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Chapter 7: Problem 17

In each of Exercises \(17-24,\) calculate the length of the given parametric curve. $$ x=3 t-7 \quad y=5-4 t \quad-1 \leq t \leq 1 $$

Short Answer

Expert verified
The length of the parametric curve is 10.

Step by step solution

01

Understand the Curve

The given parametric equations are \(x = 3t - 7\) and \(y = 5 - 4t\). These equations define a straight line in the plane, as both \(x\) and \(y\) are linear functions of \(t\). We need to find the length of this line segment as \(t\) varies between \(-1\) and \(1\).
02

Find Derivatives

Calculate the derivatives of \(x\) and \(y\) with respect to \(t\). These are \(\frac{dx}{dt} = 3\) and \(\frac{dy}{dt} = -4\). These derivatives represent the rates of change of \(x\) and \(y\) with respect to \(t\).
03

Set Up Arc Length Integral

The formula for the arc length of a parametric curve from \(t=a\) to \(t=b\) is \(L = \int_a^b \sqrt{\left(\frac{dx}{dt}\right)^2 + \left(\frac{dy}{dt}\right)^2} \, dt\). Substitute the derivatives found in Step 2 into this formula. Since \(a = -1\) and \(b = 1\), we get \(L = \int_{-1}^{1} \sqrt{3^2 + (-4)^2} \, dt\).
04

Simplify the Integrand

Simplify the expression under the square root: \(\sqrt{9 + 16} = \sqrt{25} = 5\). This means the integral simplifies to \(L = \int_{-1}^{1} 5 \, dt\).
05

Evaluate the Integral

Evaluate the integral \(L = \int_{-1}^{1} 5 \, dt\). This integral is \(5\times [t]_{-1}^{1} = 5 \times (1 - (-1)) = 5 \times 2 = 10\). This is the length of the curve.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Parametric Equations
Parametric equations are a way to express two or more variables in terms of a single parameter, often denoted by \( t \). This method is particularly useful in describing curves and paths that an object may follow.
  • For a two-dimensional curve, the position of a point is given by two parametric equations, \( x = f(t) \) and \( y = g(t) \).
  • Each value of \( t \) gives a specific point \((x, y)\) on the curve.
  • This representation is different from the usual \( y = f(x) \) equation because it defines both \( x \) and \( y \) independently based on the parameter \( t \).
In the example problem, parametric equations \( x = 3t - 7 \) and \( y = 5 - 4t \) define a straight line as \( t \) ranges from \(-1\) to \(1\), highlighting the ease with which linear paths can be defined using parametric equations.
Arc Length Integral
The arc length of a curve can be calculated using the arc length integral. This method gives the total distance traveled along a curve between two points.
  • The formula for the arc length of a parametric curve \( x = f(t) \), \( y = g(t) \), from \( t = a \) to \( t = b \) is \[ L = \int_a^b \sqrt{\left(\frac{dx}{dt}\right)^2 + \left(\frac{dy}{dt}\right)^2} \, dt \].
  • This expression under the integral sign combines the rates of change of both \( x \) and \( y \) to find the path taken by the curve.
In the given solution, this integral was specifically used to determine the length of the straight line defined by the parametric equations. The calculation involved evaluating a definite integral over the interval from \(-1\) to \(1\). This interval reflects the specific range of \( t \) given in the exercise.
Derivatives of Parametric Equations
Derivatives with respect to the parameter \( t \) play a critical role in analyzing parametric equations. They help us understand the curve's behavior.
  • The derivative \( \frac{dx}{dt} \) is the rate at which \( x \) changes as \( t \) changes.
  • Similarly, \( \frac{dy}{dt} \) is the rate at which \( y \) changes with \( t \).
By differentiating the parametric equations \( x = 3t - 7 \) and \( y = 5 - 4t \), we get \( \frac{dx}{dt} = 3 \) and \( \frac{dy}{dt} = -4 \).These values indicate how steeply \( x \) and \( y \) change, respectively. More importantly, they are plugged into the arc length integral to measure the curve's total length.
Straight Line Segment
A straight line segment is the simplest type of curve and often serves as an introduction to more complex curve calculations.
  • It is defined by two endpoints and shows a constant slope between those points.
  • The length of a straight line segment can often be calculated directly using the distance formula without integration.
In this exercise, even though parametric integration was used to find the arc length, recognizing the parametric equations define a straight line shows that simpler calculations might suffice. The understanding of parametric equations still forms the basis for working comfortably with more intricate curves.

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Most popular questions from this chapter

For what a, \(0 \leq a \leq \pi,\) does the function \(f(x)=\sin (x)-\) \(\cos (x)\) have the greatest average over the interval \([a, a+\pi]\) of length \(\pi ?\)

In each of Exercises \(65-74\) calculate the expectation of a random variable whose probability density function is given. $$ \frac{3}{16} \sqrt{4-x} \quad 0 \leq x \leq 4 $$

If \(P(X \leq m)=P(X \geq m)=1 / 2,\) then we say that \(m\) is the median of a random variable \(X .\) In each of Exercises \(75-78\), calculate the median of a random variable whose probability density function \(f\) is given. $$ f(x)=e^{1-x} /(e-1) \quad 0 \leq x \leq 1 $$

If \(P(X \leq m)=P(X \geq m)=1 / 2,\) then we say that \(m\) is the median of a random variable \(X .\) In each of Exercises \(75-78\), calculate the median of a random variable whose probability density function \(f\) is given. $$ f(x)=\cos (x) \quad 0 \leq x \leq \pi / 2 $$

In each of Exercises \(89-92,\) a function \(f\) and an interval \(I\) are given. Calculate the average \(f_{\text {avg }}\) of \(f\) over \(I,\) and find a value \(c\) in \(I\) such that \(f(c)=f_{\text {avg. }}\) State your answers to three decimal places. $$ f(x)=\sin \left(\pi\left(x^{2}-x^{3}\right)\right) \quad I=[0,1] $$
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