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aznshark4
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Homework Statement
"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 [tex]\theta[/tex] is changing when the angle is 30[tex]\circ[/tex]"
variables:
- x = ground distance of plane from the observer.
- [tex]\theta[/tex] = angle of elevation from observer to plane.
given:
- altitude of plane is 5 miles.
- [tex]\frac{dx}{dt}[/tex] = -600mi/h (because the plane is traveling towards observer, distance between them decreases).
Homework Equations
logic:
the situation creates a right triangle with base x and height 5 mi; hypotenuse is not necessary in this problem. Angle of elevation is [tex]\theta[/tex], so the base equation for this problem is:
tan [tex]\theta[/tex] = [tex]\frac{5}{x}[/tex]
The Attempt at a Solution
I first found what x was, since I would need x to solve my problem:
- tan [tex]\theta[/tex] = [tex]\frac{5}{x}[/tex]
- tan 30[tex]\circ[/tex] = [tex]\frac{5}{x}[/tex]
- x = 5[tex]\sqrt{3}[/tex]
Then, I found the derivative of both sides of the base equation, so my work looks like this:
- tan [tex]\theta[/tex] = [tex]\frac{5}{x}[/tex]
- sec2 [tex]\theta[/tex] [tex]\frac{d\theta}{dt}[/tex] = 5 (-x-2) [tex]\frac{dx}{dt}[/tex]
- [tex]\frac{d\theta}{dt}[/tex] = [tex]\frac{-5cos^2\theta}{x^2}[/tex] [tex]\frac{dx}{dt}[/tex]
plugging in, the equation would turn into:
- [tex]\frac{d\theta}{dt}[/tex] = [tex]\frac{-5cos^230}{(5\sqrt{3})^2}[/tex] * -600 = 30
[tex]\frac{d\theta}{dt}[/tex] = [tex]\frac{miles}{miles^2}[/tex] * (miles/hour) = hour-1
the answer I had was in "nothing per hour" when the answer should be in "radians per hour".
What did I do wrong in this problem? Thanks in advance!