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Gabrielle's question at Yahoo! Answers regarding related rates

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MarkFL

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Feb 24, 2012
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Here is the question:

This is a calculus I question in the section about related rates.?

This is a calculus I question in the section about related rates.
A street light is at the top of a 13.000 ft. tall pole. A man 6.100 ft tall walks away from the pole with a speed of 6.000 feet/sec along a straight path. How fast is the tip of his shadow moving when he is 33.000 feet from the pole?
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This is a calculus I question in the section about related rates.? - Yahoo! Answers

I have posted a link there to this topic so the OP can find my response.
 
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MarkFL

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Feb 24, 2012
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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:

\(\displaystyle \frac{L}{x+s}=\frac{M}{s}\)

Cross-multiply:

\(\displaystyle Ls=M(x+s)\)

Solve for $s$:

\(\displaystyle (L-M)s=Mx\)

\(\displaystyle s=\frac{M}{L-M}x\)

Now, differentiate with respect to time $t$:

\(\displaystyle \frac{ds}{dt}=\frac{M}{L-M}\cdot\frac{dx}{dt}\)

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 \(\displaystyle \frac{dx}{dt}\) as this growth is relative to the man's position:

\(\displaystyle \frac{dT}{dt}=\frac{M}{L-M}\cdot \frac{dx}{dt}+\frac{dx}{dt}= \frac{dx}{dt}\left(\frac{M}{L-M}+1 \right)= \frac{dx}{dt}\cdot\frac{L}{L-M}\)

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:

\(\displaystyle \frac{dT}{dt}=\left(6\,\frac{\text{ft}}{\text{s}} \right)\left(\frac{13\text{ ft}}{(13-6.1)\text{ ft}} \right)=\frac{260}{23}\,\frac{\text{ft}}{\text{s}}\)

Since the given data is accurate to 3 decimal places, we round the final result to:

\(\displaystyle \frac{dT}{dt}\approx11.304\,\frac{\text{ft}}{\text{s}}\)

To Gabrielle and any other guests viewing this topic, I invite and encourage you to post other related rates problems in our Calculus forum.

Best Regards,

Mark.