Hi, I am doing a practice exam, and I came across this problem and I am have difficultly starting it.
A UFO flies horizontally at a constant speed at an altitude of 15 km and passes directly over a tracking telescope on the ground. When the angle of elevation is pi/3, this angle is decreasing at a rate of 0.1 rad/ min. How fast is the UFO flying?
I need help formulating the equation, the hardest part for me. Because I know after I have the equation, I just need to take the derivative implicitly and plug in values into the equation.
Do not attempt any of these types without a diagram.
Draw the ground and the horizontal path of the UFO 15 km above the ground.
Label the tracking station as T, the point directly above T as A, and the position of the UFO as B
(AB will be parallel to the ground, 15 km above it, right ?)
Let AB = x and TB = y
Let Ø be the angle of elevation, the angle ABT = Ø
then x^2 + 15^2 = y^2
tanØ = 15/x or x/15 = cotØ
plug in Ø = pi/3 to find x, then you can find y
you will have to know how to take the derivative of cotØ which is csc^2 Ø, and that is why we also have to know the value of y.
(I got 2 km/min for the speed of the UFO, check that)
In my 3rd last line, if left out a negative sing
d(cotx)/dx = -csc^2 x , as I worked out my 2 km/min, I did use the negative )
Yes! Thank you very much! I got the exact answer. ;DD!
To solve this problem, we need to relate the variables involved and formulate an equation. Let's start by visualizing the situation.
Consider a right triangle with the angle of elevation \(\theta\) and the altitude of the UFO as the two sides. The distance covered by the UFO is the hypotenuse. Let's denote this distance as \(d\).
From the problem statement, we know that the altitude of the UFO is 15 km. Let's call this side of the triangle \(a\), and we can write \(a = 15\) km.
We are given that the angle of elevation is \(\frac{\pi}{3}\), and it is decreasing at a rate of 0.1 rad/min. Let's call this angle \(\theta\) and the rate of change \(\frac{d\theta}{dt}\).
Now, let's establish a relationship between \(d\), \(a\), \(\theta\), and \(\frac{d\theta}{dt}\). In a right triangle, we have the trigonometric relationship:
\(\tan(\theta) = \frac{a}{d}\)
Taking the derivative implicitly with respect to time \(t\), we get:
\(\sec^2(\theta) \cdot \frac{d\theta}{dt} = \frac{-a}{d^2} \cdot \frac{dd}{dt}\)
We can simplify this expression using trigonometric identities:
\(\frac{d\theta}{dt} = \frac{-a}{d^2} \cdot \cos^2(\theta) \cdot \frac{dd}{dt}\)
Since we are interested in finding the speed of the UFO, we need to find \(\frac{dd}{dt}\).
To isolate \(\frac{dd}{dt}\), we rearrange the equation:
\(\frac{dd}{dt} = \frac{d\theta}{dt} \cdot \frac{-d^2}{a \cdot \cos^2(\theta)}\)
We can now substitute the given values into the equation:
\(a = 15\) km
\(\theta = \frac{\pi}{3}\)
\(\frac{d\theta}{dt} = -0.1\) rad/min
Plugging these values into the equation will give you the rate of change of \(d\), which represents the speed of the UFO.
Remember to convert units where necessary and pay attention to the signs in the final answer.
I hope this explanation helps you formulate the equation and solve the problem. Let me know if you need any further assistance!