Problem 43 - Essential Calculus

Question

Write an integral that represents the arc length of the curve on the given interval. Do not evaluate the integral. $$ x=e^{t}+2, \quad y=2 t+1 \quad-2 \leq t \leq 2 $$

Step-by-Step Solution

Answer

The integral that represents the arc length of the curve over the interval from -2 to 2 is \(L = \int_{-2}^{2} \sqrt{e^{2t} + 4} dt\)

Step 1: Calculate the derivatives of x(t) and y(t)

The function \(x(t) = e^{t} + 2\), differentiating it with respect to \(t\) gives \(\frac{dx}{dt} = e^{t}\). The function \(y(t) = 2t + 1\), and its derivative with respect to \(t\) is \(\frac{dy}{dt} = 2\).

Step 2: Insert the derivatives into the arc length formula

Place these derivative values into the arc length formula, i.e., \(L = \int_{a}^{b} \sqrt{(\frac{dx}{dt})^{2} + (\frac{dy}{dt})^{2}} dt\), where the interval is \(t = -2\) to \(t = 2\). So we get \(L = \int_{-2}^{2} \sqrt{(e^{t})^{2} + (2)^{2}} dt\)

Step 3: Simplify the expression under the square root

Simplifying, \(L = \int_{-2}^{2} \sqrt{e^{2t} + 4} dt\) would be the integral that represents the arc length of the curve

Key Concepts

Differentiation

Differentiation is a fundamental concept in calculus, involving the computation of the derivative of a function. The derivative represents the rate at which the function's value is changing at any given point. In our exercise, we are given parametric equations:\[ x(t) = e^t + 2 \] and \[ y(t) = 2t + 1. \] To find how these curves change, we compute their derivatives with respect to \(t\).

• For \(x(t) = e^t + 2\), the derivative \( \frac{dx}{dt} = e^t \).
• For \(y(t) = 2t + 1\), the derivative \( \frac{dy}{dt} = 2 \).

Calculating these derivatives helps us find the rate of change of each variable concerning the parameter \(t\). These rates of change are crucial for further calculations, like finding arc length.
Differentiation is not only limited to parametric equations but applies across all functional forms, helping us understand and predict behavior in various fields, from physics to economics.

Integral Calculus

Integral calculus deals with the accumulation of quantities and the spaces under and between curves. In our context, we use integral calculus to determine the arc length of a curve.
The arc length formula for parametric equations \( x(t) \) and \( y(t) \) over an interval \([a, b]\) is expressed as:
\[ L = \int_{a}^{b} \sqrt{\left(\frac{dx}{dt}\right)^2 + \left(\frac{dy}{dt}\right)^2} \, dt. \]
In the given exercise, we substitute the calculated derivatives:
\[ \frac{dx}{dt} = e^t \] and \[ \frac{dy}{dt} = 2 \], resulting in the integral:
\[ L = \int_{-2}^{2} \sqrt{(e^t)^2 + 2^2} \, dt. \]
This integral calculates the distance along the curve between \(t = -2\) and \(t = 2\), without directly evaluating it. Integral calculus not only helps in finding arc lengths but also in computing areas, volumes, and other quantities where accumulation is needed.

Parametric Equations

Parametric equations express the coordinates of the points on a curve as functions of a parameter, often denoted as \(t\). This is useful when dealing with curves in which \(x\) and \(y\) aren't easily related by a simple function.
In the given exercise, the equations \(x(t) = e^t + 2\) and \(y(t) = 2t + 1\) define the path of a curve as \(t\) changes. Instead of expressing \(y\) solely as a function of \(x\), each coordinate is given by its own equation in terms of the parameter \(t\).
Using parametric equations provides flexibility and is particularly useful in modeling realistic situations where different components have their own behavior over time or another variable. They are essential in calculus for computing concepts like arc length, as seen in this exercise, because they allow us to break down the behavior of curves in terms of simple parameterized changes.