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The parametric equations describe the motion of a particle. Find an equation of the curve along which the particle moves. $$\begin{aligned} &x=t^{2}+3\\\ &y=t^{2}-2 \end{aligned}$$

Short Answer

Expert verified
The Cartesian equation of the curve is \( y = x - 5 \).

Step by step solution

01

Introduction to Parametric Equations

Parametric equations express variables such that both coordinates, usually defined as \( x \) and \( y \), are an outcome of a third variable, typically \( t \). Here, \( x = t^2 + 3 \) and \( y = t^2 - 2 \). We need to find a relationship between \( x \) and \( y \) that doesn't include \( t \).
02

Express \( t^2 \) from \( x \)

From \( x = t^2 + 3 \), solve for \( t^2 \): \[ t^2 = x - 3 \]
03

Express \( t^2 \) from \( y \)

From \( y = t^2 - 2 \), solve for \( t^2 \): \[ t^2 = y + 2 \]
04

Equate the expressions for \( t^2 \)

Now that \( t^2 \) can be represented in terms of \( x \) and \( y \), equate the two expressions: \[ x - 3 = y + 2 \]
05

Solve the equation for \( y \)

Rearrange the equation \( x - 3 = y + 2 \) to express \( y \) in terms of \( x \): \[ y = x - 5 \]
06

Write the Cartesian Equation

The equation \( y = x - 5 \) is the Cartesian equation representing the path of the particle. It describes a straight line with a slope of 1 and a y-intercept at -5.

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Key Concepts

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

Understanding Motion of a Particle
The motion of a particle can be represented using parametric equations, where a third variable, often denoted as \( t \), dictates the position of the particle over time. In this context, think of \( t \) as time. As time progresses, the values of \( x \) and \( y \) change, plotting a trajectory in the plane. This method helps in illustrating complex movements simply by expressing \( x \) and \( y \) as functions of \( t \).
For example, consider the parametric equations given in the exercise:
  • \( x = t^2 + 3 \)
  • \( y = t^2 - 2 \)
This setup reveals how both \( x \) and \( y \) are interconnected through the square of \( t \). As \( t \) varies, the particle moves along a path defined by these equations.
Parametric equations are particularly useful because they can describe a wide range of movements, including linear, circular, and more complicated paths.
Creating the Cartesian Equation
A Cartesian equation is a familiar form of representing curves, usually in the format \( y = f(x) \). To convert the parametric form into a Cartesian one involves eliminating the parameter \( t \). This allows us to describe the path taken by the particle in terms of only \( x \) and \( y \).
In our exercise, we've derived expressions from the parametric equations by isolating \( t^2 \):
  • From \( x = t^2 + 3 \), we find \( t^2 = x - 3 \).
  • From \( y = t^2 - 2 \), we find \( t^2 = y + 2 \).
With these, we equate \( x - 3 \) and \( y + 2 \) because both equal \( t^2 \):
\( x - 3 = y + 2 \)
By rearranging this equation, we convert it into the Cartesian form:
\( y = x - 5 \)
This equation is straightforward and represents a line, giving a clear path for the particle in Cartesian coordinates.
Exploring Curve Tracing
Curve tracing involves mapping out the path of the particle from its parametric description to a recognizable curve. Understanding the path is crucial for visualizing the motion.
Our Cartesian equation \( y = x - 5 \) now describes a straight line in the coordinate plane. By interpreting this:
  • The slope is \( 1 \), indicating a diagonal ascent in the positive direction.
  • The y-intercept is \(-5\), showing where the line crosses the y-axis.

To trace the particle's motion along this line, consider how the values of \( t \) influence the graph:
  • As \( t \) increases, \( x \) and \( y \) increase, with the curve moving diagonally upwards.
  • When \( t \) is negative, both \( x \) and \( y \) decrease but remain on the same linear trajectory.
This translation from parametric form to Cartesian allows us to easily visualize and comprehend the path of motion in a familiar coordinate system.

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Most popular questions from this chapter

Give an example of a function \(f\) that makes the statement true, or say why such an example is impossible. Assume that \(f^{\prime \prime}\) exists everywhere. \(f(x) f^{\prime \prime}(x)<0\) for all \(x\).

Explain what is wrong with the statement. Every function of the form \(f(x)=a / x+b x,\) where \(a\) and \(b\) are non-zero constants, has two critical points.

(a) Derive formulas for the first and second derivatives of the logistic function: \(y=\frac{L}{1+A e^{-k t}} \quad\) for \(L, A,\) and \(k\) positive constants. (b) Derive a formula for the \(t\) value of any inflection \(\operatorname{point}(\mathrm{s})\) (c) Use the second derivative to determine the concavity on either side of any inflection points.

A gas station stands at the intersection of a north-south road and an east- west road. A police car is traveling toward the gas station from the east, chasing a stolen truck which is traveling north away from the gas station. The speed of the police car is 100 mph at the moment it is 3 miles from the gas station. At the same time, the truck is 4 miles from the gas station going 80 mph. At this moment: (a) Is the distance between the car and truck increasing or decreasing? How fast? (Distance is measured along a straight line joining the car and the truck.) (b) How does your answer change if the truck is going 70 mph instead of 80 mph?

Let \(h(x)=e^{-x}+k x,\) where \(k\) is any constant. For what value(s) of \(k\) does \(h\) have (a) No critical points? (b) One critical point? (c) A horizontal asymptote?

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