Chapter 4.2, Problem 18E

a.

To determine

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

Answer to Problem 18E

The equation of tangent is $4y-3x+25=0$ .

Explanation of Solution

Given information:The equation of the curve is $x^2+y^2=25$ and the point is $\left(x_1,y_1\right)=\left(3,-4\right)$ .

Formula used:

Slope-point form of line is

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

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

Calculation:

The curve equation is $x^2+y^2=25$.............. (1)

And the point at which the tangent equation is to be determined is $\left(x_1,y_1\right)=\left(3,-4\right)$ .

By differentiating equation (1) we get

  $\begin{array}{l}{x}^2+y^2=25\\ \frac{d{x}^2}{dx}+\frac{d{y}^2}{dx}=0\\ 2x+2y\frac{dy}{dx}=0\mathrm{......}(2)\\ \text{Now}\text{put}\text{the}\text{given}\text{point}\left(x_1,y_1\right)=\left(3,-4\right)\text{in}\text{equation}\text{(2)}\\ 2\left(3\right)+2\left(-4\right)\frac{dy}{dx}=0\\ \frac{dy}{dx}\left(-8\right)=-6\\ \frac{dy}{dx}=m=\frac{3}{4}\end{array}$

Here $\frac{dy}{dx}=m=\frac{3}{4}$ is the slope of the tangent.

The slope point form of the tangent line is

  $\begin{array}{l}\left(y-y_1\right)=m\left(x-x_1\right)\text{,where}\left(x_1,y_1\right)=\left(3,-4\right)\text{and}m=\frac{3}{4}\\ \left(y-\left(-4\right)\right)=\frac{3}{4}\left(x-3\right)\\ 4\left(y+4\right)=3\left(x-3\right)\\ 4y+16=3x-9\\ 4y-3x+25=0\end{array}$

Hence the equation of tangent is $4y-3x+25=0$ .

b.

To determine

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

Answer to Problem 18E

The equation of normal to the curve is $3y+4x=0$ .

Explanation of Solution

Given information:The equation of the curve is $x^2+y^2=25$ and the point is $\left(x_1,y_1\right)=\left(3,-4\right)$ .

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:

  $\begin{array}{l}\text{Since}m=\frac{3}{4}\text{then}\text{the}\text{slope}\text{of}\text{normal}\text{is}{m}_1\text{=}-\frac{1}{m}=-\frac{4}{3}\end{array}$

Then the equation of normal at $\left(x_1,y_1\right)=\left(3,-4\right)$ is

  $\begin{array}{l}\left(y-y_1\right)=m_1\left(x-x_1\right)\\ \left(y-\left(-4\right)\right)=-\frac{4}{3}\left(x-3\right)\\ 3\left(y+4\right)=-4\left(x-3\right)\\ 3y+12=-4x+12\\ 3y+4x=0\end{array}$

Hence the equation of normal to the curve is $3y+4x=0$ .