A balloon is rising vertically above a level, straight road at a constant rate of 1 foot/second. Just when the balloon is 65 feet above the ground, a bicycle passes under it going 17 feet/sec. How fast is the distance between the bicycle and balloon increasing 3 seconds later?

let the time passed since the biker passed directly underneath be t seconds

Make a diagram showing the distance between them as d ft
I see it as
d^2 = (17t)^2 + (65+t)^2
when t=3
d^2 = 51^2 + 68^2
d = √7225 = 85

2d dd/dt = 2(17)(17t) + 2(65+t)
divide by 2 and solve for dd/dt
dd/dt = (17(17t) + 65+t)/d
when t=3 and d = 85
dd/dt = (17(51) + 68)/85 = 11 ft/s

To solve this problem, we can use the concept of related rates. Let's call the distance between the bicycle and the balloon at any time t as y, and the time as t.

The rate at which the distance between the bicycle and balloon is changing is given by dy/dt. We need to find the value of dy/dt 3 seconds later.

Given:
Rate of ascent of the balloon, dy/dt_balloon = 1 ft/s
Rate of the bicycle, dy/dt_bicycle = 17 ft/s
Initial distance between the bicycle and the balloon, y = 65 ft

To find dy/dt, we need to find an equation that relates y, the distance between the bicycle and the balloon, and t, the time passed.

As the balloon and bicycle are moving, we can use the Pythagorean theorem:
y^2 + (65 - 17t)^2 = (65 + t)^2

Now, differentiate both sides of the equation with respect to time t. We get:

2y * dy/dt + 2(65 - 17t) * (-17) = 2(65 + t) * 1

Simplify the equation:

2y * dy/dt - 34(65 - 17t) = 2(65 + t)

Now, substitute the given initial values: y = 65 and t = 0 into the equation:

2(65) * dy/dt - 34(65) = 2(65)

130 * dy/dt - 2210 = 130

130 * dy/dt = 2340

dy/dt = 2340 / 130

dy/dt = 18 ft/s

Therefore, the rate at which the distance between the bicycle and balloon is changing 3 seconds later is 18 ft/s.

To find how fast the distance between the bicycle and the balloon is increasing, we need to find the rate at which the distance is changing with respect to time. This can be determined by applying the Pythagorean theorem.

Let's denote the distance between the bicycle and the balloon as D, the vertical distance traveled by the balloon as y, and the horizontal distance traveled by the bicycle as x.

According to the Pythagorean theorem:
D^2 = x^2 + y^2

We are given that the balloon is rising at a rate of 1 foot/second, which means the vertical distance y is increasing at a rate of 1 foot/second.

The bicycle is moving horizontally at a rate of 17 feet/second, so dx/dt = 17 feet/second.

To find how fast the distance D is changing with respect to time, we need to differentiate the equation D^2 = x^2 + y^2 with respect to time (t). Differentiating both sides, we have:

2D * dD/dt = 2x * dx/dt + 2y * dy/dt

Since we are interested in finding how fast the distance D is increasing, we need to find dD/dt when t = 3 seconds.

At this moment, we have:
D = sqrt(x^2 + y^2)
y = 65 ft (given)
dx/dt = 17 ft/s (given)

To find x, we need to calculate the horizontal distance traveled by the bicycle in 3 seconds:
x = dx/dt * t = 17 ft/s * 3 s = 51 ft

Now we can calculate D:
D = sqrt(x^2 + y^2) = sqrt((51 ft)^2 + (65 ft)^2) = sqrt(2601 ft^2 + 4225 ft^2) = sqrt(6826 ft^2) ≈ 82.6 ft

Next, we need to calculate dy/dt, which is the rate at which the vertical distance is increasing. We are given that the balloon is rising at a constant rate of 1 ft/s, so dy/dt = 1 ft/s.

Let's substitute all the known values into the derived equation:
2D * dD/dt = 2x * dx/dt + 2y * dy/dt
2 * 82.6 ft * dD/dt = 2 * 51 ft * 17 ft/s + 2 * 65 ft * 1 ft/s

Simplifying:
165.2 ft * dD/dt = 1734 ft^2/s + 130 ft^2/s
165.2 ft * dD/dt = 1864 ft^2/s

To find the rate of change of D (dD/dt), divide both sides by 165.2 ft:
dD/dt ≈ 1864 ft^2/s / 165.2 ft ≈ 11.29 ft/s

Therefore, the distance between the bicycle and the balloon is increasing at a rate of approximately 11.29 feet/second.