Originally Posted by

**Jon123** Hello!

I have a question in regards below, and have answered the question but do not understand number line 5, could you please explain the reason for this.

A water tank has the shape of an inverted circular cone with base radius 3m and height 5m. If water is pumped into the tank at a rate of 1.5m3/min, find the rate at which the water level is rising when the water is 3.5m deep.

Let *t* represent time (in minutes),

*h*(*t*) represent height (in metres) of water at time *t,*

*r*(*t*) represent radius (in metres) at time *t,*

*V*(*t*) represent volume (in cubic metres) of water at time *t.*

*dV/dt = *1.5m3/min h=5m r=3m r/h=3/5 so, r=3h/5

Formula: V(t) = 1/3 \pi [r(t)]² h(t)

1. d/dt{V(t)} = d/dt{1/3 \pi [r(t)]² h(t)}

ok i am going to skip to a few steps to get to the problem i am having.

4. dV/dt= d/dt { 3 \pi h(t)3 / 25}

5. d/dv = 9 \pi h(t)² / 25 dh/dt

ok, so where did the "dh/dt" come from.. is it because we are looking for

the rate at which the water level is rising over time.. if so please advise

if that is the reason.