Question

Find the arc length of the following curves on the given interval. $$x=t^{4}, y=\frac{t^{6}}{3} ; 0 \leq t \leq 1$$

Short answer

Answer: The arc length of the curve on the interval $$0 \leq t \leq 1$$ is $$L = \frac{2}{3}(5^{\frac{3}{2}} - 8)$$.

Step-by-step solution

Calculate the derivatives of $$x(t)$$ and $$y(t)$$ with respect to $$t$$

To find the arc length, we first need the derivatives of $$x(t)$$ and $$y(t)$$ with respect to $$t$$. Find the derivative of $$x(t)=t^4$$ with respect to $$t$$: $$\frac{dx}{dt} = 4t^3$$ Find the derivative of $$y(t)=\frac{t^6}{3}$$ with respect to $$t$$: $$\frac{dy}{dt} = 2t^5$$

Write down the arc length formula for parametric curves

Now that we have the derivatives of $$x(t)$$ and $$y(t)$$, we can write down the arc length formula for parametric curves over the interval $$[0, 1]$$: $$L = \int_0^1 \sqrt{\left(\frac{dx}{dt}\right)^2 + \left(\frac{dy}{dt}\right)^2} dt$$

Substitute the derivatives into the arc length formula

Substitute the derivatives $$\frac{dx}{dt} = 4t^3$$ and $$\frac{dy}{dt} = 2t^5$$ into the formula: $$L = \int_0^1 \sqrt{(4t^3)^2 + (2t^5)^2} dt$$

Simplify the expression inside the integral

Before integrating, we need to simplify the expression inside the square root: $$L = \int_0^1 \sqrt{16t^6 + 4t^{10}} dt$$ Factor out the common term $$4t^6$$ from the expression inside the square root: $$L = \int_0^1 \sqrt{4t^6(4 + t^4)} dt$$

Evaluate the definite integral

Now we can evaluate the integral. Using substitution for \(u = 4 + t^4\), \(du = 4t^3 dt\), and integrating with respect to \(u\): $$L = \int \sqrt{4t^6 u} \frac{du}{4t^3}$$ Which simplifies to: $$L = \int \sqrt{u} du$$ Integrating with respect to u: $$L = \frac{2}{3}u^{\frac{3}{2}} + C$$ Now substitute back for \(t\): $$L = \frac{2}{3}(4 + t^4)^{\frac{3}{2}} + C$$ Finally, evaluate the definite integral from \(0\) to \(1\): $$L = \left[\frac{2}{3}(4 + t^4)^{\frac{3}{2}}\right]_0^1$$ $$L = \left[\frac{2}{3}(4 + 1)^{\frac{3}{2}}\right] - \left[\frac{2}{3}(4)^{\frac{3}{2}}\right]$$ $$L = \frac{2}{3}(5^{\frac{3}{2}} - 8)$$ Thus, the arc length of the curve on the interval $$0 \leq t \leq 1$$ is $$L = \frac{2}{3}(5^{\frac{3}{2}} - 8)$$.

Key concepts

Parametric Curves

Parametric curves offer a unique way to represent paths and shapes that are not easily captured by standard Cartesian equations. Instead of describing the curve with a single equation in terms of x and y, parametric curves use a parameter, often denoted as \(t\), to express both x and y coordinates. This means:

• \( x(t) \) describes how the x-coordinate changes as \( t \) varies.
• \( y(t) \) describes how the y-coordinate changes as \( t \) varies.

For the exercise, the curve is defined by functions \(x(t) = t^4\) and \(y(t) = \frac{t^6}{3}\). The parameter \(t\) usually represents time, but it can be any convenient variable.

In this context, parametric curves are particularly useful because they allow for a more flexible description of curves. They make it easier to compute properties like arc length, which is our goal here. By handling each coordinate separately, the calculation becomes a bit more intuitive and structured.

Derivative

The derivative is a fundamental tool in calculus that measures the rate of change of a function. When dealing with parametric curves, we are interested in the derivatives with respect to the parameter, often \(t\). In other words, we want to see how rapidly or slowly the coordinates \(x(t)\) and \(y(t)\) change as \(t\) changes.

• For \(x(t) = t^4\), the derivative \(\frac{dx}{dt} = 4t^3\) tells us the slope of the x-component of the curve at any point \(t\).
• Similarly, for \(y(t) = \frac{t^6}{3}\), the derivative \(\frac{dy}{dt} = 2t^5\) provides the rate of change of the y-component.

These derivatives are critical in calculating the arc length of the curve. They provide the necessary information on how the curve evolves, contributing to the understanding of its geometry. The derivatives are plugged into the arc length formula to account for the varying distances along the curve.

Definite Integral

The definite integral is a concept in calculus that accumulates a sum across an interval, providing the total change or area under a curve between two points. In this exercise, it helps determine the arc length of a parametric curve from \(t = 0\) to \(t = 1\).

The arc length formula for a parametric curve is \[L = \int_{a}^{b} \sqrt{\left(\frac{dx}{dt}\right)^2 + \left(\frac{dy}{dt}\right)^2} \, dt\]This formula integrates the square root of the sum of squares of the derivatives \(\frac{dx}{dt}\) and \(\frac{dy}{dt}\). This method accounts for every tiny segment of the curve's length.

In our specific problem, after substituting the derivatives, you simplify the integrand and perform the substitution to make the integral manageable. This final step requires evaluating the integral over the interval \([0, 1]\). Performing this integration helps us find the total length from start to finish. Integrals like this one transform calculus from theory into practical problem-solving by allowing the calculation of tangible measures, such as distance.