Homework Help: calculus

Posted by Susie on Monday, February 21, 2011 at 12:58pm.

Can someone tell me how to do this type of problem? A particle moves along the curve y=sqrt(1+x^3). As it reaches the point (2,3) the y coorindate is increasing at a rate os 4cm/s. How fast is the x coordinate of the point changing at that instant?

calculus - MathMate, Monday, February 21, 2011 at 1:37pm
Differentiate both sides with respect to t:
dy/dt = d(sqrt(1+x^3))/dx * dx/dt
=(3x²/(2sqrt(1+x³)))*dx/dt

So given x=2,y=3 and dy/dt=3 cm/s
substitute in formula above to solve for dx/dt.

I get dx/dt=2.

Answer this Question

Related Questions

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 ...
calculus - I need help setting up this problem: A particle moves along the curve...
calculus - A particle is moving along the curve below. y = sqrt(x) As the ...
Calculus 1 - A particle is moving along the curve y= 4 \sqrt{3 x + 1}. As the ...
PHYSICS - When the force F acting on a particle depends only on the particle'...
college calculus - 1) each side of a square is increasing at a rate of 6cm/s. At...
Calc - A particle is moving along the curve y= 3 \sqrt{3 x + 4}. As the particle...
calculus - 4. A particle starts at the point (5, 0) at t = 0 and moves along the...
Calculus - A particle is moving along the curve . As the particle passes through...

For Further Reading

First Name:
School Subject:
Answer: