# Depleting Volume

• Mar 17th 2014, 06:56 AM
VinH
Depleting Volume
Hi,

I am trying to figure out how an answer from my textbook was worked out. The textbook contains MANY ridiculous errors, so I would like to get a second opinion from someone here regarding the accuracy of the solution. It is possible that I just do not understand the logic, but I want to be sure before I go wasting any more time on it. Here is the question:

Code:

`A plastic juice bottle holds 2 L of liquid. In an experiment, a small hole is drilled in the bottom of the bottle. The volume of liquid, V, remaining after t seconds can be modelled by V(t) = 2 - t/5 + t^2/200, where t >= 0. How long does it take for the 2L of liquid to drain from the bottle?`
The solution provided is as follows:

Code:

```V(t) = 2 - t/5 + t^2/200 V(t) = 2000 - 200t + 5t^2 V(t) = 5(400 - 40t + t^2) V(t) = 5(20 - t)(20 + t) 20 - t = 0 20 = t It takes 20 seconds to drain the bottle.```
I do not understand how they are getting to step #2 in their solution and I would really appreciate some clarification. I have been racking my brain all day on other problems so it is possible I am overlooking something really obvious at this point.

Thank you.
• Mar 17th 2014, 07:35 AM
earboth
Re: Depleting Volume
Quote:

Originally Posted by VinH
Hi,

I am trying to figure out how an answer from my textbook was worked out. The textbook contains MANY ridiculous errors, so I would like to get a second opinion from someone here regarding the accuracy of the solution. It is possible that I just do not understand the logic, but I want to be sure before I go wasting any more time on it. Here is the question:

Code:

`A plastic juice bottle holds 2 L of liquid. In an experiment, a small hole is drilled in the bottom of the bottle. The volume of liquid, V, remaining after t seconds can be modelled by V(t) = 2 - t/5 + t^2/200, where t >= 0. How long does it take for the 2L of liquid to drain from the bottle?`
The solution provided is as follows:

Code:

```V(t) = 2 - t/5 + t^2/200 V(t) = 2000 - 200t + 5t^2 V(t) = 5(400 - 40t + t^2) V(t) = 5(20 - t)(20 + t) 20 - t = 0 20 = t It takes 20 seconds to drain the bottle.```
I do not understand how they are getting to step #2 in their solution and I would really appreciate some clarification. I have been racking my brain all day on other problems so it is possible I am overlooking something really obvious at this point.

Thank you.

Hello,

V(t) = 2 - t/5 + t^2/200 | *1000 to get the volume in cm³

V(t) = 2000 - 200t + 5t^2

But much more funny is this transformation:

V(t) = 5(400 - 40t + t^2)

V(t) = 5(20 - t)(20 + t) --> this is definitely wrong, it should read

V(t) = 5(20 - t)^2
• Mar 17th 2014, 11:11 AM
HallsofIvy
Re: Depleting Volume
The first step is just silly! Earboth correctly deduced that they must have converted from liters to cubic centimeters but I would object to using "V(t)" to mean both "volume in liters" and "volume in cubic centimeters". Personally, I would not have made that conversion. Given that $V(t)= 2- t/5+ t^2/200$.
When all 2 liters have drained, V(t) will be equal to 0 so we need to solve the equation $2- t/5+ t^2/200= 0$.

Now multiply both side by 200 to get rid of the fractions: $400- 40t+ t^2= (t- 20)^2= 0$ (which is what they have after factoring "5" out).