Chapter 4.2, Problem 24E

a.

To determine

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

Answer to Problem 24E

The equation of tangent is $2y+4x-2\pi =0$ .

Explanation of Solution

Given information:The equation of the curve is $x\sin 2y=y\cos 2x$ and the point is $\left(x_1,y_1\right)=\left(\frac{\pi }{4},\frac{\pi }{2}\right)$ .

Formula used:

Slope-point form of line is

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

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

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

Calculation:

The curve equation is $x\sin 2y=y\cos 2x$ ..................(1)

And the point at which the tangent equation is to be determined is $\left(x_1,y_1\right)=\left(\frac{\pi }{4},\frac{\pi }{2}\right)$ .

By differentiating equation (1) we get

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

  $\begin{array}{l}\text{Now}\text{put}\text{the}\text{given}\text{point}\left(x_1,y_1\right)=\left(\frac{\pi }{4},\frac{\pi }{2}\right)\text{in}\text{equation}\left(2\right)\\ \sin 2\left(\frac{\pi }{2}\right)-2\left(\frac{\pi }{2}\right)\sin 2\left(\frac{\pi }{4}\right)+\frac{dy}{dx}\left(2\left(\frac{\pi }{4}\right)\cos 2\left(\frac{\pi }{2}\right)+\cos 2\left(\frac{\pi }{4}\right)\right)=0\\ \left(0\right)-2\left(\frac{\pi }{2}\right)\left(1\right)+\frac{dy}{dx}\left(2\frac{\pi }{4}\left(-1\right)+\left(0\right)\right)=0\\ \frac{dy}{dx}\left(-\frac{\pi }{2}\right)=\pi \\ \frac{dy}{dx}=m=-2\end{array}$

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

The slope point form of the tangent line is

  $\begin{array}{l}\left(y-y_1\right)=m\left(x-x_1\right)\text{,where}\left(x_1,y_1\right)=\left(\frac{\pi }{4},\frac{\pi }{2}\right)\text{and}m=-2\\ \left(y-\frac{\pi }{2}\right)=-2\left(x-\left(\frac{\pi }{4}\right)\right)\\ \left(\frac{2y-\pi }{2}\right)=-2\left(\frac{4x-\pi }{4}\right)\\ \left(2y-\pi \right)=-4\left(\frac{4x-\pi }{4}\right)\\ \left(2y-\pi \right)=-\left(4x-\pi \right)\\ 2y-\pi =-4x+\pi \\ 2y+4x-2\pi =0\end{array}$

Hence the equation of tangent is $2y+4x-2\pi =0$ .

b.

To determine

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

Answer to Problem 24E

The equation of normal to the curve is $8y-4x-3\pi =0$ .

Explanation of Solution

Given information:The equation of the curve is $x\sin 2y=y\cos 2x$ and the point is $\left(x_1,y_1\right)=\left(\frac{\pi }{4},\frac{\pi }{2}\right)$ .

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:

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

Then the equation of normal at $\left(x_1,y_1\right)=\left(\frac{\pi }{4},\frac{\pi }{2}\right)$ is

  $\begin{array}{l}\left(y-y_1\right)=m_1\left(x-x_1\right)\\ \left(y-\frac{\pi }{2}\right)=\frac{1}{2}\left(x-\left(\frac{\pi }{4}\right)\right)\\ \left(\frac{2y-\pi }{2}\right)=\frac{1}{2}\left(\frac{4x-\pi }{4}\right)\\ 8\left(\frac{2y-\pi }{2}\right)=\left(4x-\pi \right)\\ 4(2y-\pi )=(4x-\pi )\\ 8y-4\pi =4x-\pi \\ 8y-4x-3\pi =0\end{array}$

Hence the equation of normal to the curve is $8y-4x-3\pi =0$ .