a balloon rises straight up to 5 ft/sec. and a crying child who lost the balloon is standing 40 ft away. how fast does the balloon move away from the child when it is 30 ft above the ground?
If the balloon is at height h, then the distance z from the child to the balloon is
z^2 = h^2 + 40^2
When h=30, z=50
So, now we want dz/dt when h=30
2z dz/dt = 2h dh/dt
Just plug in z=50 and h=30 and dh/dt=5 and solve for dz/dt
To solve this problem, we'll use related rates, which involve different rates of change with respect to time.
Let's assign variables to the quantities involved:
- Let y represent the distance between the child and the balloon (in ft), where y is changing.
- Let x represent the distance between the child and the balloon's vertical projection (in ft), where x is constant.
- Let h represent the height of the balloon above the ground (in ft), where h is changing over time t.
- Let dy/dt represent the rate at which y is changing, which is what we're looking for.
- Let dh/dt represent the rate at which the height of the balloon is changing, equal to 5 ft/sec.
Now we can set up the relationship between the variables:
y^2 + x^2 = h^2
Differentiating both sides of the equation with respect to t:
2y(dy/dt) + 2x(0) = 2h(dh/dt)
Now we can plug in the given values:
x = 40 ft (constant), dh/dt = 5 ft/sec, h = 30 ft (when we want to find dy/dt), and y is the unknown.
Substituting these values into the equation:
2y(dy/dt) + 2(40)(0) = 2(30)(5)
Simplifying:
2y(dy/dt) = 2(30)(5)
2y(dy/dt) = 300
Now, isolate dy/dt by dividing both sides of the equation by 2y:
dy/dt = 300 / (2y)
To find dy/dt when the balloon is 30 ft above the ground, substitute y = 40 - 30 = 10 into the equation:
dy/dt = 300 / (2(10))
dy/dt = 300 / 20
dy/dt = 15 ft/sec
Therefore, the balloon moves away from the child at a rate of 15 ft/sec when it is 30 ft above the ground.
To determine how fast the balloon moves away from the child when it is 30 ft above the ground, we can use the concept of similar triangles. Let's denote the height of the balloon above the ground as y and the distance between the balloon and the child as x at any given moment.
We are given that the balloon is rising straight up at 5 ft/sec. Therefore, the rate of change of y with respect to time is dy/dt = 5 ft/sec.
We need to find dx/dt, the rate of change of x with respect to time when y = 30 ft. To do this, we can set up a proportion based on similar triangles.
At the moment when y = 30 ft, the total height of the balloon from the ground is 30 ft + 40 ft (distance between the child and the balloon) = 70 ft. Let's call this height H.
Using similar triangles, we can set up the following proportion:
y / x = H / 40
Substituting y = 30 ft and H = 70 ft, we have:
30 / x = 70 / 40
Simplifying the proportion:
30 * 40 = 70 * x
1200 = 70x
x = 1200 / 70
x ≈ 17.14 ft
Therefore, when the balloon is 30 ft above the ground, x ≈ 17.14 ft. This is the distance between the balloon and the child.
To find dx/dt, we can differentiate both sides of the proportion with respect to time:
dy/dt / x = 0 / 40 - dH/dt / 40
Since dy/dt = 5 ft/sec and dH/dt is the rate of change of H with respect to time, which is the same as dy/dt (5 ft/sec) since the balloon is rising straight up, we can substitute these values:
5 / x = -5 / 40
Solving for dx/dt:
5 * 40 = -5 * x
dx/dt = (5 * 40) / (-5)
dx/dt = -200 / 5
dx/dt = -40 ft/sec
Therefore, the balloon is moving away from the child at a rate of 40 ft/sec when it is 30 ft above the ground.