Question

Kelly started at noon \((t=0)\) riding a bike from Niwot to Berthoud, a distance of \(20 \mathrm{km},\) with velocity \(v(t)=15 /(t+1)^{2}\) (decreasing because of fatigue). Sandy started at noon \((t=0)\) riding a bike in the opposite direction from Berthoud to Niwot with velocity \(u(t)=20 /(t+1)^{2}\) (also decreasing because of fatigue). Assume distance is measured in kilometers and time is measured in hours. a. Make a graph of Kelly's distance from Niwot as a function of time. b. Make a graph of Sandy's distance from Berthoud as a function of time. c. How far has each person traveled when they meet? When do they meet? d. More generally, if the riders' speeds are \(v(t)=A /(t+1)^{2}\) and \(u(t)=B /(t+1)^{2}\) and the distance between the towns is \(D,\) what conditions on \(A, B,\) and \(D\) must be met to ensure that the riders will pass each other? e. Looking ahead: With the velocity functions given in part (d), make a conjecture about the maximum distance each person can ride (given unlimited time).

Short answer

Question: Create a graph of Kelly's distance from Niwot as a function of time based on the given distance function: \(d_{K}(t) = -\frac{15}{t+1} + 15\).

Step-by-step solution

Integrate velocity functions for distances

To find the distance traveled by Kelly and Sandy, integrate their respective velocity functions with respect to time. For Kelly: \(d_{K}(t) = \int v(t) \, dt = \int \frac{15}{(t+1)^2} \, dt\) For Sandy: \(d_{S}(t) = \int u(t) \, dt = \int \frac{20}{(t+1)^2} \, dt\)

Calculate distances for Kelly and Sandy

Integrate the functions to find the distances: For Kelly: \(d_{K}(t) = -\frac{15}{t+1} + C_{1}\) For Sandy: \(d_{S}(t) = -\frac{20}{t+1} + C_{2}\) To find \(C_{1}\) and \(C_{2}\), use the initial conditions, at \(t=0\), \(d_{K}(0)=0\), and \(d_{S}(0) = 20\). This results in: \(C_{1} = 15\) and \(C_{2} = 40\). So, the distance functions are: \(d_{K}(t) = -\frac{15}{t+1} + 15\) \(d_{S}(t) = -\frac{20}{t+1} + 40\) #a. Make a graph of Kelly's distance from Niwot as a function of time.#

Key concepts

Distance and Velocity

In calculus, distance and velocity are closely related concepts. Velocity measures how quickly something is moving, while distance measures how much ground it has covered. In our example, Kelly and Sandy are riding bikes between two towns. Their velocity functions describe their speeds over time as they fatigue. Kelly's velocity function is given by \(v(t) = \frac{15}{(t+1)^2}\), which illustrates a decrease in speed over time. Similarly, Sandy's velocity function is \(u(t) = \frac{20}{(t+1)^2}\). This means:

• Velocity combines speed and direction.
• The functions show how their speeds decrease as time progresses, modeled by the inverse square of 't+1'.

Distance is then determined by integrating these velocity functions, helping us understand how far each person has traveled at any point in time. This direct relationship between velocity and distance is a key aspect of calculus problem solving in physics and engineering contexts.

Integration in Calculus

Integration is a fundamental concept in calculus that helps us find accumulations like area under a curve or, in this case, the total distance traveled. To understand the distances Kelly and Sandy have biked, we integrate their velocity functions. For Kelly, integrate the velocity function \(v(t) = \frac{15}{(t+1)^2}\), which leads to:\[d_{K}(t) = \int \frac{15}{(t+1)^2} \, dt = -\frac{15}{t+1} + C_{1} \]For Sandy, integrate \(u(t) = \frac{20}{(t+1)^2}\):\[d_{S}(t) = \int \frac{20}{(t+1)^2} \, dt = -\frac{20}{t+1} + C_{2}\]The constants \(C_{1}\) and \(C_{2}\) are calculated using initial conditions, showing where each person starts:

• For Kelly: Starts at Niwot, so \(d_{K}(0) = 0\), therefore \(C_{1} = 15\).
• For Sandy: Starts 20 km from Berthoud, so \(d_{S}(0) = 20\), therefore \(C_{2} = 40\).

Thus, integration transforms the rate of change (velocity) into a quantity traveled (distance).

Time-Dependent Functions

In this scenario, both Kelly's and Sandy's rides are described using time-dependent functions. Time-dependent functions, such as velocity functions, allow us to model real-life phenomena where quantities change over time.A time-dependent function means the output of the function changes as time changes. Here, the velocity functions:

• Kelly's function: \(v(t) = \frac{15}{(t+1)^2}\)
• Sandy's function: \(u(t) = \frac{20}{(t+1)^2}\)

illustrate that both riders slow down over time due to fatigue, commonly seen in physical activities. The behavior and analysis of these functions are essential in predicting when and where the riders meet, as well as answering questions about their travels.

Graphing Functions

Graphing functions is a powerful way to visualize relationships between variables like time, distance, and velocity. In this exercise, graphing can help us better understand how Kelly's and Sandy's distances change over time.Consider Kelly's distance function from Niwot:\[d_{K}(t) = -\frac{15}{t+1} + 15\]and Sandy's distance function from Berthoud:\[d_{S}(t) = -\frac{20}{t+1} + 40\]Plotting these functions enables us to see:

• How each rider's distance from their starting point decreases over time.
• Where and when their travel paths intersect, indicating when they meet.

Through the graph, we can visually capture the dynamics of the ride, such as the slowing rate which flattens the curve over time. Such a visual tool is crucial for deeper and practical understanding of time-dependent calculus functions.