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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ExpandBackground 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)

ExpandStep 1:
Using implicit differentiation on the equation  $x^{2}+y^{2}=25,$  we get
$2x+2y\cdot {\frac {dy}{dx}}=0.$

ExpandStep 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)

ExpandStep 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
Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \begin{array}{rcl} \displaystyle{m} & = & \displaystyle{-\bigg(\frac{4}{-3}\bigg)}\\ &&\\ & = & \displaystyle{\frac{4}{3}.} \end{array}}

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

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

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