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...

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
lemjay eNotes educator| Certified Educator

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 change of horizontal line in our triangle. The rate in which it is changing is the same as with the speed of the plane. However, the horizontal line becomes smaller as the plane approaches the observer. So the value of dx/dt is -400.

`sec^2theta (d theta)/dt =-5/x^2(-400)`

`sec^2theta (d theta)/dt =2000/x^2`

Then, isolate d theta/dt.

`(d theta)/dt=2000/(x^2sec^2theta) `

`(d theta)/dt=(2000cos^2theta)/x^2`      (Let this be EQ2.)

Use EQ1 and EQ2 to solve for the rate of change of elevation given the specified angles.

(a) `theta =30^o`

Plug-in the value of theta to EQ1 to get the value of x.

`x=5/tan (30^o)=5/(sqrt3/3)=5*3/sqrt3=5*3/sqrt3*sqrt3/sqrt3`

`x=5sqrt3`

Then, plug-in `theta =30^o` and `x=5sqrt3` to EQ2.

`(d theta)/dt=(2000cos^2(30^o))/(5sqrt3)^2`

`(d theta)/dt=20`

Hence, when `theta =30^o` , the angle of elevation is changing at 20 radian per hour.

(b) `theta =60^o`

Plug-in the value of theta to EQ1 to get the value of x.

`x=5/tan (60^o)=5/sqrt3*sqrt3/sqrt3`

`x=(5sqrt3)/3`

Then, plug-in `theta =60^o` and `x=(5sqrt3)/3` to EQ2.

`(d theta)/dt=(2000cos^2(60^o))/((5sqrt3)/3)^2`

`(d theta)/dt=60`

Hence, when `theta =60^o` , the angle of elevation is changing at 60 radian per hour.

(c) `theta =80^o`

Plug-in the value of theta to EQ1 to get the value of x.

`x=5/tan (80^o)`

`x=0.8816`

Then, plug-in `theta =80^o` and `x=0.8816` to EQ2.

`(d theta)/dt=(2000cos^2(80^o))/(0.8816)^2`

`(d theta)/dt=77.59`

Hence, when `theta =80^o` , the angle of elevation is changing at 77.59 radian per hour.

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nick-teal eNotes educator| Certified Educator

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