function of two variables, directional derivatives

**ldaniels241** · April 15th 2010, 06:47 AM

Let f(x,y) be a function that is differtiable everywhere. At a certain point P in the xy plane, the directional derivative of F in the direction of i - j is sqrt(2) and the directional derivative of f in the direction of i + j is 3sqrt(2). What is the maximum directional derivative at P?

A) 3sqrt(2) B) 2sqrt(5) C) 43sqrt(5) D) 6 E) 8

I know the directional derivative is the del of f(x,y) = <fx, fy>. I also know that the directional derivative also points in the direction of maximum increase.

So should i have <fx, fy> = i - j meaning fx = 1 and fy = -1 evaluated at P?

Same for <fx, fy> = i + j meaning fx = 1 and fy = 1 at P?

I am not sure what to do can someone help me out

**TheEmptySet** · April 15th 2010, 10:34 AM

Originally Posted by ldaniels241

Let f(x,y) be a function that is differtiable everywhere. At a certain point P in the xy plane, the directional derivative of F in the direction of i - j is sqrt(2) and the directional derivative of f in the direction of i + j is 3sqrt(2). What is the maximum directional derivative at P?

A) 3sqrt(2) B) 2sqrt(5) C) 43sqrt(5) D) 6 E) 8

I know the directional derivative is the del of f(x,y) = <fx, fy>. I also know that the directional derivative also points in the direction of maximum increase.

So should i have <fx, fy> = i - j meaning fx = 1 and fy = -1 evaluated at P?

Same for <fx, fy> = i + j meaning fx = 1 and fy = 1 at P?

I am not sure what to do can someone help me out

Since they are using an arbitary point P we might as well make it the center of our coordinate system.

This diagram may help

Attachment 16398

The two vectors in black can be used as a basis for all points on the unit circle in the picture. i.e any green line on the unit circle can be written as a linear combination of these two vectos

$x\vec{i}+y\vec{j}=\frac{c_1}{\sqrt{2}}(\vec{i}-\vec{j})+\frac{c_2}{\sqrt{2}}(\vec{i}+\vec{j})$

Provided that $c_1^2+c_2^2=1$

or in shorter notation

$(x,y)=c_1v_1+c_2v_2$

We know that

$\nabla f(0,0)\cdot \underbrace{\frac{1}{\sqrt{2}}(\vec{i}-\vec{j})}_{v_1}=\sqrt{2}$
and
$\nabla f(0,0)\cdot \underbrace{\frac{1}{\sqrt{2}}(\vec{i}+\vec{j})}_{ v_2}=3\sqrt{2}$

Now the function we want to maximize is

$M(c_1,c_2)=\nabla f(0,0)\cdot (x\vec{i}+y\vec{j})=\nabla f(0,0)\cdot (c_1v_1+c_2v_2)=$
$c_1\nabla f(0,0)\cdot v_1+c_2 \nabla f(0,0)\cdot c_2=\sqrt{2}c_1+3\sqrt{2}c_2$

Now just maximize this function with the constraint that $c_1^2+c_2^2=1$

Now when I solve for $c_1,c_2$ I get

$c_1=\frac{1}{\sqrt{10}},c_2=\frac{3}{\sqrt{10}}$

This gives the maximum of the function $M=2\sqrt{5}$

**ldaniels241** · April 15th 2010, 10:48 AM

Thank you so much. I just wish I could view the picture though.

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