# Thread:

1. ## Related Rates Problems

1. A spotlight on the ground shines on the wall 12 m away. If a man 2 m tall walks from the spotlight toward the building at a speed of 1.6 m/s, how fast is the length of his shadow on the building decreasing when he is 4 m from the building?

2. A boat is pulled into a dock by a rope attached to the bow of the boat and passing through a pulley on the dock that is m higher than the bow of the boat. If the rope is pulled in at a rate of 1 m/s, how fast is the boat approaching the dock when it is 8 m from the dock.

So far related rates problems have been pretty easy since we have been using geometric figures. I can't come up with an equation for either of these so I can solve for the change of rate. Can someone help me come up with the equations and show me how to do it in case I get something similar to these in the future?

2. Originally Posted by FalconPUNCH!
1. A spotlight on the ground shines on the wall 12 m away. If a man 2 m tall walks from the spotlight toward the building at a speed of 1.6 m/s, how fast is the length of his shadow on the building decreasing when he is 4 m from the building?

So far related rates problems have been pretty easy since we have been using geometric figures. I can't come up with an equation for either of these so I can solve for the change of rate. Can someone help me come up with the equations and show me how to do it in case I get something similar to these in the future?
1) using similar triangles, you have
$\frac{x}{2}=\frac{12}{y}$
which implies that
$xy=24$
and
$\frac{dy}{dt} x + \frac{dx}{dt} y =0$
use the given and evaluate at x=12-4=8 (note that if x=8, then y=3)

for 2) what do you min by "... m higher than..."?
Attached Thumbnails

3. Hello, FalconPUNCH!

You left out a measurement in #2.
. . I'll pick a convenient value . . .

2. A boat is pulled into a dock by a rope attached to the bow of the boat and
passing through a pulley on the dock that is 6 m higher than the bow of the boat.
If the rope is pulled in at a rate of 1 m/s,
how fast is the boat approaching the dock when it is 8 m from the dock?
Code:
```                              * P
                          *   |
                 R    *       |
                  *           | 6
              *               |
          *                   |
    B *   *   *   *   *   *   *
                  x```

The boat is at $B$, the pulley is at $P.$

The length of the rope is: . $R \,=\, BP$ . and . $\frac{dR}{dt} \,=\, -1\text{ m/s}$

From Pythagorus, we have: . $x^2 + 6^2 \:=\:R^2$

Differerentiate with respect to time: . $2x\left(\frac{dx}{dt}\right) \:=\:2R\left(\frac{dR}{dt}\right)$
. . and we have: . $\frac{dx}{dt} \:=\:\frac{R}{x}\left(\frac{dR}{dt}\right)$

Can you finish it now?

4. Oh sorry I left out 1 m. Thanks for your help. I understand now.
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