Posted by kait on Sunday, September 25, 2011 at 12:13am.
did you make a diagram?
let her distance from the lightpole be x ft
let the length of her shadow be y ft
by similar triangles ...
6/y = 10/(x+y)
6x + 6y = 10y
6x = 4y
6 dx/dt = 4 dy/dt
we are given: dx/dt + dy/dt = 2 ft/sec
dx/dt = 2 - dy/dt
so
6(2 - dy/dt) = 4 dy/dt
12 - 6dy/dt = 4dy/dt
12 = 10dy/dt
dy/dt = 12/10 = 1.2 ft/sec
Related Questions
Calculus- rates - A man 6ft. tall walks at the rate of 5 ft/sec toward a ...
Math - a man 6ft tall walks at the rate of 5ft/sec toward a streetlight that is ...
calculus - My problem is that a stft above the sidewalk. a man 6ft tall walks ...
Calculus - Shadow Length A man 6 feet tall walks at a rate of 3 ft per second ...
Calculus - A woman 5 ft tall walks at the rate of 5.5 ft/sec away from a ...
calculus - A 6ft tall woman is walking away from a srteet lamp at 4 ft/sec that ...
calculus - A woman 5.5 ft tall walks at a rate of 6 ft/sec toward a streetlight ...
Calculus - A man 2m tall walks away from a lamppost whose light is 5m above the ...
Calculus - A street light is hung 18 ft. above street level. A 6-foot tall man ...
Calculus - A street light is at the top of a 19 foot tall pole. A 6 foot tall ...
For Further Reading
