# Calculate the directional derivative

## Homework Statement

Calculate the directional derivative, direction, and rate of maximum increase in the direction of v at the given point.
f(x,y) = x^2 + y^3
v = <4,3>
P = (1,2)

## Homework Equations

I know how to calculate directional derivative but I don't know how to calculate rate of maximum increase and direction

Last edited:

## Answers and Replies

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Dembadon
Gold Member

A function increases most rapidly at P0 in the direction of grad ƒ evaluated at P0.

Given the above fact, when would you say that it deceases most rapidly? Hint: you don't need to do any other computations.

oh wait... is direction the gradient?

Dembadon
Gold Member

oh wait... is direction the gradient?
The gradient is the vector that is orthogonal (perpendicular) to the level curve at the point you gave. So yes, the gradient has direction, if that's what you meant.

For the direction: Do you know how to find a unit vector in the direction of your gradient? You can use it to find the direction in which your function is increasing and decreasing most rapidly. There should be a formula in your book somewhere, perhaps involving cosine?

Duf(P) = gradient * cos theta??
uhh-- so theta is the direction... yes?

Dembadon
Gold Member

Duf(P) = gradient * cos theta??
uhh-- so theta is the direction... yes?
Yes. If you think about it, when θ=0, your function will be increasing most rapidly. It will be decreasing most rapidly when θ=$\pi$ (the opposite direction).

Edit: Keep in mind that your final answer should be a vector.

2nd Edit: I need to go to class, so I'll check back in a little over an hour.

my directional derivative was 44/5 and my gradient was <2,12> so ||gradient|| is sqrt(148). So the direction at the given point is cos-((44/5)/sqrt(148))?

Is rate of maximum increase as same as maximum rate of increase?

Dembadon
Gold Member

Sorry if I confused you. I didn't see v at first.

Since the rate of increase is your directional derivative, the maximum rate of increase will occur when the angle between vectors v and $\nabla$ƒ is 0. Thus, the maximum rate of increase is $\frac{44}{5}$cos 0.

The direction of the maximum rate of increase is when $\nabla$ƒ is in the direction of v. This follows from the previous step because we need cos θ = 1 in order to obtain the maximum rate of increase, and since the angle between v and $\nabla$ƒ is 0, the vectors are pointing in the same direction.