Chapter 4.2, Problem 23E

a.

To determine

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

Answer to Problem 23E

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

Explanation of Solution

Given information:The equation of the curve is $2xy+\pi \sin y=2\pi$ and the point is $\left(x_1,y_1\right)=\left(1,\frac{\pi }{2}\right)$ .

Formula used:

Slope-point form of line is

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

  $\frac{d\sin y}{dx}=\cos y$

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

Calculation:

The curve equation is $2xy+\pi \sin y=2\pi$ .............. (1)

And the point at which the tangent equation is to be determined is $\left(x_1,y_1\right)=\left(1,\frac{\pi }{2}\right)$

By differentiating equation (1) we get

  $\begin{array}{l}2xy+\pi \sin y=2\pi \\ 2\frac{d\left(xy\right)}{dx}+\pi \frac{d\left(\sin y\right)}{dx}=0\\ 2\left(1\times (y)+x\times \left(\frac{dy}{dx}\right)\right)+\pi \times \left(\cos y\frac{dy}{dx}\right)=0\\ 2y+\frac{dy}{dx}\left(2x+\pi \cos y\right)=0\mathrm{......}\left(2\right)\end{array}$

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

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

The slope point form of the tangent line is

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

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

b.

To determine

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

Answer to Problem 23E

The equation of normal to the curve is $2\pi y-4x-\left({\pi }^2-4\right)=0$ .

Explanation of Solution

Given information:The equation of the curve is $2xy+\pi \sin y=2\pi$ and the point is $\left(x_1,y_1\right)=\left(1,\frac{\pi }{2}\right)$ .

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:

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

Then the equation of normal at $\left(x_1,y_1\right)=\left(1,\frac{\pi }{2}\right)$ is

  $\begin{array}{l}\left(y-y_1\right)=m_1\left(x-x_1\right)\\ \left(y-\frac{\pi }{2}\right)=\frac{2}{\pi }\left(x-\left(1\right)\right)\\ \pi \left(\frac{2y-\pi }{2}\right)=2\left(x-1\right)\\ 2\pi y-{\pi }^2=4x-4\\ 2\pi y-4x-\left({\pi }^2-4\right)=0\end{array}$

Hence the equation of normal to the curve is $2\pi y-4x-\left({\pi }^2-4\right)=0$ .