Problem 17 - Calculus Early Transcendentals: Pearson New International Edition

Question

Find the mass of a wire with the shape of the curve \(y=x^{2}\) between \((-2,4)\) and \((2,4)\) if the density is given by \(\delta(x, y)=\) \(k|x|\).

Step-by-Step Solution

Answer

The mass is \(\frac{k}{3}(17\sqrt{17} - 1)\).

Step 1: Determine the Curve Parameterization

The curve given is described by the equation \(y = x^2\). On this curve, between the points \((-2, 4)\) and \((2, 4)\), \(x\) ranges from -2 to 2, and \(y = 4\) as specified by these points.

Step 2: Calculate the Differential Arc Length

The differential arc length \(ds\) for a curve \(y = f(x)\) is given by \(ds = \sqrt{1 + \left(\frac{dy}{dx}\right)^2}dx\). Since \(y = x^2\), we have \(\frac{dy}{dx} = 2x\). Thus, \(ds = \sqrt{1 + (2x)^2}\, dx = \sqrt{1 + 4x^2}\, dx\).

Step 3: Set Up the Mass Integral

The mass \(m\) of the wire is given by \(m = \int \delta(x, y) \, ds\). Since \(\delta(x, y) = k|x|\), the mass integral becomes \[m = \int_{-2}^{2} k|x| \sqrt{1 + 4x^2} \, dx.\]

Step 4: Evaluate the Mass Integral

The integral is symmetric with respect to the y-axis because \(|x|\) and \(\sqrt{1+4x^2}\) are even functions. Thus, we can evaluate on \([0,2]\) and multiply by 2: \[m = 2k \int_{0}^{2} x \sqrt{1 + 4x^2} \, dx.\] Evaluate this integral using a u-substitution \(u = 1 + 4x^2\), \(du = 8x\, dx\), leading to \[m = \frac{k}{2} \int_{1}^{17} \sqrt{u} \, du.\] Integrate to get \[m = \frac{k}{2} \left. \frac{2}{3} u^{3/2} \right|_{1}^{17} = \frac{k}{3} \big[17^{3/2} - 1^{3/2}\big].\]

Step 5: Simplify the Result

Calculate the definite integral value: \[17^{3/2} - 1 = \sqrt{17^3} - 1 = 17 \sqrt{17} - 1.\] Plug this back into the equation for mass: \[m = \frac{k}{3}(17 \sqrt{17} - 1).\]

Key Concepts

Arc Length

In calculus, the arc length is a method that allows us to find the length of a curve. It's like stretching out a wavy line to see how long it truly is. When you have a curve given by a function, such as the parabola \(y = x^2\) in our example, the arc length can be calculated as an integral.

To find the differential arc length \(ds\), you first need the derivative of the function. The derivative \(\frac{dy}{dx}\) tells us how steep or flat our curve is. For \(y = x^2\), we have \(\frac{dy}{dx} = 2x\).

The formula for the differential arc length is \(ds = \sqrt{1 + \left(\frac{dy}{dx}\right)^2}dx\). Here, we plug in our derivative to get \(ds = \sqrt{1 + 4x^2} \, dx\). This represents a tiny piece of our curve's length.

• Used to measure the distance along a curve
• Involves taking an integral
• You'll need the derivative of the curve

Now that we have \(ds\), it can be used in further calculations, like finding the mass of a curve with varying density.

Definite Integral

In calculus, a definite integral is a powerful tool that sums up infinitely small pieces to find the total, like adding up the weight of each tiny piece of a wire. It takes a function and computes this total over a certain interval.

The notation \(\int_{a}^{b} f(x) \, dx\) represents the definite integral of function \(f(x)\) from \(x = a\) to \(x = b\). For our exercise, the integral is used to calculate the mass of a wire by summing up the contributions from each segment.

• Calculates total value between two points
• Often involves an area under a curve
• Uses certain boundaries (from \(a\) to \(b\))

In our problem, the integral \(\int_{-2}^{2} k|x| \sqrt{1 + 4x^2} \, dx\) helps us find the mass by incorporating the density function \(\delta(x, y) = k|x|\) and the differential arc length \(ds\). The symmetry of the wire helps us simplify the integral calculation. This is another awesome aspect of definite integrals, they can be adjusted for simplicity without losing accuracy.

U-Substitution

The u-substitution technique is a fundamental method for solving integration problems, especially those that seem tricky at first glance. It works by simplifying the integral into a form that is easier to solve.

In our example, we use u-substitution to evaluate \(m = 2k \int_{0}^{2} x \sqrt{1 + 4x^2} \, dx\). Here, we let \(u = 1 + 4x^2\), making the derivative \(du = 8x \, dx\). This substitution effectively changes the variable in the integral to \(u\), simplifying the process.

• Substitute a part of the integrand with a new variable
• Makes complex integrals more approachable
• Involves finding the derivative of \(u\)

After substituting, the integral becomes \(\frac{k}{2} \int_{1}^{17} \sqrt{u} \, du\). This is much more manageable and can be evaluated to find the mass. This method highlights how calculus can turn complex math problems into simpler ones.