/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 28 For each plane curve, find a rec... [FREE SOLUTION] | 91Ó°ÊÓ

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For each plane curve, find a rectangular equation. State the appropriate interval for \(x\) or \(y .\) $$x=e^{2 t}, y=e^{t}, \text { for } t \text { in }(-\infty, \infty)$$

Short Answer

Expert verified
The rectangular equation is \(x = y^2\) with \(y\) in \((0, \infty)\).

Step by step solution

01

Understand the Parameterization

The given parameterization of the curve is \(x = e^{2t}\) and \(y = e^t\). This means the curve is expressed in terms of the parameter \(t\). Our goal is to eliminate \(t\) and find a relationship directly between \(x\) and \(y\).
02

Express t in Terms of y

Since \(y = e^t\), we can take the natural logarithm on both sides to solve for \(t\): \(t = \ln(y)\).
03

Substitute t into the Equation for x

Now substitute \(t = \ln(y)\) into the equation for \(x\): \(x = e^{2t} = e^{2 \ln(y)}\).
04

Simplify the Expression for x

Using the property of exponents, \(e^{2 \ln(y)} = (e^{\ln(y)})^2 = y^2\). Thus, the rectangular equation of the curve is \(x = y^2\).
05

Determine the Appropriate Interval for y

Recall that \(y = e^t\), which is always positive for any real \(t\). Therefore, \(y\) ranges from \(0\) to \(\infty\), meaning the interval for \(y\) is \((0, \infty)\).

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Key Concepts

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

Parameterization
In mathematics, parameterization is a way to represent a curve using a parameter — often denoted as "t" — which can vary over a specific range. It allows the expression of coordinates, like
  • x = f(t)
  • y = g(t)
This provides a practical method to describe complex shapes and paths, as it simplifies the computation and understanding of the geometry of curves.
For example, in our exercise, we have the parameterization
  • \(x = e^{2t}\)
  • \(y = e^t\)
These expressions allow us to describe the path of a curve as "t" changes. The goal of parameterization is often to find a direct relationship between "x" and "y", by eliminating the parameter "t". Achieving this provides the rectangular equation, which is an equation only involving "x" and "y", making it easy to graph and interpret geometrically.
Exponential Functions
Exponential functions are fundamental in mathematics, defined by their form
  • \(f(x) = a^{(x)}\)
where "a" is a positive constant. The use of exponential functions is widespread because they describe many natural phenomena such as growth and decay.
In the given problem, both x and y are expressed using the exponential function base "e" — the natural exponential function:
  • \(x = e^{2t}\)
  • \(y = e^t\)
These expressions involve powers of "e," the Euler's number, capturing growth patterns efficiently. In particular, the expression \(x = e^{2t}\) can be simplified using logarithmic properties, transforming it into a classic relationship where \(x\) can be represented as \(y^2\), once the parameter t is eliminated and sub-operations have been applied.
This powerful transformation showcases how exponential expressions relate to one another and can be manipulated into simpler forms, aiding in the understanding and visualization of mathematical concepts.
Interval Notation
Interval notation is a systematic way to describe a set of numbers that lie within a specific range. It uses parentheses and brackets to show which of the endpoints are included in the interval. This method is concise and communicates clearly where a function or variable is valid.
In our example, we define the range for "y" by considering
  • \(y = e^t\)
Given that exponential functions with a base "e" are always positive yet unbounded as "t" ranges from negative to positive infinity, \(y\) will always be positive and increase indefinitely.
This continuous growth from a positive starting point without limit is expressed using interval notation as
  • \((0, \infty)\)
This interval notation provides a precise and efficient way to express that the values of "y" start just above zero and extend forever towards infinity, encapsulating the behavior of the function precisely.

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Most popular questions from this chapter

Solve each problem. Structure of an Atom In \(1911,\) Ernest Rutherford discovered the basic structure of the atom by "shooting" positively charged alpha particles with a speed of \(10^{7}\) meters per second at a piece of gold foil \(6 \times 10^{-7}\) meter thick. Only a small percentage of the alpha particles struck a gold nucleus head-on and were deflected directly back toward their source. The rest of the particles often followed a hyperbolic trajectory because they were repelled by positively charged gold nuclei. Thus, Rutherford proposed that the atom was composed of mostly empty space and a small, dense nucleus. The figure shows an alpha particle \(A\) initially approaching a gold nucleus \(N\) and being deflected at an angle \(\theta=90^{\circ}\) \(N\) is located at a focus of the hyperbola, and the trajectory of \(A\) passes through a vertex of the hyperbola. (a) Determine the equation of the trajectory of the alpha particle if \(d=5 \times 10^{-14}\) meter. (b) Approximate the minimum distance between the centers of the alpha particle and the gold nucleus. (GRAPH CAN'T COPY)

Graph each parabola by hand, and check using a graphing calculator. Give the vertex, axis, domain, and range. $$y=x^{2}+6 x+5$$

Find an equation for each hyperbola. Vertices \((-10,0)\) and \((10,0)\); asymptotes \(y=\pm 5 x\)

Particle When an alpha particle (a subatomic particle) is moving in a horizontal path along the positive \(x\) -axis and passes between charged plates, it is deflected in a parabolic path. If the plate is charged with 2000 volts and is 0.4 meter long, then an alpha particle's path can be described by the equation \(y=-\frac{k}{2 v_{0}} x^{2}\) where \(k=5 \times 10^{-9}\) is constant and \(v_{0}\) is the initial velocity of the particle. If \(v_{0}=10^{7}\) meters per second, what is the deflection of the alpha particle's path in the \(y\) -direction when \(x=0.4\) meter? (Source: Semat, H. and J. Albright, Introduction to Atomic and Nuclear Physics, Holt, Rinehart and Winston.)

Find an equation for each hyperbola. Vertices \((0,6)\) and \((0,-6)\); asymptotes \(y=\pm \frac{1}{2} x\)

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