A particle moves along the curve below.

Question

y =

sqrt(8 + x^3)
As it reaches the point (2, 4), the y-coordinate is increasing at a rate of 5 cm/s. How fast is the x-coordinate of the point changing at that instant?

Answer 1

y = √(8+x^3)
dy/dt = (3x^2 / 2√(8+x^3)) dx/dt

Now just plug in dy/dt=5 and x=2 to get dx/dt