009A Sample Final 1, Problem 7

A curve is defined implicitly by the equation

${\displaystyle x^{3}+y^{3}=6xy.}$

(a) Using implicit differentiation, compute  ${\displaystyle {\frac {dy}{dx}}}$.

(b) Find an equation of the tangent line to the curve  ${\displaystyle x^{3}+y^{3}=6xy}$  at the point  ${\displaystyle (3,3)}$.

Foundations:
1. What is the result of implicit differentiation of  ${\displaystyle xy?}$

It would be  ${\displaystyle y+x{\frac {dy}{dx}}}$  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  ${\displaystyle m={\frac {dy}{dx}}.}$

Solution:

(a)

Step 1:
Using implicit differentiation on the equation  ${\displaystyle x^{3}+y^{3}=6xy,}$  we get

${\displaystyle 3x^{2}+3y^{2}{\frac {dy}{dx}}=6y+6x{\frac {dy}{dx}}.}$

Step 2:
Now, we move all the  ${\displaystyle {\frac {dy}{dx}}}$  terms to one side of the equation.
So, we have

${\displaystyle 3x^{2}-6y={\frac {dy}{dx}}(6x-3y^{2}).}$

We solve to get
${\displaystyle {\frac {dy}{dx}}={\frac {3x^{2}-6y}{6x-3y^{2}}}.}$

(b)

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

${\displaystyle {\begin{array}{rcl}\displaystyle {m}&=&\displaystyle {\frac {3(3)^{2}-6(3)}{6(3)-3(3)^{2}}}\\&&\\&=&\displaystyle {-{\frac {9}{9}}}\\&&\\&=&\displaystyle {-1.}\end{array}}}$

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

${\displaystyle y\,=\,-1(x-3)+3.}$

(a)    ${\displaystyle {\frac {dy}{dx}}={\frac {3x^{2}-6y}{6x-3y^{2}}}}$
(b)    ${\displaystyle y=-1(x-3)+3}$