Course Notes for Calculus III

The \(x\) coordinate is always positive, starting at \(9\text{,}\) dropping back down to \(0\) and then rising up to \(9\) again. The \(y\) coordinates starts at \(-27\) and increases to \(27\text{.}\) Since the two coordinates have different exponents, I expect a non-linear shape following the trends I just described. This parametric curve is shown in Figure 6.3.1

Both coordinates are period functions, so I expect a shape that loops around and return to its starting point. The \(y\) coordinate has a larger amplitude, but both have the same period, so I expect this to be stretch out in the \(y\) direction. Possible an ellipse. This parametric curve is shown in Figure 6.3.2

This is hard to immediately guess the shape. Both coordinates are bounded and periodic, so I do expect a shape that loops around in some way and return to the origin. However, since we have sums of trig functions with different amplitues and periods, I expect a slightly complicated kind of looping. Possible like a sinusoidal oscilaton along a circle, but maybe with self-intersections as well. This parametric curve is shown in Figure 6.3.3

The \(y\) coordinates is just the parameter \(t\text{,}\) so I can think of this as a graph of the other coordinate (with the axes reverse, since here \(x\) will be a function of \(y\text{.}\) The \(x\) coordinate is a quadratic, so I expect a sideways parabola. This parametric curve is shown in Figure 6.3.4

This is a typical setup for a spiral, with sine and cosine terms of the same period but an amplitude which changes with the parameter. Here, the amplitude is decreasing, so I expect an curve spiraling inward. This parametric curve is shown in Figure 6.3.5

I expect a curved path out from the origin to \((10, \sqrt[3]{10}\text{.}\) This parametric curve is shown in Figure 6.3.6

Both the radius and the angle grow at the same rate. I expect a spiral with linearly growing radius. This parametric curve is shown in Figure 6.3.7

As the angle goes around, the radius oscilated between \(3\) and \(5\text{.}\) I expect a wobbly kind of circle, or perhaps more like an ellipse, since there is only one period of the oscilations as we go around the circle. This parametric curve is shown in Figure 6.3.8

The rate of revolution is increasing. At the same time, the radius is increasing. I expect a spiral into the origin. This parametric curve is shown in Figure 6.3.9

1. \begin{equation*} t = 3s \end{equation*}

   The movement will be consistently three times faster. For any value of \(s\text{,}\) the position based on the original parameter would be three times further along the curve, so movement along the curve happens three times as fast. The domain is still \(s \in [0, \infty)\text{.}\)

2. \begin{equation*} t = \frac{s}{4} \end{equation*}

   The movement will be one quarter th eoriginal speed. For any value of \(s\text{,}\) the position based on the original parameter would be only one quarter as far along the curve, so motvement along the curve happens at one quarter the original speed. The domain is still \(s \in [0, \infty)\)

3. \begin{equation*} t = s + 7 \end{equation*}

   The speed doesn't change. However, at \(s=0\) (the starting time for the new variable), the position is already partially along the curve, where the original parametrization would have been at \(t=7\text{.}\) The domain, to match the original curve, would be \(s \in [-7,\infty)\text{.}\)

4. \begin{equation*} t = s - 10 \end{equation*}

   The speed doesn't change. However, at \(s=0\) (the starting time for the new variable), the position has moved back along the curve to \(t = -10\text{.}\) If \(t\) doesn't take negative values, then the curve would have to start the new variable at \(s = 10\text{.}\) The domain, to match the original curve, would be \(s \in [10,\infty)\text{.}\)

5. \begin{equation*} t = s^2 \end{equation*}

   By squaring the new variable, the movement will be faster and faster compared with the old variable. Moving linearly in \(s\) is moving quadratically in \(t\text{.}\) The domain is still \(s\ in [0,\infty)\text{.}\)

6. \begin{equation*} t = \sqrt{s} \end{equation*}

   By taking the square root, movement along the curve will be going slower and slower compared with the old variable. Moving linearly in \(s\) is moving only \(\sqrt{s}\) in \(t\text{.}\)The domain is still \(s\ in [0,\infty)\text{.}\)

7. \begin{equation*} t = e^s \end{equation*}

   The speed will accelerate more and more compared with the old variable. This is like the quadratic change, but more dramatic, since the exponential function grows faster. The domain cannot actually match the original curve, since producing \(t = 0\) is impossible with the new variable. The closest domain is \(s \in (-\infty, \infty)\text{.}\)

8. \begin{equation*} t = -s \end{equation*}

   The speed will not change, but the direction of movement along the curve will be reversed. The domain is \(s \in (-\infty,0]\text{.}\)

I calculate the derivatives of the two coordinates.

I then put them into a pythagorian combination.

For arclength, I need to integrate this expression over the length of the curve.

The result of this integral is the arclength of the curve.

I calculate the derivatives of the two coordinates.

I then put these into a pythagorian combination.

For arclength, I need to integrate this expression over the length of the curve.

The result of this integral is the arclength of the curve.

I calculate the derivatives of the two coordinates.

I then put these into a pythagorian combination.

For arclength, I need to integrate this expression over the length of the curve.

The result of this integral is the arclength of the function.

1. Find a parametric curve which descrives an ellipse with semimajor axis \(a\) and semiminor axis \(b\text{.}\)

   I can take the standard curve for the circle and just scale the amplitude of the two coordinate functions.

   \begin{equation*} \gamma(t) = (a \cos t, b \sin t) \end{equation*}

   The shape is given by one period of the sine and cosine functions, so \(t \in [0,2\pi]\text{.}\)

2. Set up the arclength calculation to find the circumference of the ellipse.

   The derivatives are \(x^\prime = -a \sin t\) and \(y^\prime = b \cos t\text{.}\) I put these in a pythagorian combination and integrate to find the arclength.

   \begin{equation*} \int_0^{2\pi} \sqrt{a^2 \sin^2 (t) + b^2 \cos^2 (t)} dt \end{equation*}

3. Factor \(a\) out of the square root and subtract a cosine term to that you can simplify using \((\sin^2 t + \cos^2 t = 1)\text{.}\)

   \begin{align*} \amp \int_0^{2\pi} \sqrt{a^2 \sin^2 (t) + b^2 \cos^2 (t)} dt\\ \amp = \int_0^{2\pi} \sqrt{a^2 \left( \sin^2 (t) + \frac{b^2}{a^2} \cos^2 (t)\right)} dt\\ \amp = \int_0^{2\pi} a \sqrt{ \sin^2 (t) + \frac{b^2}{a^2} \cos^2 (t)} dt\\ \amp = \int_0^{2\pi} a \sqrt{ \sin^2 (t) + \cos^2 t + \left( \frac{b^2}{a^2} - 1 \right) \cos^2 (t)} dt\\ \amp = \int_0^{2\pi} a \sqrt{ 1 + \left( \frac{b^2}{a^2} - 1 \right) \cos^2 (t)} dt \end{align*}

4. The eccentricity of the ellipse is defined to be the number \(e = \sqrt{1 - \frac{b^2}{a^2}}\text{.}\) Manipulate the integral so that \(e\) shows up under the square root, but \(a\) amd \(b\) are no longer under the square root.

   \begin{gather*} \int_0^{2\pi} a \sqrt{ 1 + \left( \frac{b^2}{a^2} - 1 \right) \cos^2 (t)} dt\\ \int_0^{2\pi} a \sqrt{ 1 - \left( 1 - \frac{b^2}{a^2} \right) \cos^2 (t)} dt\\ = \int_0^{2\pi} a \sqrt{ 1 - e \cos^2 (t)} dt \end{gather*}

5. Calculate the integral in the special case \(e=0\text{.}\)

   \begin{equation*} = \int_0^{2\pi} a \sqrt{ 1 - 0 \cos^2 (t)} dt = \int_0^{2\pi} a dt = 2\pi a \end{equation*}

6. Try to solve the integral for general eccentricity \(0 \lt e \lt 1\text{.}\) Comment on the problems and barriers to integration.

   This is an impossible integral to solve with the methods in this course, or even with more advanced method. Without \(e=0\text{,}\) I can't make use of a trig identity to remove all the trig functions. I am left with this annoying \(\cos^2 t\) function inside the integral. No substitution works, since there are no trig functions outside the integral.

   This is called an elliptic integral (of the second kind, technically). It was a major motivator in 19th century mathematics, since the circumference of the ellipse was considered an important geometric problem. It led, indirectly, to the objects known as elliptic curves. These are very important geometric object which seem strangely named, since they don't look like they have anything to do with ellipses. The connection is these integrals. Elliptic curves were invented as part of some clever and complicated new methods to approach ellitpic integrals. In this sense, elliptic curves are not curves which are like ellipses, but rather, curves that solve some elliptic integrals.

First, I calculate the derivatives of the coordinate functions.

Then I calculate the pythagorian combination of these. This is a pretty long calculation, but I will be able to make use of a number of trig identities to simplify the terms inside the square root, and some terms cancel.

Then I set up the arclength function, which is the integral of this pythagorian combination.

I can use a substitution here.

Then I invert the arclength function.

Finally, I repalce \(t\) with this expression to get a parametrization in \(s\text{,}\) which is the parametrization by arclength.

First, I calculate the derivatives of the coordinate functions.

Then I calculate the pythagorian combination of these.

Then I set up the arclength function, which is the integral of this pythagorian combination.

I asked a computer algebra system for the value of this integral.

This is not algebraically invertible. All I can do is call the inverse \(t(s)\) and replace it implicitly.

First, I calculate the derivatives of the coordinate functions.

Then I calculate the pythagorian combination of these.

I'm going to complete the square here to make this more approachable.

Then I set up the arclength function, which is the integral of this pythagorian combination.

I can shift using the substitution \(v = u-2\)

This is a trig substitution. I asked a computer algebra for the solution.

This is not algebraically invertible. All I can do is implicitly write \(t(s)\) for the inverse. Then I can replace \(t\) with this \(t(s)\) in this curve.