How Fast Can A 45 Ft Sailboat Go Calculus Word Problem: A Street Light Is At The Top Of A 20 Ft Tall Pole...?
Calculus Word Problem: A street light is at the top of a 20 ft tall pole...? - how fast can a 45 ft sailboat go
A street light at the top of the mast 20 meters high. A woman of 6 feet from the pole with a speed of 8 m / sec on a straight path. What speed is the tip of his shadow moving when he is 45 m from the base of the post?
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Description of the problem
is not that big triangle on the vertical leg pole is 20 meters high and horizontal leg () D + S, where s is the length of the shadow and d is the distance the woman from the pole. The ends of the horizontal branch is the base of the mast and the tip of the shadow
There is a similar triangle, with the same top-end of the shade. The same woman creates the vertical leg of the triangle, to 6, the horizontal branch of the shadow (length s).
Items similar triangles:
Women: Pole = 6 / 20
horizontal dropouts: s / (S + D)
because the triangles are similar, the relationship becomes:
6 / 20 = S / (S + D)
3 / 10 = B / (S + D)
Cross-multiplying:
10s= 3 s (+ d)
10s = 3S + 3D
7S = 3D
s = (3 / 7) d, or the length of the shadow, if 3 / 7 by far the Women of the Pole
: different
ds / dt = (3 / 7) dd / dt, where ds / dt is the speed with which changes the length of the shadow, and dd / dt is the rate at which the distance between the light pole mutation (8 m / s is the woman walking)
ds / dt = (3 / 7) (8) = 24 / 7 m / s
Therefore, the length of its shadow 24 / 7 m / s, but the "foot" of the shadow is already on its way to 8 m / s = 56 m / sec 7
add, there are differences in the length of the shadow that moves the tip of the shadow (24 + 56) / 7 = 80 / 7 m / s
IP (AB) = 20 feet
Women (CD) = 6 feet
= EC = shadow
AC = x
tri.abe and similar tri.cde
so
AB / CD = x + y / y
20 / 6 = x + y / y
10 / 3 = (X / Y) 1
(10 / 3) -1 = x / y
3x = 7y
3dx/dt = 7dy/dt
3 (8) = 7dy/dt
dy / dt = 24 / 7
so
answer = 24 / 7
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