$f(x ,y)$ differentiable all over the plain. $g(u, v) = f(u^2 - v^2, u^2v).$ if $\nabla f (-3, 2) = 2 \vec i + \vec j$ , calculate $ \nabla g(1,2)$

Question

unfortunately, I had to miss the lecture that gradients leant and I don't know how to solve this question.

Let $f(x ,y)$ differentiable all over the plain.

Let $g(u, v) = f(u^2 - v^2, u^2v).$

if $\nabla f (-3, 2) = 2 \vec i + \vec j$ , calculate $ \nabla g(1,2)$

($\vec i$ is unit vector $[1, 0, 0]$ and $\vec j$ is unit vector $[0, 1, 0]$.

I read about gradients, but still don't know how to calculate $\nabla g(1,2).$

Can you please help me solve this question? I'd rather to get help such as hints and general explaination, than a final solution.

Thanks in advance!

$\nabla g=(\partial_x g,\partial_y g)$. What's your question on this definition? — Shuchang, Dec 29 '13 at 03:04
@Shuchang I have to substitute $(1,2)$ in the partial derivatives? — Billie, Dec 29 '13 at 03:07
Not exactly, $(1,2)$ will be plugged in after the gradient is taken. — Shuchang, Dec 29 '13 at 03:18

Kal S. · Accepted Answer · 2013-12-29T03:17:38.707

0

$g_u=2uf_u(u^2-v^2,u^2v)+2uvf_v(u^2-v^2,u^2v)$

$g_v=-2vf_u(u^2-v^2,u^2v)+u^2f_v(u^2-v^2,u^2v)$

$g_u(1,2)=2f_u(-3,2)+4f_v(-3,2)=4+4=8$

$g_v(1,2)=-4f_u(-3,2)+f_v(-3,2)=-7$

since $\nabla f(-3,2)=(2,1)$ , means that $f_u(-3,2)=2, f_v(-3,2)=1$

So $\nabla g(1,2)=(g_u(1,2),g_v(1,2))=(8,-7)$

edited Dec 29 '13 at 03:17

answered Dec 29 '13 at 03:11

Kal S.

3,781

How did you get that $f_u(-3, 2) = 1 = f_v(-3, 2)$? – Billie Dec 29 '13 at 03:15
1

I corrected my answer. – Kal S. Dec 29 '13 at 03:18

$f(x ,y)$ differentiable all over the plain. $g(u, v) = f(u^2 - v^2, u^2v).$ if $\nabla f (-3, 2) = 2 \vec i + \vec j$ , calculate $ \nabla g(1,2)$

1 Answers1