009A Sample Final 1, Problem 7

A curve is defined implicitly by the equation

$x^{3}+y^{3}=6xy$

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

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

Foundations:
1.
2.
3.

Solution:

(a)

Step 1:
Using implicit differentiation on the equation $x^{3}+y^{3}=6xy$, we get
$3x^{2}+3y^{2}{\frac {dy}{dx}}=6y+6x{\frac {dy}{dx}}$.

Step 2:
Now, we move all the ${\frac {dy}{dx}}$ terms to one side of the equation.
So, we have
$3x^{2}-6y={\frac {dy}{dx}}(6x-3y^{2})$.
We solve to get ${\frac {dy}{dx}}={\frac {3x^{2}-6y}{6x-3y^{2}}}$.

(b)

Step 1:
First, we find the slope of the tangent line at the point $(3,3)$.
We plug in $(3,3)$ into the formula for ${\frac {dy}{dx}}$ we found in part (a).
So, we get
$m={\frac {3(3)^{2}-6(3)}{6(3)-3(3)^{2}}}={\frac {9}{-9}}=-1$.

Step 2:
Now, we have the slope of the tangent line at $(3,3)$ and a point.
Thus, we can write the equation of the line.
So, the equation of the tangent line at $(3,3)$ is
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 y=-1(x-3)+3} .

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

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