# 007A Sample Midterm 3, Problem 4 Detailed Solution

Consider the circle  ${\displaystyle x^{2}+y^{2}=25.}$

(a)  Find  ${\displaystyle {\frac {dy}{dx}}.}$

(b)  Find the equation of the tangent line at the point  ${\displaystyle (4,-3).}$

Background Information:
1. What is the result of implicit differentiation of  ${\displaystyle y^{2}?}$

It would be  ${\displaystyle 2y\cdot {\frac {dy}{dx}}}$  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  ${\displaystyle m={\frac {dy}{dx}}.}$

Solution:

(a)

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

${\displaystyle 2x+2y\cdot {\frac {dy}{dx}}=0.}$

Step 2:
Now, solve for  ${\displaystyle {\frac {dy}{dx}}.}$
So, we have

${\displaystyle 2y\cdot {\frac {dy}{dx}}=-2x.}$

We solve to get
${\displaystyle {\frac {dy}{dx}}=-{\frac {x}{y}}.}$

(b)

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

${\displaystyle {\begin{array}{rcl}\displaystyle {m}&=&\displaystyle {-{\bigg (}{\frac {4}{-3}}{\bigg )}}\\&&\\&=&\displaystyle {{\frac {4}{3}}.}\end{array}}}$

Step 2:
Now, we have the slope of the tangent line at  ${\displaystyle (4,-3)}$  and a point.
Thus, we can write the equation of the line.
So, the equation of the tangent line at  ${\displaystyle (4,-3)}$  is

${\displaystyle y\,=\,{\frac {4}{3}}(x-4)-3.}$

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