Problem 56 - Calculus Early Transcendentals

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. When do they meet? How far has each person traveled when 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 \(\vec{D},\) 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).

Step-by-Step Solution

Answer

Based on the given velocity functions for Kelly and Sandy, we found their position functions and determined that they meet at 4/3 hours or 1 hour and 20 minutes, having traveled 10 kilometers each. Additionally, we discussed conditions for riders to pass each other given general velocity functions and made a conjecture about the maximum distance each person can ride based on those general functions.

Step 1: Integrate Kelly's velocity function to find position function

To find the position function \(K(t)\) for Kelly (distance from Niwot as a function of time), integrate the velocity function \(v(t) = \frac{15}{(t+1)^2}\) with respect to \(t\). Apply integration: \(K(t) = \int v(t) dt = \int \frac{15}{(t+1)^2} dt\) Let \(u = t+1\) and \(du = dt\). Then the integral becomes: \(K(t) = 15\int \frac{1}{u^2} du = -15\frac{1}{u} + C\) Now substitute back \(u = t+1\) and determine the constant \(C\) using the fact that Kelly started at Niwot at \(t=0\): \(K(0) = -15\frac{1}{(0+1)} + C = 0\) Thus, \(C=15\), and the position function for Kelly is: \(K(t) = -15\frac{1}{t+1} + 15\)

Step 2: Integrate Sandy's velocity function to find position function

To find the position function \(S(t)\) for Sandy (distance from Berthoud as a function of time), integrate the velocity function \(u(t) = \frac{20}{(t+1)^2}\) with respect to \(t\). Apply integration: \(S(t) = \int u(t) dt = \int \frac{20}{(t+1)^2} dt\) Again let \(u = t+1\) and \(du = dt\). The integral becomes: \(S(t) = 20\int \frac{1}{u^2} du = -20\frac{1}{u} + C\) Now substitute back \(u = t+1\) and determine the constant \(C\) using the fact that Sandy started at Berthoud at \(t=0\): \(S(0) = -20\frac{1}{(0+1)} + C = 20\) Thus, \(C=40\), and the position function for Sandy is: \(S(t) = -20\frac{1}{t+1} + 40\)

Step 3: Graph the position functions

Using a graphing software or calculator, graph both position functions \(K(t) = -15\frac{1}{t+1} + 15\) and \(S(t) = -20\frac{1}{t+1} + 40\) to visualize the distance traveled by Kelly and Sandy with respect to time. The graphs will illustrate parts a and b of the exercise.

Step 4: Determine when Kelly and Sandy meet

To find when they meet, we need to find when the sum of their distances from the starting points is equal to the distance between the towns, i.e., \(K(t) + S(t) = 20\). Add their position functions: \(-15\frac{1}{t+1} + 15 -20\frac{1}{t+1} + 40 = 20\) Simplify and solve for \(t\): \(-35\frac{1}{t+1} + 35 = 20\) \(\frac{1}{t+1} = \frac{3}{7}\) \(t+1 = \frac{7}{3}\) \(t = \frac{4}{3}\) So, Kelly and Sandy meet when \(t = \frac{4}{3}\) hours.

Step 5: Determine how far each person has traveled when they meet

To find the distance traveled by Kelly and Sandy when they meet, plug \(t = \frac{4}{3}\) into their position functions: \(K\left(\frac{4}{3}\right) = -15\frac{1}{\left(\frac{4}{3}+1\right)} + 15 = 10\) \(S\left(\frac{4}{3}\right) = -20\frac{1}{\left(\frac{4}{3}+1\right)}+40 = 10\) Thus, when Kelly and Sandy meet, each has traveled \(10\) kilometers.

Step 6: Determine conditions for riders to pass each other

Given the general velocity functions \(v(t) = \frac{A}{(t+1)^2}\) and \(u(t) = \frac{B}{(t+1)^2}\) and distance between towns \(D\), we need to find conditions on \(A, B,\) and \(D\) such that riders will pass each other. Integrate the general velocity functions to get the position functions: \(K(t) = -A\frac{1}{t+1} + A\) \(S(t) = -B\frac{1}{t+1} + B + D\) For them to pass each other, their combined distance should be equal to \(D\): \(K(t) + S(t) = D\) \(-A\frac{1}{t+1} + A -B\frac{1}{t+1} + B + D = D\) Now, we need to ensure that \(A - B\frac{1}{t+1}\) must be always non-negative for some \(t\). Otherwise, their combined distance can never be equal to \(D\). There is no simple expression for this condition, but this observation will be helpful when dealing with specific cases.

Step 7: Make a conjecture about the maximum distance each person can ride

Given the general velocity functions from part d, it's observed that as time approaches infinity, the riders' velocities approach zero. This suggests that they will eventually stop moving. We can conjecture that the maximum distance each person can ride (given unlimited time) is proportional to the ratio between their respective constants \(A\) and \(B\). Another approach is to integrate the velocity functions from \(0\) to \(\infty\) and find the maximum distance assuming the conditions from step 6 are met. Keep in mind that these conjectures are not proven and may not hold for all cases.

Key Concepts

Position Function

When we talk about a position function, we're discussing a mathematical expression that specifies the location of an object at any given time. In the case of our cyclists, Kelly and Sandy, the position functions represent how far each has traveled from their respective starting points as a function of time.

To arrive at these functions, one must integrate the velocity functions. In calculus, finding the position from velocity is a classic application of integration, since velocity is the derivative of position with respect to time. Consequently, integrating velocity with respect to time allows us to retrieve the position. Keep in mind, while integrating, it is crucial to consider initial conditions to determine the integration constant, so that the position function accurately reflects the starting point of the journey.

Definite Integral

A definite integral is a cornerstone concept in calculus and a key tool in finding areas, volumes, and—in our cycling scenario—the total distance traveled over a period of time. When we compute the definite integral of a velocity function over a specific interval, it gives us the change in position (or displacement) of the cyclist over that interval. In simple terms, the definite integral of velocity from the start time to any other time provides the accumulated distance covered by that time. This is represented mathematically as the area under the velocity-time curve between two points, corresponding to the start and end times of the interval considered.

Velocity-Time Graph

Analyzing a velocity-time graph provides a visual insight into how the speed of an object changes over time. For our cyclists, plotting their decreasing velocity due to fatigue would show a curve that gets closer and closer to the horizontal axis as time goes on, reflecting their slowing speeds. These graphs are especially useful in understanding the relationship between velocity and time, as the slope at any point on a position-time graph would correspond to the velocity at that instance. Furthermore, the area under the curve on a velocity-time graph up to any point gives the total distance traveled up to that time, which is exactly how we determine the position functions by integrating the velocity functions.

Related Rates

Related rates are about finding the rate at which one quantity changes with respect to another, especially when the relationship between the two is given by an equation. In the context of our cyclists' problem, we're dealing with the rates at which Kelly and Sandy approach each other. Both their velocities can be expressed as functions of time, which decrease over time due to fatigue. By integrating these functions, we obtain their position functions, which can then be used to solve for the instances at which the cyclists meet—one of the classic applications of related rates in calculus.

These concepts all tie together, demonstrating how the integration of velocity functions can lead to determining the position, visualizing this relationship through graphs, and ultimately solving real-world problems like measuring how far two cyclists have traveled over time.