Hello Gabrielle,
I like to work problems like this in general terms, derive a formula, then plug in the given data at the end. First let's draw a diagram. $L$ represents the height of the light pole, $M$ represents the height of the man, $x$ represents the distance of the man from the pole, and $s$ represents the length of the man's shadow:
By similarity, we may state:
$$\frac{L}{x+s}=\frac{M}{s}$$
Cross-multiply:
$$Ls=M(x+s)$$
Solve for $s$:
$$(L-M)s=Mx$$
$$s=\frac{M}{L-M}x$$
Now, differentiate with respect to time $t$:
$$\frac{ds}{dt}=\frac{M}{L-M}\cdot\frac{dx}{dt}$$
This is the rate of growth of the length of the shadow. To find how fast the tip $T$ of the shadow is changing, we need to add $$\frac{dx}{dt}$$ as this growth is relative to the man's position:
$$\frac{dT}{dt}=\frac{M}{L-M}\cdot \frac{dx}{dt}+\frac{dx}{dt}= \frac{dx}{dt}\left(\frac{M}{L-M}+1 \right)= \frac{dx}{dt}\cdot\frac{L}{L-M}$$
We see that this is constant, i.e., it does not depend on how far the man is from the pole. So, plugging in the given data, we find:
$$\frac{dT}{dt}=\left(6\,\frac{\text{ft}}{\text{s}} \right)\left(\frac{13\text{ ft}}{(13-6.1)\text{ ft}} \right)=\frac{260}{23}\,\frac{\text{ft}}{\text{s}}$$
Since the given data is accurate to 3 decimal places, we round the final result to:
$$\frac{dT}{dt}\approx11.304\,\frac{\text{ft}}{\text{s}}$$
To Gabrielle and any other guests viewing this topic, I invite and encourage you to post other related rates problems in our http://www.mathhelpboards.com/f10/ forum.
Best Regards,
Mark.
