Let f(x, y)=e^x y and 𝐫(t)=⟨t^3, 1+t⟩ (a) Calculate ∇ f and 𝐫^'(t) . (b) Use the Chain

Let f(x, y)=e^x y and 𝐫(t)=⟨t^3, 1+t⟩ (a) Calculate ∇ f ...

Let $f(x, y)=e^{x y}$ and $\mathbf{r}(t)=\left\langle t^{3}, 1+t\right\rangle$ $$\begin{array}{l}{\text { (a) Calculate } \nabla f \text { and } \mathbf{r}^{\prime}(t) .} \\ {\text { (b) Use the Chain Rulc for Paths to calculate } \frac{d}{d t} f(\mathbf{r}(t)) \text { . }} \\ {\text { (c) Write out the composite } f(\mathbf{r}(t)) \text { as a function of } t \text { and differentiate. }} \\ {\text { Check that the result agrees with part (b). }}\end{array}$$

Let $f(x, y)=e^{x y}$ and $\mathbf{r}(t)=\left\langle t^{3}, 1+t\right\rangle$
$$\begin{array}{l}{\text { (a) Calculate } \nabla f \text { and } \mathbf{r}^{\prime}(t) .} \\ {\text { (b) Use the Chain Rulc for Paths to calculate } \frac{d}{d t} f(\mathbf{r}(t)) \text { . }} \\ {\text { (c) Write out the composite } f(\mathbf{r}(t)) \text { as a function of } t \text { and differentiate. }} \\ {\text { Check that the result agrees with part (b). }}\end{array}$$

Key Concepts

Gradient of a Multivariable Function

The gradient is a vector that consists of all the first-order partial derivatives of a function. It points in the direction of the steepest ascent for the function and its magnitude gives the rate of increase in that direction. In the context of functions of several variables, calculating the gradient is crucial for understanding the function's behavior in different directions.

Parametric Curves and Their Derivatives

A parametric curve represents a path in space described by one or more parameters, commonly time. Differentiating the vector-valued function that defines the curve provides the tangent vector, which gives both the direction and magnitude of movement along the path. This is fundamental in analyzing motion and changes along curves in multivariable calculus.

Chain Rule for Multivariable Functions

The chain rule in the context of multivariable calculus is used to differentiate composite functions where an outer function depends on an inner function of several variables. When the inner function is a parametric curve, the chain rule allows us to compute the derivative of the composition by taking the dot product of the gradient of the outer function and the derivative of the inner function. This is essential for linking changes in different variables along a given path.

Composite Functions in Multivariable Calculus

Composite functions involve applying one function to the result of another. In multivariable calculus, this often appears when a function of several variables is evaluated along a curve or surface. Differentiating such composite functions can be approached by directly substituting the path into the function or by applying the chain rule, both of which should yield consistent results. This concept is key to connecting various methods of differentiation in higher dimensions.

Transcript

00:01 For the first part of this problem, we're asked to find the gradient of f as well as the derivative of the path r of t.

00:08 So we'll start with the gradient of f.

00:11 And the gradient of f is going to be a vector.

00:15 And the first component of the vector will be the partial derivative of f with respect to x.

00:19 And the second component of this vector will be the partial derivative of f with respect to y.

00:26 So we'll go ahead and take those partial derivatives.

00:30 It's pretty simple on this one.

00:31 We can see that for the partial with respect to x, we have a chain rule.

00:35 So the x will go to 1, and the y will come down times e to the x y.

00:42 And for the second component, the y will go to 1, the x will come down times e to the x y.

00:49 And so this will be our gradient of f.

00:52 Now for the derivative of r of t, we'll just take the component y's derivative.

00:59 So we'll have 3t squared for the first component.

01:04 For the second component, 1 will go to 0, t will go to 1.

01:07 So we'll just be left with 1.

01:09 And that is the derivative of r of t.

01:13 So that completes the first part.

01:15 We'll now go to the second part where they want us to evaluate the derivative with respect to t of f composed of r of t, which is going to be equal just to the gradient of f, of r of t times using the dot product, the derivative of r of t.

01:34 It's a bit of a mouthful, but it's not going to be too bad.

01:37 So let's take a look over here on the right -hand side.

01:40 I color -coded where we're going to be making these substitutions.

01:44 So we're going to be substituting these values of r of t into the x and y values of the gradient of f.

01:51 And when we do that, we're going to get, well, for y, we're going to have 1 plus t, and and then we're going to have e.

02:07 And then for x, we're going to have t cubed times 1 plus t for y.

02:15 And then for the second component, we're going to have x, which is going to go to t cubed times e.

02:23 And then the same thing for the exponent will be t cubed times one plus t...

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