Calculating a Fixed Distance

Benjamin Gallai and Khizer Shahid

Problem Statement

A particle \(\mathcal{P}\) is initially 1 unit away from \(\mathcal{X}\) its destination. On its path to \(\mathcal{X}\), \(\mathcal{P}\) has a velocity of \(d + 1\) units per second, where \(d\) is its distance to \(\mathcal{X}\). How long will it take for \(\mathcal{P}\) to reach \(\mathcal{X}\)?

Solution

Illustration of problem statement.

First, we will partition the interval of length \(1\) into \(n\) equally sized interval. We will assume that \(\mathcal{P}\) changes its velocity at the endpoints. Then, we may take the limit of our result as \(n \xrightarrow{} \infty\). This will allow us to simplify the problem.

Let \(d_{i}\) be the distance traveled by \(\mathcal{P}\) during the \(i\)th interval, \(v_{i}\) be its velocity at the beginning of this interval, and \(t_{i}\) be the amount of time it spends in the interval. Additionally, let \(T_{n} = t_1 + t_2 + t_3 + \cdots + t_n\) be the total amount of time that it takes \(\mathcal{P}\) to reach \(\mathcal{X}\) in a configuration with \(n\) subintervals. As we add more subintervals, the velocity will become closer to changing continuously. Hence, the total amount of time \(\mathcal{P}\) takes is \(T = \displaystyle\lim_{n\xrightarrow{} \infty} T_n\).

Now, we must compute \(T_n\) in terms of \(n\). First, we have \(T_n = \displaystyle\sum_{k = 1}^{n} t_k\). From \(d=rt\), we may conclude that \(t_k = \frac{d_k}{v_k}\). Additionally, \(d_k = \frac{1}{n}\) for all \(k\) and \(v_k = 2-\frac{k-1}{n}\) since at the \(k\)th interval it has travelled \(k-1\) subintervals of length \(\frac{1}{n}\). Hence, \[t_k = \frac{\frac{1}{n}}{2-\frac{k-1}{n}} = \frac{1}{2n-k+1} \Rightarrow T_n = \sum_{k = 1}^{n} \frac{1}{2n-k+1} = \frac{1}{2n} + \frac{1}{2n-1} + \frac{1}{2n-2} + \cdots + \frac{1}{n + 2} + \frac{1}{n+1}.\] We may further rewrite this as

\[\begin{aligned} \frac{1}{2n} + \cdots + \frac{1}{n+1} &= \left ( \frac{1}{2n} + \frac{1}{2n-1} + \cdots + \frac{1}{2} + \frac{1}{1} \right) - \left ( \frac{1}{n} + \frac{1}{n-1} + \cdots + \frac{1}{2} + \frac{1}{1}\right) \\ &= \left [ \left (\frac{1}{2n} + \frac{1}{2n-1}\right) + \left (\frac{1}{2n-2} + \frac{1}{2n-3}\right) +\cdots + \left (\frac{1}{2} + \frac{1}{1}\right)\right] - \sum_{i = 1}^{n} \frac{1}{i} \\ &= \sum_{i = 1}^{n} \left (\frac{1}{2i} + \frac{1}{2i-1}\right)- \sum_{i = 1}^{n} \frac{1}{i} \\ &= \sum_{i = 1}^{n} \left (\frac{1}{2i} + \frac{1}{2i-1} - \frac{1}{i}\right) \\ &= \sum_{i = 1}^{n} \left (\frac{1}{2i-1} - \frac{1}{2i}\right) \\ &= \left (\frac{1}{1} - \frac{1}{2} \right) + \left (\frac{1}{3} - \frac{1}{4} \right) + \cdots + \left (\frac{1}{2n-3} - \frac{1}{2n-2} \right) + \left (\frac{1}{2n-1} - \frac{1}{2n} \right) \\ &= 1-\frac{1}{2} + \frac{1}{3} - \cdots - \frac{1}{2n-2} + \frac{1}{2n-1} - \frac{1}{2n}. \end{aligned}\]

Hence, \(T = \displaystyle\lim_{n\xrightarrow{} \infty} T_n = 1 - \frac{1}{2} + \frac{1}{3} - \frac{1}{4} + \frac{1}{5} + \cdots\). However, this is not a closed form. To find the closed form, we turn to calculus.

Let the total distance \(\mathcal{P}\) has traveled at time \(t\) be \(x(t)\) and its velocity be \(v(t)\). By the problem statement, \(v(t) = (1 - x(t)) + 1 = 2 - x(t)\). Since \(x(t)\) is \(\mathcal{P}\)’s position and the rate of change of position is velocity, we have \(v(t) = x'(t)\). Now, we can simply solve for \(x(t)\), the position function, by solving the differential equation \(\frac{dx}{dt} = 2 - x\). However, we will take a different approach.

Differentiating both sides of \(v(t) = 2-x(t)\) with respect to \(t\), we get \(a(t) = -v(t)\), where \(a(t)\) is the acceleration of \(\mathcal{P}\) at time \(t\). Additionally, we have that \(a(t) = v'(t)\). Plugging this back into \(a(t) = -v(t)\), we have \(v'(t) = -v(t)\). Now, we have the differential equation \(\frac{dv}{dt} = -v\). Separating the variables and integrating both sides, we have

\[\begin{aligned} \int \frac{dv}{v} &= -\int dt \\ \Rightarrow \ln |v| = \ln v (t) &= -t + C\\ \Rightarrow v(t) = e^{-t+C} &= Ae^{-t}\end{aligned}\]

for some \(A \in \mathbb{R} ^{+}\). Note that we can remove the absolute values since \(v(t) \ge 1 > 0\). We can solve for \(A\) by plugging in \(t = 0\). Since at time \(t = 0\), \(\mathcal{P}\) is 1 unit away from \(\mathcal{X}\), its velocity is \(2\) units per second: \(v(0) = A = 2\). Hence, \(v(t) = 2e^{-t}\). Let \(\mathcal{P}\) reach \(\mathcal{X}\) after \(k\) seconds. We know that \(v(k) = 2e^{-k} = 1\). Hence, we can solve for \(k\): \[2e^{-k} = 1 \Rightarrow e^{-k} = \frac{1}{2} \Rightarrow -k = \ln \frac{1}{2} \Rightarrow k=\ln 2.\] Hence, \(\mathcal{P}\) will reach \(\mathcal{X}\) in \(\ln 2\) seconds.

Since, our answers to the problems must be the same, we have \[1 - \frac{1}{2} + \frac{1}{3} - \frac{1}{4} + \frac{1}{5} + \cdots = \ln 2\]

Generalization

Now, instead of \(\mathcal{P}\) starting \(1\) unit away from \(\mathcal{X}\), let it be \(d\) units away.

From the problem statement we have \(v(t) = (d-x(t)) + 1 = d+1-x(t)\), Taking derivative with respect to \(t\), we have \(a(t) = -v(t)\). This is the exact same equation we solved earlier. We found that \(v(t) = Ae^{-t}\) for some \(A \in \mathbb{R}\). We can analyze what happens at \(t = 0\): \(x(0) = d \Rightarrow v(0) = A = d+1\). Hence, \(v(t) = (d+1)e^{-t}\). Solving the equation \(v(t) = 1\), we have \(t = \ln(d+1)\). Hence, \(\mathcal{P}\) will reach \(\mathcal{X}\) in \(\ln(d+1)\) seconds. Plugging in \(d = 1\) from the initial problem agrees with our previous result.