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))
3 answers
am I correct
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.
It's A 10/10 cannot recommend B
2 answers