A romantic evening of physics

Set up a coordinate system. Candle 1's height is $\left(20, 15 - 2vt\right)$ and candle 2's height is $\left(50, 10 - vt\right)$ . The intersection of the line connecting those two points with the walls are where the shadows will be.

We eventually find that this line has equation $y = \dfrac{-5 + vt}{30} x + \frac{55 - 8vt}{3}$

We plug in $x = 0$ and see that candle 1's shadow moves at speed $\dfrac{8}{3}v$ . We then plug in $x = 60$ and work out the algebra and see that candle 2's shadow moves at speed $\dfrac{2}{3}v$ . Next, we divide the two and end up with $\boxed{4}$ .

1) Solve what would be the height of Shadow 1 (shadow of candle 1 on the left wall) and Shadow 2 (shadow of candle 2 on the right wall) at the given conditions.

Let

  x = height of shadow of candle 1

  y = height of shadow of candle 2

  A = angle included between the wall on the left and the line from the tip of candle 1 to the tip of its shadow on the left.

  G = distance from the bottom of the wall on the right to the tip of the constructed triangle. (Construct a triangle from the path of the ground and the shadow on the right of the wall)

Then

A = arctan (30/5) -- from the constructed triangle b/n the candles due to height difference

A = 80.53767779

tan (90-A) = 10 / (10 + G)

Solving for G, G = 50

tan (90-A) = x / 110 -- >110 = 20 + 30 + 10 +70

x = 55/3

tan (90-A) = y / 50

y = 25/3

2) Assume a change in height of 2 cm on candle 2 in 1 hr. Then candle 1 would've shrink 4 cm since the ratio of shrink rate b/n candle 1 and 2 is 2. Then the new height of candle 1 = 15 - 4 = 11 cm and candle 2 = 10 - 2 = 8 cm.

Now, a new triangle.

Let

   x' = new height of the shadow of candle 1

   y' = new height of the shadow of candle 2

   A' = new angle included

   G' = new distance from bottom of wall on the right to the vertex of the constructed triangle.

Then

A' = arctan (30/3)

tan (90-A') = 8 / (10 +G')

G' = 70

tan (90-A') = x' / (60 + 70)

x' = 13

tan (90-A') = y' / 70

y' = 7

Delta x = (55/3) - 13

Delta y = (25/3) - 7

ratio = Delta x / Delta y

ratio = 4 --> Ans.

Let the length's of Candle 1, Candle 2 and their respective shadows be $\displaystyle$ $x,y,h_1$ & $h_2$ respectively. Then from similarity we can say

$\displaystyle$ $\frac{h_2-y}{h_2-x}\,=\frac{1}{4}$ ----- $(i)$

$\displaystyle$ $\frac{h_2-x}{h_2-h_1}\,=\frac{2}{3}$ ----- $(ii)$

Differentiating both sides of $(i)$ and $(ii)$ w.r.t ' $t$ ' and setting $\frac{\mathrm d}{\mathrm d t} \left( x \right)\,=2\cdot\frac{\mathrm d}{\mathrm d t} \left( y \right)$ , we get

$\frac{\mathrm d}{\mathrm d t} \left( h_1 \right)\,=4\cdot\frac{\mathrm d}{\mathrm d t} \left( h_2 \right)$ .

So our desired answer is $\boxed{4}$ .

Idk..

Let the heights of shadows of Candle 1 and Candle 2 be $s_{1}$ and $s_{2}$ respectively.

Also, we construct a straight line joining the top of the 2 shadows and the 2 candles. Horizontal lines from the top of Candles 1 and 2 and shadow 2 to each vertical to the left respectively are also constructed.

From the problem statement, we know that $\frac{dh_{1}}{dt} = 2 \cdot \frac{dh_{2}}{dt}$ . This means that $\frac{dh_{1}}{dh_{2}} = 2$ .

Now we consider the triangles constructed. The angle of the slanted line with the verticals is constant. So we have similar triangles.

From there, we obtain $\frac {h_{1} - h_{2}} {d_{2}} = \frac {s_{1} - h_{1}} {d_{1}}$ . Substituting values gives $s_{1} = \frac{5h_{1} - 2h_{2}} {3}$ .

We can also obtain $\frac {h_{1} - h_{2}} {d_{2}} = \frac {h_{2} - s_{2}} {d_{3}}$ . Another substitution yields $s_{2} = \frac{4h_{2} - h_{1}} {3}$ .

We perform implicit differentiation on both equations.

1. Differentiating $s_{1} = \frac{5h_{1} - 2h_{2}} {3}$ gives $3 * \frac{ds_{1}}{dh_{2}} = 5 \frac{dh_{1}}{dh_{2}} - 2$ .

2. Differentiating $s_{2} = \frac{4h_{2} - h_{1}} {3}$ yields $3 * \frac{ds_{2}}{dh_{2}} = 4 - \frac{dh_{1}}{dh_{2}}$

Dividing equation 1 with equation 2 gives $\frac{ds_{1}}{ds_{2}} = \frac {5 \frac{dh_{1}}{dh_{2}} - 2}{ 4 - \frac{dh_{1}}{dh_{2}}}$ .

Substituting $\frac{dh_{1}}{dh_{2}} = 2$ yields $\frac{ds_{1}}{ds_{2}} = \boxed {4}$ .

Assuming that the shadow of one candle is only created due to the other candle

Set up the x-y coordinate system where the top of Candle 1 is at (20, 15) and Candle 2 at (30,10). Easily seen that the top of the shadows is the intersection between the line passing through these two points and the line x =0 and x = 60.

This line has the equation:

$y~=~f(x)~=~mx~+~c$

The ratio asked in the question is simply f(0)'/f(60)'.

$f{{\left( 0 \right)}^{'}}=~\frac{dm}{dt}*0~+\frac{dc}{dt}~$

$f(60)'~=\frac{dm}{dt}*60~+\frac{dc}{dt}$

So the problem now is to calculate dm/dt and dc/dt

We know that

$m~=\frac{f\left( 20 \right)-~f\left( 50 \right)}{20-50}$

Let 2a be he rate of decreasing of candle 1, thus a is the rate of decreasing of candle 2.

Hence,

$\frac{dm}{dt}=\frac{f{{\left( 20 \right)}^{'}}-~f{{\left( 50 \right)}^{'}}}{20-50}~=~-\frac{a}{30}$

We also have: $f{{\left( 20 \right)}^{'}}=~\frac{dm}{dt}*20~+\frac{dc}{dt}~=~2a$

$\to \frac{dc}{dt}=\frac{8a}{3}$

Using those value plug in

$\frac{f{{\left( 0 \right)}^{'}}}{f{{\left( 60 \right)}^{'}}}~=\frac{\frac{8a}{3}}{-2a+\frac{8a}{3}}~=~4$

The line joining the peaks of the candles will intersect the wall,say L1 for left wall & R1 for right wall. So if the speed of shadow of C1 is V1 then V1=

$\frac{(Height of Shadow of C1 at t = t1 - height at t = 0)}{t1 - 0}$

Similarly, the speed of shadow of C2 is V2 then V2=

$\frac{(Height of Shadow of C2 at t = t1 - height at t = 0)}{t1 - 0}$

We need to find $\frac{V1}{V2}$ . Initially the length of the shadow on the left wall is (10+25/3)cm, and on the right wall is (10-5/3)cm. Now the candle C1 burns at twice the rate of C2, so for after t unit time , height of shadow of C1 on left wall=15-2x height of shadow of C2 on right wall=10-x

When height of shadow on the left wall is 0, we will get similar triangles, where $\frac{15-2x}{20}$ = $\frac{10-x}{50}$ x= $\frac{55}{8}$ therefore, 15-2x=15-2 $\frac{55}{8}$ = $\frac{5}{4}$ Now, taking the height of shadow on the right wall as y, and similar triangles ratio $\frac{5/4}{20}$ = $\frac{y}{60}$

solving, y= $\frac{15}{4}$

therefore, $\frac{V1}{V2}$ = $\frac{(10+25/3)-0}{(10-5/3)-15/4}$

The time component is same for both the numerator & denominator and will be cancelled.

Solving we get $\frac{V1}{V2}$ = $\boxed{4}$