Problem 37 - Calculus Early Transcendentals

Question

Set up an integral that represents the length of the curve. Then use your calculator to find the length correct to four decimal places. $$x=t-t^{2}, \quad y=\frac{4}{3} t^{3 / 2}, \quad 1 \leqslant t \leqslant 2 $$

Step-by-Step Solution

Answer

The curve length is approximately 2.4129.

Step 1: Understand the problem

We are asked to find the length of the curve given by the parametric equations \(x = t - t^2\) and \(y = \frac{4}{3}t^{3/2}\) for \(t\) ranging between 1 and 2.

Step 2: Formula for arc length of parametric equations

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

Step 3: Find the derivatives

First, we need to find \(\frac{dx}{dt}\) and \(\frac{dy}{dt}\).- \(x(t) = t - t^2\): \(\frac{dx}{dt} = 1 - 2t\)- \(y(t) = \frac{4}{3}t^{3/2}\):\(\frac{dy}{dt} = 2t^{1/2}\)

Step 4: Set up the integral

Insert the derivatives into the arc length formula:\[ L = \int_1^2 \sqrt{(1 - 2t)^2 + (2t^{1/2})^2} \, dt \]Simplify the inside of the square root:\[ L = \int_1^2 \sqrt{1 - 4t + 4t^2 + 4t} \, dt \]Continue simplifying:\[ L = \int_1^2 \sqrt{4t^2 + 1} \, dt \]

Step 5: Calculate the integral using a calculator

Use a calculator to evaluate the integral:\[ L = \int_1^2 \sqrt{4t^2 + 1} \, dt \]Calculating this integral, we find that the length of the curve is approximately 2.4129.

Key Concepts

Parametric Equations

Parametric equations allow us to describe a curve or a geometric shape using a third variable, usually denoted by \( t \). This is different from the traditional way of expressing functions with symbols like \( y = f(x) \). Each of the coordinates, \( x \) and \( y \), is expressed as a separate function of the parameter \( t \). This method is particularly useful when the relationship between \( x \) and \( y \) cannot be easily expressed as a single function. For example, Lissajous curves and the trajectory of projectiles are best modeled using parametric equations. In this exercise, the given equations \( x = t - t^2 \) and \( y = \frac{4}{3}t^{3/2} \) describe a path by connecting points as \( t \) varies from 1 to 2. This allows a more comprehensive description for complex and smooth curves.

Integral Calculus

Integral calculus is a branch of mathematics that deals with the accumulation of quantities and the areas under or between curves. In the context of this exercise, we're using integral calculus to find the length of a curve represented by parametric equations. By setting up an integral, we can calculate the entire length of the curve between two points. If we think of derivatives as the rate of change, in integral calculus, we're trying to "add up" small bits of changes to find a total. When faced with a parametric curve, integrating gives us the total distance along the curve by "adding" up all the infinitesimally small segments of the curve between the bounds \( t = 1 \) and \( t = 2 \). This technique shows how powerful integral calculus is in understanding and solving real-world problems.

Derivatives

Derivatives measure how a function changes as its input changes—essentially giving us the "slope" or "rate of change" at any point along the curve. In this exercise, our task is to find the derivatives of the parametric functions \( x(t) = t - t^2 \) and \( y(t) = \frac{4}{3}t^{3/2} \). The derivative for \( x \) with respect to \( t \) is calculated as \( \frac{dx}{dt} = 1 - 2t \), which tells us how quickly the \( x \)-coordinate is changing as \( t \) changes. Similarly, the derivative \( \frac{dy}{dt} = 2t^{1/2} \) informs us about the rate of change of the \( y \)-coordinate. These derivatives are crucial because they are used in the arc length formula to determine the overall change and shape of the curve.

Arc Length Formula

The arc length formula helps us determine the length of a curve defined by parametric equations. The formula is given by:

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

This equation takes into account the changes in both the \( x \) and \( y \) direction as \( t \) varies from \( a \) to \( b \). By finding the square root of the sum of the squares of the derivatives \( \frac{dx}{dt} \) and \( \frac{dy}{dt} \), we essentially find the hypotenuse of tiny right triangles formed by horizontal and vertical changes along the curve. Then, integrating over an interval \([a, b]\) sums up these infinitesimally small changes to compute the total length of the curve. In this specific exercise, the formula is simplified and evaluated over the interval from 1 to 2, using numerical methods or a calculator to approximate the length of the curve.