Problem 32 - Applied Mathematics: For the Managerial, Life, and Social Sciences

Question

A hot-air balloon rises vertically from the ground so that its height after \(t\) sec is \(h=\frac{1}{2} t^{2}+\frac{1}{2} t \mathrm{ft}(0 \leq t \leq 60)\). a. What is the height of the balloon at the end of \(40 \mathrm{sec}\) ? b. What is the average velocity of the balloon between \(t=0\) and \(t=40\) ? c. What is the velocity of the balloon at the end of \(40 \mathrm{sec} ?\)

Step-by-Step Solution

Answer

a. At 40 seconds, the hot-air balloon's height is 820 ft. b. The average velocity of the balloon between \(t=0\) and \(t=40\) is \(20.5\, \frac{ft}{s}\). c. The velocity of the balloon at the end of 40 seconds is \(40.5\, \frac{ft}{s}\).

Step 1: (a) Finding the height of the balloon at t=40 sec

Plug in 40 as the value of t in the height function to get the height at 40 seconds. \(h(40) = \frac{1}{2}(40)^2 + \frac{1}{2}(40) = 820 ft\) At 40 seconds, the hot-air balloon's height will be 820 feet.

Step 2: (b) Average Velocity of the Balloon

The average velocity (Vave) can be calculated using the formula: \(V_{ave} = \frac{h(40) - h(0)}{40 - 0}\) First, let's find \(h(0)\): \(h(0) = \frac{1}{2}(0)^2 + \frac{1}{2}(0) = 0\) Now calculate the average velocity: \(V_{ave} = \frac{820 - 0}{40} = 20.5\, \frac{ft}{s}\) The average velocity of the balloon between \(t=0\) and \(t=40\) is \(20.5\, \frac{ft}{s}\).

Step 3: (c) Finding the Velocity of the Balloon at t=40 sec

First, we need to find the velocity function, which is the derivative of the height function. The height function: \(h(t) = \frac{1}{2}t^2 + \frac{1}{2}t\). Taking the derivative we have: \(v(t) = \frac{d}{dt} \left(\frac{1}{2}t^2 + \frac{1}{2}t\right) = t + \frac{1}{2}\). Now, plug in 40 as the value of t in the velocity function: \(v(40) = 40 + \frac{1}{2} = 40.5\, \frac{ft}{s}\) The velocity of the balloon at the end of 40 seconds is \(40.5\, \frac{ft}{s}\). To summarize: a. The height of the balloon at \(t=40\) is 820 ft. b. The average velocity of the balloon between \(t=0\) and \(t=40\) is \(20.5\, \frac{ft}{s}\). c. The velocity of the balloon at the end of \(t=40\) is \(40.5\, \frac{ft}{s}\).

Key Concepts

Velocity Calculation

Velocity describes how fast something is moving. More specifically, it shows both speed and direction. In our exercise, we talk about the velocity of a hot-air balloon rising. There are two kinds of velocity we're interested in: average velocity and instantaneous velocity.
To calculate the velocity, we sometimes use derivatives of a function. This will help us understand how quick the motion is at a specific point in time.

Derivative

A derivative is a powerful tool in calculus. It tells us the rate at which a function is changing at any given point. By finding the derivative of a function, we can determine instantaneous rates of change, like velocity.

• In our balloon example, we have a height function given as \( h(t) = \frac{1}{2}t^2 + \frac{1}{2}t \).


• To find the velocity at a specific time, we compute the derivative of this function, resulting in \( v(t) = t + \frac{1}{2} \).

The derivative thus becomes a new function which we can evaluate at any point to find how swiftly the balloon is rising right then.

Average Velocity

Average velocity gives us the overall change in position over a range of time. It's like calculating the average speed of a journey. You take the total distance covered and divide it by the total time taken.
For the hot-air balloon, the formula for average velocity is:
\[ V_{ave} = \frac{h(40) - h(0)}{40 - 0} \]

• Find \( h(0) \) by substituting 0 into the height function, giving us \( h(0) = 0 \).


• Next, calculate \( h(40) \), or the height after 40 seconds, which is 820 feet.


• Therefore, \( V_{ave} = \frac{820 - 0}{40} = 20.5 \frac{ft}{s} \).

This tells us, on average, the balloon travels 20.5 feet every second over the 40-second interval.

Height Function

The height function helps us understand how the height of an object changes over time. It is a mathematical expression which describes how high above the ground something is at different moments.

• In our problem, the height of the balloon is given by \( h(t) = \frac{1}{2}t^2 + \frac{1}{2}t \).


• This function shows that the height increases as time goes on, which makes sense for a balloon rising from the ground.


• To find how high the balloon is at any specific time, simply plug that time into the function.

This specific height function shows both a parabolic and linear growth, reflecting the balloon's accelerating ascent.