Definition4.1.1
If each of \(f\) and \(g\) are continuous functions, then the curve in the plane defined by \(x=f(t),y=g(t)\) is called a parametric curve, and the equations \(x=f(t),y=g(t)\) are called parametric equations for the curve.
Section4.1Parametric Equations
In middle school, you learned to write an equation of a line as \(y=mx+b\text{.}\) In the vector unit, we learned to write this in vector form as:
\begin{equation*} (x,y)=(1,m)t+(0,b) \end{equation*}This style of equation is called a vector equation. It is equivalent to writing the two equations
\begin{equation*} x=1t+0,y=mt+b, \end{equation*}which we call the parametric equations of the line. We were able to quickly develop equations of lines in space, by just adding a third equation for \(z\text{.}\)
Parametric equations provide us with a way of specifying the location \((x,y,z)\) of an object by giving an equation for each coordinate. We will use these equations to model motion in the plane and in space. In this section we'll focus mostly on planar curves.
You can generalize this definition to 3D and beyond by just adding more variables.
Exercise4.1.1
By plotting points, construct graphs of the three parametric curves given below (just make a \(t,x,y\) table, and then plot the \((x,y)\) coordinates). Place an arrow on your graph to show the direction of motion.
(a)
\(x=\cos t, y=\sin t\text{,}\) for \(0\leq t\leq 2\pi\text{.}\)
(b)
\(x=\sin t, y=\cos t\text{,}\) for \(0\leq t\leq 2\pi\text{.}\)
(c)
\(x=\cos t, y=\sin t, z=t\text{,}\) for \(0\leq t\leq 4\pi\text{.}\)
Exercise4.1.2
For the parametric curve \(x=1+2\cos t, y=3+5\sin t\text{:}\)
(a)
Plot the path traced out by the curve.
(b)
Use the trig identity \(\cos^2t+\sin^2t=1\) to give a Cartesian equation of the curve (an equation that only involves \(x\) and \(y\)).
(c)
What are the foci of the resulting object (it's a conic section)?
Exercise4.1.3
What we did in the vector unit should help here.
Find parametric equations for a line that passes through the points \((0,1,2)\) and \((3,-2,4)\text{.}\)
What we did in the vector unit should help here.
Exercise4.1.4
For the parametric curve \(\vec r(t)= (t^2+1, 2t-3)\text{:}\)
(a)
Plot the path traced out by the curve.
(b)
Give a Cartesian equation of the curve (eliminate the parameter \(t\))
Hint
Set up a system of equations then use substitution.
(c)
Now find the focus of the resulting curve.
Exercise4.1.5
Consider the parametric curve given by \(x=\tan t, y=\sec t\text{.}\) Plot the curve for \(-\pi/2\lt t\lt \pi/2\text{.}\) Give a Cartesian equation of the curve (a trig identity will help). Then find the foci of the resulting conic section.
Hint
this problem will probably be easier to draw if you first find the Cartesian equation, and then plot the curve.
Subsection4.1.1Derivatives and Tangent lines
We're now ready to discuss calculus on parametric curves. The derivative of a vector valued function is defined using the same definition as first semester calculus.
Definition4.1.2
If \(\vec r(t)\) is a vector equation of a curve (or in parametric form just \(x=f(t), y=g(t)\)), then the derivative is defined as:
\begin{equation*} \frac{d\vec r}{dt}=\ds\lim_{h\to 0}\frac{\vec r(t+h)-\vec r(t)}{h}. \end{equation*}The subtraction above requires vector subtraction. The following problem will provide a simple way to take derivatives which we will use all semester long.
Exercise4.1.6
Show that if \(\vec r(t) = (f(t),g(t))\text{,}\) then the derivative is just \(\frac{d\vec r}{dt} = (f'(t),g'(t))\text{.}\)
Hint
[The definition above says that \(\frac{d\vec r}{dt}=\ds\lim_{h\to 0}\frac{\vec r(t+h)-\vec r(t)}{h}\text{.}\) We were told \(\vec r(t) = (f(t),g(t))\text{,}\) so use this in the derivative definition. Then try to modify the equation to obtain \(\frac{d\vec r}{dt} = (f'(t),g'(t))\text{.}\)]
The previous problem shows you can take the derivative of a vector valued function by just differentiating each component separately. The next problem shows you that velocity and acceleration are still connected to the first and second derivatives.
Exercise4.1.7
Consider the parametric curve given by \(\vec r(t)=( 3\cos t, 3\sin t )\text{.}\)
(a)
Graph the curve \(\vec r\text{,}\) and compute \(\frac{d\vec r}{dt}\) and \(\frac{d^2\vec r}{dt^2}\text{.}\)
(b)
On your graph, draw the vectors \(\frac{d\vec r}{dt}\left(\frac{\pi}{4}\right)\) and \(\frac{d^2\vec r}{dt^2}\left(\frac{\pi}{4}\right)\) with their tail placed on the curve at \(\vec r\left(\frac{\pi}{4}\right)\text{.}\) These vectors represent the velocity and acceleration vectors.
(c)
Give a vector equation of the tangent line to this curve at \(t=\frac{\pi}{4}\text{.}\) (You know a point and a direction vector.)
Definition4.1.3
If an object moves along a path \(\vec r(t)\text{,}\) we can find the velocity and acceleration by just computing the first and second derivatives. The velocity is \(\frac{d\vec r}{dt}\text{,}\) and the acceleration is \(\frac{d^2\vec r}{dt^2}\text{.}\) Speed is a scalar, not a vector. The speed of an object is just the length (or magnitude) of the velocity vector.
Exercise4.1.8
Consider the curve \(\vec r(t) = (2t+3, 4(2t-1)^2)\text{.}\)
(a)
Construct a graph of \(\vec r\) for \(0\leq t\leq 2\text{.}\)
(b)
If this curve represented the path of a horse running through a pasture, find the velocity of the horse at any time \(t\text{,}\) and then specifically at \(t=1\text{.}\) What is the horse's speed at \(t=1\text{?}\)
(c)
Find a vector equation of the tangent line to \(\vec r\) at \(t=1\text{.}\) Include this on your graph.
(d)
Show that the slope of the line is
\begin{equation*} \ds \frac{dy}{dx}\big|_{x=5} = \frac{ (dy/dt)\big|_{t=1} }{ (dx/dt)\big|_{t=1} }. \end{equation*}
Hint
How can you turn the direction vector, which involves \((dx/dt)\) and \((dy/dt)\) into a slope \((dy/dx)\text{?}\)
Subsection4.1.2Arc Length
This section covers:
finding rates of change along space curves
If an object moves at a constant speed, then the distance travelled is
\begin{equation*} \text{ distance } = \text{ speed } \times\text{ time } . \end{equation*}This requires that the speed be constant. What if the speed is not constant? Over a really small time interval \(dt\text{,}\) the speed is almost constant, so we can still use the idea above. The following problem will help you develop the key formula for arc length.
Exercise4.1.9Derivation of the arc length formula
Suppose an object moves along the path given by:
\begin{equation*} \vec r(t)=(x(t),y(t)) \end{equation*}for
\begin{equation*} a\leq t\leq b \end{equation*}(a)
Show that the object's speed at any time \(t\) is \(\ds\sqrt{\left(\frac{dx}{dt}\right)^2+\left(\frac{dy}{dt}\right)^2}\text{.}\)
(b)
If you move over a really small time interval, say of length \(dt\text{,}\) then the speed is almost constant. If you move at a constant speed of \(\ds\sqrt{\left(\frac{dx}{dt}\right)^2+\left(\frac{dy}{dt}\right)^2}\) for a time length of \(dt\text{,}\) what's the distance \(ds\) you have traveled.
Hint
This should be really fast to write down
(c)
This is the arc length formula. Ask me in class for an alternate way to derive this formula.
Explain why the length of the path given by \(\vec r(t)\) for \(a\leq t\leq b\) is
\begin{equation*} \text{ Arc Length } =s=\int ds=\int_a^b \left|\frac{d\vec r}{dt}\right| dt=\int_a^b \sqrt{\left(\frac{dx}{dt}\right)^2+\left(\frac{dy}{dt}\right)^2}dt. \end{equation*}Exercise4.1.10
Use the equation from Task 4.1.9.c to find the length of the curve \(\ds \vec r(t) = \left(t^3,\frac{3t^2}{2}\right)\) for \(t\in[1,3]\text{.}\) The notation \(t\in[1,3]\) means \(1\leq t\leq 3\text{.}\) Be prepared to show us your integration steps in class (you'll need a \(u\)-substitution).
Exercise4.1.11
For each curve below, set up an integral formula which would give the length, and sketch the curve. Do not worry about integrating them.
The reason I don't want you to actually compute the integrals is that they will get ugly really fast. Try doing one in Wolfram Alpha and see what the computer gives.
The reason I don't want you to actually compute the integrals is that they will get ugly really fast. Try doing one in Wolfram Alpha and see what the computer gives.
(a)
The parabola \(\vec p(t) = (t,t^2)\) for \(t\in[0,3]\text{.}\)
(b)
The ellipse \(\vec e(t) = (4\cos t,5\sin t)\) for \(t\in[0,2\pi]\text{.}\)
(c)
The hyperbola \(\vec h(t) = (\tan t,\sec t)\) for \(t\in[-\pi/ 4,\pi/4]\text{.}\)
Subsection4.1.3Computational Practice
There are multiple copies of several exercises for your practice. You don't need to do them all. Do as many as it takes to get one right quickly.
Paramterizing Lines
Un-parameterizing lines
Computing Arc Lengths -- There are two types here