Chapter 3.1, Problem 65E

To determine

To show : the curves are orthogonal to each other at the origin.

Explanation of Solution

Given information :

Curves $y=\sin \left(2x\right)$ and $y=-\sin \left(x/2\right)$ .

Formula needed :

Chain rule of derivative: $\frac{d}{dx}\left(f\left(v\right)\right)=f^{\prime }\left(v\right)\cdot \frac{dv}{dx}$ .

For orthogonal, the product of slopes is minus 1.

Differentiate $y=\sin \left(2x\right)$ with respect to $x$ ,

  $\begin{array}{c}\frac{d}{dx}\left(y\right)=\frac{d}{dx}\sin \left(2x\right)\\ =\cos \left(2x\right)\cdot \frac{d}{dx}\left(2x\right)\\ =\cos \left(2x\right)\cdot \left(2\right)\\ =2\cos \left(2x\right)\end{array}$

Differentiate $y=-\sin \left(x/2\right)$ with respect to $x$ ,

  $\begin{array}{c}\frac{d}{dx}\left(y\right)=\frac{d}{dx}\left(-\sin \left(x/2\right)\right)\\ =-\cos \left(x/2\right)\cdot \frac{d}{dx}\left(x/2\right)\\ =-\cos \left(x/2\right)\cdot \left(1/2\right)\\ =-\frac{1}{2}\cos \left(x/2\right)\end{array}$

The slope is given by $m={\left.\frac{dy}{dx}\right|}_{\left(0,0\right)}$ .

The slope of function $y=\sin \left(2x\right)$ is

  $\begin{array}{c}{m}_1=2\cos \left(2\cdot 0\right)\\ =2\end{array}$

The slope of function $y=-\sin \left(x/2\right)$ is

  $\begin{array}{c}{m}_2=-1/2\cos \left(0/2\right)\\ =-\frac{1}{2}\end{array}$

The product is

  $\begin{array}{c}{m}_1{m}_2=2\cdot \left(-\frac{1}{2}\right)\\ =-1\end{array}$

Hence curves are orthogonal to each other at origin.

The graph of curves is

  

Curve in red is $y=\sin \left(2x\right)$ , curve in blue is $y=-\sin \left(x/2\right)$ .