Question

I have a calculus question about related rates, pic included

Answer 1

40807 cm³/min

Step-by-step explanation

The tank has the shape of a cone, with a total height of 9 meters and a diameter of 3.5 m - so the radius, which is half the diameter, is 1.75 m. As we can see, the relationship between the height of the cone and the radius is,

`(r)/(h)=(1.75m)/(9m)=(7)/(36)\Rightarrow r=(7)/(36)h`

So the volume of water will be given by,

`V(h)=(1)/(3)(\pi r^2)h=(1)/(3)\cdot\pi\cdot(7^2)/(36^2)h^2\cdot h=(49\pi)/(3888)h^3`

Where h is the height of the water (not the tank).

If we derive this equation, we will find the rate at which the volume of water is changing with time,

`(dV)/(dt)=(49\pi)/(3888)\cdot3h^(3-1)=(49\pi)/(3888)\cdot3h^2=(49\pi)/(1296)h^2`

We want to know what is the change of volume with respect to time, and this is,

`(dV)/(dt)=(dV)/(dt)\cdot(dh)/(dt)`

Because the height also changes with time. We know that this change is 24 cm per minute when the height of the water in the tank is 1 meter (or 100 cm), so we have,

`(dV)/(dt)=(49\pi)/(1296)h^2\cdot(dh)/(dt)=(49\pi)/(1296)\cdot100^2cm^2\cdot(24cm)/(1min)\approx28507cm^3/min`

This is the rate at which the water is increasing in the tank. However, we know that there is a leak at a rate of 12300 cm³/min, which means that in fact the water is being pumped into the tank at a rate of,

`28507cm^3/min+12300cm^3/min=40807cm^3/min`

Hence, the water is being pumped into the tank at a rate of 40807 cm³/min, rounded to the nearest whole cm³/min.