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.
make a sketch to get a right-angled triangle
Let Ø be the angle of elevation, the height is 2,
the horizontal distance is x and the hypotenuse is h
tanØ = 2/x
xtanØ = 2
xsec^2 Ø dØ/dt + tanØ dx/dt = 0
when h = 3
x^2 + 4 = 9
x = √5 , and dx/dt = -600 , (x is decreasing)
also when h = 3, tanØ = 2/√5 ,
cosØ = √5/3
secØ = 3/√5
sec^2 Ø = 9/5
plug into the derivative equation:
√5(9/5) dØ/dt + (2/√5)(-600) = 0
dØ/dt = 1200/√5) (5/(9√5) )
= 1200/9 radians/hr
= 20/9 radians/min
Thank you
To find the rate at which the angle of elevation is increasing, we need to consider the relationship between the altitude, the distance from the kangaroo to the plane, and the angle of elevation.
Let's assume that the distance from the kangaroo to the plane is represented by the variable x, and the angle of elevation is represented by θ.
We can form a right triangle with the hypotenuse being the distance from the kangaroo to the plane (x), the side adjacent to the angle of elevation as the altitude of the plane (2 miles), and the side opposite to the angle of elevation as the height of the kangaroo.
Using trigonometry, the tangent of the angle of elevation is given by:
tan(θ) = height of kangaroo / altitude of the plane
Taking the derivative of both sides with respect to time (t), we get:
sec^2(θ) * dθ / dt = d(height of kangaroo) / dt / altitude of the plane
We are given that the altitude of the plane is a constant 2 miles.
Let's differentiate the equation representing the distance between the kangaroo and the plane with respect to time (t):
x^2 = (altitude of plane)^2 + (height of kangaroo)^2
Differentiating both sides, we have:
2x * dx/dt = 0 + 2(height of kangaroo) * d(height of kangaroo) / dt
Simplifying, we get:
x * dx/dt = (height of kangaroo) * d(height of kangaroo) / dt
Dividing this equation by (altitude of plane)^2 (which is 4):
(x * dx/dt) / 4 = (height of kangaroo) * d(height of kangaroo) / dt / 4
Since x represents the distance from the kangaroo to the plane, we have x = 3 miles when the rate is being measured. Let's substitute this value into the equation:
(3 * dx/dt) / 4 = (height of kangaroo) * d(height of kangaroo) / dt / 4
We know that the rate of change of the distance from the kangaroo to the plane is given by the velocity of the plane, which is 600 mph:
(3 * 600) / 4 = (height of kangaroo) * d(height of kangaroo) / dt / 4
Simplifying further:
1800 / 4 = (height of kangaroo) * d(height of kangaroo) / dt / 4
450 = (height of kangaroo) * d(height of kangaroo) / dt
Finally, we can substitute the given altitude of the plane (2 miles) for the height of the kangaroo:
450 = 2 * d(height of kangaroo) / dt
Simplifying, we get:
d(height of kangaroo) / dt = 225
Therefore, the rate at which the height of the kangaroo is changing is 225 miles per hour.
Since the question asks for the rate in radians per minute, we need to convert the units.
1 hour = 60 minutes
1 mile = 5280 feet
1 radian = 180/π degrees
Converting the units:
225 miles per hour * 1 hour / 60 minutes * 5280 feet / mile * π / 180 degrees * 1 degree / 60 minutes
Simplifying, we get:
= 18π / π radians per minute
= 18 radians per minute
Therefore, the angle of elevation of the kangaroo's line of sight is increasing at a rate of 18 radians per minute.
To find the rate at which the angle of elevation of the kangaroo's line of sight is changing, we need to use trigonometry and related rates.
Let's assume:
- The position of the kangaroo is point K.
- The position of the airplane is point A.
- The altitude of the airplane is 2 miles.
- The speed of the airplane is 600 miles per hour.
- The distance between the kangaroo and the airplane is given by the variable x.
We are looking for the rate at which the angle of elevation (θ) is changing when the distance from the kangaroo to the plane (x) is 3 miles.
First, let's draw a diagram:
K
\
\
\
A
We have a right triangle where the hypotenuse is the line of sight between the kangaroo and the airplane and the opposite side is the altitude of the airplane (2 miles). The adjacent side is the distance between the kangaroo and the airplane (x).
Now, let's use the tangent function to relate the angle of elevation (θ) to the sides of the triangle:
tan(θ) = opp/adj
tan(θ) = 2/x
Next, let's differentiate both sides of this equation with respect to time (t), treating θ and x as functions of t:
d/dt[tan(θ)] = d/dt[2/x]
To find d/dt[tan(θ)], we'll use the chain rule. Let's set y = tan(θ):
d/dt[y] = d/dt[tan(θ)] = d[tan(θ)]/dθ * dθ/dt
The derivative of tangent is:
d/dt[y] = sec^2(θ) * dθ/dt
Now, let's differentiate the right side of the equation, d/dt[2/x]:
d/dt[2/x] = -2/x^2 * dx/dt
Now, we can substitute these derivatives into our original equation:
sec^2(θ) * dθ/dt = -2/x^2 * dx/dt
We know the altitude of the airplane is 2 miles and the distance from the kangaroo to the airplane is 3 miles. Therefore, x = 3 and substituting these values into the equation:
sec^2(θ) * dθ/dt = -2/3^2 * dx/dt
sec^2(θ) * dθ/dt = -2/9 * dx/dt
We are given the speed of the airplane is 600 miles per hour, which means dx/dt = 600 mph. Substituting this value:
sec^2(θ) * dθ/dt = -2/9 * 600 mph
sec^2(θ) * dθ/dt = -400/9 mph
We need to find the rate of change in radians per minute, so let's convert hour to minute:
-400/9 mph * 1 hr/60 min = -400/540 radians per minute
dθ/dt = -400/540 radians per minute
dθ/dt ≈ -0.741 radians per minute
Hence, the angle of elevation of the kangaroo's line of sight is decreasing at a rate of approximately -0.741 radians per minute when the distance from the kangaroo to the plane is 3 miles.