Problem 3 - Thomas Calculus

Question

Find the lengths of the curves in Exercises \(1-6\) $$ x=t^{3}, \quad y=3 t^{2} / 2, \quad 0 \leq t \leq \sqrt{3} $$

Step-by-Step Solution

Answer

The length of the curve is 7.

Step 1: Identify the Formula for Arc Length

The formula for the arc length of a parametric curve defined by \( x = f(t) \) and \( y = g(t) \) from \( t = a \) to \( t = b \) is \( L = \int_a^b \sqrt{\left( \frac{dx}{dt} \right)^2 + \left( \frac{dy}{dt} \right)^2} \, dt \).

Step 2: Differentiate the Parametric Equations

Differentiate \( x = t^3 \) and \( y = \frac{3t^2}{2} \) with respect to \( t \):- \( \frac{dx}{dt} = 3t^2 \)- \( \frac{dy}{dt} = 3t \)

Step 3: Plug Derivatives into Arc Length Formula

Substitute the derivatives into the arc length formula: \[ L = \int_0^\sqrt{3} \sqrt{(3t^2)^2 + (3t)^2} \, dt \]

Step 4: Simplify the Expression Under the Square Root

Calculate inside the square root:\[(3t^2)^2 = 9t^4, \quad (3t)^2 = 9t^2\]Thus, the integral becomes:\[\int_0^\sqrt{3} \sqrt{9t^4 + 9t^2} \, dt = \int_0^\sqrt{3} \sqrt{9t^2(t^2 + 1)} \, dt\]

Step 5: Simplify the Square Root Expression

Factor \( 9t^2 \) out of the square root:\[= \int_0^\sqrt{3} 3t \sqrt{t^2 + 1} \, dt \]

Step 6: Solve the Integral

To solve \( \int 3t \sqrt{t^2 + 1} \, dt \), use substitution: let \( u = t^2 + 1 \), then \( du = 2t \, dt \), or \( t \, dt = \frac{du}{2} \).When \( t = 0, \ u = 1 \) and when \( t = \sqrt{3}, \ u = 4 \).Thus, the integral becomes:\[ \int_1^4 \frac{3}{2} \sqrt{u} \, du \]Integrate:\[ \frac{3}{2} \int_1^4 u^{1/2} \, du = \frac{3}{2} \left[ \frac{2}{3} u^{3/2} \right]_1^4 \]\[= \left[ u^{3/2} \right]_1^4 = 4^{3/2} - 1^{3/2} = 8 - 1 = 7 \]

Key Concepts

Parametric Equations

Parametric equations are a way to represent mathematical curves using a parameter, usually denoted as \( t \). These equations give us the coordinates \( (x, y) \) as functions of \( t \), which is useful for describing the path of an object in motion.
In the given exercise, we see the parametric equations \( x = t^3 \) and \( y = \frac{3t^2}{2} \).
These specify the position of a point as \( t \) changes from 0 to \( \sqrt{3} \). Instead of expressing \( y \) directly as a function of \( x \), parametric equations capture a more intuitive understanding of motion.
They allow for paths that loop back or self-intersect, unlike standard Cartesian equations.

• \( x = t^3 \) tells us how the \( x \)-coordinate changes with \( t \).
• \( y = \frac{3t^2}{2} \) describes the corresponding \( y \)-coordinate.

Differentiation

Differentiation is the process of finding a derivative, which represents the rate of change of a function with respect to its variable.
This concept is crucial when working with parametric equations to find the velocity components of the curve.
For the given parametric equations, we differentiate each with respect to \( t \).

• For \( x = t^3 \), the derivative is \( \frac{dx}{dt} = 3t^2 \).
• For \( y = \frac{3t^2}{2} \), the derivative is \( \frac{dy}{dt} = 3t \).

These derivatives provide us the components required for calculating the arc length.
They give the instantaneous rate of change of \( x \) and \( y \) as \( t \) changes.
This enables us to comprehend how the curve evolves in its path.

Integration

Integration is the reverse process of differentiation, often used to determine the area under a curve or the total accumulation of quantities over a given interval.
In the context of finding arc lengths, integration helps us sum up tiny segments of the curve over the interval.
After finding the derivatives \( \frac{dx}{dt} \) and \( \frac{dy}{dt} \), we use them in the arc length integral formula.

• \( L = \int_0^\sqrt{3} \sqrt{(3t^2)^2 + (3t)^2} \, dt \)

To solve this, we further simplify inside the square root and then integrate.
Using a substitution method, \( u = t^2 + 1 \), it transforms the integral into something manageable:

• \( \int_1^4 \frac{3}{2} \sqrt{u} \, du \)

Integration then gives us the total arc length, \( L = 7 \).
This process involves understanding both the algebraic manipulation and conceptual application of integration.

Arc Length Formula

The arc length formula provides a method for finding the length of a curve defined by parametric equations over a specified interval.
This is especially important for determining distances traveled along curved paths.
The formula is:

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

It is derived from the Pythagorean Theorem, accounting for the infinitesimal distance changes in \( x \) and \( y \) components as \( t \) changes.
In our example, we identified the derivatives \( \frac{dx}{dt} = 3t^2 \) and \( \frac{dy}{dt} = 3t \), then substituted them into the formula.
This gives an integral that computes the curve's length from \( t = 0 \) to \( t = \sqrt{3} \).
The resulting arc length of 7 units encompasses the distance along the curve from the beginning to the end of the parameter \( t \).