Chapter 4.2, Problem 25E

a.

To determine

To calculate:The equation of line that is tangent to the given curve at the given point.

Answer to Problem 25E

The equation of tangent is $y+2\pi x+2\pi =0$ .

Explanation of Solution

Given information:The equation of the curve is $y=2\sin \left(\pi x-y\right)$ and the point is $\left(x_1,y_1\right)=\left(1,0\right)$ .

Formula used:

Slope-point form of line is

  $\left(y-y_1\right)=m\left(x-x_1\right),\text{where}\left(x_1,y_1\right)\text{is}\text{the}\text{intercept}\text{and}m\text{is}\text{the}\text{slope}\text{of}\text{the}\text{line}$.

  $\begin{array}{l}\frac{d\sin y}{dx}=\cos y\\ \frac{d\cos y}{dx}=-\sin y\end{array}$

Also product, quotient and chain rule of derivatives are used.

Calculation:

The curve equation is $y=2\sin \left(\pi x-y\right)$ ...................(1)

And the point at which the tangent equation is to be determined is $\left(x_1,y_1\right)=\left(1,0\right)$

By differentiating equation (1) we get

  $\begin{array}{l}y=2\sin \left(\pi x-y\right)\\ \frac{dy}{dx}=2\frac{d\left(\sin \left(\pi x-y\right)\right)}{dx}\\ \frac{dy}{dx}=2\left(\cos \left(\pi x-y\right)\times \left(\pi -\frac{dy}{dx}\right)\right)\\ \frac{dy}{dx}\left(1+2\cos \left(\pi x-y\right)\right)=2\pi \cos \left(\pi x-y\right)\mathrm{...............}\left(2\right)\end{array}$

  $\begin{array}{l}\text{Now}\text{put}\text{the}\text{given}\text{point}\left(x_1,y_1\right)=\left(1,0\right)\text{in}\text{equation}\left(2\right)\\ \frac{dy}{dx}\left(1+2\cos \left(\pi \left(1\right)-0\right)\right)=2\pi \cos \left(\pi \left(1\right)-0\right)\\ \frac{dy}{dx}\left(1+2\cos \pi \right)=2\pi \cos \left(\pi \right)\\ \frac{dy}{dx}\left(1-2\right)=-2\pi \\ \frac{dy}{dx}\left(-1\right)=-2\pi \\ \frac{dy}{dx}=m=2\pi \end{array}$

Here $\frac{dy}{dx}=m=2\pi$ is the slope of the tangent.

The slope point form of the tangent line is

  $\begin{array}{l}\left(y-y_1\right)=m\left(x-x_1\right)\text{,where}\left(x_1,y_1\right)=\left(1,0\right)\text{and}m=2\pi \\ \left(y-0\right)=2\pi \left(x-1\right)\\ y-0=2\pi x-2\pi \\ y+2\pi x+2\pi =0\end{array}$

Hence the equation of tangent is $y+2\pi x+2\pi =0$ .

b.

To determine

To calculate:The equation of line that is normal to the given curve at the given point.

Answer to Problem 25E

The equation of normal to the curve is $2\pi y+x-1=0$ .

Explanation of Solution

Given information:The equation of the curve is $y=2\sin \left(\pi x-y\right)$ and the point is $\left(x_1,y_1\right)=\left(1,0\right)$ .

Formula used:

The slope of normal to the curve at a given point is

  $m_1=-\frac{dx}{dy}=-\frac{1}{m}$

Where m is the slope of the tangent to the curve at that point.

Calculation:

  $\begin{array}{l}\text{Since}m=2\pi \text{then}\text{the}\text{slope}\text{of}\text{normal}\text{is}{m}_1\text{=}-\frac{1}{m}=-\frac{1}{2\pi }\end{array}$

Then the equation of normal at $\left(x_1,y_1\right)=\left(1,0\right)$ is

  $\begin{array}{l}\left(y-y_1\right)=m_1\left(x-x_1\right)\\ \left(y-0\right)=-\frac{1}{2\pi }\left(x-1\right)\\ 2\pi \left(y-0\right)=-1\left(x-1\right)\\ 2\pi y=-x+1\\ 2\pi y+x-1=0\end{array}$

Hence the equation of normal to the curve is $2\pi y+x-1=0$ .