Q. 24 - Calculus

Question

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

$\frac{dy}{dt} when y=r \sin \theta , r=t^3, and \theta =\sqrt{t}$

Step-by-Step Solution

Answer

The value is $\frac{dy}{dt}=t^2(\sqrt{t}·\cos \sqrt{t}+2\sin \sqrt{t})$

Step 1. Given Information:

Given:

$$\begin{gathered} y=r \sin \theta , \\ r=t^3, and \\ \theta =\sqrt{t} \end{gathered}$$

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

Step 2. Solution:

$$\begin{gathered} Using r=t^3 and \theta =\sqrt{t} in y=r \sin \theta we get \\ y=t^3 \sin \sqrt{t} \\ Diff. w.r.t. t we get \\ \frac{dy}{dt}=t^3\frac{d}{dt}\sin \sqrt{t}+\sin \sqrt{t}\frac{d}{dt}{t}^3 \\ \frac{dy}{dt}=t^3·\cos \sqrt{t}·\frac{1}{2\sqrt{t}}+\sin \sqrt{t}·2t^2 \\ \frac{dy}{dt}=t^2(\sqrt{t}·\cos \sqrt{t}+2\sin \sqrt{t}) \end{gathered}$$