Wednesday

July 23, 2014

July 23, 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

**Related Questions**

Calculus - A particle moves along the curve y = sqr(1+x^3). As it reaches the ...

calculus - Can someone tell me how to do this type of problem? A particle moves ...

Calculus - A particle moves along the curve y = sqr(1+x^3). As it reaches the ...

calculus - A particle moves along the curve y=square sqrt 1+x^3. As it reaches ...

math - a particle moves along the curve y= sqrt 1+x cubed. As it reaches the ...

college calculus - 1) each side of a square is increasing at a rate of 6cm/s. At...

calculus (related rates) - A particle moves in the plane along the curve y=xln(x...

calculus - a particle is moving counterclockwise around the circular path whose ...

Calculus - A particle is moving along the curve whose equation is (xy^3)/(1+y^2...

Calculus - A particle is moving along the curve . As the particle passes through...