A street light is mounted at the top of a 15-ft tall pole. A man 6 ft tall walks away from the pole with a speed of 5 ft/s along a straight path. How fast is the tip of his shadow moving when he is 40 ft from the pole?

---- i drew the picture and got the equation 15/6=(x+y)/y

(y=the shadow)

Question

A street light is mounted at the top of a 15-ft tall pole. A man 6 ft tall walks away from the pole with a speed of 5 ft/s along a straight path. How fast is the tip of his shadow moving when he is 40 ft from the pole?

---- i drew the picture and got the equation 15/6=(x+y)/y

(y=the shadow)

anonymous · Answer

tough question

working:

40/ 5 = 8seconds to walk 40 ft

9 ft difference between man and light

angle from ground to top of pole at 40 ft mark = arctan (15/40) = 35.8 degrees

angle from man height to top of light pole at 40 ft mark = arctan(9/40) = 22.1 degrees

anonymous · Answer

is there an equation for this or do you have to create one as you have?

amistre64 · Answer

light is mounted at the top of a 15-ft tall pole. 

  man 6 ft tall walks away from the pole with a 
                              speed of 5 ft/s along a straight path. 

How fast is the tip of his shadow moving when he is 40 ft from the pole?

this is a similar triangles problem:

|dw:1319286771599:dw|

amistre64 · Answer

almost sensible ... that needs to be refined

amistre64 · Answer

in my drawing the he has (w)alked a distance of 40 so...

         15         is to         6
                       as
 40+(s)hadow  is to   (s)hadow

amistre64 · Answer

15s = 6(40+s)

15s = 240+6s

9s = 240

s = 240/9
........................

that is the value of s at w=40 (if we need it), but that is just a position and not a speed. speed is calculated from:

 15        6
---- = --- ; and get the derivative implicity
w+s       s

[15s = 6w+6s]' :: 15 s' = 6 w' + 6 s'
                   15 s'- 6 s' = 6 w' 
                     s'(15- 6) = 6 w' 
                                s' = 6 w'/9 

and it says that walking speed: w' = 5

s' = 6(5)/9 
   = 2(5)/3 
   = 10/3  ft/sec

maybe

anonymous · Answer

Not quite, amistre.

amistre64 · Answer

im open to being wrong :)

anonymous · Answer

You're very, very close though. The end is a tad off...

|dw:1319293480551:dw|

My drawing's pretty ugly, but we'll work with it. The little point on the right of the triangle is the tip of his shadow. Let x be the man's distance from the lamppost and s be the length of his shadow, as shown. Note that he's moving to the right.

In order to calculate the rate of the tip of the shadow, we need to find 

d(x+s)/dt at x=40. 

We are given that he's walking at 5 ft/sec. This is dx/dt (the rate at which his distance from the lamp is changing). So dx/dt=5. 

Setting up similar triangles yields

6/15=s/(x+s)

6(x+s)=15s
6x+6s=15s
6x=9s
s=(2/3)x
and of course ds/dt=(2/3)dx/dt

We want d(x+s)/dt, which of course is just a sum of two derivatives...

d(x+s)/dt=dx/dt+ds/dt=5+(2/3)dx/dt

=5+(2/3)(5)=5+10/3=8 1/3 ft/sec