An airplane flies at an altitude of 5 miles toward a point directly over an observer. The speed of the plane is 600 miles per hour. Find the rate at which the angle of elevation tetra is changing when the angle is 30 degrees
would you just replace H with 5 miles since that's the height the plane is flying at?
This is where you need to do some work.
The formula is general, but the numerical values you substitute have to be consistent in units.
For H=5 miles, and dx/dt=600 miles/hour, it would give dθ/dt in radians/hour.
You need to substitute values in appropriate units of your choice. Finally, you can also change radians to degrees using degrees=radians*180/π.
To find the rate at which the angle of elevation is changing, we need to use trigonometry and related rates. Let's assume that the observer is standing at point O and the airplane is at point A.
We are given that the airplane flies at an altitude of 5 miles, so OA = 5 miles. We also know that the speed of the plane is 600 miles per hour. Let's call the point where the airplane is directly over the observer at a specific time t, point B.
Now, we need to find the rate at which the angle of elevation (θ) is changing when the angle is 30 degrees. Let's denote this rate as dθ/dt.
Step 1: Find the distance OB.
We can use the definition of sine to relate the angle θ and the sides of the right triangle OBA.
sin(θ) = O/BA
Since BA is the same as OB, we have sin(θ) = 5/OB.
Step 2: Differentiate with respect to time.
To find the rate of change of sin(θ) with respect to time, we differentiate the equation with respect to t.
d/dt(sin(θ)) = d/dt(5/OB)
Step 3: Substitute known values and solve for dθ/dt.
We are given that sin(θ) = 5/OB and θ = 30 degrees. We need to find dθ/dt when θ = 30 degrees.
d/dt(sin(θ)) = d/dt(5/OB)
Using the chain rule, we have:
cos(θ) * dθ/dt = 0
Since θ = 30 degrees, we have:
cos(30) * dθ/dt = 0
(√3/2) * dθ/dt = 0
To solve for dθ/dt, we divide both sides by (√3/2):
dθ/dt = 0 / (√3/2)
dθ/dt = 0
Therefore, the rate at which the angle of elevation is changing when the angle is 30 degrees is 0. The angle does not change at this specific angle.
Let
angle of elevation=θ
height of plane = H
horizontal distance from observer = x
tan(θ)=x/H
Use implicit differentiation
d(tan(θ))/dt = d(x/H)/dt
sec²(θ)dθ/dt = (dx/dt)/H
dθ/dt=(1/(Hsec²(θ))(dx/dt)
dθ/dt=(cos²(θ)/H)*(dx/dt)
Can you take it from here?