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N

NYC

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I really need some help with my calc 3 hw, it's review but I don't understand how to complete the problem. Well, here it is:

There are 2 buildings A&B, each 75 feet tall are 50 ft apart. A wire is strung between the buildings 40 feet above the ground. A light is hung at the top of building A illuminating the space between the buildings. A tight rope walker starts walking from building A to building B at 3ft/ min. As he walks, his shadow, a dot, moves along the ground then up the wall of building B.
1. what rate does the shadow move across the ground
2 At what rate does it move up the building
3. when will his shadow be at the spot where building B meets the ground

I drew a diagram, and got the rate of the shadow's movement across the ground to be 24/7 ft per min. Otherwise, as far as applying derivatives, I am rather confused. Any help would be greatly appreciated.
 
In my diagram, A is at P(0,0) with a line to P(0,75) where the Light is.
B is at P(40,0)
The Walker is at P(W,40)
His Shadow is at P(S,0)
The triangle with the hypotenuse LW is similar to the triangle with the hypotenuse LS so we have
35/W=75/S
S=75W/35 = 15W/7
dS/dt = (15/7)dW/dt
dW/dt= 3 ft/min: given

1) dS/dt = (15/7)*3 = 45/7 ft/min till S=50

New triangles:
S is now the height of the shadow above the ground.
W is still the distance from A
The triangle with the hypotenuse LW is similar to the triangle with the hypotenuse WB so we have
35/W=(50-W)/(40-S)
For 2) do the same calculation as for 1)

3) Distance/velocity = time.

Comment: It takes him 50/3 = 16 2/3 minutes to get across. You had the shadow taking 50/(24/7) = 14.5+ minutes to cross the courtyard. Suspicious :?:
The time for the shadow to cross 50 ft + the time for it to climb 40 ft must = 16 2/3 so you can check your answer.
 
Thanks

I really do appreciate the help, but let me tell you what I did to get 24/7

in my diagram, I established that the distance covered by the walker is z, and the rate of change for z, dz/dt=3
I labeled y as the distance traveled by his shadow, so the thing in question was dy/dt
z is also in the courtyard, and is still the distance covered by the walker.
To give you a better idea of what I did, when the shadow reaches the end of the courtyard, I said (z+y)=50
I used a ratio, that being: 35/z=40/y or (75)/(z+y)=35/z
Using that, I took the derivative, and got 7*dy/dt=8*dz/dt
I solved for dy/dt and got 24/7

If I went wrong somewhere, please let me know
 
I see, but your dy/dt isn't relative to the ground. It is relative to z, which is also moving. To change it to be relative to the ground (the usual reference) you have to add dz/dt.
24/7 + dz/dt = 45/7
Does that make sense to you?
 
Aye

Ok, I understand that part, but in regard to finding the rate of the shadow moving up the wall, we were told to use the shadow's position 2 min before the walker finished if it became necessary. I tried it the way you suggested, and got 4.5 feet/min by doing it the way one would "check" their answer.
My problem now is getting 4.5 as the rate at which the shadow travels up the wall. Sorry to bug you, but I'm still confused.
 
Oh....wait....

Wait, is the speed of the shadow up the wall variable? At 2 minutes before the walker is done, I calculated a speed of 2.711 ft/min.
The walker has covered 44 feet, and the shadow has covered 775/22 feet on the wall.
 
Shame on me!!!

Mia culpa! :oops: :oops: :oops:
I goofed on which legs of the new triangle were similar. :oops:

35/W=(50-W)/(40-S)
should be
35/W=(40-S)/(50-W).
35(50-W)=W(40-S)
35(50-W)/W=40-S
S=40-35(50-W)/W = 40-(1750-35W)/W =
(40W-1750+35W)/W =
(75W-1750)/W =
75-1750/W

dS/dt = 1750/W²*dW/dt = 5250/W²

You are right, definitly non-linear. I should have looked closer. :oops:
 
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