Q. 22 - Calculus

Question

Use Theorem 12.32 to find the indicated derivatives in Exercises 21–26. Express your answers as functions of a single variable.

$\frac{dz}{dt}when z = x^3{e}^y, x = \sin t, and y = \cos t$

Step-by-Step Solution

Answer

The value is $\frac{dz}{dt}=e^{\cos t}·{\sin }^{2}t [3\cos t+{\sin }^{2}t]$

Step 1. Given Information:

Given:

$$\begin{gathered} z=x^3{e}^y, \\ x = \sin t, and \\ y = \cos t \end{gathered}$$

We have to find the indicated derivatives and express your answers as functions of a single variable.

Step 2. Solution:

We know that Theorem 12.32 state that

Given functions $z=f(x, y), x=u\left(t\right), and y=v\left(t\right)$, for all values of t at which u and v are differentiable, and if f is differentiable at $\left(u\right(t), v(t\left)\right)$, then $\frac{dz}{dt}=\frac{\partial z}{\partial x}·\frac{dx}{dt}+\frac{\partial z}{\partial y}·\frac{dy}{dt}$

$$\begin{gathered} Using x=\sin t and y = \cos t in z=x^3{e}^y we get \\ z={(\sin t)}^3·e^{\cos t} z=e^{\cos t}·{\sin }^{3}t \\ Diff. w.r.t. t we get \\ \frac{dz}{dt}=e^{\cos t}\frac{d}{dt}{\sin }^{3}t+{\sin }^{3}t\frac{d}{dt}{e}^{\cos t} \\ \frac{dz}{dt}=e^{\cos t}(3{\sin }^{2}t·\cos t)+{\sin }^{3}t (e^{\cos t}·\sin t) \\ \frac{dz}{dt}=e^{\cos t}·{\sin }^{2}t [3\cos t+{\sin }^{2}t] \end{gathered}$$