Question about level curves

Question

find the slope of the tangent of the level curve y^3+3x^2y=-13 at an arbitrary point on the curve, and find in particular the equation of the tangent at the point( 2,-1)

c) examine whether the level curve in (b) is convex or concave around the point (2,-1) by computing Y" at this point

Answer 1

First, verify if the point is on that curve.

 

-1 + 3(4)(-1) = -13

-1 - 12 = -13

 

 

It verifies.

 

Now we find the slope of the tangent by implicit differentiation.

 

3y²y' + 6xy + 3x² y' = 0

 

3y²y' + 3x²y' = -6xy

 

3y'(y² + 3x²) = -6xy

 

  y' = -2xy / (y² + 3x²)

 

  y' = -2(2)(-1) / ((-1)² + 3(2²))

 

  y' = 4 / 13

 

 

Then the tangent line using point-slope form is

 

y = (4/13)x - (4/13)(2) - 1

 

Simplify this equation.

 

 

To find y'', take the derivative of y'.  Starting from the first line of the first derivative, we use implicit differentiation.

 

note that derivative of y' will be y''.

 

3y²y' + 6xy + 3x² y' = 0

 

(6y(y') ² + 3y²y'') + (6y + 6xy') + (6xy' + 3x² y'') = 0

 

 

Using the fact that x=2 , y=-1, and y'= 4/13, plug in those values into the above equation.  After you do that, solve for y''.