An airplane flys at a constant altitude of 2 miles and a constant speed of 600 miles per hour on a straight course that will take it directly over a kangaroo on the ground. How fast is the angle of elevation of the kangaroo's line of sight increasing when the distance from the kangaroo to the plane is 3 miles? Give your answer in radians per minute.
Why are you posting this again ?
I answered your same question about 3 hours ago.
Always check back on your posts, mark their time, and they are easy to find that way.
http://www.jiskha.com/display.cgi?id=1412721979
The answer was wrong!
Did you try to find an error in my solution?
How do you know it is wrong?
The math looks good to me, but the negative answer is strange because the question was, how fast is the angle increasing?
That is because dx/dt = -600; the plane is approaching the kangaroo.
Hailey, did you do as Reiny asked, and check his math?
Yup, Steve is right, I had the plane flying away from the kangaroo, should have read it more carefully.
The numerical answer would simply change to a positive rate of change of the angle.
This answer is still incorrect. The 3 miles from the airplane to the kangaroo is the hypotenuse of the triangle. It is the shortest distance from the airplane to the kangaroo, not the horizontal distance.
To solve this problem, we can use trigonometry and rates of change.
Let's begin by drawing a diagram to visualize the situation. We have a right triangle where the hypotenuse represents the distance from the kangaroo to the plane, one side represents the altitude of the plane (2 miles), and another side represents the height of the kangaroo's line of sight.
Since we know the distance from the kangaroo to the plane is 3 miles, and the altitude of the plane is 2 miles, we can label these sides accordingly in our diagram.
Now, let's define some variables:
- Let x represent the height of the kangaroo's line of sight.
- Let θ represent the angle of elevation.
Using trigonometry, we know that tan(θ) = x/3.
To determine how fast the angle of elevation is changing, we need to find dθ/dt (the rate of change of the angle) when the distance from the kangaroo to the plane is 3 miles.
To do this, we differentiate both sides of the equation tan(θ) = x/3 with respect to t (time):
(d/dt) [tan(θ)] = (d/dt) [x/3]
Next, we use the chain rule. Since dθ/dt represents the rate of change of the angle and dx/dt represents the rate of change of the height, we can rewrite the equation as:
sec^2(θ) * dθ/dt = (1/3) * (dx/dt)
Now, let's substitute the given values into the equation:
- We know the altitude of the plane is constant at 2 miles, so dx/dt = 0.
- The constant speed of the airplane is given as 600 miles per hour.
Now, we can solve for dθ/dt:
sec^2(θ) * dθ/dt = (1/3) * 0
Since dx/dt = 0, the right side of the equation becomes zero.
Therefore, we can conclude that dθ/dt = 0, which means the angle of elevation is not changing.
Thus, when the distance from the kangaroo to the plane is 3 miles, the angle of elevation of the kangaroo's line of sight is not increasing or decreasing.