Differentiation help plz!!!!!!!!!!

The volume, V cm^3, of a cone of height H is πh^3/12. If h increases at a constant rate of 0.2cm per second and the initial height is 2cm express, V in terms of T and find the rate of change of V at time T

answer for part 1 is 2π /3( T/10 + 1) ^3

answer for part b is π/5(T/10 +1)^2


thank u in advance :)

Given:

d[h(t)]/dt = 0.2 cm/s h(0) = 2 cm

Integrating both sides:

h(t) = (0.2 cm/s)∫dt

This is the results with a constant of integration:

h(t) = (0.2 cm/s)t + C

Initial conditions will give us the value of C:

C = h(0) = 2 cm

h(t) = (0.2 cm/s)t + 2 cm

Substitute for h:

V(t) = π{0.2 cm/s(t) + 2cm}³/12

Factor out a 2 cubed:

V(t) = π{2}³{0.1 cm/s(t) + 1cm}³/12

Divide by the factor of 4 within the 12:

V(t) = (2π/3){t/10 + 1}³

There is your desired answer.

To find d[V(t)]/dt, you must use the chain rule

d[V(t)]/dt = {d[V(u)]/du}{du/dt}

let u = t/10 + 1, then du/dt = 1/10, V(u) = (2π/3)u³, and d[V(u)]/du = 3(2π/3)u²

Substituting the above values into the chain rule:

d[V(t)]/dt = {3(2π/3)u²}{1/10}

Canceling 3/3:

d[V(t)]/dt = {(2π)u²}{1/10}

Divide by the factor of 2 within the 10:

d[V(t)]/dt = {(π/5)u²}

Reversing the "u" substitution:

d[V(t)]/dt = (π/5)(t/10 + 1)²

I tried to solve it but having doubt in my answer, but I can tell you about an online tutor who is helping students regarding subject issues and assignments etc. So, if you have time, just contact him at 1classhelp@gmail.com or wait for other guys to answer.

I am sure your problems will be sorted out. : )