# Calculate the directional derivative

• DrunkApple
In summary, the directional derivative at the given point is 44/5 and the direction of maximum increase is in the direction of v. The rate of maximum increase is the same as the maximum rate of increase, and it occurs when the gradient is in the direction of v.
DrunkApple

## 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

## The Attempt at a Solution

Last edited:

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?

DrunkApple said:
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?

DrunkApple said:
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?

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.

## What is the directional derivative?

The directional derivative is a measure of how a function changes in a specific direction. It is the rate of change of the function in the direction of a given vector.

## How do you calculate the directional derivative?

The directional derivative can be calculated using the gradient vector and the direction of the desired vector. It is the dot product of the gradient vector and the unit vector in the direction of the desired vector.

## What is the significance of the directional derivative?

The directional derivative is important in understanding how a function changes in a specific direction. It is useful in optimization problems and in determining the slope of a surface in a particular direction.

## Can the directional derivative be negative?

Yes, the directional derivative can be negative. It indicates that the function is decreasing in the direction of the given vector.

## How is the directional derivative related to the partial derivatives?

The directional derivative is related to the partial derivatives through the gradient vector. The gradient vector is a vector containing all the partial derivatives of a function, and the directional derivative is the dot product of this vector with a unit vector in the desired direction.

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