Chapter 4.1, Problem 59E

To determine

To show: the given curves are orthogonal at the origin and draw both graphs and both tangents in a square viewing window.

Answer to Problem 59E

Derivative of $y=\sin 2x$ is $y^{\prime }=2\cos 2x$ and the equation of the tangent line is $y=2x$ .

Derivative of $y=-\sin \frac{x}{2}$ is $y^{\prime }=-\frac{1}{2}\cos \frac{x}{2}$ and the equation of the tangent line is $y=-\frac{1}{2}x$ .

  

Explanation of Solution

Given information:

The curves $y=\sin 2x$ and $y=-\sin \frac{x}{2}$ crossing at a right angle and the tangents are perpendicular at the crossing point.

Calculation:

The curve $y=\sin 2x$ ,

  $\begin{array}{c}{y}^{\prime }=2\cos 2x\\ y^{\prime }\left(0\right)=2\cos 0\\ y^{\prime }\left(0\right)=2\end{array}$

Finding the equation of the tangent line with the slope $2$ ,

  $\begin{array}{c}y-0=2\left(x-0\right)\\ y=2x\end{array}$

The curve $y=-\sin \frac{x}{2}$ ,

  $\begin{array}{c}{y}^{\prime }=-\frac{1}{2}\cos \left(\frac{x}{2}\right)\\ y^{\prime }\left(0\right)=-\frac{1}{2}\cos \left(\frac{0}{2}\right)\\ y^{\prime }\left(0\right)=-\frac{1}{2}\end{array}$

Finding the equation of the tangent line with the slope $-\frac{1}{2}$ ,

  $\begin{array}{c}y-0=-\frac{1}{2}\left(x-0\right)\\ y=-\frac{1}{2}x\end{array}$

Thus we can find the derivative of $y=\sin 2x$ is $y^{\prime }=2\cos 2x$ and the equation of the tangent line is $y=2x$ .

Thus we can find the derivative of $y=-\sin \frac{x}{2}$ is $y^{\prime }=-\frac{1}{2}\cos \frac{x}{2}$ and the equation of the tangent line is $y=-\frac{1}{2}x$ .