Calculus help--- related rates? - range finder rating
The hot air balloon rides and a level domain is followed by a search range of 500 meters from the starting point. The balloon rose at a rate of 15 meters per second.
A) How fast is the direct distance between the rangefinder and the ball change in 5 seconds
B) velocity change of elevation angle when the balloon is 500 meters high
He gave us a chart trianle hypotenuse of a right side up, with the D and H and the bottom labeled equivalent to 500
Range Finder Rating Calculus Help--- Related Rates?
1 comments:
A)
D = (500 ^ 2 + H ^ 2) ^ (1 / 2)
dd / dt = 1 / 2 ((500 ^ 2 + H ^ 2) ^ (-1 / 2)) * 2H * dh / dt
/ DH Dt = 15 m / s
t = h * dh / dt = 5 * 15 = 75 feet
dd / dt = 1 / 2 ((500 ^ 2 + 75 ^ 2) ^ (-1 / 2)) * 2 * 75 * 15
/ JJ Dt = 2.23 m / s
B)
Tan (H) = h/500
Take derivatives
s ^ 2 (H) dh / dt = (1 / 500) dh / dt
dH / dt = cos ^ 2 (H) (1 / 500) dh / dt
when balloon is 500 meters high, the angle is 45 degrees
Therefore, cos (H) = sqrt (2) / 2
dH / dt = ((sqrt (2) / 2) ^ 2) (1 / 500) (15)
dH / dt = (1 / 2) (1 / 500) (15)
dH / dt = 0.015 rad / sec
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