A kite 80 feet above the ground moves horizontally at a speed of 8 ft/sec. At what rate is the angle between the string and the horizontal decreasing when 100 ft of string have been let out?
Draw your little triangle.
cos a = x/s
where s is the string length (the hypotenuse) and x is the horizontal distance from the flyer.
x = s*cos a
dx/dt = ds/dt cos a - s*sin a da/dt
At our moment, s=100, so
8 = ds/dt cos a - 100 * sin a da/dt
Still not home free. What are ds/dt and a?
a little examination shows that we have a 60-80-100 triangle, so
8 = ds/dt * .6 - 100 * .8 da/dt
8 = .6 ds/dt - 80 da/dt
What's ds/dt?
well, s^2 = x^2 + 80^2
2s ds dt = 2x dx/dt
200 ds/dt = 120 * 8 = 160
ds/dt = .8
8 = .6*.8 - 80 da/dt
8 = .48 - 80 da/dt
7.52/-80 = da/dt
-.094 = da/dt
That's in radians, so at the moment in question,
a = 53.1° and
da/dt = -5.38°/sec
To solve this problem, we will use the concept of related rates.
Let's assume that the angle between the string and the horizontal is represented by θ at a given time t, and the length of the string let out is represented by x at the same time t.
We are given that the kite is 80 feet above the ground, so the height of the kite above the ground can be represented by h. We also know that the kite moves horizontally at a speed of 8 ft/sec.
First, let's find an equation that relates the variables h, x, and θ.
Using trigonometry, we can see that tan(θ) = h / x.
Taking the derivative of both sides with respect to t, we get:
sec^2(θ) * dθ/dt = (dh/dt * x - h * dx/dt) / x^2.
Since we are interested in finding the rate at which the angle between the string and the horizontal is decreasing (dθ/dt), we need to rearrange the equation to solve for it.
We know that the length of the string let out is 100 ft (x = 100 ft), and the height of the kite above the ground is 80 ft (h = 80 ft).
Given that the kite moves horizontally at a speed of 8 ft/sec (dx/dt = 8 ft/sec), we can substitute these values into the equation to solve for dθ/dt.
sec^2(θ) * dθ/dt = (dh/dt * x - h * dx/dt) / x^2
sec^2(θ) * dθ/dt = (0 - 80 * 8) / 100^2
sec^2(θ) * dθ/dt = -640 / 10000
dθ/dt = -640 / (10000 * sec^2(θ))
Now, we need to find the value of sec(θ). To do that, we can use the Pythagorean theorem.
Using the Pythagorean theorem, we can see that h^2 + (x^2 - 80^2) = x^2.
Simplifying this equation, we have h^2 + x^2 - 80^2 = x^2,
h^2 = 80^2.
h = 80.
So, sec(θ) = hypotenuse / adjacent side = h / x = 80 / 100 = 4 / 5.
Now we can substitute the value of sec(θ) into the equation:
dθ/dt = -640 / (10000 * (4/5)^2)
dθ/dt = -640 / (10000 * 16/25)
dθ/dt = -640 * 25 / (10000 * 16)
dθ/dt = -1 / 4 rad/sec.
Therefore, the angle between the string and the horizontal is decreasing at a rate of 1/4 radians per second when 100 ft of string have been let out.
To find the rate at which the angle between the string and the horizontal is decreasing, we need to use calculus and apply the chain rule. Here's how you can solve this problem step-by-step:
1. Draw a diagram representing the situation. Draw a right-angled triangle where the string is the hypotenuse, and one side is the height above the ground (80 ft) and the other side is the horizontal distance (let's call it x).
2. Let's denote the changing angle between the string and the horizontal as θ (theta). So, we want to find dθ/dt, the rate at which θ is changing with respect to time.
3. We are given that dx/dt = 8 ft/sec, which means the horizontal distance x is changing at a constant rate of 8 ft/sec.
4. The length of the string is the hypotenuse of the triangle, which is given by the Pythagorean theorem as follows:
(string length)^2 = (height)^2 + (horizontal distance)^2
Let's call the length of the string L:
L^2 = 80^2 + x^2
Differentiating both sides of this equation with respect to time t:
2L(dL/dt) = 2(80)(0) + 2x(dx/dt)
Simplifying:
dL/dt = x(dx/dt)/L
5. We know that when 100 ft of string have been let out, the length of the string L is 100 ft:
L = 100
So, we want to find dθ/dt when L = 100.
6. We can find dx/dt when L = 100 by substituting L = 100 into the equation we obtained in step 4:
dL/dt = x(dx/dt)/L
dx/dt = (dL/dt)(L/x)
dx/dt = (dL/dt)(100/x)
7. Now, we need to find dL/dt when L = 100. Differentiating the equation L^2 = 80^2 + x^2 with respect to time t:
2L(dL/dt) = 2(0) + 2x(dx/dt)
dL/dt = x(dx/dt)/L
dL/dt = x(dx/dt)/100
8. Substituting dx/dt = 8 ft/sec (from the given information) into the equation obtained in step 7:
dL/dt = (x)(8)/100
dL/dt = (x/12.5) ft/sec
9. Now, we can find dθ/dt by substituting dx/dt = 8 ft/sec and dL/dt = x/12.5 ft/sec into the equation we obtained in step 4:
dθ/dt = (dx/dt)(x/L)
dθ/dt = (8)(x)/100
dθ/dt = (2x)/25 radians/sec
10. Since we are interested in the angle decreasing, the rate will be negative. Therefore, dθ/dt = -(2x)/25 radians/sec.
So, when 100 ft of string have been let out, the angle between the string and the horizontal is decreasing at a rate of -(2x)/25 radians per second, where x is the horizontal distance (which is changing at a constant rate of 8 ft/sec in this case).