67.9k views
3 votes
Standing 500 feet away, you watch a hot air balloon lift off and the rise at a constant rate of 2ft/s you continue to watch it as it rises slowly leaning your head back to keep your gaze fixed on it as it dwindles into a small dot against the blue sky…

Let theta denote your angle of elevation to the balloon. Draw a diagram modeling the situation. Define variables to represent the varying quantities and then label your diagram with all quantities. (Constant and varying)

How fast is your angle of elevation to the balloon changing when the balloon is 75 feet above the ground?

How fast is your angle of elevation to the balloon changing when the balloon is 150 feet above it the ground?

2 Answers

4 votes

Answer: Let's set up a diagram and define the variables to represent the varying quantities:

Let:

H = Height of the hot air balloon above the ground (in feet),

D = Distance from you to the hot air balloon (in feet),

θ = Angle of elevation from you to the hot air balloon (in radians).

Initially, when the balloon is at the ground, we have:

H = 0 feet (since it just lifted off),

D = 500 feet (the initial distance between you and the balloon).

As the balloon rises, H and D will change with time. When the balloon is at a certain height H above the ground, and you are still looking at it with an angle of elevation θ, we want to find how fast the angle θ is changing.

Since the angle of elevation is the angle between the line of sight (your gaze) and the horizontal plane, we have a right triangle formed between you, the hot air balloon, and a point on the ground directly beneath the balloon. Let's label the right triangle with sides and angles:

/

/ H

/ |

/ |

/ |θ

/______|

D

We have the following trigonometric relationship:

tan(θ) = H / D.

Differentiate both sides with respect to time t:

d(tan(θ)) / dt = d(H / D) / dt.

Using the chain rule, we have:

sec^2(θ) * dθ / dt = (dH / dt * D - H * dD / dt) / D^2.

Now, we need to find expressions for dH/dt (the rate at which the height of the balloon is changing) and dD/dt (the rate at which the distance from you to the balloon is changing).

Given:

dH/dt = 2 ft/s (the rate at which the balloon rises),

D = 500 ft (the distance between you and the balloon remains constant).

We can also find dD/dt using the Pythagorean theorem. When the balloon is at height H, the distance from you to the balloon is given by:

D^2 = H^2 + 500^2.

Differentiate both sides with respect to time t:

2D * dD/dt = 2H * dH/dt.

Now, solve for dD/dt:

dD/dt = (H * dH/dt) / D.

Now, plug the values into the expression we derived earlier:

sec^2(θ) * dθ / dt = (2 * D - H * (H * dH/dt) / D) / D^2.

At the given heights H = 75 ft and H = 150 ft, we can calculate the corresponding angles of elevation θ using the relation tan(θ) = H / D:

When H = 75 ft:

θ = arctan(75/500) ≈ 8.13 degrees.

When H = 150 ft:

θ = arctan(150/500) ≈ 16.26 degrees.

Now, substitute the corresponding values of θ, H, and dH/dt into the expression:

When H = 75 ft:

sec^2(θ) * dθ / dt ≈ (2 * 500 - 75 * (75 * 2)) / (500^2) ≈ -0.0040 rad/s ≈ -0.229 degrees/s.

When H = 150 ft:

sec^2(θ) * dθ / dt ≈ (2 * 500 - 150 * (150 * 2)) / (500^2) ≈ -0.0160 rad/s ≈ -0.918 degrees/s.

So, the rate at which your angle of elevation to the balloon is changing is approximately -0.229 degrees per second when the balloon is 75 feet above the ground and approximately -0.918 degrees per second when the balloon is 150 feet above the ground. The negative sign indicates that your angle of elevation is decreasing as the balloon rises.

User Guy Mograbi
by
8.1k points
2 votes

The rate at which the angle of elevation changes as the hot air balloon lifts off can be determined using trigonometry by creating a right triangle with the height of the balloon and the distance from the observer to the balloon. The derivative of the tangent function of theta with respect to time provides the rate of change of the angle, which can be calculated at any given height such as 75 feet or 150 feet above the ground.

The student's angle of elevation to the hot air balloon can be modeled as a right triangle, where the adjacent side (the horizontal ground distance) is a constant 500 feet.

The opposite side (the height of the balloon) and the angle theta are the varying quantities. As the balloon ascends, we want to find out how fast the angle of elevation theta is changing, which is given by the rate d(theta)/dt.

Using trigonometry, tan(theta) = opposite/adjacent, or in this case, tan(theta) = height/500. To find d(theta)/dt, we take the derivative with respect to time t of both sides, applying the chain rule to the left side (d/dt of tan(theta) = sec^2(theta)*d(theta)/dt) and the right side (d/dt of height/500 is just the rise rate, which is 2ft/s).

At the instant when the balloon is 75 feet above the ground, using the right triangle we have tan(theta) = 75/500.

Plugging that into the differentiated equation and solving for d(theta)/dt, we can find the rate at which theta is changing at that moment. To find the rate of change of theta at 150 feet, we repeat the process with the new height.

User Hkdalex
by
8.8k points