A kite is 40ft high, with 50ft of cord out. If the kite moves horizontally at 5min per hour directly away from the boy flying it, how fast is the cord being paid out.
Well, if I were to take a stab at this question, I'd say that the cord is probably being paid out at a speed of "reel-y fast"! After all, the boy flying the kite wouldn't want to lose it in the sky now, would he?
To find the rate at which the cord is being paid out, we can use the concept of similar triangles.
Let's define the variables:
h = height of the kite (40 ft)
d = distance between the kite and the boy (which is changing)
First, we need to find the rate at which the distance between the kite and the boy is changing. We are given that the kite moves horizontally at a rate of 5 ft/min per hour. Since there are 60 minutes in an hour, the horizontal speed of the kite is 5 ft/60 min = 1/12 ft/min.
Next, we can set up a ratio of similar triangles to relate the changing distance between the kite and the boy (d) to the length of the cord out (50 ft).
Using the similarity of triangles, we have:
d / h = (d + x) / (h + 50),
where x is the additional distance between the kite and the boy due to the cord being paid out.
Simplifying this equation, we get:
d / 40 = (d + x) / 90.
Now, let's solve for x:
Cross-multiply the equation:
90d = 40(d + x).
Expand and simplify:
90d = 40d + 40x.
Rearrange the equation to solve for x:
90d - 40d = 40x.
50d = 40x.
Divide both sides by 40:
x = (50d) / 40.
Finally, we can differentiate both sides with respect to time (t) to find the rate at which the cord is being paid out:
dx/dt = (50d / 40) * (dd/dt).
The left side represents the rate at which the cord is being paid out (dx/dt), and the right side represents the rate of change of the distance between the kite and the boy (dd/dt).
Given that the horizontal speed of the kite is 1/12 ft/min, we have:
dx/dt = (50d / 40) * (1/12).
Simplifying further,
dx/dt = (5/8) * (d/12) = (5d) / 96.
Therefore, the rate at which the cord is being paid out is (5d) / 96 ft/min.
To find the speed at which the cord is being paid out, we need to use the concept of related rates.
Let's denote the height of the kite as h and the length of the cord as c. We are given that h = 40 ft and c = 50 ft. We need to find how fast the cord is being paid out, which is the rate of change of c with respect to time.
We are also given that the kite moves horizontally at a rate of 5 ft per hour away from the boy. Let's denote this horizontal distance as x.
The key here is to relate the variables using a right triangle. The length of the cord is the hypotenuse of the right triangle, and the height of the kite is one of the legs.
Using the Pythagorean theorem, we have x^2 + h^2 = c^2.
Differentiating both sides of this equation with respect to time (t), we get:
2x(dx/dt) + 2h(dh/dt) = 2c(dc/dt).
Since we are interested in finding dc/dt, we can solve for it by isolating this term:
dc/dt = (x(dx/dt) + h(dh/dt))/c.
Substituting the given values, we have x = 5 ft/hour, h = 40 ft, and c = 50 ft.
Now let's find dx/dt, the rate at which x is changing. Since we are given that x is changing at a rate of 5 ft per hour, dx/dt = 5 ft/hour.
Next, let's find dh/dt, the rate at which h is changing. In the given problem, we are not explicitly given how the height of the kite is changing with time. Therefore, we assume that the height remains constant.
Plugging these values back into the equation for dc/dt, we get:
dc/dt = (5 ft/hour * 5 ft + 40 ft * 0 ft/hour) / 50 ft.
Simplifying this expression, we have:
dc/dt = (25 ft + 0 ft) / 50 ft.
Finally, evaluate the expression:
dc/dt = 25 ft / 50 ft = 0.5 ft/hour.
Therefore, the cord is being paid out at a rate of 0.5 ft per hour.
let the length of the chord be y
let the horizontal distance of the kite be x ft
given: dx/dt = 5 ?? (looks like a typo in the units)
I will assume you meant:
5 mph = 5280 ft/60 min = 88 ft/min
y^2 = x^2 + 40^2
when y = 50
50^2 = x^2 + 40^2
x = 30
2y dy/dt = 2x dx/dt
dy/dt = 2x dx/dt / 2y
= (x/y)dx/dt
= (30/50)(5) ft/min
= 3 ft/min
( looks quite slow, check my arithmetic)