Difference between revisions of "009A Sample Final 3, Problem 5"

Revision as of 11:57, 6 March 2017

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:

Final Answer:

@@ Line 15: / Line 15: @@
 !Step 1: &nbsp;
 |-
-|
+|We use implicit differentiation to find the derivative of the given curve.
 |-
-|
+|Using the product and chain rule, we get
 |-
-|
+|&nbsp; &nbsp; &nbsp; &nbsp; <math>3x^2+3y^2y'=2y+2xy'.</math>
 |-
-|
+|We rearrange the terms and solve for &nbsp;<math style="vertical-align: -5px">y'.</math>
+|-
+|Therefore,
+|-
+|&nbsp; &nbsp; &nbsp; &nbsp; <math>3x^2-2y=2xy'-3y^2y'</math>
+|-
+|and
+|-
+|&nbsp; &nbsp; &nbsp; &nbsp; <math>y'=\frac{3x^2-2y}{2x-3y^2}.</math>
 |}