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.)

sadly, it didn't work :( thanks for trying

To find the speed of the plane at that time, we can use trigonometry and related rates.

Let's denote the distance between the tracking telescope and the plane by d, and the angle of elevation by θ. We are given that the plane is flying horizontally at an altitude of 5 km, so the height of the plane is constant.

From trigonometry, we know that the tangent of the angle of elevation is equal to the height of the plane divided by the distance from the tracking telescope to the plane. So we have:

tan(θ) = 5/d

Now, we need to find the rate at which the angle of elevation is decreasing, dθ/dt. We are given that dθ/dt = -π/4 rad/min. This negative sign indicates that the angle of elevation is decreasing.

To find the rate at which the distance between the tracking telescope and the plane is changing, dd/dt, we can differentiate both sides of the equation tan(θ) = 5/d with respect to time t:

sec^2(θ) * dθ/dt = -5/dd/dt

Since we are given the value of dθ/dt, we can now solve for dd/dt:

dd/dt = -5/(sec^2(θ) * dθ/dt)

We are interested in finding the speed of the plane, which is the magnitude of dd/dt. The magnitude of sec^2(θ) is simply the reciprocal of the cosine of θ:

dd/dt = -5/(1/cos^2(θ) * dθ/dt) = -5*cos^2(θ)/dθ/dt

Substituting the given values, we have:

dd/dt = -5*cos^2(π/3)/(-π/4) = (-5)*(1/4) = -5/4 km/min

Since the speed cannot be negative, the magnitude of the speed of the plane at that time is 5/4 km/min.

Therefore, the plane is traveling at a speed of 1.25 km/min at that time.

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.

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!)