Chapter 4.2, Problem 26E

a.

To determine

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

Answer to Problem 26E

The equation of tangent is $y=\pi$ .

Explanation of Solution

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

Formula used:

Slope-point form of line is

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

  $\begin{array}{l}\frac{d\sin y}{dx}=\cos y\\ \frac{d\cos y}{dx}=-\sin y\\ \sin 2y=2\sin y\cos y\end{array}$

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

Calculation:

The curve equation is $x^2{\cos }^{2}y-\sin y=0$ ................(1)

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

By differentiating equation (1) ,

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

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

Here $\frac{dy}{dx}=m=0$ is the slope of the tangent.

The slope point form of the tangent line is

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

Hence the equation of tangent is $y=\pi$ .

b.

To determine

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

Answer to Problem 26E

The equation of normal to the curve is $y=\pi$ .

Explanation of Solution

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

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:

  $\begin{array}{l}\text{Since}m=0\text{then}\text{the}\text{slope}\text{of}\text{normal}\text{is}{m}_1\text{=}-\frac{1}{m}=0\end{array}$

Here $m_1=m=0$

Then the equation of normal at $\left(x_1,y_1\right)=\left(0,\pi \right)$ is same as the equation of tangent

Hence the equation of normal to the curve is $y=\pi$ .