Let f(x, y)=x y^2 and 𝐫(t)=⟨(1)/(2) t^2, t^3⟩ (a) Calculate ∇ f and 𝐫^'(t) . (b) Use t

step 1 Okay, so for this problem we're asked a couple of questions. First, we need to find the gradient

Let f(x, y)=x y^2 and 𝐫(t)=⟨(1)/(2) t^2, t^3⟩ (a)...

Let $f(x, y)=x y^{2}$ and $\mathbf{r}(t)=\left\langle\frac{1}{2} t^{2}, t^{3}\right\rangle$ $$ \begin{array}{l}{\text { (a) Calculate } \nabla f \text { and } \mathbf{r}^{\prime}(t) .} \\ {\text { (b) Use the Chain Rulc for Paths to evaluate } \frac{d}{d t} f(\mathbf{r}(t)) \text { at } t=1 \text { and }} \\ {t=-1 .}\end{array} $$

Let $f(x, y)=x y^{2}$ and $\mathbf{r}(t)=\left\langle\frac{1}{2} t^{2}, t^{3}\right\rangle$
$$
\begin{array}{l}{\text { (a) Calculate } \nabla f \text { and } \mathbf{r}^{\prime}(t) .} \\ {\text { (b) Use the Chain Rulc for Paths to evaluate } \frac{d}{d t} f(\mathbf{r}(t)) \text { at } t=1 \text { and }} \\ {t=-1 .}\end{array}
$$

Key Concepts

Gradient

The gradient of a scalar function is a vector that contains all the function's partial derivatives. It points in the direction of the steepest increase of the function and its magnitude indicates the rate of change in that direction. This concept is essential in multivariable calculus for analyzing how functions change in different directions.

Chain Rule for Paths

The chain rule for paths is used when a scalar function is composed with a vector-valued function. It explains how to differentiate a composition by taking the dot product of the gradient of the scalar function and the derivative (or tangent vector) of the path. This allows us to track how changes along a curve in the domain affect the overall output of the scalar function.

Parametrized Paths in Multivariable Functions

A parametrized path represents a curve within the domain of a multivariable function, expressed as a function of a single variable (often time). This concept is crucial in vector calculus as it helps in understanding the behavior of functions along curves, making it possible to compute derivatives and integrals along those paths.

Transcript

00:01 Okay, so for this problem we're asked a couple of questions.

00:04 First, we need to find the gradient of f as well as the derivative of the path function r.

00:13 So we'll start with finding the gradient of f, which is going to be composed of the partial derivatives with respect to x and y.

00:30 So if we take the partial derivative of f with respect to x, we're going to get y squared.

00:43 And then if we take the partial derivative with respect to y, we're just using the power rule and treating x as a constant, just as a reminder.

00:51 So we'll get 2xy.

00:56 And that is the gradient of f.

01:00 Now, to get the derivative of the path function r, we're just going to take the component -wise derivative as usual.

01:10 So this is just going to be some power rules here for one -half t squared, bringing the two down and subtracting one from the power, we're just going to get t.

01:23 And then for t -cubed, we're going to get 3t squared.

01:28 So that is the derivative of the path function r.

01:33 And now they want us to take the derivative of the path function r composed with f.

01:42 So to do this, we can use the formula.

01:45 We can take the gradient of f and just multiply it by the derivative of the path function r of t.

01:56 So we already have the, excuse me, and we're going to be evaluating the gradient of f at r of t.

02:09 So we'll be composing the function first and then taking the gradient of it.

02:16 So i'll go ahead and show that right now.

02:17 So we already have the gradient, and the gradient of f is y squared and 2xy...

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