# An airplane flies at an altitude of y = 5 miles toward a point directly over an observer. The speed of the plane is 400 miles per hour. Find the rates at which the angle of elevation θ is changing when the angle is θ = 30°, θ = 60°, and θ = 80°.

## Expert Answers You are exactly right!

There are 60 minutes in an hour.  So instead of 77.59 Radians per 1 hour, its 77.59 radians per 60 minutes.

So we divide 77.59 by 60 to get our answer in terms of radians per minutes.  1.293 Radians/Minute

The attachment is the figure that represents the given in our problem.

To solve for the rate of change of angle of elevation, let's apply the formula of tangent function.

`tan theta=(o p p o s i te) / (a d ja cent)`

For our triangle, the tangent is

`tan theta =5/x`

Then, isolate the x.

`x=5/tan theta `          (Let this be EQ1).

Next, lets take the derivative of our tangent with respect to t.

`d/dt (tan theta)=d/dt (5/x)`

`sec^2 theta (d theta)/dt=-5/x^2 dx/dt`

Take note that dx/dt is the rate of...

(The entire section contains 2 answers and 396 words.)

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