A simple derivative question

The height if a cone is decreeing at 3cm/s while its radius is increasing at 2cm/s . When the radius is 4cm and the height is 6cm , is the volume of the cone increasing or deceasing? At what rate is the volume changing then?

We know that for a cone, V = (π/3)*(r^2*h) To find dv/dt (the rate of change of volume), we can differentiate the volume equation. Since you're given r, dr/dt, h, and dh/dt, we can just differentiate both sides of the equation to find dV/dt. For the right side of the equation, you can use product rule on r^2*h:

dV/dt = (π/3) (2r*dr/dt*h + dh/dt*r^2) = (π/3) (8*2*6 + -3*16) = 48π/3 cm^3/s

Since dV/dt is positive, the volume of the cone is increasing with respect to time at 48π/3 cm^3/s.

This is a "related rates" question. You are given information about the radius and height of a cone (with their rates of change), and you are being asked about the rate of change in the cone's volume. Connect the variables using a known equation:
V=(pi/3) r^2 h.
V, r, and h are functions of time t; take the derivative with respect to t of both sides of the equation, using the product rule and the chain rule on the RHS:
dV/dt=(pi/3)[2r(dr/dt)h+r^2 dh/dt].
We are given that dh/dt=-3, dr/dt=2, r=4, and h=6. So
dV/dt=(pi/3)[2*4*2*6+(4^2)*(-3)]=
=(pi/3)16[6-3]=
=16pi.
Since this is positive, the volume of the cone is increasing at the given instant in time; the volume is changing at a rate of 16pi cm^3/s at that point in time.

V = (1/3) * π * r^2 * h =====> taking deerivative

V ' = (1/3) * π * ( r^2 * dh/dt + h * 2 * r * dr/dt )

V ' = (1/3) * π * ( r^2 * dh/dt + h * 2 * r * dr/dt )

having:

dh/dt = -3cm/s , dr/dt = +2cm/s , r = 4 cm , h = 6cm

V ' = (1/3) * π * ( 4^2 * (-3) + 6 * 2 * 4 * 2 )

V ' = (1/3) * π * ( 16 * (-3) + 96 )

V ' = (1/3) * π * ( -48 + 96 )

V ' = (1/3) * π * ( 48 )

V ' = 16π

dV/dt ≈ 50.3 cm^3/s =====> incresaing