A balloon is rising at a constant speed of 5 ft/sec. A boy is cycling along a straight road at a constant speed of 15 ft/sec. When he passes under the balloon, it is 5 feet above him. Determine how fast the distance between the boy and the balloon is increasing 3 seconds after he has passed underneath it.

x = 15 t

h = 5 t + 5 = 5(t+1)

y^2=x^2+h^2 = 225t^2 + 25t^2 + 50 t + 25

y^2 = 250 t^2 + 50 t + 25

2 y dy/dt = 500 t + 50

now at t = 3
y^2 = 250(9)+150 +25 = 2425
y = 49.2
so
2(49.2) dy/dt = 1500 + 50 = 1550
dy/dt = 15.8 m/s

make a sketch putting the boy somewhere on the road

let the time passes since the boy was directly under the ballon be t seconds
the the boy went 15t ft and the balloon is at a height of (3 + 5t) ft
let the distance between them be d ft
so we have a right-angled triangle with base 15t, height of 3+5t and hypotenuse d

d^2 = (15t)^2 + (3+5t)^2
2d dd/dt + 2(15t)(15) + 2(3+5t)(5)
dd/dt = (225t + 15 +25t)/d
= (250t + 15)/d

when t=3
d^2 = 45^2 + 18^2 = 2349
d = √2349

dd/dt = (250(3) + 15)/√2349
= appr 15.78 ft/s

Go with Damon's solution, I somehow read the balloon as 3 ft instead of 5 ft above the boy

correction:
d^2 = (15t)^2 + (5+5t)^2
2d dd/dt + 2(15t)(15) + 2(5+5t)(5)
dd/dt = (225t + 25 +25t)/d
= (250t + 25)/d

when t=3
d^2 = 45^2 + 20^2 = 2425
d = √2425

dd/dt = (250(3) + 25)/√2425
= appr 15.738 ft/s

Well, this situation sounds like a real balancing act! Let's break it down.

We know the balloon is rising at a constant speed of 5 ft/sec, and the boy is cycling along at a constant speed of 15 ft/sec. When the boy passes under the balloon, the distance between them is 5 feet.

To find out how fast the distance between the boy and the balloon is increasing 3 seconds after he has passed underneath it, we can use the rates at which both the boy and the balloon are moving.

The boy is cycling at 15 ft/sec, so after 3 seconds he would have traveled a distance of 3 seconds × 15 ft/sec = 45 feet.

During these 3 seconds, the balloon has been rising at a constant speed of 5 ft/sec, so it would have gone up by 3 seconds × 5 ft/sec = 15 feet.

Now, we need to calculate the hypotenuse of the right triangle formed by the boy's distance and the balloon's height. Using the Pythagorean theorem, we have:

Distance^2 = (45 ft)^2 + (15 ft)^2 = 2025 ft^2 + 225 ft^2 = 2250 ft^2

Take the square root of both sides to find the distance:

Distance = √2250 ft^2 = 47.43 ft (approximately)

Now that we have the distance, we can differentiate it with respect to time to find how fast it's changing:

d(Distance)/dt = d/dt(√2250 ft^2) = (1/2√2250) × (d(2250)/dt) = (1/2√2250) × 0

Since the distance is constant and not changing with time, the speed at which it's changing is 0 ft/sec.

So, after 3 seconds, the distance between the boy and the balloon is not changing – it remains at 47.43 feet. It seems like they've found the perfect balance!

To determine how fast the distance between the boy and the balloon is increasing 3 seconds after he has passed underneath it, we can use the concept of related rates.

Let's denote the distance between the boy and the balloon as "d" (in feet) at time "t" (in seconds) after the boy has passed underneath.

Here are the known rates:
- The balloon is rising at a constant speed of 5 ft/sec. Therefore, the rate of change of the balloon's height can be expressed as dh/dt = 5 ft/sec.
- The boy is cycling at a constant speed of 15 ft/sec. Therefore, the rate of change of the boy's position can be expressed as dx/dt = 15 ft/sec.

Now, we can set up a right triangle to represent the situation. The horizontal leg of the triangle represents the distance the boy has traveled, and the vertical leg represents the height of the balloon. The hypotenuse of the triangle represents the distance d between the boy and the balloon.

Using the Pythagorean theorem, we have:
d^2 = x^2 + h^2

Differentiating both sides of the equation with respect to time t, we get:
2d * dd/dt = 2x * dx/dt + 2h * dh/dt

Since we want to find the rate of change of distance (dd/dt) at t = 3 seconds, we can plug in the given values:
dh/dt = 5 ft/sec
dx/dt = 15 ft/sec
h = 5 ft

At t = 3 seconds, the boy has already passed underneath the balloon, so x = 15 ft/sec * 3 sec = 45 ft.

Plugging these values into the equation, we have:
2d * dd/dt = 2(45 ft) * (15 ft/sec) + 2(5 ft) * (5 ft/sec)

Simplifying:
2d * dd/dt = 1350 ft + 50 ft^2/sec

Now we can solve for dd/dt:
dd/dt = (1350 ft + 50 ft^2/sec) / (2d)

We still need to find the value of d at t = 3 seconds. To do this, we can use the Pythagorean theorem once again:
d^2 = x^2 + h^2
d^2 = (45 ft)^2 + (5 ft)^2
d^2 = 2025 ft^2 + 25 ft^2
d^2 = 2050 ft^2
d ≈ 45.35 ft (rounded to two decimal places)

Now we can substitute this value of d into the equation for dd/dt:
dd/dt = (1350 ft + 50 ft^2/sec) / (2 * 45.35 ft)

Calculating the result gives us:
dd/dt ≈ 16.47 ft/sec

Therefore, the distance between the boy and the balloon is increasing at a rate of approximately 16.47 ft/sec, 3 seconds after the boy has passed underneath the balloon.