Question

a point moves around the circle x^2+y^2=9. When the point is at (-sqrt3, sqrt6), the x coordinate is increasing at the rate of 20 units per second. How fast is its y coordinate at that instant?

Answer 1

To answer, subtract x² from both sides of the equation to give us,
                                   y² = -x² + 9
Then, we derive the equation in terms t.
                             2y(dy/dt) = -2x(dx/dt) 
Substituting the known values from the given,
                           2(sqrt6)(dy/dt) = -2(-sqrt3)(20)
The value of dy/dt is 14.14.