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  $(4,-3).$

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Background Information:
1. What is the result of implicit differentiation of  $y^{2}?$
It would be  $2y{\frac {dy}{dx}}$  by the Product 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
$2x+2y{\frac {dy}{dx}}=0.$

Step 2:
Now, solve for  ${\frac {dy}{dx}}.$
So, we have
$2y{\frac {dy}{dx}}=-2x.$
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
${\begin{array}{rcl}\displaystyle {m}&=&\displaystyle {\frac {-4}{-3}}\\&&\\&=&\displaystyle {{\frac {4}{3}}.}\end{array}}$

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

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

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