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Eliminate the parameter \(t\) to find an equivalent equation with \(y\) in terms of \(x\). Give any restrictions on \(x\). Sketch the corresponding graph, indicating the direction of in- creasing \(t\). $$x=e^{-t}, \quad y=e^{t}, t \geq 0$$

Short Answer

Expert verified
The equivalent equation in terms of \(x\) is \(y= \frac{1}{x}\). The restriction on \(x\) is that \(x \neq 0\) and \(x \leq 1\). The graph is a hyperbola opening sideways and exists only in the interval (0,1], and the direction of increasing \(t\) is from right to left.

Step by step solution

01

Rewrite the equation in terms of \(t\)

The given equations are \(x=e^{-t}\) and \(y=e^{t}\). Start by rewriting one of the equations in terms of \(t\). By taking the natural logarithm of both sides in \(x=e^{-t}\), we can isolate \(t\) as follows: \(t = - \ln(x)\).
02

Substitute \(t\) in \(y=e^{t}\)

Now, replace \(t\) in the given second equation with the expression we found: \(y=e^{t}=e^{-\ln(x) } = \frac{1}{x}\). Hence, \(y=\frac{1}{x}\). This is the equivalent equation in terms of \(x\) instead of \(t\).
03

Identify restrictions on \(x\)

Looking at the derived function, \(y=\frac{1}{x}\), it's clear that \(x\) cannot be zero because division by zero is undefined. Therefore, the restriction on \(x\) is that \(x \neq 0\). Also, since \(t \geq 0\), we have \(x = e^{-t} \leq 1\). This means that \(x\) must be in the open interval (0, 1], not including 0 but including 1.
04

Sketch the graph

Sketch the graph of \(y=\frac{1}{x}\) on the interval (0,1] only. The direction of increasing \(t\) corresponds to moving from right to left because \(t\) increases as \(x\) decreases (since \(x=e^{-t}\)).

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Key Concepts

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

Understanding Parametric Equations
Parametric equations are a set of equations in which the system’s coordinates are expressed as functions of one or more independent variables, called parameters. Think of them as a way to describe a path or a curve in space by dictating how each point on the curve is placed based on a separate variable, typically time.

In our exercise example, the equations given are in parametric form, where both the x-coordinate and the y-coordinate of a point on the graph are described in terms of the parameter 't'. To create a regular equation with y in terms of x, you usually need to solve one of the parametric equations for the parameter and then substitute this into the other equation. This process is called 'eliminating the parameter'.

In practice, understanding parametric equations can be key for analyzing motion, as in physics, or for creating curves in computer graphics. By mastering this concept, students can tackle a wide variety of problems spanning multiple disciplines.
The Natural Logarithm's Role in Mathematics
The natural logarithm, written as \(\ln(x)\), is the logarithm to the base \(e\), where \(e\) is an irrational and transcendental constant approximately equal to 2.71828. This special logarithm is the inverse operation of the exponential function with base \(e\), which means that if you have \(e^y = x\), then you can express \(y\) as \(\ln(x)\).

In the step-by-step solution of our exercise, we used the natural logarithm to isolate the parameter \(t\) from the expression \(e^{-t} = x\). Taking the natural logarithm of both sides of this equation allowed us to solve for \(t\) and then substitute it back into the other parametric equation to eliminate the parameter.

Why Use Natural Logarithm?

The natural logarithm is particularly useful in calculus because it simplifies the differentiation and integration of exponential functions, which often arise in the modelling of real-world phenomena, such as population growth, compound interest, and radioactive decay.
Exploring Exponential Functions
Exponential functions are a class of mathematical functions denoted generally by \(f(x) = a^x\), where \(a\) is a positive constant (the base of the exponent), and \(x\) is the variable. These functions are characterized by a rate of growth that is proportional to the value of the function itself - the larger the value, the faster it grows.

Our exercise involves the specific exponential function \(e^{t}\), which is a function with the base \(e\). Its remarkable utility in mathematics stems from its rate of growth: the function grows at the same rate as its current value, hence it is exponentially increasing. In the case of \(e^{-t}\), the negative sign indicates an exponentially decreasing function.

Exponential Functions in Real Life

The concept of exponential functions extends far beyond the classroom. From computing complex compound interest to describing the spread of a virus in epidemiology, and even analyzing sound decibel levels, exponential functions are fundamental in describing scenarios where change occurs at a rate proportional to the current value.

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Most popular questions from this chapter

Graph each equation using a graphing utility. $$r=\frac{\sqrt{2}}{1+\sin \theta}$$

An elliptical track is used for training race horses and their jockeys. Under normal circumstances, two coaches are stationed in the interior of the track, one at each focus, to observe the races and issue commands to the jockeys. One day, when only one coach was on duty, a horse threw its rider just as it reached the vertex closest to the vacant observation post. The coach at the other post called to the horse, which dutifully came running straight toward her. How far did the horse run before reaching the coach if the minor axis of the ellipse is 600 feet long and each observation post is 400 feet from the center of the interior of the track?

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In this set of exercises, you will use hyperbolas to study real-world problems. Video Games A video game designer is creating a video game about a bank robbcry. In the game, a person robs a bank and then flecs. The robber's path forms one branch of a hyperbola. At the time of the robbery, the bank's security officer is stationed at one focus of the hyperbola and a police cruiser is parked at the opposite vertex (the vertex closer to the opposite focus). If the length of the transverse axis of the hyperbola is 70 millimeters and the foci are 120 millimeters apart, how far from the police cruiser was the bank's security officer when the robbery occurred?

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