Candle's Height

  • Thread starter Bashyboy
  • Start date
  • #1
1,421
5

Homework Statement


In the problem, we are to consider two candles, call them C1 and C2, with different heights and different thicknesses. Call the height of C1 H1, and for C2, call it H2. The taller candle burns can burn for 7/2 hours, and the short one, 5 hours. After two hours lapses, the candles have equal heights. Two hours ago, what fraction of the candle's height gave the short candle's height?


Homework Equations





The Attempt at a Solution


Here some observations, of which I am not certain are true, that I made. Assuming the candles burn at a constant rate, the rate of C1 is H1/3.5, and the rate of C2 is H2/5.

After two hours of burning has lapsed, h1/h2 = 1.

Now, I am going to assume the candles are cylindrical.

[itex]V_1 = \pi r^2_1H_1[/itex] I am also going to assume that the burning process is such that the radius is not altered.

[itex]\frac{dV_1}{dt} = \pi r^2_1 \frac{dH_1}{dt}[/itex]. I know how the height changes with time--it's constant. [itex]\frac{dH_1}{dt} = \frac{H_1}{3.5}[/itex] If I write it in the form [itex]\frac{dH_1}{dt} = (H_1)(3.5)^{-1}[/itex], this is a differential equation in which I can use the separation of variables method.

[itex]\int H_1 dH_1 = (3.5)^{-1} \int dt[/itex]

which comes out to be [itex]H_1 = \sqrt{\frac{t}{3.5} + C}[/itex]

----------------------------------------------------------------

There are a few details I am confused about. Did I properly apply the separation of variables method? If so, what does the constant C represent, and can it be set to zero? Will this solution lead anywhere?

Thank you, and please don't give every detail of the problem away.
 
Last edited:

Answers and Replies

  • #2
Borek
Mentor
28,635
3,106
I have a feeling most of the things you did were unnecessary. The only thing you need is to express H(t) as a function of initial H and the burning speed (for each candle separately) and apply condition H1(2hours)=H2(2hours) - most likely it will be of the form that can be easily solved for the initial ratio of heights.
 
  • #3
223
10
This is an interesting one, although could you please explain the problem:
Two hours ago, what fraction of the candle's height gave the short candle's height?
It is confusing to say the least. 2 hours ago, I understand, means before they were both lit, but "what fraction of the (taller?) candle's height gave short candle's height"?

Were I to pose a problem with this information I would want to find the relation between thicknesses or heights.
 
Last edited:
  • #4
pasmith
Homework Helper
1,965
599

Homework Statement


In the problem, we are to consider two candles, call them C1 and C2, with different heights and different thicknesses. Call the height of C1 H1, and for C2, call it H2. The taller candle burns can burn for 7/2 hours, and the short one, 5 hours. After two hours lapses, the candles have equal heights. Two hours ago, what fraction of the candle's height gave the short candle's height?
I take it that we are assuming that C1 is the taller candle.

The Attempt at a Solution


Here some observations, of which I am not certain are true, that I made. Assuming the candles burn at a constant rate, the rate of C1 is H1/3.5, and the rate of C2 is H2/5.
Correct, if H1 and H2 are the initial heights of the candles.

After two hours of burning has lapsed, h1/h2 = 1.

Now, I am going to assume the candles are cylindrical.
I don't think you need assume anything about the shape or material of each candle. You need only assume that the rate of change of height is constant.

[itex]V_1 = \pi r^2_1H_1[/itex] I am also going to assume that the burning process is such that the radius is not altered.

[itex]\frac{dV_1}{dt} = \pi r^2_1 \frac{dH_1}{dt}[/itex]. I know how the height changes with time--it's constant. [itex]\frac{dH_1}{dt} = \frac{H_1}{3.5}[/itex]
No. When you said that the constant rate of burning of C1 was [itex]H_1/3.5[/itex], you must have been taking [itex]H_1[/itex] to mean the initial height (because otherwise that statement would be false). You now appear to be using [itex]H_1[/itex] both for initial height (which is a constant, so it's derivative with respect to time is zero) and for height at time t (which varies with t). This has led you to an incorrect ODE.

If I write it in the form [itex]\frac{dH_1}{dt} = (H_1)(3.5)^{-1}[/itex], this is a differential equation in which I can use the separation of variables method.

[itex]\int H_1 dH_1 = (3.5)^{-1} \int dt[/itex]
You're dividing both sides by [itex]H_1[/itex] so you should have
[tex]
\int \frac1{H_1}\,dH_1 = \frac1{3.5}\int\,dt
[/tex]
which you may go on to solve as practice in integration, but it doesn't solve the original problem.


To answer the actual question: You are entitled to assume that the rate of loss of height is constant, but different for each candle. You have already calculated the rates in terms of the initial heights: the rate of C1 is [itex]H_1/3.5[/itex], and the rate of C2 is [itex]H_2/5[/itex].

Now you need to work out the height of each candle after two hours of burning at those constant rates, and use the condition that after two hours the heights are the same.
 
  • #5
223
10
Would still like to know what we are to determine in this assignment. Is it about determining the relation of the heights before lighting the candles?
 
Last edited:
  • #6
1,421
5
pasmith, all right. I made a notation error; however, wouldn't my method still work? The volume function should then be [itex]V_1(t) = \pi r^2_1 h(t)[/itex]. Would that then work?
 
  • #7
Ray Vickson
Science Advisor
Homework Helper
Dearly Missed
10,706
1,728
pasmith, all right. I made a notation error; however, wouldn't my method still work? The volume function should then be [itex]V_1(t) = \pi r^2_1 h(t)[/itex]. Would that then work?
Why are you trying to make a very simple problem complicated? Just do what others have already suggested to you.
 

Related Threads on Candle's Height

  • Last Post
Replies
3
Views
793
  • Last Post
Replies
2
Views
1K
  • Last Post
Replies
12
Views
1K
  • Last Post
Replies
3
Views
731
  • Last Post
Replies
8
Views
2K
  • Last Post
Replies
3
Views
1K
  • Last Post
Replies
1
Views
923
  • Last Post
Replies
1
Views
2K
  • Last Post
Replies
8
Views
2K
  • Last Post
Replies
7
Views
3K
Top