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October 24, 2014

October 24, 2014

Posted by **XCS** on Friday, May 8, 2009 at 9:16am.

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 -
**drwls**, Friday, May 8, 2009 at 11:10amBoth 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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