Q. 26 - Calculus

Question

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

$\frac{d\theta }{dt} when \theta ={\tan }^{-1}\frac{y}{x} , x=e^t , and y=e^{2t}$

Step-by-Step Solution

Answer

The value is $\frac{d\theta }{dt}=\frac{e^t}{1+e^{2t}}$

Step 1. Given Information:

Given:

$$\begin{gathered} \theta =ta{n}^{-1}\frac{y}{x} , \\ x=e^t and \\ y=e^{2t} \end{gathered}$$

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

Step 2. Solution:

$$\begin{gathered} U\sin g x=e^t , and y=e^{2t} in \theta ={\tan }^{-1}\frac{y}{x} we get \\ \theta ={\tan }^{-1}\frac{e^{2t}}{e^t} \\ \theta ={\tan }^{-1}{e}^t \\ Diff. w.r.t.t we get \\ \frac{d\theta }{dt}=\frac{1}{1+e^{2t}}·e^t \\ \frac{d\theta }{dt}=\frac{e^t}{1+e^{2t}} \end{gathered}$$