Derivative Applications - Trapezoidal Trough

[pahuja]

Water is leaking from a trough at the rate of 0.8 L/s. The trough has a trapezoidal cross section, where the width at the bottom is 55 cm, at the top is 85cm, and the height is 25 cm. The length of the trough is 3 m.Find the rate at which the height is changing when the depth of water is 11 cm.

I have already solved the question, my final answer was approx. 0.039cm/s. If someone could take a look at the steps I attached and confirm that I did it correctly that would be appreciated.

 

[Dr.Peterson]

A few words of explanation about how you are finding the area formula would have made it much easier to be sure you were right, but you are. The answer appears to be correct.

Was there any specific point at which you were unsure?

 

[MarkFL]

When I was a student, I liked to work such problems in general terms so that I could obtain a formula to apply to all similar problems.

The variable width \(w\) of the volume of water at the top would increase linearly as the height \(h\). Let \(w_B\) be the fixed width at the bottom and \(w_T\) be the fixed width of the trough at the top. Let \(h_T\) be the height of the trough. Let \(\ell\) be the length of the trough. All linear measures will be in cm, and time will be measured in seconds. We know:

[MATH]w(0)=w_B[/MATH]
[MATH]w(h_T)=w_T[/MATH]
Hence:

[MATH]w(h)=\frac{w_T-w_B}{h_T}h+w_B[/MATH]
Hence the volume of the water at time \(t\) is

[MATH]V(t)=\frac{h\ell}{2}\left(\frac{w_T-w_B}{h_T}h+2w_B\right)[/MATH]
And so we may compute:

[MATH]V'(t)=\frac{h\ell}{2}\cdot\frac{w_T-w_B}{h_T}\cdot\frac{dh}{dt}+\frac{\ell}{2}\left(\frac{w_T-w_B}{h_T}h+2w_B\right)\frac{dh}{dt}[/MATH]
There results:

[MATH]\frac{dh}{dt}=\frac{h_T}{\ell\left(h(w_T-w_B)+w_Bh_T\right)}V'(t)[/MATH]
Now we may plug in the given values:

[MATH]\frac{dh}{dt}=\frac{25\text{ cm}}{(300\text{ cm})\left((11\text{ cm})(85-55)\text{ cm}+(55\text{ cm})(25\text{ cm})\right)}\left(-800\frac{\text{cm}^3}{\text{s}}\right)=-\frac{40}{1023}\,\frac{\text{cm}}{\text{s}}\approx-0.039100684261974585\,\frac{\text{cm}}{\text{s}}[/MATH]
It appears our results do agree.