- #1

brotherbobby

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- 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?

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