Question

The volume of oil in a cylindrical container is increasing at a rate of 150 cubic inches per second. The height of the cylinder is approximately ten times the radius. At what rate is the height of the oil changing when the oil is 35 inches high?

Answer 1

`\dot h = 1.299\,(in)/(s)`

Explanation:

The volume of the cylindrical container is:

`V = \pi\cdot r^(2)\cdot h`

Where:

`r` - Radius, in inches.

`h` - Height, in inches.

The rate of change of the volume is:

`\dot V = 2\pi\cdot r \cdot h \cdot \dot r + \pi\cdot r^(2)\cdot \dot h`

`\dot V = \pi \cdot r \cdot (2\cdot h \cdot \dot r + r \cdot \dot h)`

`\dot V = 0.1\pi \cdot h \cdot (0.2 \cdot h \cdot \dot h + 0.1\cdot h \cdot \dot h)`

`\dot V = 0.03\pi\cdot h^(2) \cdot \dot h`

The rate of change of the height is:

`\dot h = (\dot V)/(0.03\pi\cdot h^(2))`

`\dot h = (150\,(in^(3))/(s) )/(0.03\pi\cdot (35\,in)^(2))`

`\dot h = 1.299\,(in)/(s)`

Answer 2

Answer: The height of the oil changing at the rate of
`(200)/(49\pi)\ inches/sec`

Explanation:

Since we have given that

volume is increasing at a rate of 150 cubic inches per second.

Height of cylinder = 10 times of radius.

As we know the formula for "Volume":

`V=\pi r^2h\\\\V=\pi ((h)/(10))^2* 10r\\\\V=(1)/(100)\pi h^3`

We will derivative it w.r.t 't'.

So, it becomes,

`(dv)/(dt)=(3h^2\pi)/(100)* (dh)/(dt)\\\\150=(3h^2\pi)/(100)(dh)/(dt)\\\\(150* 100)/(3* 35* 35* \pi)=(dh)/(dt)\\\\(200)/(49\pi)=(dh)/(dt)`

Hence, the height of the oil changing at the rate of
`(200)/(49\pi)\ inches/sec`