Q. 25 - Calculus

Question

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

$\frac{dr}{dt} when r=\sqrt{x^2+y^2} , x=\sqrt{t} and y=t^2$

Step-by-Step Solution

Answer

The value is $\frac{dr}{dt}=\frac{1+4t^3}{2\sqrt{t+t^4}}$

Step 1. Given Information:

Given:

$$\begin{gathered} r=\sqrt{x^2+y^2} , \\ x=\sqrt{t} and \\ y=t^2 \end{gathered}$$

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

Step 2. Solution:

$$\begin{gathered} U\sin g x=\sqrt{t} and y = t^2 in r=\sqrt{x^2+y^2} we get \\ r=\sqrt{{\left(\sqrt{t}\right)}^2+{\left(t^2\right)}^2} \\ r=\sqrt{t+t^4} \\ Diff. w.r.t.t we get \\ \frac{dr}{dt}=\frac{1}{2\sqrt{t+t^4}}·(1+4t^3) \\ \frac{dr}{dt}=\frac{1+4t^3}{2\sqrt{t+t^4}} \end{gathered}$$