# Velocity and Rate of Change of Shadow

A candle burns at a speed of −4/3 inches per minute. A 16-inch stick is placed upright between the candle and a wall such that the base of the stick is 2 inches away from the wall and 3 inches away from the base of the candle. We also note that the stick casts a shadow on the wall due to the light of the candle. a. At the time when the candle is 20.8 inches tall, what is the instantaneous velocity of the tip of the shadow? b. At the time when the candle is 20.8 inches tall, what is the instantaneous rate of change of the distance between the tip of the candle and the tip of the shadow?

This is my drawing
Is this correct:

1. $\displaystyle \frac{d c}{ d t} = \frac{-4}{3}$ in/min when $c=20.8$ inches ($c$ is candle height )
2. Using similar triangles:
• $\displaystyle \frac{20.8-s}{5} = \frac{(20.8-s)-(16-s)}{3}$
• $\displaystyle \frac{20.8-s}{5}= \frac{4.8}{3}$
• $\displaystyle 3(20.8-s)=5(4.8)$
• $\displaystyle 62.4-3s=24$
• $\displaystyle 38.4=3s$
• $\displaystyle s= 12.8$
3. $\displaystyle ab^2 = 5^2 + (20.8-s)^2$

• $\displaystyle ab^2 = 5^2 + 8^2$
• $\displaystyle ab^2 = 25 + 64$
• $\displaystyle ab = \sqrt{89}$

I'm stuck, what should I do next? I hope you could explain it simply.

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Good for you for drawing a picture. Your evaluation that $s=12.8$ is correct for the time $c=20.8$. What you need to do is write an equation $s=f(c)$. You incorporated the data $c=20.8$ too early. Just follow the same calculation you did, but leave $c$ as a variable. Then you can take $\frac {ds}{dc}$ and evaluate that at $c=20.8$ For b, it is the same. You need to evaluate $(a+b)=g(c)$, then take $\frac {d(a+b)}{dc}$ and evaluate it at $c=20.8$
No. You will use the chain rule $\frac {ds}{dt}=\frac{ds}{dc}\frac {dc}{dt}$ You are given that $\frac {dc}{dt}=-\frac 43$ –  Ross Millikan Jan 8 '14 at 16:17
You dropped a sign. It should be $s=\frac 13(80-2c)$ Then to get $\frac {ds}{dt}$ you use the chain rule. Note that $\frac {ds}{dt}$ should be positive, because as the candle burns down the top of the shadow goes up. –  Ross Millikan Jan 8 '14 at 17:07
I get $\frac {ds}{dc}=-\frac 23, \frac {ds}{dt}=\left(-\frac 23 \right)\left(-\frac 43 \right)=\frac 89$ –  Ross Millikan Jan 8 '14 at 17:41