# Two candles making shadows of one another on opposite walls

• brotherbobby
brotherbobby
Homework Statement
Two candles of equal height ##h## at the initial moment are at a distance ##a## from each other. The distance between each candle and the nearest wall is also ##a## (see figure below). With what speed will the shadows of the candles move along the walls if one candles burns down during a time ##t_1## and the other during a time ##t_2##?
Relevant Equations
1. (Uniform) Velocity of a point along a line : ##v = \dfrac{dx}{dt}##

2. For two similar right angled triangles of heights ##h_1## and ##h_2## and bases ##b_1## and ##b_2## : ##\quad\dfrac{h_2}{b_2}=\dfrac{h_2}{b_2}##

I copy and paste the problem as it appears in the text along with the diagram alongside. It is clear from the image that the base of the shadows don't move - only their tops would move either up or down. Let us assume the time ##t_1>t_2##, implying that candle 1 burns at a slower rate. How would things appear after a time of burning ##t## for both candles?

I draw my own diagram, showing the situation after a time ##t##. The rate at which the candles burn ##\small{r_i=\dfrac{h}{t_i}}##. The height of candle 1 after a time ##t## : ##\small{h_1 = h-r_1t=h-\dfrac{h}{t_1}t\Rightarrow h_1=h\left(1-\dfrac{t}{t_1}\right)\quad (1)}.## Likewise, ##\small{h_2= h\left(1-\dfrac{t}{t_2}\right)\quad (2)}.##

If the shadows have lengths ##s_1, s_2##, I have to find their velocities as the candles burn, i.e. ##\boxed{\dot{s_1}=?\;,\;\dot{s_2}=?}##.

In ##\triangle's## CDO and EFO, we have (by similarity) : ##\small{\dfrac{h_1-s_2}{2a}=\dfrac{h_2-s_2}{a}\Rightarrow h_1-s_2 = 2(h_2-s_2)\Rightarrow s_2=2h_2-h_1}=2h\left(1-\dfrac{t}{t_2}\right)-h\left(1-\dfrac{t}{t_1}\right)## upon using equations ##(1),(2)## above.
This simplifies to ##\small{s_2=h+h\left(\dfrac{1}{t_1}-\dfrac{2}{t_2}\right)t\Rightarrow \dot{s}_2=h\left(\dfrac{1}{t_1}-\dfrac{2}{t_2}\right)}\Rightarrow \boxed{\boldsymbol{\dot{s}_2=\dfrac{h}{t_1t_2}(t_2-2t_1)}}\quad (3)##
Since ##t_1>t_2,\; \dot{s}_2=\text{-ve}##.

From similar ##\triangle's## ABO and EFO, ##\small{\dfrac{s_1-s_2}{3a}=\dfrac{h_2-s_2}{a}\Rightarrow s_1-s_2=3(h_2-s_2)}##
##\small{\Rightarrow s_1=3h_2-2s_2=3\left(1-\dfrac{t}{t_2}\right)-2h-2h\left(\dfrac{1}{t_2}-\dfrac{2}{t_2}\right)t}## after using equations ##(2),(3)##.

This reduces to ##\small{s_1=h-\dfrac{2h}{t_1}t+\dfrac{h}{t_2}t\Rightarrow \dot{s}_1=-\dfrac{2h}{t_1}+\dfrac{h}{t_2}}\Rightarrow \boxed{\boldsymbol{\dot{s}_1=\dfrac{h}{t_1t_2}(t_1-2t_2)}}\quad (4)##

My answers miss those of the text by a change of sign.

Request : Where do you think am mistaken?

Last edited:
You have defined the velocity in the opposite direction to the text. This will make it differ by a sign.

The problem also asks for speed, not velocity, so technically the correct answer is the absolute value of these expressions (which make them the same). The difference being that your expressions will always have (at least) one negative value.

Orodruin said:
You have defined the velocity in the opposite direction to the text. This will make it differ by a sign.
I think I should look into first myself and then post the text solution. I measured distances from the bottom up, taking the base as point 0.

Yes you were right. They are taking distances from the top down.

Orodruin said:
The difference being that your expressions will always have (at least) one negative value.

Yes that is what am not sure of. Let us have the expressions again, given my conventions.

I assumed ##t_1>t_2##. However, if ##t_1<2t_2\Rightarrow \dot{s}_1=-ve##. Either way, ##\dot{s}_2=-ve##.

So it could be that both shadows are moving down?

brotherbobby said:
So it could be that both shadows are moving down?
Of course, consider the case when both candles burn at the same speed. Then both shadows will move down at that speed.

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