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?
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.
