Motivating Questions
-
What is a vector-valued function? What do we mean by the graph of a vector-valued function?
-
What is a parameterization of a curve in \(\R^2\text{?}\) In \(\R^3\text{?}\) What can the parameterization of a curve tell us?
Section 10.1 Vector-Valued Functions of One Variable
Subsection 10.1.1 Introduction
In our previous work, we have seen several examples of curves in space, such as lines in three dimensions and conic sections (in two dimensions). For a line through a fixed point \(\vr_0\) in the direction of vector \(\vv\), we can express the line parametrically through the single vector equation
\begin{equation*}
\vr(t) = \vr_0 + t \vv.
\end{equation*}
We call \(\vr(t)\) a vector-valued function of one variable (or simply a vector-valued function, when the the single variable aspect is clear in context) because there is one scalar input (\(t\)) and the output of \(\vr\) is a vector that will vary based on the value of the scalar input. When the output vectors are graphed with initial point at the origin, the terminal points of these vector outputs trace out the line in space. Figure 9.5.5 and Figure 9.5.9 provide illustrations of this idea.
Similar to lines, other curves in space are classified as one-dimensional objects because there is only one way to move along this object. We refer to this direction as forward/backward depending on whether motion is toward larger (forward) or smaller (backward) values of the input variable. We will similarly express the coordinates of points on a curve in terms of a single variable. This expression will describe movement along this path. For instance, the graph of an ellipse, such as the one shown in Figure 10.1.1, cannot be expressed with \(y\) as a function of \(x\text{.}\) However, we still classify an ellipse as a one-dimensional graph because at any point on the graph, there is only one dimension to move along the graph.
Vector-valued functions of one variable are the perfect vehicle for describing curves in general. We use vectors starting at the origin to identify points in space and the terminal points of these vectors trace out a curve in space. This approach allows us to draw a wide variety of graphs in two- and three-dimensional space. We can also describe curves in \(\R^n\) for any \(n\text{.}\) This same approach to drawing curves will also allow us to represent slices of surfaces in space, which we will discuss more in Activity 10.1.4 and many times in Chapter 11.
Preview Activity 10.1.1.
After graduating, you decide to start a self-driving car company because it didn’t look too hard based on a few articles you scrolled past online. You decide to call your company Steer Clear and start working on the self-driving part of the car. Your first task is to understand how the location tracking equipment you bought online works. There are two parts to your location tracking system: a receiver box and a tracker that you put in the object you want to track.
(a)
After putting the tracker in your pocket, you get in a car and drive around the receiver box. When you download the data about your drive, you notice that the software is outputting a vector from the receiver box to the tracker. The software also seems to love math, because these vectors are written in terms of trig functions.
Below are some of the output vectors from the software. Evaluate each of these vectors and draw these vectors with initial point at the origin on the axes in Figure 10.1.2.
-
\(\displaystyle \langle \cos(0), \sin(0) \rangle\)
-
\(\displaystyle \langle \cos\left(\frac{\pi}{2}\right), \sin\left(\frac{\pi}{2}\right)\rangle\)
-
\(\displaystyle \langle \cos(\pi), \sin(\pi) \rangle \)
-
\(\displaystyle \langle \cos\left(\frac{3\pi}{2}\right), \sin\left(\frac{3\pi}{2}\right)\rangle\)
(b)
Below are a few more output vectors from your software. Evaluate each of these vectors and draw each of these vectors with initial point at the origin on the axes in Figure 10.1.2.
-
\(\displaystyle \langle \cos\left(\frac{\pi}{4}\right), \sin\left(\frac{\pi}{4}\right) \rangle\)
-
\(\displaystyle \langle \cos\left(\frac{3\pi}{4}\right), \sin\left(\frac{3\pi}{4}\right) \rangle\)
-
\(\displaystyle \langle \cos\left(\frac{5\pi}{4}\right), \sin\left(\frac{5\pi}{4}\right) \rangle\)
-
\(\displaystyle \langle \cos\left( \frac{7\pi}{4}\right), \sin\left(\frac{7\pi}{4}\right) \rangle\)
(c)
Based on the data you are seeing from the two sets of vector outputs of your software, it appears that the output of your software is the set of vectors of the form \(\langle \cos(t), \sin(t) \rangle\text{,}\) where \(t\) goes from \(0\) to \(2 \pi\text{.}\) Add to Figure 10.1.2 a sketch the set of terminal points (only) of vectors \(\langle \cos(t), \sin(t) \rangle\) plotted in standard position (with initial point at the origin). You should allow \(t\) to assume values from \(0\) to \(2 \pi\text{.}\) Write a couple sentences about what path your drive took.
(d)
What part of the path you drove is described by \(\langle \cos(t), \sin(t) \rangle\text{,}\) where \(t\) goes from 0 to \(\pi\text{?}\) What would the path be if you drove along \(\langle \cos(t), \sin(t) \rangle\) for \(t\) from 0 to \(4 \pi\text{?}\)
In the preview activity, we saw how location data for how an object is moving along a curve can be given by vectors with a common initial point. In this section, we will continue to look at the implications of using this as an algebraic description of movement along a curve. In particular, we will look at what algebraic rules and definitions for functions using the output vectors as points and we will conclude this section by generating several families of examples.
Subsection 10.1.2 Vector-Valued Functions
As in Preview Activity 10.1.1, we can think of a point on the curve shown in Figure 10.1.3 as the terminal point of a vector starting at the origin. As the point we are looking at moves along the curve, the vector changes to terminate at the desired point. Figure 10.1.3 shows a curve in space with a vector from the origin to a highlighted point on the curve. Use the slider to change the location on the curve to see how changing the scalar input gives different outputs with terminal points on the curve.
This correspondence between a location and a vector starting at the origin will allow us to think of a curve as a collection of terminal points of vectors emanating from the origin. We can describe an object traveling along this curve by defining a function \(\vr\) whose input is the variable \(t\) and whose output is the vector from the origin to the point on the curve at time \(t\text{.}\) In so doing, we have introduced a new type of function, one whose input is a scalar and whose output is a vector.
The terminal points of the output vectors of \(\vr\) will trace out the curve in space. From this perspective, the \(x\text{,}\) \(y\text{,}\) and \(z\) coordinates of the points on the curve are functions of time, \(t\text{,}\)
\begin{equation*}
x = x(t), \quad y = y(t), \quad \text{ and } \quad z = z(t)\text{,}
\end{equation*}
and thus we have three coordinate functions that represent the collection of points on the curve. The variable \(t\) is called a parameter and the equations \(x = x(t)\text{,}\) \(y = y(t)\text{,}\) and \(z = z(t)\) are called parametric equations (or a parameterization of the curve). The function \(\vr\) whose output is the vector from the origin to a point on the curve is defined by
\begin{equation*}
\vr(t) = \langle x(t), y(t), z(t) \rangle\text{.}
\end{equation*}
Definition 10.1.4.
A vector-valued function of one variable is a function whose input is a real parameter \(t\) and whose output is a vector that depends on \(t\text{.}\) The graph of a vector-valued function is the set of all terminal points of the output vectors with their initial points at the origin.
Parametric equations for a curve in space are equations of the form
\begin{equation*}
x = x(t), \quad y = y(t), \quad \text{ and } \quad z = z(t)
\end{equation*}
that describe the \((x,y,z)\) coordinates of a point on a curve in \(\R^3\text{.}\)
Note that every set of parametric equations determines a vector-valued function of the form
\begin{equation*}
\vr(t) = \langle x(t), y(t), z(t) \rangle
\end{equation*}
and every vector-valued function defines a set of parametric equations for a curve. Moreover, we can consider vector-valued functions as parameterizations in \(\R^2\text{,}\) \(\R^3\text{,}\) or any dimension.
As a reminder, in Section 9.5, we determined the parametric equations of a line in space using a point and a direction vector. For a nonlinear example, the curve in Figure 10.1.3 has the parametric equations
\begin{equation*}
x(t) = \cos(t), \quad y(t) = \sin(t), \quad \text{ and } \quad z(t)
= \cos(t) \sin(t)\text{.}
\end{equation*}
We represent this as the vector-valued function \(\vr\) given by
\begin{equation*}
\vr(t) = \langle \cos(t), \sin(t), \cos(t) \sin(t) \rangle\text{.}
\end{equation*}
Before we start working on parameterizations of curves, you may be asking yourself “Why can’t we just have the output of our function be a point? Why do vectors need to be involved?!?” This is a good question because simpler tends to be better. However, using vectors means we will have significantly more tools at our disposal. These vector tools will make calculations easier and help us make geometric sense of measurements throughout the rest of this course. For example, we used a constant vector to indicate the direction of a line and this vector is related to the location vectors along the path. Remember that we already have vector tools that allow us to separate measurements into different pieces (like magnitude or direction), determine how much of a vector is in a particular direction, or determine how much of a vector is orthogonal to another direction.
Activity 10.1.2.
The same curve can be represented with different parameterizations. Use appropriate technology to plot each of the curves generated by the following vector-valued functions for values of \(t\) from \(0\) to \(2 \pi \text{.}\) For each example, you should write a few sentences to explain how the graphs are alike and how they are different. Be sure to pay attention to how the parameter values are related to different points on the curve.
(a)
\(\vr(t) = \langle \sin(t), \cos(t) \rangle\)
(b)
\(\vr(t) = \langle \sin(2t), \cos(2t) \rangle\)
(c)
\(\vr(t) = \langle \cos(t+\pi), \sin(t+\pi) \rangle\)
(d)
\(\vr(t) = \langle \cos(t^2), \sin(t^2) \rangle\)
A mantra we will use when parameterizing curves is “There is more than one way to walk the same path”, which means there is not a unique way to parameterize a curve in space. The examples in Activity 10.1.2 illustrate that a parameterization allows us not only to describe a curve in space as the graph of a vector-valued function of one variable but also to specify the direction and speed at which the graph is traversed as \(t\) changes. In the different parameterizations of the circle, we see that we can start at different points and move around the circle in either direction. In Section 10.2, we will begin to investigate the calculus of vector-valued functions. This will enable us to precisely quantify the direction, speed, and acceleration of an object moving along a curve in space. As such, describing curves parametrically will allow us not only to indicate the points on the curve itself but also to describe motion along the curve.
While the ideas of a path or curve in space are closely related to a parameterization, we will take a moment here to clarify the precise meaning of these terms. A curve in space refers to the collection of points on the path. A curve does not have a specific direction to be traveled or even a particular way to move along points. A parameterization is an explicit description for how to travel along the points on the curve in terms of the parameter \(t\text{.}\) A common analogy we will use is that a curve in space would be represented by a bike path or road while a parameterization would be a description of how you took a trip along this bike path or road as a function of time.
In general, we will avoid having parameterizations that pause at any point on the curve or reverse their direction along the curve because many of the measurements we consider in this chapter will be undefined when this occurs. It is common in practice to split parameterizations up into separate pieces that correspond to the intervals between where these pauses or direction changes happen.
The graph of a vector-valued function of one variable is the curve traced out by the endpoints of the output vectors, but this graph does not typically have parameter values labeled. This means that there is significant information lost when considering only the graph of a vector-valued function because no information about how the path was traveled/parameterized is available.
In the next subsection, we will cover some ideas that will be useful in parameterizing curves. First, however, we will address a couple of ideas related to the new category of functions we will be studying. Remember that the key idea behind functions is that for each input there is exactly one output assigned to that input. The domain of a function is the set of allowed inputs for the function. Sometimes a restricted set will be given that limits what inputs should be considered. Unless otherwise stated, however, the domain of a function will be the largest set of inputs for which the function is defined. Common ways for scalar functions to not be defined for a particular input include:
-
dividing by zero,
-
taking the square root of negative numbers, and
-
restricted domains from other functions like \(\arcsin\) or logarithms.
The domain of a vector-valued function of one variable will be the subset of \(\R\) where all of the component functions are defined. In other words, the domain of a vector-valued function of one variable is the intersection of the domains for the component functions, because that is the set of inputs for which the vector output will be defined.
Example 10.1.5.
Let
\begin{equation*}
\vr(t) = \left\langle \frac{-1}{t-2},\, \sqrt{25-t^2},\, \ln(t+1) \right\rangle\text{.}
\end{equation*}
The domains of the component functions are \(\{x \in \R | x\neq 2\}\) for the \(x\)-component, \([-5,5]\) for the \(y\)-component, and \((-1,\infty)\) for the \(z\)-component. The function \(\vr\) has an output for only the parameter values for which all three component functions are defined. Thus, the domain of \(\vr(t)\) is \((-1,2)\cup(2,5]\text{.}\)
Subsection 10.1.3 Parameterizing Curves
We now discuss how to parameterize several types of curves in space. An important idea that will be used throughout the rest of the text is the way that different relationships of between coordinates can be expressed or understood, such as expressing one coordinate as a function of the other or relationships such as trig identities between the coordinate values.
Using parametric equations to define vector-valued functions in two dimensions is a more versatile way of describing curves than defining \(y\) as a function of \(x\text{.}\) In fact, if \(y = f(x)\) is a function of \(x\text{,}\) then we can parameterize the graph of \(f\) by
\begin{equation*}
\vr(t) = \langle t, f(t) \rangle\text{.}
\end{equation*}
Hence, the graph of every single-variable function can be described parametrically. In addition, as we saw in Preview Activity 10.1.1 and Activity 10.1.2, we can use vector-valued functions to represent curves in the plane that cannot be described with \(y\) as a function of \(x\) (or \(x\) as a function of \(y\)).
Example 10.1.6. Parameterizing \(y=f(x)\).
Consider the parabola given by \((x-2)^2=8(y+1)\text{.}\) From Figure 10.1.7, you can see that the graph of \((x-2)^2=8(y+1)\) passes the vertical line test. Therefore, we can write the \(y\)-coordinate as a function of the \(x\)-coordinate. Specifically, we isolate \(y\) and obtain \(y=f(x)=\frac{(x-2)^2}{8}-1\text{.}\) Using the idea above, we set \(x=t\) to be our parameter. Now we can express both coordinates of points on our curve as functions of \(t\text{.}\) In particular, setting \(x(t)=t\) means that we can write \(y\) as a function of \(t\) because \(x=t\text{.}\) Thus \(x(t)=t\) and \(y(x)=y(t)=\frac{(t-2)^2}{8}-1\) will parameterize the parabola given by \((x-2)^2=8(y+1)\text{.}\) This corresponds to the vector-valued function \(\vr(t) = \langle t , f(t) \rangle=\langle t,\frac{(t-2)^2}{8}-1 \rangle\text{.}\)
A parabola with vertex indicated by a point at \((2,-1)\text{.}\) It is plotted for \(-4\leq x\leq 6\) and also passes through the points \((-2,1)\) and \((6,1)\text{.}\)
Remember that there is not a unique way to parameterize a curve. This means that the choices we made above are not the only way to describe motion on this parabola. You could move along the curve with double the speed (relative to our original parameterization). In this case, the parameterization would be \(\vr(t) = \langle 2t,\frac{(2t-2)^2}{8}-1 \rangle\text{.}\)
Other choices for the parameterization might include wanting to have the vertex \((2,-1)\) of the parabola correspond to a specific parameter value, such as \(t=0\text{.}\) One parameterization with this property would be \(\vr(t) =\langle t-2 , \frac{t^2-2}{2} \rangle\text{.}\) Remember that there is more than one way to walk the same path.
The next activity has you consider a few examples of curves generated by vector-valued functions, including in three dimensions.
Activity 10.1.3.
Vector-valued functions can be used to generate many interesting curves. Graph each of the following using an appropriate technological tool, and then write one sentence for each function to describe the behavior of the resulting curve.
(a)
\(\vr(t) = \langle t\cos(t), t\sin(t) \rangle\)
(b)
\(\vr(t) = \langle \sin(t)\cos(t), t\sin(t) \rangle\)
(c)
\(\vr(t) = \langle \sin(5t), \sin(4t) \rangle\)
(d)
\(\vr(t) = \langle t^2\sin(t)\cos(t), 0.9t\cos(t^2), \sin(t) \rangle\text{.}\) (Note that this defines a curve in three-dimensional space.)
Experiment with different formulas for \(x(t)\) and \(y(t)\) and ranges for \(t\) to see what other interesting curves you can generate. Share your best results with peers.
The next example helps illustrate how thinking of curves as graphs of vector-valued functions can help make it easier to algebraically address some ideas about the graphs of functions that are tricker when working from the perspective of \(y=f(x)\text{.}\)
Example 10.1.8.
(a)
Consider a one-to-one function, such as \(f(x) = x^2+1\) on \([0,2]\text{.}\) The graph of this function is shown in Figure 10.1.9. In this scenario, we can also find the inverse function \(f^{-1}\) and graph it as well. Algebraically, this can sometimes be rather complex. However, when we parameterize the function as we did in Example 10.1.6, plotting the inverse function using a vector-valued function becomes routine. The method for doing this will rely on the fact that the graph of the inverse \(y=f^{-1}(x)\) the reflection of the graph of \(y=f(x)\) across the line \(y=x\text{.}\)
The portion of a parabola with vertex \((0,1)\) lying to the right of the \(y\)-axis is shown with label \(y=f(x)\text{.}\) The line \(y=x\) is shown as a dashed line. The reflection of the parabola across the line \(y=x\) is illustrated as well and labeled \(y=f^{-1}(x)\text{.}\)
As in Example 10.1.6, we parameterize the graph of \(f(x)=x^2+1\) on \([0,2]\) as \(\vr(t) = \left\langle t,f(t)\right\rangle = \left\langle t,t^2+1\right\rangle\) for \(0\leq t\leq 2\text{.}\) The points on the graph of the inverse function are determined by swapping the coordinates of the points on the graph of \(y=f(x)\text{,}\) so we swap the components of our vector-valued function that has parameterized the graph and consider \(\vs(t) = \left\langle f(t),t)\right\rangle = \left\langle t^2+1,t\right\rangle\text{.}\) Notice here that while we do not have an algebraic formula that allows us to take a given \(x\) and find \(f^{-1}(x)\text{,}\) we do have a nice way of graphing the inverse function with absolutely no algebra required.
(b)
Now suppose that we need to think about the parabola with equation \(x=y^2+1\text{.}\) If we wanted to graph this using familiar tools from precalculus or single-variable calculus, we would need to graph two functions, since this curve does not pass the vertical line test. However, if we think parametrically, we can view \(y\) as the parameter here. This is similar to what we did in Example 10.1.6, but now we are thinking of a vector-valued function \(\vr(t) = \langle x(t),y(t)\rangle\) where \(y(t)=t\text{.}\) With that settled, we then have that \(x(t) = \left(y(t)\right)^2 + 1 = t^2+1\text{.}\) This leads us to \(\vr(t) = \langle t^2+1,t\rangle\text{.}\)
The intersection of surfaces in space are often curves. Thus, we may determine parameterizations for these curves defined by satisfying multiple equations. For example, if we want to describe the intersection of the surface \(z = \cos(x^2 + y^2)\) and the fundamental plane \(y = 1\) , then we can set \(y = 1\) in \(z = \cos(x^2 + y^2)\text{.}\) We then let \(x\) be parameterized by the variable \(t\text{.}\) This tells us that the intersection of \(z = \cos(x^2 + y^2)\) and the fundamental plane \(y = 1\) is the curve parameterizated as \(\langle t, 1, \cos(t^2 + 1)\rangle\text{.}\) The next activity guides you through similar questions for a more familiar surface. You may find it useful to use the idea of using a variable other than \(x\) as the parameter, as we did in the second part of Example 10.1.8
Activity 10.1.4.
For this activity we will consider the paraboloid with equation \(z = x^2+y^2\text{.}\)
(a)
Parameterize the intersection of the paraboloid with the fundamental plane \(x=2\text{.}\) What type of curve is the intersection?
(b)
Parameterize the intersection of the paraboloid with the fundamental plane \(y = -1\text{.}\) What type of curve is the intersection?
(c)
Parameterize the intersection of the paraboloid with the fundamental plane \(z = 25\text{.}\) What type of curve is the intersection?
(d)
How do your responses change to all three of the preceding questions if you instead consider the surface defined by \(z = x^2 - y^2\text{?}\)
As you saw in Activity 10.1.4, it is convenient to parameterize conic sections and other curves defined by common relationships between coordinates. In Exercise 13, we will generalize our work for parameterizing circles earlier to give a general parameterized of a translated ellipse. These ideas are extended in Exercise 18, which asks you to parameterize hyperbolas.
Summary
-
A vector-valued function is a function whose input is a real parameter \(t\) and whose output is a vector that depends on \(t\text{.}\) The graph of a vector-valued function is the set of all terminal points of the output vectors with their initial points at the origin.
-
Every vector-valued function provides a parameterization of a curve. In \(\R^2\text{,}\) a parameterization of a curve is a pair of equations \(x = x(t)\) and \(y = y(t)\) that describes the coordinates of a point \((x,y)\) on the curve in terms of a parameter \(t\text{.}\) In \(\R^3\text{,}\) a parameterization of a curve is a set of three equations \(x = x(t)\text{,}\) \(y=y(t)\text{,}\) and \(z = z(t)\) that describes the coordinates of a point \((x,y,z)\) on the curve in terms of a parameter \(t\text{.}\)
Exercises 10.1.4 Exercises
The WeBWorK problems are written by many different authors. Some authors use parentheses when writing vectors, e.g., \((x(t),y(t),z(t))\) instead of angle brackets \(\langle x(t),y(t),z(t) \rangle\text{.}\) Please keep this in mind when working WeBWorK exercises.
1.
Find the domain of the vector function
\begin{equation*}
\mathbf r(t) = \left \langle \ln(7 t), \sqrt{t + 15}, \frac{1}{\sqrt{14 - t}}
\right \rangle
\end{equation*}
using interval notation.
Domain:
2.
3.
4.
Find a vector parametric equation \(\vec{r}(t)\) for the line through the points \(P = \left(2,-3,3\right)\) and \(Q = \left(3,-7,5\right)\) for each of the given conditions on the parameter \(t\text{.}\)
(a) If \(\vec{r}(0) = \left\lt 2,-3,3\right>\) and \(\vec{r}(7) = \left\lt 3,-7,5\right>\text{,}\) then
\(\vec{r}(t) =\)
\(\vec{r}(t) =\)
(c) If the points \(P\) and \(Q\) correspond to the parameter values \(t = 0\) and \(t = -3\text{,}\) respectively, then
\(\vec{r}(t) =\)
5.
Suppose parametric equations for the line segment between \((7, 8)\) and \((0, -5)\) have the form:
\begin{equation*}
\begin{array}{r@{\,}c@{\,}l}
x \amp = \amp a + bt \cr
y \amp = \amp c + dt
\end{array}
\end{equation*}
If the parametric curve starts at \((7, 8)\) when \(t=0\) and ends at \((0, -5)\) at \(t=1\text{,}\) then find \(a\text{,}\) \(b\text{,}\) \(c\text{,}\) and \(d\text{.}\)
\(a =\) ,
\(b =\) ,
\(c =\) ,
\(d =\) .
6.
7.
8.
Are the following statements true or false?
-
The parametric curve \(x = (3t+4)^2, y = 5(3t+4)^2-9\text{,}\) for \(0 \leq t \leq 3\) is a line segment.
-
A parametrization of the graph of \(y = \ln(x)\) for \(x > 0\) is given by \(x = e^t, y = t\) for \(-\infty \lt t \lt \infty\text{.}\)
-
The line parametrized by \(x = 7, y = 5t, z = 6 + t\) is parallel to the x-axis.
9.
10.
A bicycle wheel has radius R. Let P be a point on the spoke of a wheel at a distance d from the center of the wheel. The wheel begins to roll to the right along the the x-axis. The curve traced out by P is given by the following parametric equations:
\(x = 10 \theta - 8 \sin(\theta)\)
\(y = 10 - 8 \cos(\theta)\)
What must we have for R and d?
R=
d =
11.
A standard parameterization for the unit circle is \(\langle \cos(t), \sin(t) \rangle\text{,}\) for \(0 \le t \le 2\pi\text{.}\)
-
Find a vector-valued function \(\vr\) that describes a point traveling along the unit circle so that at time \(t=0\) the point is at \(\left(\frac{\sqrt{2}}{2}, \frac{\sqrt{2}}{2} \right)\) and travels clockwise along the circle as \(t\) increases.
-
Find a vector-valued function \(\vr\) that describes a point traveling along the unit circle so that at time \(t=0\) the point is at \(\left(\frac{\sqrt{2}}{2}, \frac{\sqrt{2}}{2} \right)\) and travels counter-clockwise along the circle as \(t\) increases.
-
Find a vector-valued function \(\vr\) that describes a point traveling along the unit circle so that at time \(t=0\) the point is at \(\left(-\frac{\sqrt{2}}{2}, \frac{\sqrt{2}}{2} \right)\) and travels clockwise along the circle as \(t\) increases.
-
Find a vector-valued function \(\vr\) that describes a point traveling along the unit circle so that at time \(t=0\) the point is at \((0,1)\) and makes one complete revolution around the circle in the counter-clockwise direction on the interval \([0,\pi]\text{.}\)
12.
(a)
Step through the process of showing parametrize the implicit equation of a circle with center \((h,k)\) and radius \(r \gt 0\) given by
\begin{equation*}
(x-h)^2 + (y-k)^2 = r^2
\end{equation*}
we can use the relationship between \(\sin\) and \(\cos\text{.}\) We found a parametric equation for the circle can be expressed by
\begin{equation*}
x(t) =r \cos(\theta) + h \qquad y(t) = r \sin(\theta) + k
\end{equation*}
13.
Earlier we saw that to parametrize the implicit equation of a circle with center \((h,k)\) and radius \(r \gt 0\) given by
\begin{equation*}
(x-h)^2 + (y-k)^2 = r^2
\end{equation*}
we can use the relationship between \(\sin\) and \(\cos\text{.}\) We found a parametric equation for the circle can be expressed by
\begin{equation*}
x(t) =r \cos(\theta) + h \qquad y(t) = r \sin(\theta) + k
\end{equation*}
The conic section most closely related to the circle is the ellipse. The general equation of an ellipse (centered at the origin) is given by
\begin{equation*}
\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1.
\end{equation*}
We can continue to make use of the relationship between \(\sin\) and \(\cos\) to discover parametric equations for an ellipse. In fact, without the \(a\) and \(b\) in the equation things would work perfectly. To remedy this (that is, get rid of \(a,b\)) we simply multiply by \(\sin\) and \(\cos\) by these values. Consider the equations
\begin{equation*}
x(t) = a \cos(t) \qquad y(t) = b \sin(t).
\end{equation*}
(a)
Substitute the equations for \(x(t), y(t)\) above into
\begin{equation*}
\frac{x^2}{a^2} + \frac{y^2}{b^2}
\end{equation*}
and simplify.
(b)
Consider the ellipse given by \(\frac{x^2}{9} + \frac{y^2}{4} = 1.\) What are the parametric equations for this ellipse? Graph them below to ensure you obtain the exact same graph.
(c)
In Figure 10.1.11, update the equation in \(x\) and \(y\) as well as the parametric equations for the ellipse \(\frac{x^2}{16} + \frac{y^2}{25} = 1.\)
(d)
Write a few sentences about the orientation of the parametric equations. In particular, you should talk about the following: at what point do you "start", i.e. what point corresponds to \(t=0\text{?}\) How do you move around the ellipse (clockwise or counterclockwise)?
(e)
Update the equation in \(x\) and \(y\) as well as the parametric equations so that the ellipse has a center of \((-2,1)\text{.}\)
(f)
Since we know that
\begin{equation*}
(\cos(t))^2 +(\sin(t))^2=1 \quad \text{for every }t\in\mathbb{R}
\end{equation*}
and the form of a translated ellipse is given by
\begin{equation*}
\frac{(x-h)^2}{a^2} + \frac{(y-k)^2}{b^2} = 1,
\end{equation*}
set
\begin{equation*}
\frac{(x-h)^2}{a^2}=(\cos(t))^2
\end{equation*}
and solve for x.
(g)
Set up a similar equation involving \(y\) and \(\sin(t)\) then solve for \(y\) to get a general set of parametric equations for the translated ellipse.
14.
Let \(a\) and \(b\) be positive real numbers. You have probably seen the equation \(\frac{(x-h)^2}{a^2} + \frac{(y-k)^2}{b^2} = 1\) that generates an ellipse, centered at \((h,k)\text{,}\) with a horizontal axis of length \(2a\) and a vertical axis of length \(2b\text{.}\)
-
Explain why the vector function \(\vr\) defined by \(\vr(t) = \langle a\cos(t), b\sin(t) \rangle\text{,}\) \(0 \le t \le 2\pi\) is one parameterization of the ellipse \(\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1\text{.}\)
-
Find a parameterization of the ellipse \(\frac{x^2}{4} + \frac{y^2}{16} = 1\) that is traversed counterclockwise.
-
Find a parameterization of the ellipse \(\frac{(x+3)^2}{4} + \frac{(y-2)^2}{9} = 1\text{.}\)
-
Determine the \(xy\)-equation of the ellipse that is parameterized by\begin{equation*} \vr(t) = \langle 3 + 4\sin(2t), 1 + 3\cos(2t) \rangle. \end{equation*}
15.
Consider the two-variable function \(z = f(x,y) = 3x^2 + 4y^2 - 2\text{.}\)
-
Determine a vector-valued function \(\vr\) that parameterizes the curve which is the \(x = 2\) trace of \(z = f(x,y)\text{.}\) Plot the resulting curve. Do likewise for \(x = -2, -1, 0,\) and \(1\text{.}\)
-
Determine a vector-valued function \(\vr\) that parameterizes the curve which is the \(y = 2\) trace of \(z = f(x,y)\text{.}\) Plot the resulting curve. Do likewise for \(y = -2, -1, 0,\) and \(1\text{.}\)
-
Determine a vector-valued function \(\vr\) that parameterizes the curve which is the \(z = 2\) contour of \(z = f(x,y)\text{.}\) Plot the resulting curve. Do likewise for \(z = -2, -1, 0,\) and \(1\text{.}\)
-
Use the traces and contours you’ve just investigated to create a wireframe plot of the surface generated by \(z = f(x,y)\text{.}\) In addition, write two sentences to describe the characteristics of the surface.
16.
Recall that any line in space may be represented parametrically by a vector-valued function.
-
Find a vector-valued function \(\vr\) that parameterizes the line through \((-2,1,4)\) in the direction of the vector \(\vv = \langle 3, 2, -5 \rangle\text{.}\)
-
Find a vector-valued function \(\vr\) that parameterizes the line of intersection of the planes \(x + 2y - z = 4\) and \(3x + y - 2z = 1\text{.}\)
-
Determine the point of intersection of the lines given by\begin{equation*} x = 2 + 3t, \ y = 1 - 2t, \ z = 4t, \end{equation*}\begin{equation*} x = 3 + 1s, \ y = 3-2s, \ z = 2s. \end{equation*}Then, find a vector-valued function \(\vr\) that parameterizes the line that passes through the point of intersection you just found and is perpendicular to both of the given lines.
17.
For each of the following, describe the effect of the parameter \(s\) on the parametric curve for \(t\) in the interval \([0,2 \pi]\text{.}\)
-
\(\displaystyle \vr(t) = \langle \cos(t), \sin(t) + s \rangle\)
-
\(\displaystyle \vr(t) = \langle \cos(t)-s, \sin(t) \rangle\)
-
\(\displaystyle \vr(t) = \langle s\cos(t), \sin(t) \rangle\)
-
\(\displaystyle \vr(t) = \langle s\cos(t), s\sin(t) \rangle\)
-
\(\displaystyle \vr(t) = \langle \cos(st), \sin(st) \rangle\)
18. Parameterizing a Translated Hyperbola.
Another conic section that will be useful to quickly parameterize is the hyperbola. The translated hyperbola is given by
\begin{equation}
\frac{(x-h)^2}{a^2} - \frac{(y-k)^2}{b^2} = 1\tag{10.1.1}
\end{equation}
for horizontal orientation and
\begin{equation}
\frac{(y-k)^2}{a^2} - \frac{(x-h)^2}{b^2} = 1\tag{10.1.2}
\end{equation}
for vertical orientation.
(a)
Find a trig identity of the form
\begin{equation*}
(\underline{\hspace{0.5in}})^2-(\underline{\hspace{0.5in}})^2=1
\end{equation*}
(b)
Set the first blank element of your trig identity from the previous part equal to the following expression, and then solve for \(x\text{.}\)
\begin{equation*}
\frac{(x-h)^2}{a^2}
\end{equation*}
(c)
Set up a similar equation involving \(y\) and the trig function from the second blank of part 10.1.4.18.a then solve for \(y\) to get a general set of parametric equations for the translated hyperbola.
(d)
Subsitute in your parametric equations for the translated hyperbola into the Desmos Interactive below to check that your equations trace the same graph as the translated hyperbola. You should move the sliders for \(h\text{,}\) \(k\text{,}\) \(a\text{,}\) and \(b\) to convince yourself that this set of parametric equations works for all cases (of the horizontal orientation).
19.
For each of the following equations, give a parameterization and be sure state the bounds of your parameterization.
-
\(\displaystyle \frac{(x-2)^2}{9}+\frac{(y+1)^2}{16}=1\)
-
\(\displaystyle \frac{(y+3)^2}{4}-\frac{(x+1)^2}{9}=1\)
-
\(\displaystyle (y-7)^2=-4(x+1)\)
