calculus problem... help??

Question

A spotlight is on the ground 100ft from a building that has vertical sides. A person 6 ft tall starts at the spotlight and walks directly toward the building at a rate of 5ft/sec.

a) how fast is the top of the persons shadow moving down the building when the person is 50 feet away from the building?

b) how fast is the top of the shadow moving when the person is 25 feet away?

thank you

Answer 1

What a fun problem!

First, the slope of the shadow is 6 divided by the distance between the man the the spotlight.

Then, the height on the building is the slope times 100 feet, or, to substitute from above, 6/(distance between man and spotlight)*100, or 600/distance of man from spoptlight

Distance of the man from the spotlight is 5*time

so the height of the shadow is now:
120/time

the derivative with time of this is -1*120/(time-squared).
The -1 means the shadow is moving "down" which makes sense.

The man is 50 feet from the building 10 seconds after he starts walking, so at this time the shadow is moving down at 120/100, or 1.2 feet per second.

When the person is 25 feet from the wall, he has been walking for 15 seconds, so the shadow is moving down at 120/225 feet per second

Answer 2

Draw the picture and see the 2 similar triangles. Call the distance from the person to the building d and call the shadow's length s. The bottom of the little triangle is 100 - d.
By similar triangles, (100-d)/6 = 100/s

600 = s(100-d)
600 = 100s - sd
600 = s(100-d)

s = 600/(100-d)

You can find ds/dd. You are given dd/dt. You want ds/dt.

ds/dd times dd/dt = ds/dt

Answer 3

Since the point of the spotlite to the shadow on the wall makes a triangle, Let:
Z= angle of triangle at spotlite
x= distance from spotlite to person
y= distance from shadow to floor

For the first problem:
x=50
dx/dt=5

equation for angle:
tanZ=6/x

solve for angle:
tanZ=6/50
Z=6.843 degrees (approximately)

Find rate of change of angle with derivative of the first equation:
(secZ)^2*(dZ/dt)= (-6/x^2)*(dx/dt)
(sec(6.843))^2*(dZ/dt)= (-6/50^2)*(5)
dZ/dt= -.01183 degrees per second (appr)

Now you use this data for the rate of change of shadow:
tanZ=y/100

solve for derivative:
(secZ)^2*(dZ/dt)=1/100*(dy/dt)
dy/dt= -1.19 ft per sec (approximately) or 1.19 ft downwards per sec


that's only the first problem.
The second one is the same thing just a slight change in numbers. Do it by yourself for practice! Good luck

Answer 4

Wow, your fresh out of luck with me but you can try different math websites like http://www.math.com/homeworkhelp/Calculus.html or just search calculus on google (or Yahoo).

Best of luck,
Steamhead10