W13 - Notes

More calculus with parametric curves

01 Theory - Arclength

Theory 2

Arclength formula

The arclength of a parametric curve with coordinate functions $x(t)$ and $y(t)$ is:

$$L={\int }_a^b\sqrt{{(x^{\prime })}^2+{(y^{\prime })}^2}dt$$

This formula assumes the curve is traversed one time as $t$ increases from $a$ to $b$.

Counts total traversal

This formula applies when the curve image is traversed one time by the moving point.

Sometimes a parametric curve traverses its image with repetitions. The arclength formula would add length from each repetition!

Extra - Derivation of arclength formula

The arclength of a parametric curve is calculated by integrating the infinitesimal arc element:

$$\begin{array}{c}ds=\sqrt{d{x}^2+d{y}^2}\\ \\ L={\int }_a^{b}ds\end{array}$$

In order to integrate $ds$ in the $t$ variable, as we must for parametric curves, we convert $ds$ to a function of $t$:

$$\begin{array}{c}ds=\sqrt{d{x}^2+d{y}^2}\gg \gg \sqrt{\frac{1}{d{t}^2}\cdot (d{x}^2+d{y}^2)\cdot d{t}^2}\\ \\ \gg \gg \sqrt{\frac{d{x}^2}{d{t}^2}+\frac{d{y}^2}{d{t}^2}}\cdot \sqrt{d{t}^2}\gg \gg \sqrt{{\left(\frac{dx}{dt}\right)}^2+{\left(\frac{dy}{dt}\right)}^2}dt\\ \\ \gg \gg ds=\sqrt{x^{\prime }{(t)}^2+y^{\prime }{(t)}^2}dt\end{array}$$

So we obtain $ds=\sqrt{{(x^{\prime })}^2+{(y^{\prime })}^2}dt$ and the arclength formula follows from this:

$$L={\int }_a^b\sqrt{{(x^{\prime })}^2+{(y^{\prime })}^2}dt$$

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02 Illustration

Example - Perimeter of a circles

Perimeter of a circle

(1) The perimeter of the circle $(R\cos t,R\sin t)$ is easily found. We have $(x^{\prime },y^{\prime })=(-R\sin t,R\cos t)$, and therefore:

$$\begin{array}{c}{(x^{\prime })}^2+{(y^{\prime })}^2={(-R\sin t)}^2+{(R\cos t)}^2\\ \\ \gg \gg R^2{\sin }^{2}t+R^2{\cos }^{2}t\gg \gg R^2\\ \\ ds=\sqrt{{(x^{\prime })}^2+{(y^{\prime })}^2}dt=Rdt\end{array}$$

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(2) Integrate around the circle:

$$\begin{array}{c}\text{Perimeter}={\int }_0^{2\pi }ds\gg \gg {\int }_0^{2\pi }Rdt\\ \\ \gg \gg Rt{|}_0^{2\pi }=2\pi R\end{array}$$ Link to original

Example - Perimeter of an asteroid

Perimeter of an asteroid

Find the perimeter length of the ‘asteroid’ given parametrically by $(x,y)=(a{\cos }^3\theta ,a{\sin }^3\theta )$ for $a=2$.

Solution

(1) Notice: Throughout this problem we use the parameter $\theta$ instead of $t$. This does not mean we are using polar coordinates!

Compute the derivatives in $\theta$:

$$(x^{\prime },y^{\prime })=(3a{\cos }^2\theta \sin \theta ,3a{\sin }^2\theta \cos \theta )$$

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(2) Compute the infinitesimal arc element.

$$\begin{array}{c}{(x^{\prime })}^2+{(y^{\prime })}^2\gg \gg 9a^2{\cos }^4\theta {\sin }^2\theta +9a^2{\sin }^4\theta {\cos }^2\theta \\ \\ \gg \gg 9a^2{\sin }^2\theta {\cos }^2\theta ({\cos }^2\theta +{\sin }^2\theta )\\ \\ \gg \gg 9a^2{\sin }^2\theta {\cos }^2\theta \end{array}$$

Plug into the arc element, simplify:

$$\begin{array}{c}ds=\sqrt{{(x^{\prime })}^2+y^{\prime }{)}^2}d\theta \\ \\ \gg \gg \sqrt{9a^2{\sin }^2\theta {\cos }^2\theta }d\theta \\ \\ \gg \gg ds=3a\left|\sin \theta \cos \theta \right|d\theta \end{array}$$

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(3) Bounds of integration?

Easiest to use $\theta \in [0,\pi /2]$. This covers one edge of the asteroid. Then multiply by 4 for the final answer.

On the interval $\theta \in [0,\pi /2]$, the factor $3a\sin \theta \cos \theta$ is positive. So we can drop the absolute value and integrate directly.

Absolute values matter!

If we tried to integrate on the whole range $\theta \in [\mathrm{0,2}\pi ]$, then $3a\sin \theta \cos \theta$ really does change sign.

To perform integration properly with these absolute values, we’d need to convert to a piecewise function by adding appropriate minus signs.

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(4) Integrate the arc element:

$$\begin{array}{ccc}& {\int }_0^{\pi /2}ds\gg \gg {\int }_0^{\pi /2}3a\sin \theta \cos \theta d\theta & \\ & & \\ & \gg \gg 3a{\int }_{u=0}^{1}udu& \text{(}u=\sin \theta \text{)}\\ & & \\ & \gg \gg 3a{\frac{u^2}{2}|}_0^1\gg \gg \frac{3a}{2}& \end{array}$$

Finally, multiply by 4 to get the total perimeter: $L=6a$

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03 Theory - Distance, speed

Theory 3

Distance function

The distance function $s(t)$ returns the total distance traveled by the particle from a chosen starting time $t_0$ up to the (input) time $t$:

$$s(t)={\int }_{t_0}^{t}ds={\int }_{t_0}^t\sqrt{x^{\prime }{(u)}^2+y^{\prime }{(u)}^2}du$$

We need the dummy variable $u$ so that the integration process does not conflict with $t$ in the upper bound.

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Speed function

The speed of a moving particle is the rate of change of distance:

$$v(t)=s^{\prime }(t)=\sqrt{x^{\prime }{(t)}^2+y^{\prime }{(t)}^2}$$

This formula can be explained in either of two ways:

1. Apply the Fundamental Theorem of Calculus to the integral formula for $s(t)$.
2. Consider $ds=\sqrt{x^{\prime }{(t)}^2+y^{\prime }{(t)}^2}dt$ for a small change $dt$: so the rate of change of arclength is $\frac{ds}{dt}$, in other words $s^{\prime }(t)$.

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04 Illustration

Example - Speed, distance, displacement

Speed, distance, displacement

The parametric curve $(t,\frac{2}{3}{t}^{3/2})$ describes the position of a moving particle ($t$ measuring seconds). (a) What is the speed function?

Suppose the particle travels for $8$ seconds starting at $t=0$. (b) What is the total distance traveled? (c) What is the total displacement?

Solution

(a)

Compute derivatives:

$$(x^{\prime },y^{\prime })=(1,t^{1/2})$$

Now compute the speed:

$${(x^{\prime })}^2+{(y^{\prime })}^2=1+{(t^{1/2})}^2=1+t\gg \gg v(t)=\sqrt{{(x^{\prime })}^2+{(y^{\prime })}^2}=\sqrt{1+t}$$

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(b)

Distance traveled by using speed.

Compute total distance traveled function:

$$s(t)={\int }_{u=0}^t\sqrt{1+u}du$$

Substitute $w=1+u$ and $dw=du$. New bounds are $1$ and $1+t$. Calculate:

$$\begin{array}{c}\gg \gg {\int }_1^{1+t}\sqrt{w}dw\\ \\ \gg \gg \frac{2}{3}{w}^{3/2}{|}_1^{1+t}\gg \gg \frac{2}{3}({(1+t)}^{3/2}-1)\end{array}$$

The distance traveled up to $t=8$ is:

$$s(8)=\frac{2}{3}(9^{3/2}-1)\gg \gg \frac{2}{3}(27-1)\gg \gg \frac{52}{3}$$

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(c)

Displacement formula: $d=\sqrt{{(x_1-x_0)}^2+{(y_1-y_0)}^2}$

Now compute starting and ending points.

For starting point, insert $t=0$:

$$(x(t),y(t)){|}_{t=0}\gg \gg (t,\frac{2}{3}{t}^{3/2}){|}_{t=0}\gg \gg (\mathrm{0,0})$$

For ending point, insert $t=8$:

$$\begin{array}{c}(x(t),y(t)){|}_{t=8}\gg \gg (t,\frac{2}{3}{t}^{3/2}){|}_{t=8}\\ \\ \gg \gg (8,\frac{2}{3}{8}^{3/2})\gg \gg (8,\frac{32\sqrt{2}}{3})\end{array}$$

Insert $(\mathrm{0,0})$ and $(\mathrm{8,32}\sqrt{2}/3)$:

$$\begin{array}{c}\gg \gg \sqrt{8^2+{\left(\frac{32\sqrt{2}}{3}\right)}^2}\gg \gg \sqrt{64+\frac{2048}{9}}\\ \\ \gg \gg \frac{\sqrt{2624}}{3}\end{array}$$ Link to original

05 Theory - Surface area of revolutions

Theory 4

Surface area of a surface of revolution: thin bands

Suppose a parametric curve $(x(t),y(t))$ is revolved around the $x$-axis or the $y$-axis.

The surface area is:

$$A={\int }_a^{b}2\pi R(t)\sqrt{{(x^{\prime })}^2+{(y^{\prime })}^2}dt$$

The radius $R(t)$ should be the distance to the axis:

$$\begin{array}{cc}R(t)& =y(t)\text{revolution about }x\text{-axis}\\ R(t)& =x(t)\text{revolution about }y\text{-axis}\end{array}$$

This formulas adds the areas of thin bands, but the bands are demarcated using parametric functions instead of input values of a graphed function.

The formula assumes that the curve is traversed one time as $t$ increases from $a$ to $b$.

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06 Illustration

Example - Surface of revolution - parametric circle

Surface of revolution - parametric circle

By revolving the unit upper semicircle about the $x$-axis, we can compute the surface area of the unit sphere.

Parametrization of the unit upper semicircle:

$$\begin{array}{c}(x,y)=(\cos t,\sin t)\\ \\ \gg \gg (x^{\prime },y^{\prime })=(-\sin t,\cos t)\end{array}$$

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Therefore, the arc element:

$$\begin{array}{c}ds=\sqrt{{(x^{\prime })}^2+{(y^{\prime })}^2}dt\\ \\ \gg \gg \sqrt{{(-\sin t)}^2+{(\cos t)}^2}dt\gg \gg dt\end{array}$$

Now for $R(t)$ we choose $R(t)=y(t)=\sin t$ because we are revolving about the $x$-axis.

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Plugging all this into the integral formula and evaluating gives:

$$A={\int }_0^{\pi }2\pi \sin tdt\gg \gg -2\pi \cos t{|}_0^{\pi }\gg \gg 4\pi$$

Notice: This method is a little easier than the method using the graph $y=\sqrt{1-x^2}$.

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Example - Surface of revolution - parametric curve

Surface of revolution - parametric curve

Set up the integral which computes the surface area of the surface generated by revolving about the $x$-axis the curve $(t^3,t^2-1)$ for $0\le t\le 1$.

Solution

For revolution about the $x$-axis, we set $R=y(t)=t^2-1$.

Then compute $ds$:

$$\begin{array}{c}ds=\sqrt{{(x^{\prime })}^2+{(y^{\prime })}^2}\gg \gg \sqrt{{(3t^2)}^2+{(2t)}^2}\gg \gg \sqrt{9t^4+4t^2}\\ \\ \gg \gg \sqrt{t^2(9t^2+4)}\gg \gg t\sqrt{9t^2+4}\end{array}$$

Therefore the desired integral is:

$$A={\int }_0^{1}2\pi Rds\gg \gg {\int }_0^{1}2\pi (t^2-1)t\sqrt{9t^2+4}dt$$Link to original

Polar curves

Videos

Review Videos

Videos, Organic Chemistry Tutor

• Polar coordinates intro
• Graphing polar curves

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07 Theory - Polar points, polar curves

Theory 1

Polar coordinates are pairs of numbers $(r,\theta )$ which identify points in the plane in terms of distance to origin and angle from $+x$-axis:

Converting $\mathrm{Polar}\leftrightarrow \mathrm{Cartesian}$

$$\begin{array}{c}\text{Polar}\to \text{Cartesian}\\ x=r\cos \theta \\ y=r\sin \theta \end{array}\begin{array}{c}\text{Cartesian}\to \text{Polar}\\ r=\sqrt{x^2+y^2}\\ \tan \theta ={\displaystyle \frac{y}{x}(x\ne 0)}\end{array}$$

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Polar coordinates have many redundancies: unlike Cartesian which are unique!

• For example: $(r,\theta )=(r,\theta +2\pi )$
  • And therefore also $(r,\theta )=(r,\theta -2\pi )$ (negative $\theta$ can happen)
• For example: $(-r,\theta )=(r,\theta +\pi )$ for every $r,\theta$
• For example: $(0,\theta )=(\mathrm{0,0})$ for any $\theta$

Polar coordinates cannot be added: they are not vector components!

• For example $(5,\pi /3)+(2,\pi /6)\ne (7,\pi /2)$
• Whereas Cartesian coordinates can be added: $(\mathrm{1,4})+(2,-2)=(\mathrm{3,2})$

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The transition formulas $\text{Cartesian}\to \text{Polar}$ require careful choice of $\theta$.

• The standard definition of ${\tan }^{-1}\left(\frac{y}{x}\right)$ sometimes gives wrong $\theta$
  • This is because it uses the restricted domain $\theta \in (-\pi /2,\pi /2)$; the polar interpretation is: only points in Quadrant I and Quadrant IV (SAFE QUADRANTS)
• Therefore: check signs of $x$ and $y$ to see which quadrant, maybe need $\pi$-correction!
  • Quadrant I or IV: polar angle is ${\tan }^{-1}\left(\frac{y}{x}\right)$
  • $\text{Q}\text{u}\text{a}\text{d}\text{r}\text{a}\text{n}\text{t}\text{I}\text{I}\text{o}\text{r}\text{I}\text{I}\text{I}\text{:}$ polar angle is ${\tan }^{-1}\left(\frac{y}{x}\right)+\pi$

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Equations (as well as points) can also be converted to polar.

For $\text{Cartesian}\to \text{Polar}$, look for cancellation from ${\cos }^2\theta +{\sin }^2\theta =1$.

For $\text{Polar}\to \text{Cartesian}$, try to keep $\theta$ inside of trig functions.

• For example:

$$r={\sin }^2\theta \gg \gg \sqrt{x^2+y^2}={\left(\frac{y}{\sqrt{x^2+y^2}}\right)}^2$$ Link to original

08 Illustration

Example - Converting to polar: $\pi$-correction

Converting to polar: pi-correction

Compute the polar coordinates of $(-\frac{1}{2},+\frac{\sqrt{3}}{2})$ and of $(+\frac{\sqrt{2}}{2},-\frac{\sqrt{2}}{2})$.

Solution

For $(-\frac{1}{2},\frac{\sqrt{3}}{2})$ we observe first that it lies in Quadrant II.

Next compute:

$${\tan }^{-1}\left(\frac{\sqrt{3}/2}{-1/2}\right)\gg \gg {\tan }^{-1}(-\sqrt{3})\gg \gg -\pi /3$$

This angle is in Quadrant IV. We add $\pi$ to get the polar angle in Quadrant II:

$$\theta =\pi -\pi /3\gg \gg 2\pi /3$$

The radius is of course $1$ since this point lies on the unit circle. Therefore polar coordinates are $(r,\theta )=(\mathrm{1,2}\pi /3)$.

For $(+\frac{\sqrt{2}}{2},-\frac{\sqrt{2}}{2})$ we observe first that it lies in Quadrant IV. (No extra $\pi$ needed.)

Next compute:

$${\tan }^{-1}\left(\frac{-\sqrt{2}/2}{+\sqrt{2}/2}\right)\gg \gg {\tan }^{-1}(-1)\gg \gg -\pi /4$$

So the point in polar is $(1,-\pi /4)$.

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Example - Shifted circle in polar

Shifted circle in polar

For example, let’s convert a shifted circle to polar. Say we have the Cartesian equation:

$$x^2+{(y-3)}^2=9$$

Then to find the polar we substitute $x=r\cos \theta$ and $y=r\sin \theta$ and simplify:

$$\begin{array}{c}{x}^2+{(y-3)}^2=9\\ \\ \gg \gg r^2{\cos }^2\theta +{(r\sin \theta -3)}^2=9\\ \\ \gg \gg r^2{\cos }^2\theta +r^2{\sin }^2\theta -6r\sin \theta +9=9\\ \\ \gg \gg r^2({\sin }^2\theta +{\cos }^2\theta )-6r\sin \theta =0\\ \\ \gg \gg r^2-6r\sin \theta =0\gg \gg r=6\sin \theta \end{array}$$

So this shifted circle is the polar graph of the polar function $r(\theta )=6\sin \theta$.

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