Problem 6 - Calculus: An Integrated Approach to Functions and Their Rates of Change

Question

The density of dart holes on an old dartboard is given by \(\rho(r)=\frac{1010}{\pi\left(r^{2}+1\right)^{2}}\) holes per square inch, where \(r\) is the distance, in inches, from the center of the board. If the board is a circle with diameter 20 inches, find the total number of holes in the board.

Step-by-Step Solution

Answer

The total number of holes in the dartboard is approximately 6257.

Step 1: Define the Integral

The total number of darts in the board is given by the integral of the density function over the area of the board. Using polar coordinates, the area element is \(dA = r dr dθ\). The limits of the integral correspond to the dimensions of the dartboard: \(r\) ranges from 0 to the radius of 10 inches and \(θ\) ranges from 0 to \(2\pi\). This reduces the problem to a double integral: \[\int_0^{2\pi} \int_0^{10} \rho(r) r dr d\theta. \]

Step 2: Substitute the Function

Substitute the given function \(\rho(r)\) into the integral in place of the density function: \[\int_0^{2\pi} \int_0^{10} 1010/(\pi(r^2+1)^2) r dr d\theta. \] Notice that the \(\pi\) in the numerator and denominator will cancel out.

Step 3: Evaluate the Inner Integral

First, we solve the inner integral with respect to \(r\), by making the substitution \(u = r^2 + 1\): \[1010 \int_0^{2\pi} [-1/2 * \frac{1}{u}]_{1}^{101} d\theta. \] The integral will give a value of \(\frac{-1010}{2}\) times \([ \frac{1}{101} - \frac{1}{1}]\).

Step 4: Evaluate the Outer Integral

Next, the outer integral will evaluate from 0 to \(2\pi\) which will be a multiplication of the range (i.e. \(2\pi\)) with the output of the inner integral: \[-505 \times 2\pi \times(\frac{1}{101} - 1).\]

Step 5: Simplify the Expression

Simplify the expression to get the total number of holes in the dartboard. The final answer should be a single numerical value.

Key Concepts

Density Function

A density function helps us understand how a certain quantity is distributed over a given space. In the context of this problem, we are looking at how dart holes are spread across a dartboard. The function provided, \( \rho(r)=\frac{1010}{\pi\left(r^{2}+1\right)^{2}} \), describes how many dart holes occur per square inch based on the distance \(r\) from the center.
This function is important because it allows us to integrate over the dartboard's area to find the total number of dart holes. High values of \(\rho(r)\) indicate more holes at certain distances from the center, showing that not all areas of the board have equal frequency of holes. Understanding this concept is crucial for analyzing and solving problems involving spatial distributions.

Polar Coordinates

Polar coordinates are a two-dimensional coordinate system where each point on a plane is determined by an angle and a distance, denoted by \((r, \theta)\). This system is useful when dealing with circular or radial patterns, like the distributions we find on dartboards.
In this problem, converting to polar coordinates simplifies the integration process. Instead of using Cartesian coordinates \((x, y)\), we express the area element as \(dA = r dr d\theta\), which neatly fits the circular symmetry of the dartboard.

• \( r \): distance from the origin (center of the board)
• \( \theta \): angle measured from a reference direction

Using polar coordinates allows us to more easily compute areas and densities in radial patterns.

Double Integral

A double integral extends the concept of integration to functions of two variables, allowing the calculation of a volume under a surface or an accumulated quantity over an area. Here, we use a double integral to sum the dart holes over the board's surface.
The integral we tackle is \(\int_0^{2\pi} \int_0^{10} \rho(r) r dr d\theta\). It involves integrating the density function over the angular \(\theta\) and radial \(r\) dimensions.
Evaluating such an integral typically involves two steps:

• First, perform the inner integral with respect to \(r\).
• Second, perform the outer integral with respect to \(\theta\).

This approach effectively accounts for the entire area of interest, accommodating any changes in density with distance from the center.

Substitution Method

The substitution method is a powerful technique for solving integrals. It involves transforming the variable of integration to simplify the expression, ultimately making an integral more manageable.
In this exercise, we use the substitution \(u = r^2 + 1\) to handle the inner integral. This changes the integral's variable from \(r\) to \(u\), simplifying the integration process as we solve \(\int \frac{r}{(r^2+1)^2} dr\). The relationship \(du = 2r dr\) helps replace \(r dr\) in the integrand, making it easier to evaluate.

• Transform \(r\) to \(u\).
• Adjust the limits of integration accordingly.

Substitution methods are particularly useful when dealing with complex or nested functions, as seen in our density function.