Question

A thin wire represented by the smooth curve C with a density \(\rho\) (units of mass per length) has a mass \(M=\int_{C} \rho d s\) Find the mass of the following wires with the given density. $$\text { C: }\left\\{(x, y): y=2 x^{2}, 0 \leq x \leq 3\right\\} ; \rho(x, y)=1+x y$$

Short answer

Question: Compute the mass of the wire made from the curve \(y = 2x^2\) between \(0 \leq x \leq 3\), with density function \(\rho(x, y) = 1 + xy\). Solution: For the given curve and density function, we find the arc length element \(ds = \sqrt{1 + 16t^2} dt\). Then, the mass M of the wire can be expressed as an integral in terms of the parameter t: \( M = \int_{0}^{3} (1 + 2t^3) \sqrt{1 + 16t^2} dt \). Since this integral is quite complicated, the mass of the wire can be found approximately using numerical integration techniques or a computer algebra system.

Step-by-step solution

Parametric Representation of the Curve

To find the parametric representation of the given curve, we will express both x and y in terms of a single parameter t. Since the curve is given as \(y = 2x^2\) with \(0 \leq x \leq 3\), we can use \(x = t\) with \(0 \leq t \leq 3\). Then the curve would be represented parametrically as \((t, 2t^2)\).

Finding the Arc Length Element ds

In order to find ds, we will first find the derivatives of the parametric functions with respect to t. That is, we need to find \(\dfrac{dx}{dt}\) and \(\dfrac{dy}{dt}\). For \(x(t) = t\) , we have \(\frac{dx}{dt} = 1\). For \(y(t) = 2t^2\), we have \(\frac{dy}{dt} = 4t\). Next, we can use the following formula for ds in terms of dt: \(ds = \sqrt{\left(\frac{dx}{dt}\right)^2 + \left(\frac{dy}{dt}\right)^2} dt\) Substituting the values we obtained for \(\frac{dx}{dt}\) and \(\frac{dy}{dt}\), we get: \(ds = \sqrt{(1)^2 + (4t)^2} dt\) \(ds = \sqrt{1 + 16t^2} dt\)

Set up the Integral for Mass

Now that we have ds, we can set up the integral for mass M. Having parametric representation \((t, 2t^2)\) and given density function \(\rho (x,y) = 1+xy\), we replace x with t and y with \(2t^2\), resulting in: \(\rho(t)=1+t(2t^2)\) Then, M can be expressed as follows: \( M = \int_{C} \rho ds \) Since we have converted the integral over the curve to the integral in terms of the parameter t, we can write the integral for mass M as: \( M = \int_{0}^{3} \rho(t) \sqrt{1 + 16t^2} dt \)

Compute the Integral

Now, we need to compute the integral to find the mass of the wire: \( M = \int_{0}^{3} (1 + 2t^3) \sqrt{1 + 16t^2} dt \) Unfortunately, this integral is quite complicated and cannot be solved by elementary methods or even some special functions. Therefore, the mass of the wire can be found approximately using numerical integration techniques such as Simpson's Rule, Trapezoidal Rule, or by using a computer algebra system (e.g. WolframAlpha or MathCAD).

Key concepts

Parametric Equations

When dealing with the mass of a wire or the path of an object, we often turn to parametric equations to describe the position of points along a curve in a more functional way. Parametric equations allow us to express a geometric line or shape in terms of a single parameter, usually denoted as t. This becomes incredibly useful as it converts the representation from a static form to a dynamic one, capable of encapsulating motion or growth over an interval.
For the wire in our exercise, each point can be represented as a set of two parametric equations, where x and y are both functions of t. By expressing both coordinates in terms of this common parameter, calculations, especially those involving calculus such as differentiation and integration, become much clearer and easier to handle. Parametric representations are the backbone for many applications in physics, engineering, and computer graphics.

Arc Length

The concept of arc length is crucial when determining the length of a portion of a curve or the perimeter of a more complex shape. In calculus, to find the arc length of a parametric curve, we use a formula derived from the Pythagorean theorem that includes the derivatives of the parametric equations with respect to t. This formula is represented as:
\(ds = \rva{ \frac{dx}{dt}^2 + \frac{dy}{dt}^2} dt\).
This little differential segment, ds, represents an infinitesimally small slice of the curve’s length, which, when summed up (integrated) over the appropriate interval, gives us the total arc length. In context, the arc length helps us understand not just how long the wire is, but by multiplying it by the density, we determine the wire's total mass. This is particularly important in fields like structural engineering, where the weight of materials plays a crucial role in design and safety.

Numerical Integration

Sometimes, we encounter integrals that are impossible or very challenging to solve analytically, which means we can't express their solutions in terms of elementary functions. This is where numerical integration methods come into play. They allow us to approximate the value of definite integrals using various techniques. Some common methods include the Trapezoidal Rule, Simpson's Rule, and Romberg's Method.
Numerical integration works by breaking the area under the curve into shapes for which we can calculate the area, like rectangles or trapezoids. By summing the areas of these shapes, we get an estimation of the total integral. These methods are especially helpful in real-world applications where exact solutions are not necessary, or the functions involved are too complex for standard calculus techniques. Numerical integration makes it possible to solve problems in physics, engineering, and economics that would otherwise be beyond our analytical reach.

Line Integrals in Calculus

In the context of calculus, line integrals provide a means to integrate multi-variable functions along a certain curve or path. They are an extension of the integral concept from a straight line to any curve and are utilized predominantly in vector calculus. The line integral of a scalar field, as seen in the calculation of the mass of a wire, is often represented by the integral of a function along a curve multiplied by the differential element of arc length (\(ds\)).
For our wire example, the line integral is used to accumulate the density function over the length of the wire to find its total mass. In broader uses, line integrals can also calculate work done by a force field along a path or the circulation of a field around a loop. They are pivotal in fields like electromagnetism, fluid dynamics, and in the study of conservative fields.