Objectives
In this section you will...
develop the notion of `curvature' and the associated scalar/vector values.
Section8.3Some Examples and The TNB Frame
¶Imagine that you are at a park with some kids (a nephew, a daughter, etc.). The park has a merry-go-round where the kids can sit down, hold on, and then spin in circles. As the spinning speed increases, they'll feel a greater force trying to throw them outwards. To stay on the merry-go-round, they have to counteract this outward acceleration with their own inward acceleration. For the next few examples, let's examine the connections between \(\vec v\text{,}\) \(\vec a\text{,}\) \(\vec T\text{,}\) \(d\vec T/dt\text{,}\) \(d\vec T/ds\text{,}\) and \(|d\vec T/ds|\) in the context of spinning around on a merry-go-round. These few examples are enough to develop just about the entire chapter.
Exercise8.3.1
Sammy sits on a merry go round. He sits 3 feet from the center of the merry ground, and lets his big sister spin him around. We can parameterize Sammy's path with the vector equation \(\vec r(t) = (3\cos t, 3\sin t)\text{,}\) where \(t\) is in seconds. The point \((0,3)\) corresponds to the time \(t=\pi/2\text{.}\)
(a)
Draw the curve for \(0\leq t\leq 2\pi\text{.}\)
(b)
Compute \(\vec v\) and \(\vec a\text{.}\)
(c)
At \((0,3)\) (so \(t=\pi/2\)), draw these two vectors. Are these two vectors orthogonal?
(d)
Compute \(\vec T\) and \(\dfrac{d\vec T}{dt}\text{.}\)
(e)
At \((0,3)\text{,}\) draw these two vectors. Are these two vectors orthogonal?
(f)
Compute \(\dfrac{d\vec T}{ds}\) and \(\left|\dfrac{d\vec T}{ds}\right|\text{.}\)
(g)
How is the length of \(\dfrac{d\vec T}{ds}\) related to the circle?
Exercise8.3.2
We started with the parameterization \(\vec r(t) = (3\cos t, 3\sin t)\text{,}\) now we'll replace \(t\) with \(5t\) and \(\omega t\) to understand which values can be used as descriptors of the curve vs. the motion.
(a)
Replace \(t\) with \(5t\) and compute \(\vec v\text{,}\) \(\vec a\text{,}\) \(\vec T\text{,}\) \(\dfrac{d\vec T}{dt}\text{,}\) \(\dfrac{d\vec T}{ds}\text{,}\) and \(\left|\dfrac{d\vec T}{ds}\right|\text{.}\)
(b)
If we still want to evaluate at the time corresponding to the point \((0,3)\text{,}\) what is \(t=\text{?.}\)
Hint
It should be smaller than \(\frac{\pi}{2}\)
(c)
Replace \(t\) with \(\omega t\) and find \(\vec v\text{,}\) \(\vec a\text{,}\) \(\vec T\text{,}\) \(\dfrac{d\vec T}{dt}\text{,}\) \(\dfrac{d\vec T}{ds}\text{,}\) and \(\left|\dfrac{d\vec T}{ds}\right|\text{.}\)
(d)
Explain why we should evaluate at \(t=\frac{\pi}{2\omega}\) for the point \((0,3)\text{.}\)
(e)
Evaluate each at \(t=\frac{\pi}{2\omega}\text{,}\) the time corresponding to point \((0,3)\text{.}\)
(f)
At \((0,3)\text{,}\) which of these 6 quantities remain the same?
Gauss developed and expanded many of the ideas in this section while on a map making mission for a king. He wanted to create extremely high quality 2D images of the 3D world we are in.
Subsection8.3.1Curvature
We started with motion on a circle with a relatively simple speed. We then multiplied the speed by 5, and then chose a variable speed. You should have noticed that at the same point on the curve, regardless of speed, the quantities \(\vec T\text{,}\) \(\dfrac{d\vec T}{ds}\text{,}\) and \(\left|\dfrac{d\vec T}{ds}\right|\) remained the same. Will this pattern continue if we were to change from constant speeds to variable speeds? What if we left the path of a circle, and moved to any smooth curve? How far can we take this pattern? Is it true always? If so, then the quantities \(\vec T\text{,}\) \(\dfrac{d\vec T}{ds}\text{,}\) and \(\left|\dfrac{d\vec T}{ds}\right|\) are pretty important quantities that describe the curve we're on.
Because \(\dfrac{d\vec T}{ds}\) and \(\left|\dfrac{d\vec T}{ds}\right|\) show up a lot, let's give them a definition.
Definition8.3.1
Curvature \(\kappa\) and The Curvature Vector \(\vec \kappa\)Let \(\vec r(t)\) be a smooth curve (so that \(\vec v\) is never zero).
The vector \(\vec \kappa = \dfrac{d\vec T}{ds}\) we'll call the curvature vector. It measures how quickly the unit tangent vectors changes as we increase in length (not time).
The number \(\kappa = \left|\dfrac{d\vec T}{ds}\right|\) we'll call the curvature.
Let's apply this new definition to a circle of any radius. We'll quickly see that the curvature and radius of a circle are related. The curvature vector \(d\vec T/ds\) tells us how quickly \(\vec T\) changes as we increase in length, which is a measure of how sharply we turn a corner. If our curve is a circle of radius \(3\text{,}\) then the curvature is \(1/3\text{.}\) A larger circle should result in smaller curvature.
Exercise8.3.3
Consider the curve \(\vec r(t)=(a\cos t, a\sin t)\text{.}\)
(a)
Draw the curve, and state the radius \(\rho\) of the best approximating circle.
(b)
Find the curvature vector \(\vec \kappa\) by performing a computation.
(c)
What relationship exists between \(\rho\) and \(\kappa\text{?}\) If the radius \(\rho\) were to increase, what would happen to \(\kappa\text{?}\)
For any curve, we could approximate how rapidly the curve turns at a point by drawing a circle that best approximates the curve (kind of like a Taylor polynomial, only now we'll use a circle.) We want the circle to meet the curve \(\vec r\) tangentially, and we want the curvature of the circle to match the curvature of the curve. Since the curvature of the circle must match the curvature of the curve, we know they are inverse related. This gives the following definition.
Definition8.3.2Radius of Curvature
When the curvature \(\kappa\) of a smooth curve is nonzero, we'll define the radius of curvature, written \(\rho\text{,}\) to be the reciprocal \(\rho = \dfrac{1}{\kappa}\text{.}\) The curvature and radius of curvature are inversely related.
Watch a YouTube Video.
Subsection8.3.2More on Curvature
Now let's look at what happens if we change the speed in a nonlinear way. For simplicity, let's replace \(t\) with \(t^2\text{.}\) You could pick any other change. If you aren`t confident with your product rule, then please do the first review exercise. If you need to remember how to show vectors are orthogonal, do the second.
Review8.3.4
Suppose \(f(x) = \sin(x^2)\text{.}\) Compute \(f'(x)\) and \(f''(x)\text{.}\)
Solution
We have \(f'(x) = 2x\cos(x^2)\) and \(f''(x) = -4x^2\sin(x^2)+2\cos(x^2)\text{.}\)
Review8.3.5
Show that \((-2,3,1)\) and \((4,2,2)\) are orthogonal.
Solution
The dot product of these two vectors is \((-2,3,1)\cdot(4,2,2) = -8+6+2=0\text{.}\) Because the dot product is zero, the vectors are orthogonal.
Exercise8.3.6
The big kids pushing Sammy on the merry-go-round decide to constantly increase the spinning rate. To parametrize Sammy's path, let's replace \(t\) with \(t^2\) in the original parametrization to obtain \(\vec r(t) = (3\cos t^2, 3\sin t^2)\text{.}\) Fill in the missing values below, and then answer the questions that follow.
\begin{equation*} \begin{array}{|c|c|}\hline \vec v \amp ( -6 t \sin \left(t^2\right),6 t \cos \left(t^2\right) ) \\\hline ds/dt \amp 6 t \\\hline \vec a \amp \\\hline \vec T \amp \left(-\sin \left(t^2\right),\cos \left(t^2\right)\right) \\\hline d\vec T/dt \amp \left(-2 t \cos \left(t^2\right),-2 t \sin \left(t^2\right)\right)\\\hline \vec \kappa = d\vec T/ds \amp \\\hline \kappa =|d\vec T/ds| \amp \frac{1}{3} \\\hline \end{array} \end{equation*}(a)
Show that \(\vec v\) and \(\vec a\) are not orthogonal, but that \(\vec T\) and \(d\vec T/dt\) are.
(b)
Because \(\vec r (t)\) includes periodic trig functinos at \((0,3)\) we could have \(t^2 = \frac{\pi}{2}\text{,}\) \(t^2=2\pi+\frac{\pi}{2}\) or \(t^2=2\pi k + \frac{\pi}{2}\) for any integer \(k\text{.}\)
Show that \(\vec T\) is the same at \((0,3)\text{,}\) regardless of what time we pass through \((0,3)\text{.}\)
(c)
Show the same is true for \(d\vec T/ds\text{.}\)
(d)
Let's look at some specific values...
(i)
When the speed is \(\frac{ds}{dt}=6\) (so \(t=1\)), how long are \(d\vec T/dt\) and \(d\vec T/ds\text{?}\)
(ii)
When the speed is \(\frac{ds}{dt}=12\) (so \(t=2\)), how long are \(d\vec T/dt\) and \(d\vec T/ds\text{?}\)
(iii)
When the speed is \(\frac{ds}{dt}=30\) (so \(t=5\)), how long are \(d\vec T/dt\) and \(d\vec T/ds\text{?}\)
(iv)
If the speed were some random number \(v\text{,}\) how long are \(d\vec T/dt\) and \(d\vec T/ds\text{?}\)
Observation8.3.3
The previous exercise showed the following patterns.
The vectors \(\vec v\) and \(\vec a\) are not always orthogonal.
The vectors \(\vec T\) and \(\dfrac{d\vec T}{dt}\) are always orthogonal.
The vectors \(\vec T\) and \(\dfrac{d\vec T}{ds}\) are independent of speed. They depend only on the shape of the curve, not the speed at which you traverse the curve.
If we multiply \(\dfrac{d\vec T}{ds}\) by \(\dfrac{ds}{dt}\text{,}\) we'll get \(\dfrac{d\vec T}{dt}\text{.}\) The two vectors point in the same direction, which from here on out we'll call \(\vec N\text{,}\) the direction normal to the motion.
Let's see if this pattern continues when we swap to a different curve. Rather than try to connect the curve to some physical real world example, let's look at a curve we are familiar with, a parabola. If the pattern holds there as well, then we may have found a key pattern that works with all curves. Then we can take our new knowledge and apply it to ANYTHING (like space flight, roller coasters, missiles, and anything that moves).
If you want to perform the computations below by hand, you must master the product, quotient, and chain rule, as well as working with rational exponents in algebra. Feel free to try the computations by hand (if you can get them, then well done). You may also simply use Sage to do the computations below.
Exercise8.3.7
Consider the curve \(\vec r(t) = (t, t^2)\text{.}\) The computations get intense, so let's use a computer algebra system, such as Sage (follow this link) to help us. If we don't use a computer, we could spend hours on this exercise. The computer will do all the computations, and give graphs, in seconds.
Use the Sage link.
(a)
Use the SageMathCell below to compute \(\vec v\text{,}\) \(\vec a\text{,}\) \(\dfrac{ds}{dt}\text{,}\) \(\vec T\text{,}\) \(\dfrac{d\vec T}{dt}\text{,}\) \(\dfrac{d\vec T}{ds}\text{,}\) \(\left|\dfrac{d\vec T}{ds}\right|\text{,}\) and the unit vector \(\vec N\) that is orthogonal to \(\vec T\text{.}\) Evaluate each at \(t=1\) and \(t=\sqrt{3}/2\) (use “sqrt(3)/2”). Record your answers in the provided table.
(b)
Let's now double the speed at which we traverse along the curve. Replace \(t\) with \(2t\text{,}\) and then repeat the exercise above. However, instead of putting in \(t=1\) and \(t=\sqrt{3}/2\text{,}\) we now need to use \(t=1/2\) and \(t=?\text{.}\) Use Sage to record your answers in the provided table.
(c)
After completing the table, does Observation 3 still hold?
(d)
How are \(\vec T\) and \(\vec N\) related? Conjecture a pattern.
| Function: | \(\vec r(t)=(t,t^2)\) | \(\vec r(t)=(t,t^2)\) | \(\vec r(t)=(2t,(2t)^2)\) | \(\vec r(t)=(2t,(2t)^2)\) |
| Value: | at \(t=1\) | at \(t=\sqrt3/2\) | at \(t=1/2\) | at \(t=?\) |
| \(\vec r\) | \((1,1)\) | \((3/4,3/4)\) | \((1,1)\) | \((3/4,3/4)\) |
| \(\vec v\) | \((1,2)\) | |||
| \(\vec a\) | \((0,2)\) | |||
| \(\frac{ds}{dt}\) | \(\sqrt{5}\) | |||
| \(\vec T\) | \((\frac{\sqrt{5}}{5},\frac{2\sqrt{5}}{5})\) | |||
| \(\frac{d\vec T}{dt}\) | \((-\frac{4\sqrt{5}}{25},\frac{2\sqrt{5}}{25})\) | |||
| \(\vec\kappa=\frac{d\vec T}{ds}\) | \((-\frac{4}{25},\frac{2}{25})\) | |||
| \(\kappa=\left|\frac{d\vec T}{ds}\right|\) | \(\frac{2\sqrt{5}}{25}\) | |||
| \(\vec N\) | \((-\frac{2\sqrt{5}}{5},\frac{1\sqrt{5}}{5})\) |
Our observation still holds. Let's try it on one more curve in 3D by hand.
Exercise8.3.8
Consider the helix \(\vec r(t)=(4\cos t,4\sin t, 3t)\text{.}\) Compute \(\vec v\text{,}\) \(\vec T\text{,}\) \(\dfrac{d\vec T}{dt}\text{,}\) and \(\dfrac{d\vec T}{ds}\text{.}\) Then do the following:
Because the speed is constant, many of the computations are simplified. You can use the Sage link to check your work.
(a)
Show that \(\vec T\) and \(\dfrac{d\vec T}{dt}\) are orthogonal. (Do this for any time \(t\text{.}\))
(b)
At \(t=2\pi\text{,}\) compute \(ds/dt\text{,}\) \(\vec T\text{,}\) \(\dfrac{d\vec T}{dt}\text{,}\) and \(\dfrac{d\vec T}{ds}\text{.}\)
(c)
Let's slow down how quickly we travel along the curve. Replace \(t\) with \(t/5\text{.}\) Then at \(t=10\pi\text{,}\) give \(ds/dt\text{,}\) \(\vec T\text{,}\) \(\dfrac{d\vec T}{dt}\text{,}\) and \(\dfrac{d\vec T}{ds}\) for the new curve \(\vec r_2(t) =(4\cos(\frac{t}{5}),4\sin(\frac{t}{5}), 3(\frac{t}{5}))\text{.}\)
(d)
Give a unit vector that points in the same direction as \(\dfrac{d\vec T}{dt}\) or \(\dfrac{d\vec T}{ds}\text{.}\)
Subsection8.3.3Computational Practice
These are provided to help you achieve better skills in basic computational answers.
1
2
3
The answer to \(\kappa(1)\) is incorrect. Ask me in class to work it out.
4
The answer to \(\kappa(1)\) is incorrect. Ask me in class to work it out.