Question

find the directional derivative of the function at the given point in the direction of the vector v

Answer 1

g(s,t) = s*sqrt(t) , P(2,4), v=2i-j

Answer 2

i got gs and gt. gs= sqrt(t) and gt= s/(2sqrt(t))

Answer 3

and u = <1/sqrt(5) , -1/(2sqrt(5))

Answer 4

the answer on the back of the book is 7/(2sqrt(5)) but im getting a different answer

Answer 5

@myininaya @zepdrix

Answer 6

@wio

Answer 7

Do you know how to find the gradient?

Answer 8

hey man i keep getting a different answer than the one from the book could you help me out real quick

Answer 9

yeah i multiplied the u * gs +u*gt

Answer 10

\[
g(s,t) = s\cdot \sqrt{t} \implies \nabla g = \sqrt{t}~\mathbf i+\frac{s}{\sqrt t}~\mathbf j
\]

Answer 11

\[
D_{\mathbf v}g = \nabla g\cdot \mathbf v= 2\sqrt t -\frac{s}{\sqrt t}
\]

Answer 12

yeah its \[\frac{ s }{ 2\sqrt{t} }\]

Answer 13

for the j

Answer 14

\[
\ldots = \frac{2t}{\sqrt t}-\frac{s}{\sqrt t} = \frac{2t-s}{\sqrt t}
\]

Answer 15

okay

Answer 16

Is that what you got for \(D_{\mathbf v}g(s,t)\)?

Answer 17

no i got 1/sqrt(5) * (sqrt(t)) -1/2sqrt(5)* (s/2sqrt(t))

Answer 18

What is that even supposed to be?

Answer 19

You should still have variables because you don't plug points in at this stage

Answer 20

Did you find \(\nabla g(\mathbf u)\) rfirst?

Answer 21

yeah i got \[\frac{ 2 }{ 2\sqrt{5}} and -\frac{ 1}{ 2\sqrt{5} }\]

Answer 22

is it right?

Answer 23

So you're saying \(t=5\) and \(s=1\)?

Answer 24

I think your gradient is correct, but you're not plugging \(u\) in correctly, or you aren't telling me what the correct \(u\) is.