document.write( "Question 1194694: A light is placed on the ground 30 ft. from a building. A man 6 ft. tall walks
\n" ); document.write( "from the light toward the building at the rate of 5 ft.per sec. Find the rate
\n" ); document.write( "at which the length of his shadow on the wall is changing when he is 15
\n" ); document.write( "ft. from the building. \n" ); document.write( "

Algebra.Com's Answer #826990 by KMST(5328) $\"\"$ $\"About$

You can put this solution on YOUR website!
Here is the man, 6 ft tall, walking towards the light
\n" ); document.write( "on a horizontal ground surface, shown when he is at a distance x (in ft) from the light, projecting a shadow with a length s (in ft) on the wall:
\n" ); document.write( "

\n" ); document.write( "If we count time from the moment $\"t=0\"$ when the man starts walking fron the light,
\n" ); document.write( " $\"x=5t\"$ with $\"t\"$ in seconds.
\n" ); document.write( "There are two similar right triangles, so
\n" ); document.write( " $\"s%2F30=6%2Fx\"$ --> $\"s=180%2Fx\"$
\n" ); document.write( "and if we want $\"s\"$ as a function of $\"t\"$
\n" ); document.write( " $\"x=180%2F5t\"$ --> $\"x=36%2Ft\"$
\n" ); document.write( "The rate of change of a function with its variable is the derivative of the function, so we could write
\n" ); document.write( " $\"ds%2Fdt=-36%2Ft%5E2\"$ and then
\n" ); document.write( "calculate the value of $\"ds%2Fdt\"$ when $\"5t=15\"$ <--> $\"t=3\"$ as
\n" ); document.write( " $\"-36%2F3%5E2=-36%2F9=highlight%28-4%29\"$ $\"ft%2Fs\"$
\n" ); document.write( "which means the shadow's length is decreasing at $\"highlight%284%29\"$ $\"ft%2Fsecond\"$ at the instant the man is at 15 ft from the light.
\n" ); document.write( "
\n" ); document.write( "Alternately, we could write $\"ds%2Fdt=%28ds%2Fdx%29%28dx%2Fdt%29\"$ and knowing that $\"dx%2Fdt=5\"$ as long as the man is walking between light and building
\n" ); document.write( "regardless of how we count the time we could calculate $\"ds%2Fdt\"$ ,
\n" ); document.write( "without caring how long the man has to walk to get 15 ft from the light, we calculate
\n" ); document.write( " $\"ds%2Fds=-180%2Fx%5E2\"$ and for $\"x=15\"$ $\"ds%2Fdx=-180%2F15%5E2\"$ and $\"ds%2Fdt=%28-180%2F15%5E2%29%2A5=-4\"$ to get the same result. \n" ); document.write( "

\n" );