An airplane flies at an altitude of
y = 5
miles toward a point directly over an observer (see figure). The speed of the plane is 500 miles per hour. Find the rates at which the angle of elevation π is changing when the angle is
π = 45Β°,
π = 60Β°,
and
π = 70Β°.
Really need help on this question urgently for an assignment
Well, well, well! It's time for a little math with a sprinkle of humor! Hold onto your funny bones, folks!
Using some trigonometry, we can determine the rate at which the angle of elevation is changing. Here's the deal:
When π = 45Β°, we can calculate the rate of change by differentiating the equation y = 5 with respect to time (t). So, dy/dt = 0 (since 5 is a constant). That's right, folks, for a 45Β° angle, we're not changing anything!
Now, let's ramp it up to π = 60Β°. We still differentiate y = 5 with respect to time (t). This time, we find that dy/dt = 0 (since 5 is still a constant). Looks like we're getting a bit of a repetitive pattern here!
Now, hang on tight because we're going for π = 70Β°! Differentiating y = 5 again gives us dy/dt = 0 (since 5 is a constant through and through). Oh no, it's another zero! Looks like the angle of elevation isn't changing either! It's a no-show, folks!
So, to sum it up, no matter the angle, the rate of change of the angle of elevation is a big, fat ZERO! It seems like the airplane can't make up its mind when it comes to giving us some mathematical excitement!
To find the rates at which the angle of elevation is changing, we can use trigonometry and the concept of related rates. Let's start with the given information:
The altitude of the airplane, y, is given as 5 miles. The speed of the plane is 500 miles per hour.
We want to find the rates at which the angle of elevation, π, is changing when π is 45Β°, 60Β°, and 70Β°.
First, we need to identify the relevant trigonometric relationship. In this case, we can use the tangent function. The tangent of an angle is equal to the ratio of the opposite side to the adjacent side in a right triangle.
In the diagram, the altitude of the plane forms the opposite side, and the distance the plane has traveled in the horizontal direction forms the adjacent side.
Let's define x as the horizontal distance the plane has traveled. To relate x and y, we can use the speed of the plane, which is given as 500 miles per hour. This means that dx/dt = 500 (the rate at which x is changing is 500 miles per hour).
Now, let's write down the tangent of the angle π in terms of x and y:
tan(π) = y / x
To find the rate at which π is changing, we can differentiate this equation implicitly with respect to time (t), using the chain rule:
sec^2(π) * dπ/dt = (dy/dt * x - y * dx/dt) / x^2
Note that sec^2(π) is the derivative of tan(π) with respect to π.
Since we are given the value of y as 5 miles, we can substitute it into the equation. Also, we know that dx/dt is 500 miles per hour. Thus, we have:
sec^2(π) * dπ/dt = (dy/dt * x - 5 * 500) / x^2
To find the rate at which the angle of elevation is changing, π, we need to find dπ/dt. To do this, we rearrange the equation and solve for dπ/dt:
dπ/dt = [(dy/dt * x - 5 * 500) / x^2] / sec^2(π)
Now we can plug in the given angles π = 45Β°, 60Β°, and 70Β° to find the corresponding rates at which π is changing. Let's compute them:
For π = 45Β°:
- Calculate the value of x using the given altitude y = 5 and the tangent relationship: x = y / tan(π) = 5 / tan(45) β 5 miles.
- Substitute the values into the equation: dπ/dt = [(dy/dt * x - 5 * 500) / x^2] / sec^2(π) = [(0 * 5 - 5 * 500) / 5^2] / sec^2(45) = -2000 / 25 = -80 degrees per hour.
For π = 60Β°:
- Calculate the value of x using the given altitude y = 5 and the tangent relationship: x = y / tan(π) = 5 / tan(60) β 5 * β3 β 8.66 miles.
- Substitute the values into the equation: dπ/dt = [(dy/dt * x - 5 * 500) / x^2] / sec^2(π) = [(0 * 8.66 - 5 * 500) / 8.66^2] / sec^2(60) = -2500 / 75.07 = -33.3 degrees per hour.
For π = 70Β°:
- Calculate the value of x using the given altitude y = 5 and the tangent relationship: x = y / tan(π) β 5 / tan(70) β 1.777 miles.
- Substitute the values into the equation: dπ/dt = [(dy/dt * x - 5 * 500) / x^2] / sec^2(π) = [(0 * 1.777 - 5 * 500) / 1.777^2] / sec^2(70) = -2500 / 5.02 = -498 degrees per hour.
Therefore, the rates at which the angle of elevation π is changing when π is 45Β°, 60Β°, and 70Β° are approximately -80 degrees per hour, -33.3 degrees per hour, and -498 degrees per hour, respectively.
when the plane is at a distance x miles from directly over the observer,
tanΞΈ = 5/x
sec^2ΞΈ dΞΈ/dt = -5/x^2 dx/dt
So, when ΞΈ=60Β°, 5/x = β3 and you have
4 dΞΈ/dt = -3/5 * -500
dΞΈ/dt = 75 rad/hr