Please Help me on rate of change!?

Question

volume of a cone of heigh H is (pi.H^3)/12. If H increases at a constant rate of 0.2cm/s and the initial heigh is 2cm, express V in terms of t and find the rate of change of V at a time t.

Answer 1

H = 2 + 0.2t
V = (pi* (2+0.2t)^3)/12
see? now, check:
H = 2+ 0.2cm when t = 1
H = 2.2
V = (pi*(2.2)^3)/12
= 1331/1500*pi
V = 1331/1500*pi when t = 1
H = 2+0.4 at t = 2
V = (pi*(2.4)^3)/12
= 1728/1500* pi when t= 2
Like this, V = 2197/1500 at t= 3

Answer 2

dude, it's like this:
First :
H = (2 + 0.2*t ) cm

Second : substitute the H with the formula above.
V(t) = (pi*(2+0.2*t)^3)/12

Third :for the rate of change, u find the derivative of V(t)
V(t)
= (pi/12)(4+0.8t+0.04t^2)(2+0.2t)
= (pi/12)*(8+2.4t+0.24t^2+0.008t^3)

dv(t)/dt = 0.2pi+0.04pi*t+0.002pi*t^2

(u might wanna check it again for all the result.)

Answer 3

V=(pi*H^3)/12

The rate of change of V is simply V' (or the derivative of V)
Then evaluate V'(t).

The person below has an equation regarding V'...

Answer 4

V = (π/12)h^3
h(t) = 2 + 0.2t, so
V(t) = (π/12)(0.2t + 2)^3
V(t) = (π/12)(0.008t^3 + 0.24t² + 2.4t + 8)
V(t) = (.002π/3)t^3 + 0.02πt² + 0.2πt + 2π/3
dV/dt = 0.002πt² + 0.04πt + 0.2π