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

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