Problem 12 - Calculus Early Transcendentals

Question

Find the mass of the following thin bars with the given density function. $$\rho(x)=5 e^{-2 x} ; \text { for } 0 \leq x \leq 4$$

Step-by-Step Solution

Answer

Answer: The mass of the thin bar with the given density function is approximately \(2.4992\) mass units.

Step 1: Define the mass function and intervals

We start by defining the mass function as the integral of the density function over the interval of interest \([0, 4]\). The mass function is given by: $$m = \int_a^b \rho(x) dx$$ Where \(a = 0\) and \(b = 4\).

Step 2: Insert the given density function and interval

Now we will insert the given density function, \(\rho(x)=5e^{-2x}\), and the given interval, \(0 \leq x \leq 4\), into the mass function: $$m = \int_0^4 5e^{-2x} dx$$

Step 3: Integrate and evaluate the function

To find the mass, integrate the function with respect to \(x\). Use the power rule for integration where the antiderivative of \(e^{cx}\) is \(\frac{1}{c}e^{cx}\). $$m = \left[-\frac{5}{2}e^{-2x}\right]_0^4$$ Now, evaluate the integrated function at the bounds of integration: $$m = -\frac{5}{2}e^{-8} - (-\frac{5}{2}e^{0})$$

Step 4: Simplify the expression and calculate the mass

Simplify the expression and use the properties of exponentials to find the mass: $$m = -\frac{5}{2}e^{-8} + \frac{5}{2}$$ Now, calculate the value for this expression: $$m \approx -\frac{5}{2}(0.000335) + \frac{5}{2} \approx 2.4992$$ The mass of the thin bar with the given density function is approximately \(2.4992\) mass units.

Key Concepts

Density function

A density function in mathematics is a function used to describe the distribution of a quantity throughout a space. In the context of the exercise, the density function is given as \( \rho(x) = 5e^{-2x} \). This means at each point \( x \) along the bar between 0 and 4, the density—essentially how much mass is concentrated—can be calculated using this function.

• The factor of 5 in the density function scales the density overall, giving it a base value before the exponential decay from the \(e^{-2x}\) term reduces it.
• As \( x \) increases, the \( e^{-2x} \) term decreases rapidly, meaning density decreases exponentially along the bar.
• This type of function is often used in real-world physics problems where density isn't uniform, such as in bars or beams where material distribution varies along the length.

Understanding a density function's role is crucial when calculating mass because it lets us know how the distribution of mass affects total weight or mass across an object.

Definite integral

A definite integral is a fundamental concept in calculus used to calculate the accumulation of quantities, such as area under a curve or total mass from a density function. In our exercise, we compute the definite integral of the density function \( \rho(x) = 5e^{-2x} \) over the interval \([0,4]\).

• The notation \( \int_0^4 \rho(x) \, dx \) indicates we are calculating the total mass by integrating from \( x = 0 \) to \( x = 4 \).
• This process is akin to summing up infinitesimally small elements to find a complete quantity, like the exact mass of the bar.
• Evaluating a definite integral involves finding an antiderivative, substituting the upper and lower bounds, and calculating the difference.

Appreciating how a definite integral works helps you understand problems like these because it shows how small, localized densities add up to a complete, integrated value, representing the whole physical property such as mass.

Exponential function

An exponential function is one where the variable appears in the exponent, typically showing rapid growth or decay. In the case of the density function \( \rho(x) = 5e^{-2x} \), the exponential part \( e^{-2x} \) governs how density changes with \( x \) in an exponential decay manner.

• Exponential decay functions like \( e^{-2x} \) decrease rapidly, indicating that as you move further along the bar, the density drops.
• The base \( e \) is a natural constant approximately equal to 2.718, which makes it unique as rates calculated with \( e \) are continuously compounding.
• Exponential functions are vital in modeling processes that change multiplicatively—for instance, population growth, radioactive decay, and here, decreasing density along an object's length.
• In integration, knowing how to handle exponential functions can greatly help in calculating precise areas under curves—significantly simplifying the computation of mass as was exhibited in the steps.

So, grasping how exponential functions operate within densities illustrates how physical properties like mass can vary non-linearly and be precisely quantified.