Problem 37 - Calculus Early Transcendentals: Pearson New International Edition

Question

, find the length of the parametric curve defined over the given interval. $$ x=t, y=t^{3 / 2} ; 0 \leq t \leq 3 $$

Step-by-Step Solution

Answer

The length of the parametric curve is approximately 4.63 units.

Step 1: Understand the Formula for Curve Length

The length of a parametric curve defined by \(x = f(t)\) and \(y = g(t)\) from \(t=a\) to \(t=b\) is given by the integral \( L = \int_{a}^{b} \sqrt{ \left( \frac{dx}{dt} \right)^2 + \left( \frac{dy}{dt} \right)^2 } \, dt \). In this case, \(x = t\) and \(y = t^{3/2}\). The interval given is \(0 \leq t \leq 3\).

Step 2: Compute Derivatives

Calculate the derivatives \(\frac{dx}{dt}\) and \(\frac{dy}{dt}\). We have \(\frac{dx}{dt} = 1\) since \(x = t\). For \(\frac{dy}{dt}\), compute the derivative of \(y = t^{3/2}\), which is \(\frac{dy}{dt} = \frac{3}{2}t^{1/2}\).

Step 3: Substitute Derivatives into Formula

Substitute \(\frac{dx}{dt} = 1\) and \(\frac{dy}{dt} = \frac{3}{2}t^{1/2}\) into the length integral. This gives: \[ L = \int_{0}^{3} \sqrt{1^2 + \left(\frac{3}{2}t^{1/2}\right)^2} \, dt = \int_{0}^{3} \sqrt{1 + \frac{9}{4}t} \, dt. \]

Step 4: Simplify the Integral

To evaluate \(\int_{0}^{3} \sqrt{1 + \frac{9}{4}t} \, dt\), use a substitution if needed, or solve directly if it matches a standard form. Let \(u = 1 + \frac{9}{4}t\), then \(du = \frac{9}{4} dt\) which means \(dt = \frac{4}{9}du\).

Step 5: Evaluate the Integral

Under the substitution, when \(t = 0\), \(u = 1\); when \(t = 3\), \(u = 1 + \frac{9}{4} \times 3 = \frac{21}{4}\). The integral becomes \( L = \frac{4}{9} \int_{1}^{\frac{21}{4}} \sqrt{u} \, du = \frac{4}{9} \left[ \frac{2}{3} u^{3/2} \right]_{1}^{\frac{21}{4}} \).

Step 6: Perform Final Calculations

Evaluate \(\frac{4}{9} \times \frac{2}{3} \left( \left(\frac{21}{4}\right)^{3/2} - 1^{3/2} \right)\). Simplify each part to find the result: \(L = \frac{8}{27} \left( \left(\frac{21}{4}\right)^{3/2} - 1 \right)\). Calculate the numerical value for the final answer.

Key Concepts

Parametric Curves

When we talk about parametric curves, we are dealing with a way to represent curves through equations involving a third variable, usually denoted as "t." These equations express the coordinates \(x\) and \(y\) as functions of \(t\). In the given exercise, you have \(x = t\) and \(y = t^{3/2}\). Here, \(t\) is known as the parameter, and it dictates how the point moves along the curve.

• Independent Path: Parametric equations allow you to control both coordinates independently of each other.
• Flexibility: You can easily describe motion in a plane with parametric equations, including loops, tight curves, and complex motions not possible with standard \(y = f(x)\) representations.

This form of representing curves is powerful, especially when dealing with non-straight paths or dynamic processes.

Curve Length

The challenge with parametric curves is finding the length of the curve over a given interval. Let's take a closer look at this concept. The curve's length is essentially a measure of how far you would travel if you walked along the curve from one point to another.
To calculate this, we use a specific integral formula: \[ L = \int_{a}^{b} \sqrt{ \left( \frac{dx}{dt} \right)^2 + \left( \frac{dy}{dt} \right)^2 } \, dt \]Here, \(a\) and \(b\) are the interval limits. It's important to square the derivatives before adding them under the square root. This formula derives from the Pythagorean theorem, representing the incremental change in distance traveled as "you move slightly along the curve."

• Precision: It calculates the actual path walked along, including tiny turns and twists.
• Applications: Curve length is useful in physics, engineering, and any field dealing with real-world trajectory or path calculations.

Integral Calculus

Integral calculus allows us to calculate the total accumulation of quantities. In the context of this exercise, we use integral calculus to compute the curve's length by summing up infinitesimally small pieces of the curve.
The integral used in finding curve lengths is definite, meaning it has bounds: from \(t=a\) to \(t=b\). You adjust these bounds to find the curve's length over a specific interval.

• Definite Integrals: These provide the accumulated value, crucial when finding lengths, areas, and volumes.
• Substitution: Frequently used to simplify or solve integrals by changing variables.

In the exercise, substitution was employed to evaluate the integral, involving a strategic change of variable, making the integral more manageable.

Derivatives

Derivatives play a vital role in calculating the curve length of parametric curves. They tell us how \(x\) and \(y\) change with respect to the parameter \(t\). These changes are essential in understanding the curve's geometry over the interval.
In the given problem, we compute two derivatives:- \( \frac{dx}{dt} = 1 \) because \(x = t\) changes uniformly with \(t\).- \( \frac{dy}{dt} = \frac{3}{2}t^{1/2} \) gives the rate at which \(y\) changes with respect to \(t\). This derivative shows how \(y\) grows more rapidly as \(t\) increases, represented in the exercise where \(y = t^{3/2}\).

• Rate of Change: Derivatives measure instantaneous rates of change, crucial for understanding dynamics.
• Applications: Beyond mathematics, derivatives apply to velocity, acceleration, and many real-world changing phenomena.

Accurate derivative calculations are essential for plugging into the curve length formula and ensuring precision in the result.