Problem 48 - Calculus Early Transcendentals

Question

Two bars of length \(L\) have densities \(\rho_{1}(x)=4 e^{-x}\) and \(\rho_{2}(x)=6 e^{-2 x},\) for \(0 \leq x \leq L\) a. For what values of \(L\) is bar 1 heavier than bar \(2 ?\) b. As the lengths of the bars increase, do their masses increase without bound? Explain.

Step-by-Step Solution

Answer

#Answer# a) For bar 1 to be heavier than bar 2, the length \(L\) must be less than \(-\ln{(\frac{1+\sqrt{5}}{2})}\). b) The masses of the bars do not increase without bound as the lengths increase. The masses have limits: the mass of bar 1 approaches 4 and the mass of bar 2 approaches 3 as the lengths increase.

Step 1: Calculate the mass of each bar

To calculate the mass of each bar, we integrate the density function over the length of the bar (from \(0\) to \(L\)): Mass of bar 1: \(m_{1}(L) = \int_{0}^{L} \rho_{1}(x) dx = \int_{0}^{L} 4e^{-x} dx\) Mass of bar 2: \(m_{2}(L) = \int_{0}^{L} \rho_{2}(x) dx = \int_{0}^{L} 6e^{-2x} dx\) Now, we integrate each function: For mass of bar 1: \(m_{1}(L) = [-4e^{-x}]_{0}^{L}= -4e^{-L} - (-4e^{0})\) \(m_{1}(L) = 4(1-e^{-L})\) For mass of bar 2: \(m_{2}(L) = [-3e^{-2x}]_{0}^{L}= -3e^{-2L} - (-3e^{0})\) \(m_{2}(L) = 3(1-e^{-2L})\)

Step 2: Compare the masses of the bars

To find when bar 1 is heavier than bar 2, we need to find the values of \(L\) when \(m_{1}(L) > m_{2}(L)\): \(4(1-e^{-L}) > 3(1-e^{-2L})\) Now, we can reorganize the inequality: \(4 - 4e^{-L} > 3 - 3e^{-2L}\) Moving all terms to one side, we get: \(e^{-2L} - e^{-L} > 1\) Let's substitute \(u = e^{-L}\) to make this inequality easier to work with. Therefore, \(u^2 - u > 1\) Rearrange the inequality into a quadratic inequality: \(u^2 - u - 1 > 0\) At this point, we can find the real roots of the equation if any: \(u^2 - u - 1 = 0\) Using the quadratic formula, we find that the roots are \(u = \frac{1 \pm \sqrt{5}}{2}\). Since we want the inequality \(u^2 - u - 1 > 0\), we need to choose the range of \(u\) where the quadratic is positive. This is when \(u < \frac{1-\sqrt{5}}{2}\) or \(u > \frac{1+\sqrt{5}}{2}\). Now, recall our substitution: \(u = e^{-L}\). We can now revert back to \(L\) and present the answer: \(L < - \ln{(\frac{1+\sqrt{5}}{2})}\) for bar 1 to be heavier than bar 2.

Step 3: Analyze mass behavior as length increases

To determine if the masses of the bars increase without bound when the lengths increase, we will analyze the mass functions as \(L \to \infty\): For mass of bar 1: \(\lim_{L \to \infty} m_1(L) = \lim_{L \to \infty} 4(1-e^{-L}) = 4\) For mass of bar 2: \(\lim_{L \to \infty} m_2(L) = \lim_{L \to \infty} 3(1-e^{-2L}) = 3\) As the lengths of the bars increase, their masses do not increase without bound. In fact, their masses reach a limit of \(4\) and \(3\) respectively.