Derivatives, rates of change (trapezoidal prism)

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Homework Help Overview

The problem involves a water trough with an isosceles trapezoidal cross-section, where participants are exploring how the water level changes over time as the trough is filled at a specified rate. The context includes concepts from calculus, specifically derivatives and rates of change.

Discussion Character

  • Exploratory, Assumption checking, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss the relationship between the volume of water in the trough and the height of the water level, with attempts to derive the rate of change of height with respect to time. There are questions about how to express the top width of the trapezoid in terms of the water height and how to differentiate the volume formula correctly.

Discussion Status

Some participants have suggested alternative approaches to express the top width as a function of height and to differentiate the volume formula. There is ongoing confusion regarding the correct application of formulas and unit conversions, indicating a lack of consensus on the method to find the desired rate of change.

Contextual Notes

Participants note the importance of consistent units, particularly the need to convert the fill rate from cubic meters to cubic centimeters to match the dimensions used in the problem.

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1. A water trough is 10 m long and a cross-section has the shape of an isosceles trapezoid that is 30 cm wide at the bottom, 80 cm wide at the top, and has height 50 cm. If the trough is being filled with water at the rate of 0.2 m^3/min, how fast is the water level rising when the water is 30 cm deep?

Homework Equations


$$V=\frac{(a+b)h}{2}×L$$

The Attempt at a Solution



Given (I converted all to cm):
$$a=30$$ $$\frac{da}{dt}=0$$ $$base=80$$ $$height=50$$ $$\frac{dV}{dt}=0.2$$ $$h=30$$ $$L=10$$ $$\frac{dL}{dt}=0$$

I need to find $$\frac{dh}{dt}$$

After I find the derivative of V using natural log.
$$lnV=\frac{1}{2}ln(a+b)hL$$
$$lnV=\frac{1}{2}ln(a+b)+lnh+lnL$$
$$\frac{dV}{dt}=\frac{1}{2}(\frac{db/dt}{a+b}+\frac{dh/dt}{h})×V$$

I don't know how I'm supposed to find $$\frac{db}{dt}$$ i get b=60.

Any help is much appreciated.
 

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This is the same as the other one - you don't need to deal with all those logs.

(a+b)L/2 is just a constant that you are given. Put k=2/(a+b)L
Lynchpin You need to express b in terms of h.

Think it through - let a is the bottom, and c is the top, the overall height is H.
0<h<H
when h=0, what is b(h) equal to?
when h=H, what is b(h) equal to?
what formula makes that happen? b(h)=

* Rearrange your formula to make h the subject.
* differentiate both sides to get dh/dt in terms of dV/dt.
* You are given dV/dt.
 
Last edited:
Thanks, but I still get this.

$$2V=(a+b)hL$$ $$h=\frac{2V}{(a+b)L}=Vk$$ $$\frac{dh}{dt}=\frac{dV}{dt}k=0.2×\frac{2}{(30+60)1000}$$

I don't get the right answer, which is 10/3.
 
Still the wrong formula.
I misread you - I thought you put b as the top width - my bad.
I have rewritten post #2 to reflet this.

Think it through - let a is the bottom, and c is the top, the overall height is H.
0<h<H
when h=0, what is b(h) equal to?
when h=H, what is b(h) equal to?
what formula makes that happen? b(h)=
 
Still the wrong formula.
I misread you - I thought you put b as the top width - my bad.
I have rewritten post #2 to reflet this.

Think it through - let a is the bottom, and c is the top, the overall height is H.
0<h<H
when h=0, what is b(h) equal to?
when h=H, what is b(h) equal to?
what formula makes that happen? b(h)=
 
Watch your units!

If you are going to use cm for the dimensions of the trough, then the fill rate of 0.2 m^3/min should also be converted to units of cm^/min.
 

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