# calculus

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I need help setting up this problem:

A particle moves along the curve
y=sqr(1+x^3). As it reaches the point (2,3) the y-cordinate is increasing at a rate of 4cm/s. How fast is the x-coordinate of the point changing at that instant.

I could only come up with this:

dy/dx=4

• calculus -

Both y and x are functions of t.
dy/dt = (dy/dx)*(dx/dt)

You are given that dy/dt = 4 and, according to my calculations,
dy/dx = [(1/2)/sqr(1+x^3)]*3x^2
= [(1/2)/3]*12 = 2

That means dx/dt = 4/2 = 2

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