Directional derivative formula

[mit_hacker]

Homework Statement

(Q) The derivative of f(x,y) at Po(1,2) in the direction i + j is 2sqrt(2) and in the direction of -2j is -3. What is the derivative of f in the direction of -i - 2j? Give reasons for your answers.

Homework Equations

The directional derivative is given by the formula:

∂f/∂x i+∂f/∂y j

The Attempt at a Solution

You get simultaneous equations when you apply the above equation and you find that

∂f/∂y = 3/2.
And ∂f/∂x = [4sqrt(2) - 3] / 2.

Then applying the dot product of this and -i - 2j, you get [-3-4sqrt(2)] / 2 but the answer is supposed to be -7/sqrt(5). How did they get that??

 

[HallsofIvy]

Homework Statement

(Q) The derivative of f(x,y) at Po(1,2) in the direction i + j is 2sqrt(2) and in the direction of -2j is -3. What is the derivative of f in the direction of -i - 2j? Give reasons for your answers.

Homework Equations

The directional derivative is given by the formula:

∂f/∂x i+∂f/∂y j

The Attempt at a Solution

You get simultaneous equations when you apply the above equation and you find that

∂f/∂y = 3/2.

Yes, that's true.

And ∂f/∂x = [4sqrt(2) - 3] / 2.

No, that's not true. "The derivative of f(x,y) at Po(1,2) in the direction i + j is 2sqrt(2)" tells you that f_x/\sqrt{2}+ f_y/\sqrt{2}= 2\sqrt{2} (dividing by the length of i+ j) or that f_x+ f_y=4. Since f_y= 3/2, that gives f_x= 5/2

Then applying the dot product of this and -i - 2j, you get [-3-4sqrt(2)] / 2 but the answer is supposed to be -7/sqrt(5). How did they get that??

No, take the dot product of (5/2)i+ (3/2)j with the unit vector in the direction of -i- 2j.

Remember that the derivative in the direction of vector v is \nabla f \cdot v/||v||.

You keep forgetting to divide by the length of v.

 

[mit_hacker]

Eye opener!

Thank-you very much for explicitly exposing my weakness! I really mean it. Now, I'll never forget to divide by the length!