Helpp!!! This question is crazy =(

[Queenisabella87]

A tank is shaped like an inverted cone with a 10 foot height and 4 ft radius at the top.

Water is being pumped into it at a rate of 2 pi ft^3 / min.

how fast is the depth of the water increasing when it is 5ft deep full???


pi = the pi sign = ~3.14125

 

[tkhunny]

You'll need the formula for the volume of a right circular cone, given its height and base radius.
You'll need a concept of similar triangles.

Let's see your plan.

 

[galactus]

Actually, this problem is not as crazy as it is cliche. This is a typical related rates problem one encounters in Calc I.

Do some Googling and you can find plenty about related rates.

A little trick I like to use when dealing with cross sectional area, is to note that

the rate of change of the volume is equal to the cross sectional area at that instant times the rate of change of the height.

You are given the rate of change of the volume, dV/dt

and you can find the surface area of the water, A(t), when it is 5 feet deep by using similar triangles and the area of a circle formula.

You are asked to find dh/dt.

\(\displaystyle \frac{dV}{dt}=A(t)\cdot \frac{dh}{dt}\)

\(\displaystyle \frac{dh}{dt}=\frac{\frac{dV}{dt}}{A(t)}\)

 

[Deleted member 4993]

A tank is shaped like an inverted cone with a 10 foot height and 4 ft radius at the top.

Water is being pumped into it at a rate of 2 pi ft^3 / min.

how fast is the depth of the water increasing when it is 5ft deep full???


pi = the pi sign = ~3.14125

On the top right hand corner of this page - there is a search button - type in "inverted cone" and hit "search" - you'll find many similar problems solved in this forum.

This problem is very similar to that of you had posted sometime ago:

viewtopic.php?f=3&t=42474&p=165282#p165282

By the way, the value of ? you have posted is wrong - but that should not matter because the final answer for "this" problem does not have ? as a factor.

Alternative way to Galactus's method is to follow tk's method:

1) Find the equation of the volume of a cone in terms of radius (r) and height (h) - these are functions of time.

2) Find the relation between 'r' and 'h' from your given geometry (and similar triangles).

3) Rewrite the equation in (1) in terms of 'h' only - using relation in (2)

4) Differentiate and solve.

Please share your work with us, indicating exactly where you are stuck - so that we may know where to begin to help you.

 

[soroban]

Hello, Queenisabella87!

A tank is shaped like an inverted cone with a 10-ft height and 4-ft radius.

Water is being pumped into it at a rate of \(\displaystyle 2\pi\text{ ft}^3\text{/min.}\)

How fast is the depth of the water increasing when it is 5 ft deep?

      : - 4 - : - 4 - :
    - * - - - * - - - *
    :  \      |      /
    :   \     |  r  /
    :    \- - + - -/
   10     \:::|:::/
    :      \:h|::/
    :       \:|:/
    :        \|/
    -         *

\(\displaystyle \text{We are told that: }\:\frac{dV}{dt} \,=\,2\pi\text{ ft}^3\text{/min.}\)

\(\displaystyle \text{The volume of the water is: }\:V \:=\:\frac{\pi}{3}r^2 h\) .[1]

\(\displaystyle \text{From similar right triangles, we have: }\:\frac{h}{r} \,=\,\frac{10}{4} \quad\Rightarrow\quad r \,=\,\frac{2}{5}h\) .[2]


Substitute [2] into [1]:

. . \(\displaystyle V \;=\;\frac{\pi}{3}\left(\frac{2}{5}h\right)^2h \;=\;\frac{4\pi}{75}h^3\)


\(\displaystyle \text{Differentiate with respect to time: }\;\frac{dV}{dt} \;=\;\frac{4\pi}{25}h^2\cdot\frac{dh}{dt}\)

\(\displaystyle \text{When }h = 5\text{, we have: }\;2\pi \;=\;\frac{4\pi}{25}(5^2)\,\frac{dh}{dt} \quad\Rightarrow\quad 4\pi \frac{dh}{dt} \:=\:2\pi \quad\Rightarrow\quad \frac{dh}{dt} \:=\:\frac{1}{2}\)


\(\displaystyle \text{The water is rising at }\tfrac{1}{2}\text{ ft/min.}\)

 

[galactus]

I will go ahead and show you what I was getting at.

Note: the rate of change of the volume is equal to the cross sectional area at that instant times the rate of change of the height.

This means \(\displaystyle \frac{dh}{dt}=\frac{\frac{dV}{dt}}{A(t)}\)

We are given \(\displaystyle \frac{dV}{dt}=2{\pi} \;\ ft^{3}\)

We need \(\displaystyle \frac{dh}{dt}\)

The only thing we need at this point is A(t).

It can be found by using the same similar triangle reasoning Soroban used. \(\displaystyle \frac{r}{h}=\frac{4}{5}\Rightarrow r=\frac{2h}{5}\)

In other words, the radius is 2/5th the height. But, you can see that right off by just looking at radius 4 and height 10. 4 is 2/5th of 10.

Anyway, sub this into the area of a circle formula. \(\displaystyle A(h)={\pi}\left(\frac{2}{5}h\right)^{2}=\frac{4{\pi}h^{2}}{25}\)

We want this when h=5, so \(\displaystyle A=\frac{4{\pi}(5)^{2}}{25}=4{\pi}\)

Now, just use \(\displaystyle \frac{dh}{dt}=\frac{\frac{dV}{dt}}{A(h)}=\frac{2\pi}{4\pi}=\frac{1}{2}\)