Draw a diagram. If x is the distance from the kangaroo, then
tanθ = 2/x
sec^2θ dθ/dt = -2/x^2 dx/dt
so, when x=3, using dx/dt = -10 mi/min
13/3 dθ/dt = -2/9 * -10 = 20/9
dθ/dt = 2/3 rad/min
tanθ = 2/x
sec^2θ dθ/dt = -2/x^2 dx/dt
so, when x=3, using dx/dt = -10 mi/min
13/3 dθ/dt = -2/9 * -10 = 20/9
dθ/dt = 2/3 rad/min
Let's define the following variables:
- Let "θ" be the angle of elevation of the kangaroo's line of sight.
- Let "x" be the horizontal distance between the kangaroo and the plane.
- Let "y" be the vertical distance between the kangaroo and the plane.
We are given the following information:
- The plane is flying at a constant altitude of 2 miles.
- The plane is flying at a constant speed of 600 miles per hour.
- We need to find the rate at which the angle of elevation (θ) is changing when the distance "x" is 3 miles.
First, let's draw a diagram to illustrate the situation:
/|
/ |
/ | y
/ |
/ θ |
/____|
x
From the diagram, we can see that:
- tan(θ) = y/x
- y = 2 miles (constant altitude of the plane)
Differentiating both sides of the equation with respect to time (t), we get:
sec^2(θ) * dθ/dt = (1/x) * dx/dt
Now, we need to find dx/dt. Since the plane is traveling at a constant speed of 600 miles per hour, dx/dt = 600 mph.
Substituting the given values into the equation, we have:
sec^2(θ) * dθ/dt = (1/3) * (600/60) [converting mph to miles per minute]
Simplifying the equation, we get:
sec^2(θ) * dθ/dt = 10
To find dθ/dt, we need to find sec^2(θ):
From the right triangle, we have:
sec(θ) = hypotenuse (h) / adjacent side (x)
Using the Pythagorean theorem, we have:
h^2 = x^2 + y^2
h^2 = x^2 + (2 miles)^2
h^2 = x^2 + 4 miles^2
Taking the square root of both sides, we get:
h = sqrt(x^2 + 4 miles^2)
Substituting the values into the equation, we have:
sec(θ) = sqrt(x^2 + 4 miles^2) / x
Taking the reciprocal of both sides, we have:
cos(θ) = x / sqrt(x^2 + 4 miles^2)
Finally, substituting cos(θ) into the previous equation, we have:
(1 + tan^2(θ)) * dθ/dt = 10
Simplifying further, we get:
1 + tan^2(θ) = 10 / dθ/dt
tan^2(θ) = 10 / dθ/dt - 1
Now, we can substitute the value of θ when x = 3 miles into the equation and solve for dθ/dt.
Once we have dθ/dt, we will have the rate at which the angle of elevation of the kangaroo's line of sight is changing, given in radians per minute.