,t)=s\sqrt t, (2,4), \vec{v}=2\hat{i} - \hat{j}$$

Homework Equations



$$\nabla g(s,t) = <g_s(s,t), g_t(s,t)>\\
\vec{u} = \vec{v}/|\vec{v}|\\
D_u g(s,t) = \nabla g(s, t) \cdot \vec{u}$$

The Attempt at a Solution


I found $$\nabla g(s, t) =<\sqrt{t}, s/(2\sqrt{t})>$$ which gives $$\nabla g(2,4) = <2, 1/2>$$ and the directional vector to be $$<2/\sqrt{5}, -1/\sqrt{5}>$$ Which gives a dot product of $$5/2\sqrt{5}$$ but my book says that it should be $$7/2\sqrt{5}$$.
 

Answers and Replies

  • #2
12,653
9,172

Homework Statement


[/B]
Find the directional derivative of the function at the given point in the direction of the vector v.

$$g(s,t)=s\sqrt t, (2,4), \vec{v}=2\hat{i} - \hat{j}$$

Homework Equations



$$\nabla g(s,t) = <g_s(s,t), g_t(s,t)>\\
\vec{u} = \vec{v}/|\vec{v}|\\
D_u g(s,t) = \nabla g(s, t) \cdot \vec{u}$$

The Attempt at a Solution


I found $$\nabla g(s, t) =<\sqrt{t}, s/(2\sqrt{t})>$$ which gives $$\nabla g(2,4) = <2, 1/2>$$ and the directional vector to be $$<2/\sqrt{5}, -1/\sqrt{5}>$$ Which gives a dot product of $$5/2\sqrt{5}$$ but my book says that it should be $$7/2\sqrt{5}$$.
What is ##\langle (2,1/2)\,,\,(2,-1)\rangle \,?##
 
  • #3
33,270
4,967

Homework Statement


[/B]
Find the directional derivative of the function at the given point in the direction of the vector v.

$$g(s,t)=s\sqrt t, (2,4), \vec{v}=2\hat{i} - \hat{j}$$

Homework Equations



$$\nabla g(s,t) = <g_s(s,t), g_t(s,t)>\\
\vec{u} = \vec{v}/|\vec{v}|\\
D_u g(s,t) = \nabla g(s, t) \cdot \vec{u}$$

The Attempt at a Solution


I found $$\nabla g(s, t) =<\sqrt{t}, s/(2\sqrt{t})>$$ which gives $$\nabla g(2,4) = <2, 1/2>$$ and the directional vector to be $$<2/\sqrt{5}, -1/\sqrt{5}>$$ Which gives a dot product of $$5/2\sqrt{5}$$
Check your work on the dot product. I think you missed that the second fraction has a 2 in the denominator.
betamu said:
but my book says that it should be $$7/2\sqrt{5}$$.
I get this, as well, but it should be written as ##.7/(2\sqrt{5})## or better, as $$\frac 7 {2\sqrt 5}$$
 
  • #4
12
3
What is ##\langle (2,1/2)\,,\,(2,-1)\rangle \,?##
I'm unsure what you mean. You're putting the gradient vector and v into one vector together?
 
  • #5
12
3
Check your work on the dot product. I think you missed that the second fraction has a 2 in the denominator. I get this, as well, but it should be written as ##.7/(2\sqrt{5})## or better, as $$\frac 7 {2\sqrt 5}$$
Haha wow. Yeah you're right. Well at least I gained some experience in writing out latex code in posting this. Thanks!
 
  • #6
12,653
9,172
I'm unsure what you mean. You're putting the gradient vector and v into one vector together?
Yes, it is ##\nabla f \cdot \vec{v}##, I simply left out the norm ##\sqrt{5}## since your mistake was the dot product, not the factor.
 
  • #7
33,270
4,967
I'm unsure what you mean. You're putting the gradient vector and v into one vector together?
That should be the dot product of two vectors. fresh_42's notation might be for the inner product, a generalization of the dot product.
 
  • #8
12
3
Oh alright, I'd never seen that notation before. Thank you both!
 

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