Chapter 4, Problem 47RE

(a)

To determine

To calculate: To find an equation for the tangent to the curve at the given point.

Answer to Problem 47RE

The equation of the line of tangent is $y-2=-\frac{1}{4}\left(x-1\right)$ .

Explanation of Solution

Given Information: The given equation is $x^2+2y^2=9$ and the point is $\left(1,2\right)$ .

Calculation:

The slope of the tangent line is calculated by finding the derivative of the given equation at the given point.

The derivative of the function is,

  $\begin{array}{c}{x}^2+2y^2=9\\ 2x+2\times 2y\frac{dy}{dx}=\frac{d}{dx}\left(0\right)\\ 2\left(x+2y\frac{dy}{dx}\right)=0\left(\because \frac{d}{dx}\left(\text{constant}\right)=0\right)\\ x+2y\frac{dy}{dx}=0\\ \frac{dy}{dx}=-\frac{x}{2y}\end{array}$

Derivative of the function at $\left(1,2\right)$ is,

  $\begin{array}{c}\frac{dy}{dx}=-\frac{x}{2y}\\ =-\frac{1}{4}\left(\because x=1,y=2\right)\end{array}$

That is,

  $m_{\text{tangent}}=-\frac{1}{4}$

• Here, $m_{\text{tangent}}$is the slope of the tangent.

Thus, equation of the line of tangent using the point-slope form is,

  $\begin{array}{c}y-y_1=m_{\text{tangent}}\left(x-x_1\right)\\ y-2=-\frac{1}{4}\left(x-1\right)\end{array}$

The equation of the line of tangent is $y-2=-\frac{1}{4}\left(x-1\right)$ .

(b)

To determine

To calculate: To find an equation for the normal to the curve at the given point.

Answer to Problem 47RE

The equation of the line of normal is $y-2=4\left(x-1\right)$ .

Explanation of Solution

Given Information: The given equation is $x^2+2y^2=9$ and the point is $\left(1,2\right)$ .

Calculation:

The slope of the tangent line is calculated by finding the derivative of the given equation at the given point.

The derivative of the function is,

  $\begin{array}{c}{x}^2+2y^2=9\\ 2x+2\times 2y\frac{dy}{dx}=\frac{d}{dx}\left(0\right)\\ 2\left(x+2y\frac{dy}{dx}\right)=0\left(\because \frac{d}{dx}\left(\text{constant}\right)=0\right)\\ x+2y\frac{dy}{dx}=0\\ \frac{dy}{dx}=-\frac{x}{2y}\end{array}$

Derivative of the function at $\left(1,2\right)$ is,

  $\begin{array}{c}\frac{dy}{dx}=-\frac{x}{2y}\\ =-\frac{1}{4}\left(\because x=1,y=2\right)\end{array}$

That is,

  $m_{\text{tangent}}=-\frac{1}{4}$

The slope of the normal line is perpendicular to the slope of the tangent line.

That is,

  $\begin{array}{c}{m}_{\text{normal}}\times m_{\text{tangent}}=-1\\ m_{\text{normal}}=\frac{-1}{m_{\text{tangent}}}\\ m_{\text{normal}}=\frac{-1}{\frac{-1}{4}}\\ =4\end{array}$

• Here, $m_{\text{normal}}$ is the slope of the normal.

Thus, equation of the line of normal using the point-slope form is,

  $\begin{array}{c}y-y_1=m_{\text{normal}}\left(x-x_1\right)\\ y-2=4\left(x-1\right)\end{array}$

The equation of the line of normal is $y-2=4\left(x-1\right)$ .