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- stanschimLv 71 month ago
At any time, t, as the plane passes over Marvin, up to and including the point where the airplane is over the tower, the distance between Marvin and the plane, D, obeys the Pythagorean Theorem.

D^2 = 5000^2 + x^2, where x is how far the plane is from Marvin, horizontally.

Taking derivatives with respect to time gives:

2D(dD/dt) = 0 + 2x(dx/dt)

Now we know from the problem statement that dx/dt = 800 ft/sec.

Furthermore, at the point the plane is over the tower, by the Pythagorean Theorem, the distance from Marvin to the plane is 13000 feet. This is the hypotenuse of a right triangle with legs 5000 feet and 12000 feet.

Substituting values:

2(13000)(dD/dt) = 2(12000)(800)

26000(dD/dt) = 19200000

dD/dt = about 738.46 feet per second.

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- TomVLv 71 month ago
s² = 5000² + x²

2sds/dt = 2xdx/dt

ds/dt = (x/s)dx/dt

Given:

x = 12000 ft

dx/dt = 800 f/sec

At x = 12000 ft,

s = √(5000² + 12000²) = √169000000 = 13000 ft

ds/dt = (12000/13000)(800) = 738.5 ft/sec

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- Anonymous1 month ago
related rates solution

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