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 ycorrdinate is increasing at a rate of 4cm/s. How fast is the xcoordinate 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...
 Calculus  A particle moves along the curve y = sqr(1+x^3). As it reaches the ...
 college calculus  1) each side of a square is increasing at a rate of 6cm/s. At...
 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...
 calulus  a particle is moving along the cure y=sqrt x. as the particle passes ...
More Related Questions