Posted by **Tay** on Tuesday, January 17, 2012 at 8:34pm.

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

- math -
**bobpursley**, Tuesday, January 17, 2012 at 8:37pm
y= sqrt(1+x^3)

dy/dt= 1/2 * 1/sqrt(1+x^3)* 3x^2 dx/dt

dy/dt= 3x^2/2y * dx/dt

dx/dt= 2y dy/dt * 1/3x^2

you are given x, y, dy/dt, solve fod dx/dt.

check my work, I did it in a hurry

- math -
**Tay**, Tuesday, January 17, 2012 at 9:43pm
hey thanks we are checking it now:)

## Answer this Question

## Related Questions

- calculus - A particle moves along the curve y=square sqrt 1+x^3. As it reaches ...
- 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 ...
- calc - A particle moves along the curve below. y = sqrt(8 + x^3) As it reaches ...
- calculus - I need help setting up this problem: A particle moves along the curve...
- college calculus - 1) each side of a square is increasing at a rate of 6cm/s. At...
- Calculus - A particle moves along the curve y = sqr(1+x^3). As it reaches the ...
- calculus - A particle is moving along the curve below. y = sqrt(x) As the ...
- Calculus - A particle is moving along the curve whose equation is (xy^3)/(1+y^2...
- DIFF. CALCULUS - A particle is moving along the parabola 4y = (x + 2)^2 in such ...