Problem 38 - Advanced Engineering Mathematics

Question

Consider the three curves between \((0,0)\) and \((2,4)\) : $$ \begin{array}{lll} C_{1}: x=t, & y=2 t, & 0 \leq t \leq 2 \\ C_{2}: x=t, & y=t^{2}, & 0 \leq t \leq 2 \\ C_{3}: x=2 t-4, & y=4 t-8, & 2 \leq t \leq 3 \end{array} $$ Show that \(\int_{C_{1}} x y d s=\int_{C_{3}} x y d s\), but \(\int_{C_{1}} x y d s \neq \int_{C_{2}} x y d s\).

Step-by-Step Solution

Answer

Integrals on C1 and C3 are equal: \(\frac{16 \sqrt{5}}{3}\); not equal for C2.

Step 1: Parametrize Curve C1

For the curve \( C_1 \), we have \( x = t \) and \( y = 2t \). The parametric equation is given as \( t ext{ varies from } 0 ext{ to } 2 \). We need the differential arc length \( ds \), which is given by \( ds = \sqrt{(\frac{dx}{dt})^2 + (\frac{dy}{dt})^2} \, dt \). Here, we compute \( \frac{dx}{dt} = 1 \) and \( \frac{dy}{dt} = 2 \), so \( ds = \sqrt{1^2 + 2^2} \, dt = \sqrt{5} \, dt \).

Step 2: Compute Integral for C1

Now compute the integral for \( C_1 \):\[ \int_{C_1} xy \, ds = \int_{0}^{2} t (2t) \times \sqrt{5} \, dt = \sqrt{5} \int_{0}^{2} 2t^2 \, dt \].Evaluating the integral: \[ \sqrt{5} (2) \left[ \frac{t^3}{3} \right]_{0}^{2} = \sqrt{5} (2) \left[ \frac{8}{3} \right] = \frac{16 \sqrt{5}}{3} \].

Step 3: Parametrize Curve C2

For the curve \( C_2 \), we have \( x = t \) and \( y = t^2 \). Again we find \( ds \) as \( \sqrt{(\frac{dx}{dt})^2 + (\frac{dy}{dt})^2} \, dt \). Here, \( \frac{dx}{dt} = 1 \) and \( \frac{dy}{dt} = 2t \), hence \( ds = \sqrt{1 + 4t^2} \, dt \).

Step 4: Compute Integral for C2

Now compute the integral for \( C_2 \):\[ \int_{C_2} xy \, ds = \int_{0}^{2} t (t^2) \times \sqrt{1 + 4t^2} \, dt = \int_{0}^{2} t^3 \sqrt{1 + 4t^2} \, dt \].This integral is not equal to \( \frac{16 \sqrt{5}}{3} \).

Step 5: Parametrize Curve C3

For \( C_3 \), notice it's a line, so check if it coincides with \( C_1 \) as it is a line extension:Let \( t = \frac{x+4}{2} \), then \( y = 4 \left(\frac{x+4}{2}\right) - 8 \), hence \( x = t \) and \( y = 2t \) hold as coordinates parameterizing it from a different viewpoint,showing \( C_3 \) is a linear transformation of \( C_1 \). Thus, the same integral holds.

Key Concepts

Parametric Equations

Understanding parametric equations is essential when dealing with curves described by two functions, where both coordinates depend on a parameter, typically denoted by \( t \). In our example, we have three curves described using parametric equations:

• For \( C_1 \), \( x = t \) and \( y = 2t \).
• For \( C_2 \), \( x = t \) and \( y = t^2 \).
• For \( C_3 \), \( x = 2t - 4 \) and \( y = 4t - 8 \).

Each pair of equations describes how the \( x \) and \( y \) coordinates of a curve evolve as \( t \) changes. The parametric approach is particularly useful in visualizing and calculating various properties of curves without relying solely on the Cartesian coordinate system. Here, the parameter \( t \) provides a more dynamic view of how these curves behave over a specific interval. This understanding is vital for calculating subsequent properties like arc length and integrals.

Arc Length

Calculating the arc length of a parametric curve involves determining the infinitesimally small segments along the curve and summing them up to get the total length. For a curve represented by \((x(t), y(t))\), the differential arc length \( ds \) can be expressed as:\[ ds = \sqrt{\left(\frac{dx}{dt}\right)^2 + \left(\frac{dy}{dt}\right)^2} \, dt \]Let's see how this plays out for our curves:- For \( C_1 \), we found: \( ds = \sqrt{5} \, dt \).- For \( C_2 \), it was a bit more complex: \( ds = \sqrt{1 + 4t^2} \, dt \).The arc length helps in integral calculations, as the integration over the curve \( \int xy \, ds \) requires \( ds \) in its computation. This step subtly captures how a function changes along a curve, summarizing it through a single integrated value across a defined path.

Integral Calculation

The task involves calculating the integral \( \int xy \, ds \) over given curves using the arc length derived previously. The method begins with substituting parametric expressions for \( x \) and \( y \), multiplying them together, and then incorporating \( ds \):- For curve \( C_1 \): \[ \int_{C_1} xy \ ds = \int_{0}^{2} t(2t) \times \sqrt{5} \, dt = \sqrt{5} \int_{0}^{2} 2t^2 \, dt \] After evaluating, this yields: \( \frac{16\sqrt{5}}{3} \).- For curve \( C_2 \): \[ \int_{C_2} xy \ ds = \int_{0}^{2} t(t^2) \times \sqrt{1 + 4t^2} \, dt \] This integral, due to its complexity, would not simplify to the result from the previous curve.These calculations showcase the application of differential elements in evaluating geometric properties of curves, providing insights into their relative magnitudes.

Curve Parametrization

Curve parametrization allows different expressions for the same geometric path, offering flexibility in how a curve is described. For instance, curve \( C_3 \) shows a different parametrization while remaining equivalent in path to \( C_1 \):- \( C_3 \) \((x = 2t - 4, y = 4t - 8)\): By substituting and simplifying \( t \equiv \frac{x+4}{2} \), we achieve: \[ y = 2\left(\frac{x+4}{2}\right) = 2x \]Thus, it matches the parametric form of \( C_1 \) when transformed. This indicates curve \( C_3 \) is an extension of the line formed by \( C_1 \), confirming their integral values match.Understanding curve parametrization in parametric form is crucial for interpreting equivalent paths, especially when using different variable ranges or expressions, reinforcing the unification of varying equations and their geometric representations.