# 007A Sample Midterm 3, Problem 4 Detailed Solution

Consider the circle  $x^{2}+y^{2}=25.$ (a)  Find  ${\frac {dy}{dx}}.$ (b)  Find the equation of the tangent line at the point  $(4,-3).$ Background Information:
1. What is the result of implicit differentiation of  $y^{2}?$ It would be  $2y\cdot {\frac {dy}{dx}}$ by the Chain Rule.

2. What two pieces of information do you need to write the equation of a line?

You need the slope of the line and a point on the line.

3. What is the slope of the tangent line of a curve?

The slope is  $m={\frac {dy}{dx}}.$ Solution:

(a)

Step 1:
Using implicit differentiation on the equation  $x^{2}+y^{2}=25,$ we get

$2x+2y\cdot {\frac {dy}{dx}}=0.$ Step 2:
Now, solve for  ${\frac {dy}{dx}}.$ So, we have

$2y\cdot {\frac {dy}{dx}}=-2x.$ We solve to get
${\frac {dy}{dx}}=-{\frac {x}{y}}.$ (b)

Step 1:
First, we find the slope of the tangent line at the point  $(4,-3).$ We plug  $(4,-3)$ into the formula for  ${\frac {dy}{dx}}$ we found in part (a).
So, we get

${\begin{array}{rcl}\displaystyle {m}&=&\displaystyle {-{\bigg (}{\frac {4}{-3}}{\bigg )}}\\&&\\&=&\displaystyle {{\frac {4}{3}}.}\end{array}}$ Step 2:
Now, we have the slope of the tangent line at  $(4,-3)$ and a point.
Thus, we can write the equation of the line.
So, the equation of the tangent line at  $(4,-3)$ is

$y\,=\,{\frac {4}{3}}(x-4)-3.$ (a)    ${\frac {dy}{dx}}=-{\frac {x}{y}}$ (b)    $y\,=\,{\frac {4}{3}}(x-4)-3$ 