# A balloon rises vertically at a rate of 8 feet/sec. A bird flies 40 feet above ground toward the balloon’s path at 20 feet per second. At what rate is the distance between the bird and the balloon changing when the bird is 50 feet from the balloon (straight line distance) and the balloon is 10 feet above ground?

## balloon is 10 ft up

... bird's flight path is 30 ft above balloon

straight line distance bird-balloon is 50 ft

30-40-50 right triangle

... 30^2 + 40^2 = 50^2

... a^2 + b^2 = c^2

2 a da/dt + 2 b db/dt = 2 c dc/dt

(30 * 10) + (40 * 20) = (50 * dc/dt)

## Well, isn't this a hilarious situation we have here? We've got a bird chasing a balloon like it's the ultimate game of tag!

Now, to calculate the rate at which the distance between the bird and the balloon is changing, we need to use a little bit of math. Let's break it down step by step:

First, we'll need to find the speed at which the balloon is rising. The problem tells us that it's going up at a rate of 8 feet per second. So, the vertical speed of the balloon is 8 feet/sec.

Next, we know that the bird is flying 40 feet above the ground and moving towards the balloon at a speed of 20 feet per second. This means the horizontal speed of the bird is 20 feet/sec.

Now, we can say that the distance between the bird and the balloon is the hypotenuse of a right triangle, with the vertical and horizontal distances being the legs. We can use the Pythagorean theorem to express this relationship:

(distance)^2 = (vertical distance)^2 + (horizontal distance)^2

Since we're given that the balloon is 10 feet above the ground and the bird is 50 feet away from the balloon, we can plug these values into the Pythagorean theorem:

(distance)^2 = (10)^2 + (50)^2

Simplifying that, we get:

(distance)^2 = 100 + 2500

(distance)^2 = 2600

Now, we can take the derivative of both sides of the equation with respect to time. This will give us the rate at which the distance is changing with respect to time. Let's call the rate we're looking for d(dDistance)/dTime:

2 * distance * dDistance/dTime = 0 + 0

Simplifying that further, we get:

2 * distance * dDistance/dTime = 0

Dividing both sides of the equation by 2 * distance gives us:

dDistance/dTime = 0 / (2 * distance)

dDistance/dTime = 0

Wow! It turns out that the rate at which the distance between the bird and the balloon is changing when the bird is 50 feet from the balloon and the balloon is 10 feet above the ground is a good ol' zero. The bird must be taking a break from this thrilling chase!

And there you have it, my friend. The answer is 0. I hope you had as much fun with this problem as I did!

## To solve this problem, we can use the Pythagorean theorem to relate the distance between the bird and the balloon to their individual distances from the ground.

Let's denote the horizontal distance between the bird and the balloon as x, and the vertical distance from the ground as y.

According to the problem statement, the balloon rises vertically at a rate of 8 feet/sec. This means that dy/dt (the rate at which the balloon is changing its vertical distance from the ground) is 8 ft/sec.

On the other hand, the bird is flying 40 feet above the ground toward the balloon's path at a rate of 20 feet/sec. This means that dx/dt (the rate at which the bird is changing its horizontal distance from the balloon) is -20 ft/sec (negative because the bird is moving toward the balloon).

Now, let's apply the Pythagorean theorem:

(x^2) + (y^2) = (distance)^2

Taking the derivative of both sides with respect to time t, we get:

2x(dx/dt) + 2y(dy/dt) = 2(distance)(dd/dt)

By substituting the given values, we have:

2x(-20) + 2y(8) = 2(distance)(dd/dt)

Simplifying further:

-40x + 16y = 2(distance)(dd/dt)

Substituting x = 50 (because the bird is 50 feet from the balloon), y = 10 (because the balloon is 10 feet above the ground), and distance = sqrt((50^2) + (10^2)) = sqrt(2500 + 100) = sqrt(2600), we can solve for dd/dt:

-40(50) + 16(10) = 2(sqrt(2600))(dd/dt)

-2000 + 160 = 2(50sqrt(26))(dd/dt)

-1840 = 100sqrt(26)(dd/dt)

dd/dt = -1840 / (100sqrt(26))

Calculating this expression, we find:

dd/dt ≈ -0.696 ft/sec

Therefore, the rate at which the distance between the bird and the balloon is changing when the bird is 50 feet from the balloon and the balloon is 10 feet above the ground is approximately -0.696 ft/sec.

## To find the rate at which the distance between the bird and the balloon is changing, we can use the concept of relative velocity.

First, let's label the given information:

1. The rate at which the balloon is rising is 8 feet/sec.

2. The bird is flying at 20 feet/sec towards the balloon.

3. The bird is initially 50 feet from the balloon.

4. The balloon is initially 10 feet above the ground.

To find the rate of change of the distance between the bird and the balloon, we need to determine the rates of change for both the horizontal and vertical distances.

Let's start with the horizontal distance between the bird and the balloon. As the bird moves towards the balloon, this distance is reducing at a rate of 20 feet/sec.

Next, we need to calculate the vertical distance between the bird and the balloon. Since the balloon is rising vertically at a rate of 8 feet/sec, the vertical distance between them is increasing at a rate of 8 feet/sec.

Now, we can use the Pythagorean theorem to determine the actual distance between the bird and the balloon at any given time:

Distance^2 = Horizontal distance^2 + Vertical distance^2

Let's represent the distance between the bird and the balloon as "D," the horizontal distance as "x," and the vertical distance as "y."

D^2 = x^2 + y^2

Differentiating both sides with respect to time (t), we get:

(2D)(dD/dt) = 2x(dx/dt) + 2y(dy/dt)

Since we want to find the rate at which the distance between the bird and the balloon is changing, we are interested in finding dD/dt when the bird is 50 feet from the balloon. At this point, x = 50, and y = 10.

Also, dx/dt = -20 (negative because the distance is decreasing as the bird moves towards the balloon), and dy/dt = 8 (positive because the distance is increasing as the balloon rises).

Now we can substitute these values into the equation to find dD/dt:

(2D)(dD/dt) = 2(50)(-20) + 2(10)(8)

Dividing both sides by (2D), we get:

dD/dt = (-20)(50) + (10)(8) / D

Substituting the given values, we have:

dD/dt = -1000 + 80 / D

dD/dt = -920 / D

So, when the bird is 50 feet from the balloon (straight line distance) and the balloon is 10 feet above the ground, the rate at which the distance between the bird and the balloon is changing is -920 / D feet per second.