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Chapter 6: Problem 50

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.

Short Answer

Expert verified
Answer: Bar 1 is heavier than bar 2 for values of L such that \(0<e^{-L}<\frac{1-\sqrt{\frac{1}{3}}}{2}\) or \(\frac{1+\sqrt{\frac{1}{3}}}{2}<e^{-L}<1\). The masses of the bars do not increase without bound as their lengths increase, because the exponential terms in their density functions go to 0.

Step by step solution

01

1. Find the mass of bar 1 for any L

To find the mass of bar 1, integrate the density function \(\rho_{1}(x)=4 e^{-x}\) along its length \(0 \leq x \leq L\). \(\text{Mass}_{1}(L) = \int_{0}^{L} 4 e^{-x} dx\) Now, integrate the function: \(\text{Mass}_{1}(L) = -4\left[e^{-x}\right]_{0}^{L} = -4(e^{-L} - 1)\)
02

2. Find the mass of bar 2 for any L

Similarly, to find the mass of bar 2, integrate the density function \(\rho_{2}(x) = 6 e^{-2x}\) over the length \(0 \leq x \leq L\). \(\text{Mass}_{2}(L) = \int_{0}^{L} 6 e^{-2x} dx\) Now, integrate the function: \(\text{Mass}_{2}(L) = -3\left[e^{-2x}\right]_{0}^{L} = -3(e^{-2L} - 1)\)
03

3. Compare the masses to find the values of L for which bar 1 is heavier

We want to find the values of \(L\) such that \(\text{Mass}_{1}(L) > \text{Mass}_{2}(L)\). Plug in the expressions for each mass: \(-4(e^{-L} - 1) > -3(e^{-2L} - 1)\) Simplify the inequality: \(e^{-2L}-e^{-L} < \frac{1}{3}\) Let \(y = e^{-L}\), so the inequality becomes: \(y^2 - y < \frac{1}{3}\) Add \(\frac{1}{3}\) to both sides of the inequality: \(y^2 - y + \frac{1}{3} < 1\) This inequality represents a quadratic function with a positive leading coefficient, which means it is a parabola opening upwards. We are interested in the range of \(y\) values for which the inequality holds true. Since we let \(y = e^{-L}\), it's important to note that the range of the exponential function is \((0,1)\). Therefore, \(0<y<1\), and we need to find where the inequality holds for this range. The roots of the quadratic equation \(y^2 - y + \frac{1}{3} = 0\) can be found using the quadratic formula: \(y = \frac{1 \pm \sqrt{1^2 - 4\left(\frac{1}{3}\right)}}{2}\) \(y = \frac{1 \pm \sqrt{\frac{1}{3}}}{2}\) Note that the roots are within the range \((0,1)\). So, the inequality holds true for \(L\) values that correspond to \(y\) values in the range: \(0<e^{-L}<\frac{1-\sqrt{\frac{1}{3}}}{2}\) or \(\frac{1+\sqrt{\frac{1}{3}}}{2}<e^{-L}<1\)
04

4. Determine if the masses increase without bound

As \(L\) increases, the exponential term in both mass integrals goes to 0. Therefore, both bars have finite masses and their masses do not increase without bound. In conclusion, for values of \(L\) such that \(0<e^{-L}<\frac{1-\sqrt{\frac{1}{3}}}{2}\) or \(\frac{1+\sqrt{\frac{1}{3}}}{2}<e^{-L}<1\), bar 1 is heavier than bar 2. As the lengths of the bars increase, their masses do not increase without bound, because the exponential terms in their density functions go to 0.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Density Functions
Density functions are a fundamental concept in calculus, especially when dealing with physical applications such as mass distribution. These functions represent the density of a substance at a given point. For a one-dimensional object like a bar, the density function \( \rho(x) \) gives the mass per unit length at a point \( x \). It's crucial to note that the density can vary along the length of the object.For example, in the exercise, the density functions \( \rho_{1}(x)=4e^{-x} \) and \( \rho_{2}(x)=6e^{-2x} \) describe how the density of each bar changes along their length, indicating that both bars have densities that decrease exponentially from one end to the other. Understanding how to work with these functions is key to solving problems related to mass and distribution.
Definite Integrals
The definite integral of a function represents the signed area under the curve between two limits. In practical scenarios like the given exercise, the definite integral is used to calculate total quantities, such as the mass of an object with a variable density.By integrating the density function over the length of the object, we obtain the total mass. This is what's happening when we calculate \( \text{Mass}_{1}(L) \) and \( \text{Mass}_{2}(L) \) by integrating their respective density functions from \( 0 \) to \( L \) as shown in the steps. The concept of definite integrals is a pillar in calculus and appears in various real-world applications.
Exponential Functions
Exponential functions are characterized by a constant base raised to a variable exponent, typically written as \( a^x \). These functions grow or decay at rates proportional to their current value, which makes them extremely important in describing processes like radioactive decay or population growth.In our case, the exercise features the exponential functions \( e^{-x} \) and \( e^{-2x} \) in the density expressions. Such decreasing exponential functions approach zero as \( x \) tends to infinity. The bases of these functions are irrational numbers approximately equal to \( 2.718 \) (Euler's number, \( e \) ), which is a fundamental constant in mathematics.
Mass Comparison
Comparing masses involves finding relationships between different mass distributions, a common task in physics and engineering. In our exercise, we're tasked with comparing the masses of two bars of equal length but with different exponential density functions.By calculating the definite integrals of the density functions, we determine the mass for each bar over its length. We then set up an inequality to compare the masses. Understanding how to properly set up and solve these inequalities is critical when comparing quantities that vary across different conditions or objects.
Limit of a Function
The concept of the limit of a function is critical in calculus, and it concerns what value a function approaches as the input approaches a certain point. Limits are the foundation of calculus, as they are required to formally define both the derivative and the integral.Referring to the given exercise, we analyze the limit of the exponential functions involved in the mass calculation of the bars as the length \( L \) increases. Specifically, we observe in Step 4 that as \( L \) increases without bound, the exponential terms \( e^{-L} \) and \( e^{-2L} \) approach zero, thus affecting the limits of the mass of the bars. In this case, it indicates that the masses of the bars do not increase without bound despite increasing lengths, due to the properties of exponential decay within the density functions.

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Most popular questions from this chapter

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