Posted by yolana on .
A plane flies horizontally at an altitude of 5 km and passes directly over a tracking telescope on the ground. When the angle of elevation is π/3, this angle is decreasing at a rate of π/4 rad/min. How fast is the plane traveling at that time? (Round your answer to two decimal places.)

Calculus some1 plz!!!! 
Brooke & Ashland,
x = 12
x' = 13/10
y' = ? when x = 4
m = 2
y = ?
Use proportiions.
12 / 2 = 4 / y
6 = 4 / y
2/3 = y
12 / x = 4 / y
Differentiating both sides with respect to time, we get:
12 / x^2 * x' = 4 / y^2 * y'
12 / 4^2 * x' = 4 / (2/3)^2 * y'
12 / 16 * x' = 9 y'
3/4 * 13/10 = 9 y'
39 / 40 = 9 y'
39 / 360 = y'
0.11 m/s = y'
…
y = 5
a = pi/4
a' = pi/3
x' = ? when a = pi/4
x = ?
Let's first find x.
tan(a) = opposite / adjacent
tan(a) = y / x
tan(pi/4) = 5 / x
1 = 5/x
5 = x
Now we can rewrite as:
tan(a) = 5 / x
Now let's differentiatite both sides with respect to time. Doing so gives:
a' * sec^2(a) = 5 / x^2 * x'
pi/3 * (1+tan^2a) = 5 / 25 * x'
pi/3 * (1 + tan^2(pi/4)) = 1/5 * x'
pi/3 * 2 = 1/5 * x'
2pi/3 * 5 = x'
10pi/3 = x'
10.47 km/min = x'
Hope this helped. 
Calculus some1 plz!!!! 
yolana,
sadly, it didn't work :( thanks for trying

Calculus some1 plz!!!! 
MathMate,
x=h*cot(θ)
dx/dθ=h*csc²(θ)
Given
h=5 km
dθ/dt = π/4
θ=π/3
dx/dt=(dx/dθ)*d(θ/dt)
=h*csc²(θ)*dθ/dt
=5csc²(π/3)*(π/4)
=5*(√3)/2*(π/4)
=1.083π/min.
=3.40 km/min (not a fast plane!)