A man 6 feet tall walks at a rate of 5 feet per second away from a light that is

protisvitfc 2021-11-23 Answered
A man 6 feet tall walks at a rate of 5 feet per second away from a light that is 15 feet above the ground. (a) When he is 10 feet from the base of the light, at what rate is the tip of his shadow moving? (b) When he is 10 feet from the base of the light, at what rate is the length of his shadow changing?

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pseudoenergy34
Answered 2021-11-24 Author has 9410 answers
Look at the ilustration in the book.
The lamppost is 15 ft. tall, and the man is 6 ft. tall. We use a proportion to find the shadow. s is the length of the base of the lamppost to the shadow, x is the length of the base of the lamppost to the man, so the length of the shadow is s - x. The lamppost is 15 ft. tall, and the man is 6 ft. tall. We use a proportion to find the shadow. s is the length of the base of the lamppost to the shadow, x is the length of the base of the lamppost to the man, so the length of the shadow is s - x.
\(\displaystyle{\frac{{{15}}}{{{6}}}}={\frac{{{s}}}{{{s}-{x}}}}\)
Cross multiply and distribute
\(\displaystyle{15}{s}-{15}{x}={6}{s}\)
Isolate the variable s, and simplify the fraction.
\(\displaystyle{s}={\frac{{{5}}}{{{3}}}}{x}\)
Take the derivative according to time.
\(\displaystyle{\frac{{{d}{s}}}{{{\left.{d}{t}\right.}}}}={\frac{{{5}{\left.{d}{x}\right.}}}{{{3}{\left.{d}{t}\right.}}}}\)
Given that \(\displaystyle{\frac{{{\left.{d}{x}\right.}}}{{{\left.{d}{t}\right.}}}}\) is 5 ft/sec, multply. This answer for a.
\(\displaystyle{\frac{{{d}{s}}}{{{\left.{d}{t}\right.}}}}={\frac{{{5}}}{{{3}}}}\times{5}={\frac{{{25}}}{{{3}}}}\) ft/sec
\(\displaystyle{\frac{{{d}{s}}}{{{\left.{d}{t}\right.}}}}\) we found in part a, \(\displaystyle{\frac{{{\left.{d}{x}\right.}}}{{{\left.{d}{t}\right.}}}}\) is given in the problem.
\(\displaystyle{\frac{{{d}{s}}}{{{\left.{d}{t}\right.}}}}={\frac{{{25}}}{{{3}}}},{\frac{{{\left.{d}{x}\right.}}}{{{\left.{d}{t}\right.}}}}={5}\)
Subtract. This is the answer to part b.
\(\displaystyle{\frac{{{25}}}{{{3}}}}-{5}={\frac{{{10}}}{{{3}}}}\) ft/sec
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