Related Rates: Equilateral Triangle Trough

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ChaoticLlama

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Dec 11, 2004
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The cross section of a water trough is an equilateral triangle with its top edge horizontal . If the trough is 5m long and 25cm deep, and if water is flowing in at a rate of 0.25m³/min, how fast is the water level rising when the water is 10cm deep at the deepest point?

What I know:
After the equation relating volume at any time has been differentiated, then
dV/dt = .25m³/min and I want to solve for dh/dt

However, I do not know how to set up the equation because I do not know which variables are dependent on time.

The equation I am using for the Volume is V = bhw <-- which of these -the base, height, or width- can be numerical, and which must be left as variables with respect to time?

The answer is √(3) / 4 m/s

Thanks.
 
We want \(\displaystyle \frac{dh}{dt}\) when h=10 cm

We know \(\displaystyle \frac{dV}{dt}\)=.25

Using triangles, the volume of water in the trough can be given by:

\(\displaystyle V=\frac{1}{2}(b)(h)(5)\)

\(\displaystyle V=\frac{5}{2}bh\)

Half of the trough forms a right triangle with side \(\displaystyle (\frac{1}{4})tan(\frac{\pi}{6})=\frac{sqrt{3}}{12}\)

and hypoteneuse \(\displaystyle \frac{1}{4cos(\frac{\pi}{6})}=\frac{sqrt{3}}{6}\)

We can use similar triangles \(\displaystyle \frac{\frac{b}{2}}{h}=\frac{1}{sqrt{3}}\)

\(\displaystyle b=\frac{2h\sqrt{3}}{3}\).

Sub into volume equation:

\(\displaystyle V=\frac{5}{2}(\frac{2h\sqrt{3}}{3})h=\frac{5h^{2}sqrt{3}}{3}\)

Differentiate with respect to t:

\(\displaystyle \frac{dV}{dt}=\frac{10h\sqrt{3}}{3}\frac{dh}{dt}=.25\)

\(\displaystyle .25=\frac{10sqrt{3}}{3}(.10)\frac{dh}{dt}\)

Solving for\(\displaystyle \frac{dh}{dt}=\frac{sqrt{3}}{4}=.433 m/min.\)

Is this the answer you're looking for?. Check it out. Easy to make mistakes.
 
Hello, ChaoticLlama!

This one takes a LOT of explaining . . .

The cross section of a water trough is an equilateral triangle with its top edge horizontal.
If the trough is 5m long and 25cm deep, and if water is flowing in at a rate of 0.25m³/min,
how fast is the water level rising when the water is 10cm deep at the deepest point?
Click to expand...
Very sneaky . . . they have meters and centimeters!

The length is: 500 cm.
The volume is changing at: 250,000 cm\(\displaystyle ^3\)/min.

Code: 
              25/√3
    *-------+-------* 
     \      |      /
      \     |  r  /
       \  25+----/
        \   |   / 50/√3
         \  |h /
          \ | /
           \|/
            *
If an equilateral triangle has an altitude of \(\displaystyle 25\), its side is \(\displaystyle \frac{50}{\sqrt{3}}\)

From the similar right triangles: .\(\displaystyle \frac{r}{h}\,=\,\frac{25/\sqrt{3}}{25}\,=\,\frac{1}{\sqrt{3}}\;\;\Rightarrow\;\;r\,=\,\frac{h}{\sqrt{3}}\) [1]

The area of the triangle (of water) is: .\(\displaystyle A\:=\:\frac{1}{2}(\text{base})(\text{height})\:=\:\frac{1}{2}(2r)(h)\:=\:rh\)

Substitue [1]: .\(\displaystyle A\:=\:\left(\frac{h}{\sqrt{3}}\right)(h)\:=\:\frac{h^2}{\sqrt{3}}\)

Volume is Area times Length: .\(\displaystyle V\:=\:\left(\frac{h^2}{\sqrt{3}}\right)\cdot500\:=\:\frac{500}{\sqrt{3}}h^2\)


Differentiate with respect to time: .\(\displaystyle \frac{dV}{dt}\;=\;\frac{1000}{\sqrt{3}}h\left(\frac{dh}{dt}\right)\)

At that instant: .\(\displaystyle \frac{dV}{dt}\,=\,250,000,\;h\,=\,10\)

We have: . \(\displaystyle 250,000\:=\:\frac{1000}{\sqrt{3}}(10)\left(\frac{dh}{dt}\right)\)

Therefore: .\(\displaystyle \frac{dh}{dt}\;=\;25\sqrt{3}\;\approx\;43.3\) cm/min

. . . Someone check my work . . . please!
 
Thanks a lot, you have helped a great deal.

\(\displaystyle \
\int_0^\pi {\frac{1}{{1 + x^2 }}}
\\)
 
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