1) A street light is at the top of a 19 ft tall pole. A woman 6 ft tall walks aw
ID: 2847290 • Letter: 1
Question
1) A street light is at the top of a 19 ft tall pole. A woman 6 ft tall walks away from the pole with a speed of 5 ft/sec along a straight path. How fast is the tip of her shadow moving when she is 35 ft from the base of the pole?
answer:_____ft/sec
answer:_____dollars per day
3) A rotating light is located 11 feet from a wall. The light completes one rotation every 4 seconds. Find the rate at which the light projected onto the wall is moving along the wall when the light's angle is 15 degrees from perpendicular to the wall.Answer is a positive value. Give your answer accurate to at least one decimal place.
answer:_______feet per second
Explanation / Answer
1)Height of Pole, p = 13 ft.
Height of Woman, w = 6 ft.
Distance Between Woman and Pole, s = 35 ft.
Speed of Woman, ds/dt = 5 ft./sec
Length of Shadow = L
Distanc Between Woman and End of Shadow = x
Find dL/dt:
See diagram: http://s533.photobucket.com/user/revo_emag/media/image623.jpg.html
p / (s + x) = w / x
13 / (35 + x) = 6 / x
13x = 6(35 + x)
13x = 210 + 6x
13x - 6x = 210
7x = 210
x = 210 / 7
x = 30
L = s + x
L = 35 + 30
L = 65
p / L = w / (L - s)
13 / L = 6 / (L - s)
13(L - s) = 6L
13L - 13s = 6L
13L - 6L = 13s
7L = 13s
Differentiate Implicitly:
7(dL/dt) = 13(ds/dt)
7(dL/dt) = 13(5)
7(dL/dt) = 65
dL/dt = 65 / 7
dL/dt = 9.3
The tip of the shadow is moving at a rate of 9.3 ft./sec.
2)
You are given the equation and one rate. The rate is that the change in sales per day is 45. This is the same as the change in x with respect to time. We can write this as the derivative dx/dt. The question is asking for the change in revenue with respect to time at 190 units. This can be written as the derivative dR/dt. Now, taking our equation, differentiate both sides with respect to time (i.e. implicitly):
R = 1700x - 3x
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