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A 25 ft ladder is leaning against a vertical wall. At what rate (with respect to time) is the angle theta between the ground an...Asked by lilac
A 25ft ladder is leaning against a vertical wall. At what rate (with respect to time) is the angle between the ground and the ladder changing, if the top of the ladder is sliding down the wall at the rate of r inches per second, at the moment that the top of the ladder is h feet from the ground. (equation in terms of h and the angle)
a. r / (300√ (1 - (h/25)^2))
b. 1 / (25√ (1 - (h/25)^2))
c. r / (12√ (1 - (h/25)^2))
d. r / (25√ (1 + (h/25)^2))
e. 1 / (25√ (1 + (h/25)^2))
a. r / (300√ (1 - (h/25)^2))
b. 1 / (25√ (1 - (h/25)^2))
c. r / (12√ (1 - (h/25)^2))
d. r / (25√ (1 + (h/25)^2))
e. 1 / (25√ (1 + (h/25)^2))
Answers
Answered by
anonymous
am I correct
Answered by
oobleck
sinθ = h/25
cosθ dθ/dt = 1/25 dh/dt
dθ/dt = -r/25 * 25/√(625-h^2) = -r/√(625-h^2) = -r/(25√(1 - (h/25)^2)
Looks like B, except that the numerator should be -r, not 1, since clearly the angle is decreasing as the ladder slides down.
cosθ dθ/dt = 1/25 dh/dt
dθ/dt = -r/25 * 25/√(625-h^2) = -r/√(625-h^2) = -r/(25√(1 - (h/25)^2)
Looks like B, except that the numerator should be -r, not 1, since clearly the angle is decreasing as the ladder slides down.
Answered by
Jonathon
It's A 10/10 cannot recommend B
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