Posted by **yolana** on Monday, March 14, 2011 at 7:47pm.

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**, Monday, March 14, 2011 at 7:51pm
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**, Monday, March 14, 2011 at 7:53pm
sadly, it didnt work :( thanks for trying

- Calculus some1 plz!!!! -
**MathMate**, Monday, March 14, 2011 at 8:20pm
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!)

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