Related Rates

Question

A 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?

Answer 1

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/cos²(x))(dx/dt)=(1/d)(dh/dt).

 

(dx/dt)=(cos²(x)/d)(dh/dt)

 

You are told that (dh/dt)=2.5

 

From the triangle,  cos²(x)=(d²)/(d²+h²)=(5.75²)/(5.75²+18²)=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 

Answer 2

Hi Lisa, I got something different. The tricky part of these sorts of questions is in the set-up. Let's define three variables: h is the height of the balloon, d is the horizontal distance to where the boy let go of the balloon and \theta is "angle of observation." Did you start the question by writing a trigonometric relation between these three variables?