Robert F. answered • 07/21/15

A Retired Professor to Tutor Math and Physics

^{2}(x))(dx/dt)=(1/d)(dh/dt).

^{2}(x)/d)(dh/dt)

^{2}(x)=(d

^{2})/(d

^{2}+h

^{2})=(5.75

^{2})/(5.75

^{2}+18

^{2})=0.0926

Lisa W.

asked • 07/21/15A child's balloon rises into the air at a rate of 2.5 ft/sec. The child runs over to his mom (5.75 feet from where he lost it) and watches the balloon ascend into the sky. How fast is the angle of observation, Θ, changing (in radians per second) at the moment when the balloon is 18 feet above the ground?

I found that the angle of observation is changing at 0.14 rad/sec. Could somebody please confirm this for me?

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Robert F. answered • 07/21/15

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The height of the child is not give, so the problem is ambiguous, since the angle of observation is not given. Therefore, assume the child's height is zero.

Construct a right triangle mother at the origin, release point d=5.75 feet to the right, and current balloon position h=18 feet above the release point. Denote the angle of observation by x.

When the balloon is 18 feet high, tan(x)=h/d.

d(tan(x))/dt=(1/cos^{2}(x))(dx/dt)=(1/d)(dh/dt).

(dx/dt)=(cos^{2}(x)/d)(dh/dt)

You are told that (dh/dt)=2.5

From the triangle, cos^{2}(x)=(d^{2})/(d^{2}+h^{2})=(5.75^{2})/(5.75^{2}+18^{2})=0.0926

(dx/dt)=(0.0926/5.75)(2.5)=0.040 radians/sec

The height of the child is not give, so the problem is ambiguous, since the angle of observation

Aaron M. answered • 07/21/15

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Experienced physics, math and scientific writing tutor

Lisa W.

Yes, I got tan(θ)=h/5.75

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07/21/15

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Lisa W.

07/21/15