Computing directional derivatives with the gradient Compute...

Computing directional derivatives with the gradient Compute the directional derivative of the following functions at the given point $P$ in the direction of the given vector. Be sure to use a unit vector for the direction vector. $$f(x, y)=\sqrt{4-x^{2}-2 y} ; P(2,-2) ;\left\langle\frac{1}{\sqrt{5}}, \frac{2}{\sqrt{5}}\right\rangle$$

Computing directional derivatives with the gradient Compute the directional derivative of the following functions at the given point $P$ in the direction of the given vector. Be sure to use a unit vector for the direction vector.
$$f(x, y)=\sqrt{4-x^{2}-2 y} ; P(2,-2) ;\left\langle\frac{1}{\sqrt{5}}, \frac{2}{\sqrt{5}}\right\rangle$$

Key Concepts

Directional Derivative

A directional derivative measures the rate at which a function changes at a point as one moves in a specific direction. It is computed by taking the dot product of the function's gradient and a unit vector in the desired direction. This concept is central in multivariable calculus as it generalizes the idea of a derivative to multiple dimensions.

Gradient

The gradient is a vector that comprises the partial derivatives of a function with respect to its variables. It points in the direction of the steepest ascent of the function. In the context of directional derivatives, the gradient plays a crucial role because its dot product with the unit direction vector gives the rate of change of the function in that direction.

Partial Derivatives

Partial derivatives are the derivatives of a multivariable function with respect to one variable while holding the other variables constant. They are used to form the gradient vector, which is essential in evaluating both the directional derivatives and understanding how functions change in different directions.

Unit Vector

A unit vector has a magnitude of one and indicates direction. When computing a directional derivative, it is important to use a unit vector to ensure that the calculated rate of change accurately reflects the function's behavior in that specific direction without being affected by the length of the vector.

Transcript

00:01 They give us the function f of xy is equal to the square root of, let's see, 4 minus x squared, minus 2y.

00:15 And they're going to want us to compute a directional derivative, so let's compete the gradient for anything else.

00:21 We'll have one -half of that input for minus x squared minus 2y to the negative 1 -half times using the chain rule, negative 2x.

00:32 The partial derivative with respect to x and in the y component we're going to have the same first factor because we're using chain row except the different variable is being taken the partial derivative of when we compute the derivative of the inside we'll have a negative 2 in the front here yeah let's see specifically we're going to want to compete the gradient at the given point 1 over square root of 5 2 over square root of 5 let's see i know that's not the point.

01:08 The point they want us to compute it at is 2 negative 2.

01:15 Okay.

01:19 Let's plug in for this little expression right here because it appears in both.

01:24 And in fact, the one half in front is canceled out in both.

01:30 So the one half is also canceled out.

01:32 We end up with 4 minus 4.

01:35 Well, that's a 0.

01:36 And then minus a 2 times a negative 2.

01:40 So it's 1 over square root of 4.

01:43 It's really just a 1 half here.

01:45 And there's also a 1 half here...

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