Q. 21. - 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 $z=\sin x\cos y, x=e^t and y=t^3$

Step-by-Step Solution

Answer

The required single variable function is

$\frac{dz}{dt}=\left(\cos \left(e^t\right)\cos \left(t^3\right)\right)\left(e^t\right)+\left(-\sin \left(e^t\right)\sin \left(t^3\right)\right)\left(3t^2\right)$

Step 1: Given information

Think about the following function.

$z=\sin x\cos y,x=e^t and y=t^3$

Step 2: The objective is to find out the single variable function.

By using the chain rule,

$$\begin{gathered} \frac{dz}{dt}=\frac{dz dx}{dx dt}+\frac{dz dy}{dy dt} \\ z=\sin x \cos y \\ \frac{dz}{dx}=\cos x \cos y \\ \frac{dz}{dy}=-\sin x \sin y \end{gathered}$$

Before moving on to the next stage,

$$\begin{gathered} x=e^t \\ \Rightarrow \frac{dx}{dt}=e^t \\ y=t^3 \\ \Rightarrow \frac{dy}{dt}=3t^2 \end{gathered}$$

Then,

$$\begin{gathered} \frac{dz}{dt}=\frac{dz dx}{dx dt}+\frac{dz dy}{dy dt} \\ \frac{dz}{dt}=(\cos x \cos y)\left(e^t\right)+(-\sin x \sin y)\left(3t^2\right) \end{gathered}$$

The single variable function is

$\frac{dz}{dt}=\left(\cos \left(e^t\right)\cos \left(t^3\right)\right)\left(e^t\right)+\left(-\sin \left(e^t\right)\sin \left(t^3\right)\right)\left(3t^2\right)$