AP CALCULUS exam 1987 #2Equation is y^3+3x^2y+13=0.Find the minimum y-coordinate of any point on the curve?

Answer 1

Veggies> asked how to find dy/dx for this expression three days ago so you must be working on the same problem! The answer was

dy/dx= -2xy/(x^2+y^2)

and we know a minimum value for y occurs when dy/dx=0. This means x=0 or y=0, but y=0 doesn't satisfy the original equation so x=0 is where the minimum (or maximum) y-value occurs. Putting x=0 into the original equation gives y^3+13=0 so y=(-13)^1/3 is the minimum y-coordinate.

To show this is the minimum rather than maximum y-coordinate a graph of the curve helps. Because x only appears as an x^2 term the function is even (i.e. symmetric about the y-axis). Also, we have already seen the curve never crosses the x-axis (no solution when y=0) and dy/dx is always positive for x>0, y<0 so the value of y=(-13)^1/3 is a minimum.

Answer 2

3y^2*y´+6xy+3x*^2y´= 0
y´(3y^2+3x^2)=-6xy y´= 0 ,implies x=0 or y=0
if x=0
y^3+13 =0 so y=-(13)^1/3
If y=0 (not possible)
3x^2=(-13-y^3)/y >= 0 so the minimum y is -(13)^1/3

Answer 3

Differentiate the equation implicitly with respect to x. y will be minimized when at a point where y' = 0 or y' does not exist. Good luck.