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.

tan(Θ) = 2/x

dx/dt = -600
y = 2

Start with tan(Θ) = 2/x.Taking the derivative leaves you with:

sec^2(Θ)*dΘ/dt = -2/x^2*dx/dt

Solve for dΘ/dt.

dΘ/dt = -2/x^2 * dx/dt * 1/sec^2(Θ)

when the distance from the plane to the kangaroo is 3, x = sqrt(5); so Θ = arcsin(2/3).

Use the equation above and plug in dx/dt, x, and theta, then divide by 60 to get units to rads/min.

the answer should be about 2.22222 rad/min.

Well, isn't this a "high"-flying problem! Let's tackle it with a pinch of humor.

First things first, we need to find the angle of elevation of the kangaroo's line of sight. So, picture this: the airplane is flying above, and the kangaroo is chilling on the ground, minding his own business.

Now, imagine a right triangle formed by the airplane, the kangaroo, and a point directly below the kangaroo. The hypotenuse represents the line of sight between the airplane and the kangaroo. The height of the triangle is the altitude of the airplane, which is 2 miles. The distance from the kangaroo to the point directly below is 3 miles (now, let's hope the kangaroo doesn't get "hoppy" during our calculations!).

To find the angle of elevation, we can use trigonometry. The tangent of the angle is equal to the opposite side (2 miles) divided by the adjacent side (3 miles). So, the tangent of the angle is 2/3.

Now, let's calculate the derivative of the tangent of the angle with respect to time. We are given that the airplane's speed is constant at 600 miles per hour. Since we are interested in the rate of change of the angle with respect to time, we need to find dtan(theta)/dt.

Using the chain rule, dtan(theta)/dt = sec^2(theta) * d(theta)/dt.

But wait, we don't know d(theta)/dt directly -- we need to find it in terms of known values. Thankfully, there is a relationship between theta and the tangent:
tan(theta) = 2/3.

Differentiating both sides with respect to time gives us:
sec^2(theta) * d(theta)/dt = 0.

Now, let's substitute the value of tangent from our earlier calculation:
sec^2(theta) * d(theta)/dt = 0.

Since sec^2(theta) is always positive, this implies that d(theta)/dt must be zero. In other words, the angle of elevation isn't changing with time!

So, to answer your question, the angle of elevation of the kangaroo's line of sight is NOT increasing when the distance from the kangaroo to the plane is 3 miles. It's as "stale" as an old joke!

Remember, math and physics might not always take you on a "funny" journey, but understanding them can be a real "triumph"! Keep on learning and keep on laughing!

To find how fast the angle of elevation of the kangaroo's line of sight is increasing, we need to use the concept of trigonometry.

Let's denote the distance from the kangaroo to the airplane as "x" (in miles) and the angle of elevation of the kangaroo's line of sight as "θ" (in radians).

We know that the altitude of the airplane is 2 miles and the distance traveled by the airplane is equal to its speed multiplied by time. Therefore, the distance traveled by the airplane in terms of x can be written as 600 * (t - x/600) (in miles).

Now, we can form a right triangle with the kangaroo at the base, the airplane at the vertical side, and the line connecting the airplane to the kangaroo as the hypotenuse.

Using trigonometry, we can relate the angle θ to the distance x and the altitude of the airplane. We have:

tan(θ) = 2 / x

We can differentiate both sides of this equation with respect to time (since we want to find the rate of change of θ with respect to time), and use the chain rule:

sec^2(θ) * dθ/dt = (-2 / x^2) * dx/dt

To find dθ/dt, we need to solve for it. Since we are given that x = 3 miles, we can substitute the given values into the equation, and rearrange it:

sec^2(θ) * dθ/dt = (-2 / 3^2) * dx/dt

sec^2(θ) * dθ/dt = (-2 / 9) * dx/dt

dθ/dt = (-2 / 9) * dx/dt * 1 / sec^2(θ)

Now, we need to find dx/dt, the rate at which x is changing with respect to time. Since we know the speed of the airplane is 600 miles per hour, dx/dt is equal to -600.

If we substitute this value and the given distance x = 3 miles into the equation, we get:

dθ/dt = (-2 / 9) * (-600) * 1 / sec^2(θ)

Now, we only need to find sec(θ) at x = 3 miles. From the given triangle, we know that sec(θ) is equal to the hypotenuse divided by the adjacent side, which is:

sec(θ) = sqrt(3^2 + 2^2) / 3

sec(θ) = sqrt(13) / 3

Substituting this value into the equation, we have:

dθ/dt = (-2 / 9) * (-600) * 1 / (sqrt(13) / 3)^2

Simplifying further:

dθ/dt = 400 * 3 / 13

dθ/dt = 1200 / 13

Therefore, the rate of change of the angle of elevation of the kangaroo's line of sight is increasing at a rate of approximately 92.31 radians per minute.

To solve this problem, we can use the concept of similar triangles. Let's denote the angle of elevation of the kangaroo's line of sight as θ, the distance from the kangaroo to the plane as x, and the altitude of the plane as h.

Since the plane is flying at a constant altitude of 2 miles, the height of the right triangle formed by the plane, the kangaroo, and the line of sight is also 2 miles.

Using the concept of similar triangles, we can write the equation:

tan(θ) = h/x

Now, we need to find how fast the angle of elevation is changing when the distance from the kangaroo to the plane is 3 miles. This can be expressed as dθ/dt, the rate of change of θ with respect to time.

Differentiating both sides of the equation with respect to time, we get:

sec^2(θ) * dθ/dt = -h/x^2 * dx/dt

Since we are given the altitude (h) and the speed of the plane (dx/dt = 600 mph), we can substitute these values into the equation.

sec^2(θ) * dθ/dt = -2/3^2 * 600

Simplifying, we get:

sec^2(θ) * dθ/dt = -400

To find dθ/dt, we need to determine the value of sec^2(θ). Since we know that tan(θ) = h/x, we can use the Pythagorean identity to find sec^2(θ):

sec^2(θ) = 1 + tan^2(θ) = 1 + (h^2/x^2)

Substituting the given values, we have:

sec^2(θ) = 1 + (2^2/3^2) = 1 + 4/9 = 13/9

Now, we can solve for dθ/dt:

(13/9) * dθ/dt = -400

Dividing both sides by 13/9, we get:

dθ/dt = -400 * 9/13 ≈ -276.92 radians per hour

Since the question asks for the answer in radians per minute, we convert the rate to minutes by dividing by 60:

dθ/dt ≈ -276.92/60 ≈ -4.62 radians per minute

Therefore, the angle of elevation of the kangaroo's line of sight is decreasing at a rate of approximately 4.62 radians per minute.

I made a sketch showing the distance covered by the plane after it passed over the kangaroo as x miles

let the angle of elevation be Ø

I have :
tanØ = 2/x
xtanØ = 2
x sec^2 Ø dØ/dt + tanØ dx/dt = 0

when the distance between the kangaroo and the plance is 3
x^2 + 2^2 = 3^2
x = √5

when x = √5 , dx/dt = 600, tanØ = 2/√5, and sec^2 Ø = 49/5
√5(49/5) dØ/dt + (2/√5)(600) = 0
dØ/dt = (-1200/√5)/(49√5/5)
= -1200/49 radians/hr
= -20/49 radians/min

notice the angle is decreasing
check my arithmetic, I should write it out on paper first.