Q. 23 - Calculus

Question

Use Theorem 12.32 to find the indicated derivatives in Exercises 21–26. Express your answers as functions of a single variable. $\frac{dx}{dt} when x=r \cos \theta , r=t^2-5, and \theta =t^3+1$

Step-by-Step Solution

Answer

The value is $\frac{dx}{dt}=2t·\cos (t^3+1)-3t^2(t^2-5)\sin (t^3+1)$

Step 1. Given information:

Given:

$$\begin{gathered} x=r \cos \theta , \\ r=t^2-5, and \\ \theta =t^3+1 \end{gathered}$$

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

Step 2. Solution:

$$\begin{gathered} U\sin g r=t^2-5 and \theta =t^3+1 in x=r \cos \theta we get \\ x=(t^2-5)·\cos (t^3+1) \\ Diff. w.r.t. t we get \\ \frac{dx}{dt}=(t^2-5)\frac{d}{dt}\cos (t^3+1)+\cos (t^3+1)\frac{d}{dt}(t^2-5) \\ \frac{dx}{dt}=(t^2-5)(-\sin (t^3+1)·3t^2)+\cos (t^3+1)·2t \\ \frac{dx}{dt}=2t·\cos (t^3+1)-3t^2(t^2-5)\sin (t^3+1) \end{gathered}$$