Problem 44 - Thomas Calculus

Question

Minimizing a moment of inertia A rectangular plate of constant density \(\delta(x, y)=1\) occupies the region bounded by the lines \(x=4\) and \(y=2\) in the first quadrant. The moment of inertia \(I_{a}\) of the rectangle about the line \(y=a\) is given by the integral $$ I_{a}=\int_{0}^{4} \int_{0}^{2}(y-a)^{2} d y d x $$ Find the value of \(a\) that minimizes \(I_{a}\)

Step-by-Step Solution

Answer

The value of \(a\) that minimizes \(I_a\) is 1.

Step 1: Setup the Problem

The given integral for the moment of inertia is a double integral:\[ I_{a} = \int_{0}^{4} \int_{0}^{2} (y-a)^{2} \, dy \, dx \]We need to find the value of \(a\) that minimizes this integral.

Step 2: Simplify the Integral

First, integrate with respect to \(y\):\[ \int_{0}^{2} (y-a)^2 \, dy \]Using the substitution \((y-a)^2 = (y^2 - 2ay + a^2)\) gives:\[ \int_{0}^{2} (y^2 - 2ay + a^2) \, dy = \left[ \frac{y^3}{3} - ay^2 + a^2y \right]_{0}^{2} \]

Step 3: Evaluate Inner Integral

Substitute the bounds 0 and 2 into the expression:\[ \left( \frac{(2)^3}{3} - a(2)^2 + a^2(2) \right) - \left( \frac{0^3}{3} - a(0)^2 + a^2(0) \right) \]Simplifying:\[ \frac{8}{3} - 4a + 2a^2 \]

Step 4: Compute the Outer Integral

Now integrate the simplified result with respect to \(x\):\[ \int_{0}^{4} \left( \frac{8}{3} - 4a + 2a^2 \right) \, dx \]This integration results in:\[ \left( \frac{8}{3} - 4a + 2a^2 \right) x \bigg|_{0}^{4} = 4 \left( \frac{8}{3} - 4a + 2a^2 \right) \]

Step 5: Differentiate the Simplified Expression

Differentiating the expression with respect to \(a\):\[ \frac{d}{da} \left[ 4 \left( \frac{8}{3} - 4a + 2a^2 \right) \right] = \frac{d}{da} [ 4(\frac{8}{3}) - 16a + 8a^2 ] \]\[ = -16 + 16a \]

Step 6: Set Derivative to Zero to Find Critical Points

Set the derivative to zero to find critical points:\[ -16 + 16a = 0 \]Solving for \(a\):\[ 16a = 16 \rightarrow a = 1 \]

Step 7: Conclude Minimum

Since we only have one critical point and the second derivative \( \frac{d^2}{da^2}(I_a) = 16 > 0 \) indicates a minimum, \(a = 1\) is the value that minimizes \(I_{a}\).

Key Concepts

Understanding Double Integrals

Double integrals are an extension of single integrals and are used to compute volume under a surface in a multi-dimensional space. They allow us to add up tiny volumes over a specified region.
These integrals have two integration signs corresponding to two variables, typically written as:
\[\int \int_R f(x, y) \, dA\]where \(R\) is the region of integration in the \(xy\)-plane.

• The first step involves integrating with respect to one variable, often the inner integral.
• Next, we integrate the resulting expression with respect to the other variable, which corresponds to the outer integral.

In the original exercise, the double integral calculates the moment of inertia of a rectangular plate, a real-life application of this concept.

The Role of Calculus

Calculus is the branch of mathematics that studies rates of change and accumulations. It's vital for solving problems involving optimization, where we aim to maximize or minimize a certain quantity.
The primary tools of calculus used in optimization are derivatives and integrals:

• Derivatives: Help us understand how a function's value changes as its inputs change. In our exercise, the derivative indicates how the moment of inertia varies with different values of \(a\).
• Integrals: Allow us to accumulate these changes over a region, which is exactly what the double integral in the exercise does for the moment of inertia.

By taking derivatives and setting them to zero, we identify critical points, allowing us to find minimum or maximum values of the function.

Optimization in Calculus

Optimization involves finding the "best" or most efficient way to achieve an objective, such as minimizing the moment of inertia in mechanical engineering.
The process generally involves:

• Identifying the function that describes the situation, which, in our case, is the integral expression for moment of inertia.
• Calculating derivatives to understand how this function changes and to identify critical points.
• Utilizing second derivatives or other tests to confirm whether these points are minima or maxima.

In our problem, we find the value of \(a\) that minimizes \(I_{a}\), verifying that the point is indeed a minimum by using second derivative analysis.

Identifying Critical Points

Critical points occur where the derivative of a function equals zero, or where it doesn't exist. They're essential for finding minimums and maximums, acting as "turning points" in the graph of a function.
In optimization:

• We set the first derivative of a function to zero to locate potential critical points.
• In our exercise, \(-16 + 16a = 0\) revealed the single critical point \(a = 1\).
• We then apply further tests, such as the second derivative test, to determine if this point is a minimum or a maximum.

For the moment of inertia, checking the second derivative confirmed \(a = 1\) as a minimum.