009A Sample Final 3, Problem 5

Calculate the equation of the tangent line to the curve defined by  $x^{3}+y^{3}=2xy$  at the point,  $(1,1).$

Foundations:
The equation of the tangent line to  $f(x)$  at the point  $(a,b)$  is
$y=m(x-a)+b$  where  $m=f'(a).$

Solution:

Step 1:
We use implicit differentiation to find the derivative of the given curve.
Using the product and chain rule, we get
$3x^{2}+3y^{2}y'=2y+2xy'.$
We rearrange the terms and solve for  $y'.$
Therefore,
$3x^{2}-2y=2xy'-3y^{2}y'$
and
$y'={\frac {3x^{2}-2y}{2x-3y^{2}}}.$

Step 2:
Therefore, the slope of the tangent line at the point  $(1,1)$  is
${\begin{array}{rcl}\displaystyle {m}&=&\displaystyle {\frac {3(1)^{2}-2(1)}{2(1)-3(1)^{2}}}\\&&\\&=&\displaystyle {\frac {3-2}{2-3}}\\&&\\&=&\displaystyle {-1.}\end{array}}$
Hence, the equation of the tangent line to the curve at the point  $(1,1)$  is
$f(x)=-1(x-1)+1.$

Final Answer:
$f(x)=-1(x-1)+1$

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