Question

Sketch the graph of the given parametric equation and find its length. \(x=t^{3} / 3, y=t^{2} / 2 ; 0 \leq t \leq 1\)

Short answer

The arc length is \(\frac{2\sqrt{2} - 1}{3}\).

Step-by-step solution

Understand the Parametric Equations

The given parametric equations describe a curve in terms of the parameter \(t\). The equations are \(x = \frac{t^3}{3}\) and \(y = \frac{t^2}{2}\). As \(t\) varies from 0 to 1, these equations determine the points \((x, y)\) that form the curve.

Plot the Curve

To sketch the curve, compute a few values of \(x\) and \(y\) for \(t\) in its range. At \(t = 0\), \(x = 0^3 / 3 = 0\) and \(y = 0^2 / 2 = 0\). At \(t = 1\), \(x = 1^3 / 3 = 1/3\) and \(y = 1^2 / 2 = 1/2\). Intermediate values can be computed to shape the curve in this range from the origin to \((1/3, 1/2)\).

Derive the Arc Length Formula

The formula for the length of a curve defined by parametric equations \(x(t)\) and \(y(t)\) over interval [a, b] is \(L = \int_a^b \sqrt{\left( \frac{dx}{dt} \right)^2 + \left( \frac{dy}{dt} \right)^2} \, dt\). First, find the derivatives: \(\frac{dx}{dt} = t^2\) and \(\frac{dy}{dt} = t\).

Set up the Integral for Arc Length

Substitute the derivatives into the arc length formula: \[ L = \int_0^1 \sqrt{(t^2)^2 + (t)^2} \, dt = \int_0^1 \sqrt{t^4 + t^2} \, dt\] This expression simplifies the calculation of the arc length from \(t = 0\) to \(t = 1\).

Simplify the Integral

Factor the expression inside the square root: \[ L = \int_0^1 \sqrt{t^2(t^2 + 1)} \, dt = \int_0^1 t\sqrt{t^2 + 1} \, dt \] This integral represents the length of the curve. We will compute it.

Compute the Integral

Perform a substitution to solve the integral. Let \(u = t^2 + 1\), then \(\frac{du}{dt} = 2t\), giving us \(dt = \frac{du}{2t}\). Change limits for \(u\): when \(t = 0\), \(u = 1\), when \(t = 1\), \(u = 2\). Thus, the integral becomes:\[ L = \int_1^2 \sqrt{u} \, \frac{du}{2} = \frac{1}{2} \int_1^2 u^{1/2} \, du \]The result of the integral is: \[ \frac{1}{2} \left[ \frac{2}{3} u^{3/2} \right]_1^2 = \frac{1}{3} \left[ 2\sqrt{2} - 1 \right] \] This yields the arc length value.

Calculate Final Length

Calculate the final numerical value of the arc length: \[ L = \frac{1}{3} (2\sqrt{2} - 1) = \frac{2\sqrt{2} - 1}{3} \] This gives the length of the curve for the given parameter range.

Key concepts

Curve Sketching with Parametric Equations

Curve sketching with parametric equations is an interesting way to describe complex curves using a parameter, often denoted as \(t\). This method allows us to map out a two-dimensional path in the \(xy\)-plane by using two separate equations, one for \(x\) and one for \(y\), both in terms of \(t\). In the given exercise, \(x = \frac{t^3}{3}\) and \(y = \frac{t^2}{2}\) form a parametric curve as \(t\) ranges from 0 to 1. By computing points for specific values of \(t\), we can see the path this curve follows.

For example, when \(t = 0\), both \(x\) and \(y\) are 0, placing the initial point at the origin. When \(t = 1\), \(x = \frac{1}{3}\) and \(y = \frac{1}{2}\), locating the curve's endpoint at \((\frac{1}{3}, \frac{1}{2})\).

Intermediary points can be calculated to see the transition. Understanding this pattern helps us visualize the curve's overall shape, which in this case, forms a path from the origin to \((\frac{1}{3}, \frac{1}{2})\) in the first quadrant.

Arc Length Calculation of Parametric Curves

Calculating the arc length of a curve defined by parametric equations can be achieved with a well-established formula. This is particularly useful for determining the distance travelled along a path. For parametric curves given by \(x(t)\) and \(y(t)\), the arc length \(L\) is calculated using

\[ L = \int_a^b \sqrt{\left( \frac{dx}{dt} \right)^2 + \left( \frac{dy}{dt} \right)^2} \, dt \]

where \(a\) and \(b\) are the limits of \(t\). This process involves deriving \(x(t)\) and \(y(t)\) to find \(\frac{dx}{dt}\) and \(\frac{dy}{dt}\).

For the given exercise, we derived \(\frac{dx}{dt} = t^2\) and \(\frac{dy}{dt} = t\). These derivatives are then substituted into the formula, resulting in:

\[ L = \int_0^1 \sqrt{t^4 + t^2} \, dt \]

Understanding how these derivatives feed into the arc length formula aids in comprehending how parametric paths can be measured.

Integration Techniques for Solving Arc Length

Integration techniques are essential when solving expressions like the arc length of a parametric curve. The challenge often lies in simplifying the integrand, which can contain square roots and polynomials. In our exercise, after forming the integral \( \int_0^1 \sqrt{t^4 + t^2} \, dt \), we simplify it to \( \int_0^1 t\sqrt{t^2 + 1} \, dt \) by factoring out \(t^2\).

To integrate, we apply a substitution technique. Letting \(u = t^2 + 1\) simplifies our problem into an easier format. With \( du = 2t \, dt \), we replace \( dt \) with \( \frac{du}{2t} \), turning the limits for \(u\) into \(1\) and \(2\), respectively. Now the integral looks like:

\[ \frac{1}{2} \int_1^2 u^{1/2} \, du \]

This simplified expression is much easier to evaluate, demonstrating how substitution can break down complex integration tasks into manageable parts.

Exploring Calculus with Parametric Curves

Calculus with parametric equations offers a different perspective compared to traditional \(y = f(x)\) functions. Using parameters like \(t\) enables the exploration of dynamic processes such as motion along a curve.

In our example, calculus helps us not just to visualize the path described by the parametric equations, but also to perform critical operations like deriving functions and calculating arc lengths. This further solidifies understanding of the curve's properties.

Derivatives tell us about the rate of change along the curve, while integrals provide us details about the whole curve, such as its length. This exercise beautifully showcases the power of calculus in comprehending and calculating fundamental aspects of curves described uniquely through parameters.