Vector analysis: Gradient of f(M_0)...

**Also sprach Zarathustra** · June 6th 2011, 08:31 AM

I will appreciate if someone could solve and explain me step by step the following problem:

Let $f:\mathbb{R}^3\to\mathbb{R}$ be a differentiable in every point and satisfies the following:

1. $f(x,y,2x^2+y^2)=3x-5y$ for all $(x,y)$ .

2. The directional derivative of $f$ at point $M_0(1,2,6)$ at direction $a=(\frac{1}{3},\frac{2}{3},\frac{2}{3})$ equals to $1$ .

Compute $\nabla f(M_0)$ .

Thank you.

**TheEmptySet** · June 6th 2011, 09:28 AM

Originally Posted by Also sprach Zarathustra

I will appreciate if someone could solve and explain me step by step the following problem:

Let $f:\mathbb{R}^3\to\mathbb{R}$ be a differentiable in every point and satisfies the following:

1. $f(x,y,2x^2+y^2)=3x-5y$ for all $(x,y)$ .

2. The directional derivative of $f$ at point $M_0(1,2,6)$ at direction $a=(\frac{1}{3},\frac{2}{3},\frac{2}{3})$ equals to $1$ .

Compute $\nabla f(M_0)$ .

Thank you.

The directional derivative in the direction u is given by

$D_u(f)=\nabla f \cdot \frac{\mathbf{u}}{||\mathbf{u}||}$ using this gives

$\nabla f(1,2,6) \cdot (\mathbf{i}+2\mathbf{j}+2\mathbf{k})=3 =f_{x}(1,2,6)+2f_{y}(1,2,6)+2f_{z}(1,2,6)$

This tells us that $f_{z}(1,2,6)=\frac{1}{2}\left[ 3-f_{x}(1,2,6)-2f_{y}(1,2,6) \right]$

Now we have that $f(x,y,z(x,y))=f(x,y,2x^2+y^2) \implies z(x,y)=2x^2+y^2$

$\nabla f=f_x\mathbf{i}+f_y\mathbf{j}+\left(\frac{\partial f}{\partial z}\frac{\partial z}{\partial x}+\frac{\partial f}{\partial z}\frac{\partial z}{\partial y } \right)\mathbf{k}$

$\nabla f=f_x\mathbf{i}+f_y\mathbf{j}+\left( 4xf_z(x,y,z)+2yf_z(x,y,z) \right)\mathbf{k}$

but we know what $f(x,y)$ is and can take the needed partial derivatives to finish.

**Also sprach Zarathustra** · June 6th 2011, 11:28 AM

Originally Posted by TheEmptySet

The directional derivative in the direction u is given by

$D_u(f)=\nabla t \cdot \frac{\mathbf{u}}{||\mathbf{u}||}$ using this gives

$\nabla f(1,2,6) \cdot (\mathbf{i}+2\mathbf{j}+2\mathbf{k})=9 =f_{x}(1,2,6)+2f_{y}(1,2,6)+2f_{z}(1,2,6)$

This tells us that $f_{z}(1,2,6)=\frac{1}{2}\left[ 9-f_{x}(1,2,6)-2f_{y}(1,2,6) \right]$

Now we have that $f(x,y,z(x,y))=f(x,y,2x^2+y^2) \implies z(x,y)=2x^2+y^2$

$\nabla f=f_x\mathbf{i}+f_y\mathbf{j}+\left(\frac{\partial f}{\partial z}\frac{\partial z}{\partial x}+\frac{\partial f}{\partial z}\frac{\partial z}{\partial y } \right)\mathbf{k}$

$\nabla f=f_x\mathbf{i}+f_y\mathbf{j}+\left( 4xf_z(x,y,z)+2yf_z(x,y,z) \right)\mathbf{k}$

but we know what $f(x,y)$ is and can take the needed partial derivatives to finish.

First, thank you very much! Now, I'm not understand how to arrive to the answer 7i-j-k (the answer appears in a book).

Thanks again!

**TheEmptySet** · June 6th 2011, 01:04 PM

Yes I can see why you were confused I wrote the incorrect magnitude for the vector. It should be

$3 =f_{x}(1,2,6)+2f_{y}(1,2,6)+2f_{z}(1,2,6)$

Now we just need to compute two other partial derivatives. Using the chain rule we get that

$\frac{\partial }{\partial x}f(x,y,z(x,y))=f_x+4xf_z=3$

$\frac{\partial }{\partial y}f(x,y,z(x,y))=f_y+2yf_z=-5$

$3 =f_{x}(1,2,6)+2f_{y}(1,2,6)+2f_{z}(1,2,6)$

This gives a system of 3 equations in three unknowns for the partial derivatives of the function at the point (1,2,6).

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