Problem 31 - Thomas Calculus

Question

Constant density Find the moment about the \(x\) -axis of a wire of constant density that lies along the curve \(y=\sqrt{x}\) from \(x=0\) to \(x=2\) .

Step-by-Step Solution

Answer

Use a numerical method to evaluate \( M_x = \sigma \int_{0}^{2} \sqrt{x} \sqrt{1 + \frac{1}{4x}} \, dx \).

Step 1: Understand the Moment Formula

The moment about the x-axis for a curve with a wire is given by the integral \( M_x = \int y \cdot \sigma \cdot ds \), where \( \sigma \) is the constant density and \( ds \) is the differential arc length on the curve. For a curve defined by \( y = f(x) \), the differential arc length is given by \( ds = \sqrt{1 + \left( \frac{dy}{dx} \right)^2} \, dx \).

Step 2: Derive the Arc Length Differential

Given the curve \( y = \sqrt{x} \), calculate \( \frac{dy}{dx} = \frac{1}{2\sqrt{x}} \). The differential arc length becomes \( ds = \sqrt{1 + \left( \frac{1}{2\sqrt{x}} \right)^2} \, dx = \sqrt{1 + \frac{1}{4x}} \, dx \).

Step 3: Set Up the Integral for the Moment

Substitute \( y = \sqrt{x} \) and \( ds \) into the moment formula: \( M_x = \int_{0}^{2} \sqrt{x} \cdot \sigma \cdot \sqrt{1 + \frac{1}{4x}} \, dx \). Factor out the constant density \( \sigma \), \( M_x = \sigma \int_{0}^{2} \sqrt{x} \cdot \sqrt{1 + \frac{1}{4x}} \, dx \).

Step 4: Simplify and Evaluate the Integral

The integral is \( \int_{0}^{2} x^{1/2} \sqrt{1 + \frac{1}{4x}} \, dx \). This integral is challenging to evaluate analytically. Use numerical integration methods or software/calculators to approximate this value.

Step 5: Incorporate Density into the Result

Multiply the result of the integral by \( \sigma \) to include the effect of constant density. This gives the moment about the x-axis for the wire along the curve.

Key Concepts

Arc Length

The concept of arc length is vital in problems involving curves and surfaces. It helps determine the distance along a curve. In integral calculus, we calculate this length using specific functions. For a curve defined by an equation like \( y = f(x) \), the arc length from point \( x=a \) to \( x=b \) is given by the integral \[L = \int_{a}^{b} \sqrt{1+\left(\frac{dy}{dx}\right)^2} \, dx.\]This formula accounts for the curve's slope at each point.

When working with curves, the differential arc length \( ds \) is a small segment's length of the arc. For example, if the curve is described as \( y = \sqrt{x} \), the derivative \( \frac{dy}{dx} \) needs to be determined first. In this case, \( \frac{1}{2\sqrt{x}} \) is the result, which is then used within the arc length formula.

• This highlights how the slope affects the total arc length.
• It's pivotal in computing various physical properties, such as moments of inertia.

Understanding arc length and its computation helps solve more complex physics or geometry problems.

Integral Calculus

Integral calculus is the branch of mathematics focusing on integrals, areas, and accumulations. It's essential in determining quantities where adding up infinitely small parts results in a whole.In problems like our exercise, we use integral calculus to find the moment about the \( x \)-axis. Defined as \( M_x = \int y \cdot \sigma \cdot ds \), this equation involves integrating the product of the curve equation and its arc length over a specified interval. Here, the integral represented by \[\int_{0}^{2} \sqrt{x} \cdot \sqrt{1 + \frac{1}{4x}} \, dx\] is tackled using integral calculus.

• Integral calculus helps calculate areas under curves and other related concepts.
• In physics, it's often employed to find centers of mass, moments, and more.

Integral calculus provides the tools for solving real-world problems by focusing on collection and accumulation over defined intervals.

Constant Density

In physics and engineering, constant density refers to a uniform distribution of mass within an object or along a path like a wire. This simplification allows for straightforward mathematical treatment, as density remains a constant factor.In our exercise, the wire's density, represented by \( \sigma \), remains constant throughout its length. This allows us to factor out \( \sigma \) from the integral expression when calculating moments, leading to \[M_x = \sigma \int_{0}^{2} \sqrt{x} \cdot \sqrt{1 + \frac{1}{4x}} \, dx.\]

• A constant density simplifies calculations by not needing to account for density variations.
• In practice, this assumption can closely approximate many real-world objects.

Understanding constant density aids in approaching physical problems with simplified, yet powerful, models.

Differential Calculus

Differential calculus analyzes how functions change and allows us to define the rate of this change — essentially, differentiation. It's indispensable in finding slopes of curves and calculating derivatives, both critical for determining properties like arc length.For our curve \( y = \sqrt{x} \), differential calculus provides \( \frac{dy}{dx} = \frac{1}{2\sqrt{x}} \). This expression is crucial for determining the differential arc length, a key component in moment calculations.

• Differential calculus is mainly about rates of change and slopes.
• It helps predict how a function behaves and transitions over small intervals.

Overall, mastering differential calculus enables tackling more advanced computational tasks and understanding dynamic changes in physical processes.