Implicit Differentiation - Calculus

Question

 Need to find slope of tangent line to curve 5(x^2 + y^2)^2 = 169(x^2-y^2) at point (3,-2) . It is implicit differentiation but I keep getting the wrong value at 3,-2. I'm not sure if I am isolating the y' correctly. Anyone want to take a shot?

Answer 1

I think the correct answer to this problem is -9/46. 

 

It is a matter of personal choice whether to expand the binomial or try some other technique.

 

I rewrote the equation as:

 

(x²+y²)²/(x² - y²) = 169/5

 

Now I used the quotient rule (differentiating implicitly). Notice that the right side becomes 0.

 

Truthfully it gets fairly messy, but after the quotient rule and multiplying both sides of the equation by the denominator on the left side, dividing out some common factors, factoring out y' etc, this is what I ended with:

 

y' = [x(x² + y²) - 2x(x² - y²)] / [2y(x²-y²) +y(x²+y²)]

 

y'_(3,-2) = -9/46 (-0.195652173913)

 

 

Answer 2

We have 5(x² + y²)² = 169 (x² - y²) so this evaluates to:

 

5(x^{4 +} 2x²y² + y⁴) = 169x² - 169y²

5x⁴ + 10x²y² + 5y⁴ = 169x² - 169y²  Now, we are ready for implicit differentiation. 

 

20x³ + 10x² (2yy') + y²(20x) + 20y³y' = 338x - 338yy'

20x²yy' + 20y³y' + 338yy' = 338x - 20x³ - 20xy²

y'(20x²y + 20y³ + 338y) = 338x - 20x³ - 20xy²

y' = (338x - 20x³ - 20xy²)/(20x²y + 20y³ + 338y)

 

When I evaluate this at the point (3, -2), I get 234/156 = 117/78 = 39/26 = 3/2. 

Answer 3

Lets take this one step at a time.

 

First, we can expand the terms to get rid of parentheses.

 

 

5(x⁴ + 2x²y² + y⁴) = 169x² - 169y²

 

5x² + 10x²y² + 5y⁴ = 169x² - 169y²

 

 

Now we can differentiate both sides of the equation.

 

10x + 20xy² + 20x²yy' + 20y³y' = 338x - 338yy'

 

 

Did you end up with this at this point?  If so, move all the y' terms to the left side of the equation.  Move all the non-y' terms to the right side.

 

20x²yy' + 20y³y' + 338yy' = -10x - 20xy² + 338x

 

 

Factor out y' on the right side.

 

y' (20x²y + 20y³ + 338y) = 338x - 20xy² - 10x

 

 

Next, divide both sides of the equation by the coefficient of y'.  After that, evaluate y' when x=3 and y=-2 as the last step.  I leave it to you to finish up here.