Question

a particle moves along curve y=x^3/2 in the first quadrant in such a way that its distance from the origin increases at the rate of 11 units per second . the value of dx/dt when x=3 is

Answer 1

The distance D of the point from the origin is given by  

D(x) = sqrt(x^2 + y^2)  

= sqrt(x^2 + x^3)  

dD/dt = dD/dx dx/dt  

So, dx/dt = (dD/dt) / (dD/dx)  

dD/dx = (2x + 3x^2) / 2(x^2 + x^3)^(1/2)  

dD/dt = 11  

So, dx/dt =  

11 / [(2x + 3x^2) / 2(x^2 + x^3)^(1/2)]  

= 22 (x^2 + x^3)^(1/2) / (2x + 3x^2)  

So, dx/dt | x = 3  

= 22 (9 + 27)^(1/2) / (6 + 27)  

= 22 x 6 / 33  

= 4 (horizontal) units per second