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Chapter 4: Problem 43

Describe in words the curve represented by the parametric equations $$x=3+t^{3}, \quad y=5-t^{3}$$

Short Answer

Expert verified
The curve is a straight line described by \(y = 8 - x\).

Step by step solution

01

Understand Parametric Equations

The given parametric equations describe the relationship between two variables, x and y, in terms of a third variable, t. These equations are: \(x = 3 + t^3\) and \(y = 5 - t^3\). The variable t is known as the parameter, and as it varies, it traces out a curve in the xy-plane.
02

Analyzing x and y Equations

Let's analyze each equation. For \(x = 3 + t^3\), as t changes, the value of x increases or decreases depending on t. Similarly, for \(y = 5 - t^3\), as t changes, y also varies correspondingly.
03

Eliminate Parameter t

To find the relation between x and y, eliminate the parameter t. From \(x = 3 + t^3\), we solve for \(t^3 = x - 3\). Similarly, from \(y = 5 - t^3\), we solve for \(t^3 = 5 - y\). Equating the two, we find: \(x - 3 = 5 - y\) or rearranging gives \(y = 8 - x\).
04

Interpret the Resulting Equation

The equation \(y = 8 - x\) is a linear equation representing a line with a slope of -1 on the xy-plane. It is a straight line that has an inverse relationship between x and y.
05

Describe the Curve

Since the parametric form can be continuously mapped to a straight line, the curve described by the given parametric equations is a line. The line passes through specific points in the plane defined by the equations, but overall it forms the line \(y = 8 - x\).

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Key Concepts

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

Parametric Equations Analysis
Parametric equations help us understand how a curve or a path is traced out by varying a parameter. In the given exercise, we have two equations: \(x = 3 + t^3\) and \(y = 5 - t^3\). Here, \(t\) is the parameter, and it plays a crucial role in plotting the path of the curve.
As \(t\) changes its value, the equations determine the corresponding x and y coordinates. This means each point on the curve is defined by a particular value of t. It's like having a set of instructions that tell you where to go as t changes.
  • For positive values of \(t\), \(x\) becomes larger than 3 while \(y\) decreases from 5.
  • For negative values of \(t\), \(x\) becomes smaller than 3 and \(y\) increases above 5.
Such equations are perfect for describing motion because they inherently include direction, making them useful in physics and engineering.
Parameter Elimination
To establish a connection directly between \(x\) and \(y\), we need to eliminate the parameter \(t\). By taking the given equations \(x = 3 + t^3\) and \(y = 5 - t^3\), we can determine the relationship between x and y.
Solve for \(t^3\) in both equations:
  • From \(x = 3 + t^3\), we get \(t^3 = x - 3\).
  • From \(y = 5 - t^3\), we find \(t^3 = 5 - y\).
By equating these two expressions for \(t^3\), we eliminate \(t\) and find the direct relationship: \[ x - 3 = 5 - y \]
After simplifying, the equation \(y = 8 - x\) is derived, showing a direct relationship excluding the parameter \(t\). Parameter elimination is essential in finding a familiar geometry form, such as a line or curve, from parametric equations.
Relationship Between x and y
Upon eliminating the parameter \(t\), we find a straightforward relationship between x and y in the form of \(y = 8 - x\). This equation represents a linear relationship, which is interpreted as a straight line when graphed on the xy-plane.
Here's what to note about this line:
  • The slope of the line is \(-1\), indicating that as \(x\) increases, \(y\) decreases. This is an inverse relationship.
  • The line crosses the y-axis at \(y = 8\), which is the intercept.
  • Every unit increase in x results in a unit decrease in y, maintaining the balance between x-offset and y-offset, which confirms the inverse nature of the relationship.
When the parametric equations describe dynamics or movements, finding the linear form like \(y = 8 - x\) helps simplify further analysis. In practical terms, understanding this relationship can aid in predicting trajectories or analyzing real-life motion scenarios.

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Most popular questions from this chapter

Every function of the form \(f(x)=a / x+b x,\) where \(a\) and \(b\) are nonzero constants, has two critical points.

(a) Find \(d^{2} y / d x^{2}\) for \(x=t^{3}+t, y=t^{2}\) (b) Is the curve concave up or down at \(t=1 ?\)

Two particles move in the \(x y\) -plane. At time \(t,\) the position of particle \(A\) is given by \(x(t)=4 t-4\) and \(y(t)=2 t-k,\) and the position of particle \(B\) is given by \(x(t)=3 t\) and \(y(t)=t^{2}-2 t-1.\) (a) If \(k=5,\) do the particles ever collide? Explain. (b) Find \(k\) so that the two particles do collide. (c) At the time that the particles collide in part (b), which particle is moving faster?

In many applications, we want to maximize or minimize some quantity subject to a condition. Such constrained optimization problems are solved using Lagrange multipliers in multivariable calculus;show an alternate method.With quantities \(x\) and \(y\) of two raw materials available, \(Q=x^{1 / 2} y^{1 / 2}\) thousand items can be produced at a cost of \(C=2 x+y\) thousand dollars. Using the following steps, find the minimum cost to produce 1 thousand items. (a) Graph \(x^{1 / 2} y^{1 / 2}=1 .\) On the same axes, graph \(2 x+y=2,2 x+y=3,\) and \(2 x+y=4\) (b) Explain why the minimum cost occurs at a point at which a cost line is tangent to the production curve \(Q=1\) (c) Using your answer to part (b) and implicit differentiation to find the slope of the curve, find the minimum cost to meet this production level.

Evaluate the limits where $$f(t)=\left(\frac{3^{t}+5^{t}}{2}\right)^{1 / t} \quad \text { for } t \neq 0$$ $$\lim _{t \rightarrow-\infty} f(t)$$
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