Homework Help: calculus

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

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 - drwls, Friday, May 8, 2009 at 11:10am
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

Answer this Question

Related Questions

calculus - Can someone tell me how to do this type of problem? A particle moves ...
calculus - A particle moves along the curve y=square sqrt 1+x^3. As it reaches ...
college calculus - 1) each side of a square is increasing at a rate of 6cm/s. At...
math - a particle moves along the curve y= sqrt 1+x cubed. As it reaches the ...
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 - 5. A particle moves along the y – axis with velocity given by v(...
calculus (related rates) - A particle moves in the plane along the curve y=xln(x...
DIFF. CALCULUS - A particle is moving along the parabola 4y = (x + 2)^2 in such ...
Calculus - A point moves along the curve y=(x-2)^2 such that the x coordinate ...

For Further Reading

First Name:
School Subject:
Answer: