Expert: Scotto - 3/22/2009

Question
I was given the following questions:
I) Find the slope of the tangent line to the curve (a lemniscate) 2*(x^2+y^2)^2=25*(x^2-y^2)at the point (3,-1)
II) If 3x^2+3x+xy=4 and y(4)= -14, find y'(4) by implicit differentiation.

Answer
I) Find the slope of the tangent line to the curve (a lemniscate)
2*(x^2+y^2)^2=25*(x^2-y^2)at the point (3,-1)

Approach 1:
If we know that (x²+y²)² = 25(x²-y²), we can implicitly differentiate and get the answer.
The derivative of x² is 2s.  The derivative of y² is 2yy’.

Therefore, the overall derivative is
2(x ² + y²)(2x + 2yy’) = 25(2x – 2yy’).

We know that x=3, y=-1, so x²=9 and y²=1.

Given this, we have 2(10)(6 – 2y’) = 25(6 + 2y’).

Multiplying it out gives us 120 – 40y’ = 150 + 50y’.

Rearranging terms gives us –90y’ = 30, so y’ = -1/3.
To take a different approach, we could find y’.


Approach 2:
The derivative is 4x(x² + y²) + 4y(x² + y²)y’ = 50x – 50yy’.

Splitting this in terms of factors times y’ and those not gives us
(4y(x² + y²) +50y)y’ = 50x – 4x(x² + y²).

Doing division of what’s times y’ gives us
y’ = 50x – 4x(x² + y²)/(4y(x² + y²) +50y).

Now putting in x=3 and y=-1 gives us
(150-12(10))/(-4(10)-50) = 30/(-90) = -1/3.

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II) If 3x^2+3x+xy=4 and y(4)= -14, find y'(4) by implicit differentiation.
6x + 3 + xy’ + y = 0, x=4, so 24  + 3 + 4y’ – 14 = 0
4y’ =-13, y’=-13/4