Problem 12 - The Calculus with Analytic Geometry

Question

The total mass of a rod of length \(L \mathrm{ft}\) is \(M\) slugs and the measure of the linear density at a point \(x \mathrm{ft}\) from the left end is proportional to the measure of the distance of the point from the right end. Show that the linear density at a point on the rod \(x \mathrm{ft}\) from the left end is \(2 M(L-x) / L^{2}\) slugs \(/ \mathrm{ft}\).

Step-by-Step Solution

Answer

\( \rho(x) = \frac{2M(L - x)}{L^2} \)

Step 1: Understand the linear density function

The linear density \(\rho(x)\) at point \(x\) from the left end is proportional to the distance from the right end \(L - x\). Thus, \(\rho(x) = k (L - x)\), where \(k\) is a constant of proportionality.

Step 2: Express the total mass

The total mass \(M\) of the rod is the integral of the linear density over the length of the rod. This gives: \[M = \int_0^L \rho(x) \, dx = \int_0^L k(L - x) \, dx\]

Step 3: Perform the integration

Compute the integral: \[M = k \int_0^L (L - x) \, dx = k \left[ Lx - \frac{x^2}{2} \right]_0^L = k \left[ L^2 - \frac{L^2}{2} \right] = k \frac{L^2}{2}\]

Step 4: Solve for the constant \(k\)

From the integral result, solve for \(k\): \[M = k \frac{L^2}{2} \Rightarrow k = \frac{2M}{L^2} \]

Step 5: Write the linear density function

Substitute \(k \) back into the density function: \[ \rho(x) = k(L - x) = \frac{2M}{L^2}(L - x) = \frac{2M(L - x)}{L^2} \]

Key Concepts

Proportionality in Calculus

Proportionality is a fundamental concept in calculus and related fields. It signifies that one quantity varies directly as another. In our exercise, this concept helps us describe how the linear density of a rod changes over its length. The linear density, \(\rho(x)\), at any point \(x\) on the rod, is proportional to the distance from the right end. This relation can be expressed mathematically as \(\rho(x) = k (L - x)\), where \(k\) is a constant of proportionality. Here, \(L\) is the total length of the rod. The distance from the right end is \(L - x\). This equation tells us that as \(x\) increases (moving from the left to the right end), the linear density decreases proportionally.

Integral of Linear Density

In this problem, we are given the total mass of the rod, which is \(M\). To find the total mass, we integrate the linear density function over the length of the rod. This integral is given by \[ M = \int_0^L \rho(x) \, dx. \] We replace \(\rho(x)\) with \(k (L - x)\) (from our proportionality equation) to get: \[ M = \int_0^L k (L - x) \, dx. \] Performing the integration yields: \[ M = k\left[ Lx - \frac{x^2}{2} \right]_0^L = k \left[ L^2 - \frac{L^2}{2} \right]. \] Simplifying this, we have: \[ M = k \frac{L^2}{2}. \] Integrating the density over the rod’s length converts a linear function of density into the total mass.

Solving for Constants

To finalize our solution, we need to solve for the constant of proportionality, \(k\). From our integrated result, we know: \[ M = k \frac{L^2}{2}. \] Rearranging to solve for \(k\) gives: \[ k = \frac{2M}{L^2}. \] Now that we have found \(k\), we substitute it back into our original linear density function. Doing this yields: \[ \rho(x) = k (L - x) = \frac{2M}{L^2} (L - x). \] Thus, the linear density at any point \(x\) from the left end is: \[ \rho(x) = \frac{2M (L - x)}{L^2} \. \] This results from incorporating the constraint that the total mass is \(M\) and simplifying the relationship between density and distance.