W13 - Examples

Parametric lines

Parametric lines

(1) Parametric coordinate functions for a line:

$$x=a+rt,y=b+st,t\in (-\infty ,+\infty )$$

Compare this to the graph of linear function:

$$y=mx+b\text{some }m,b$$

Vertical lines cannot be described as the graph of a function. We must use $x=a$.

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(2) Parametric lines can describe all lines equally well, including horizontal and vertical lines.

A vertical line $x=a$ is achieved by setting $s=0$ and $r\ne 0$.

A horizontal line $y=b$ is achieved by setting $r=0$ and $s\ne 0$.

A non-vertical line $y=mx+b$ may be achieved by setting $s=m$ and $r=1$, and $a=0$.

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(3) Assuming that $r\ne 0$, the parametric coordinate functions describe a line satisfying:

$$\begin{array}{c}y=b+s\left(\frac{x-a}{r}\right)\\ \\ \gg \gg y=\frac{s}{r}\cdot x+\left(b-\frac{s}{r}\cdot a\right)\end{array}$$

and therefore the slope is $m=\frac{s}{r}$ and the $y$-intercept is $b-\frac{s}{r}\cdot a$.

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(4) The point-slope construction of a line has a parametric analogue:

$$\begin{array}{c}\text{point-slope line:}\\ \\ y-a=m(x-b)(x,y)=(a+t,b+mt)\end{array}$$

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Parametric ellipses

Parametric ellipses

The general equation of an ellipse centered at $(h,k)$ with half-axes $a$ and $b$ is:

$${\left(\frac{x-h}{a}\right)}^2+{\left(\frac{y-k}{b}\right)}^2=1$$

This equation represents a stretched unit circle:

• by $a$ in the $x$-axis
• by $b$ in the $y$-axis

Parametric coordinate functions for the general ellipse:

$$x=h+a\cos t,y=k+b\sin t,t\in [0,2\pi )$$ Link to original

Parametric cycloids

Parametric cycloids

The cycloid is the curve traced by a pen attached to the rim of a wheel as it rolls.

It is easy to describe the cycloid parametrically. Consider the geometry of the situation:

The center $C$ of the wheel is moving rightwards at a constant speed of $1$, so its position is $(t,1)$. The angle is revolving at the same constant rate of $1$ (in radians) because the radius is $1$.

The triangle shown has base $\sin t$, so the $x$ coordinate is $t-\sin t$. The $y$ coordinate is $1-\cos t$.

So the coordinates of the point $P=(x,y)$ are given parametrically by:

$$x=t-\sin t,y=1-\cos t,t>0$$

If the circle has another radius, say $R$, then the parametric formulas change to:

$$x=Rt-R\sin t,y=R-R\cos t,t>0$$Link to original

Tangent to a cycloid

Tangent to a cycloid

Find the tangent line (described parametrically) to the cycloid $(4t-4\sin t,4-4\cos t)$ when $t=\pi /4$.

Solution

(1) Compute $x^{\prime }$ and $y^{\prime }$.

Find $x^{\prime }(t)$ and $y^{\prime }(t)$:

$$\begin{array}{rl}x(t)& =4t-4\sin t\gg \gg x^{\prime }(t)=4-4\cos t\\ & \\ y(t)& =4-4\cos t\gg \gg y^{\prime }(t)=4\sin t\end{array}$$

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(2) Plug in $t=\pi /4$:

$$\begin{array}{c}{x}^{\prime }(\pi /4)\gg \gg 4-4\cos (\pi /4)\gg \gg 4-2\sqrt{2}\\ \\ y^{\prime }(\pi /4)\gg \gg 4\sin (\pi /4)\gg \gg 2\sqrt{2}\end{array}$$

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(3) Apply formula: $\frac{dy}{dx}=\frac{y^{\prime }}{x^{\prime }}$:

Calculate $\frac{dy}{dx}$ at $t=\pi /4$:

$$\begin{array}{c}\frac{dy}{dx}(\pi /4)=\frac{y^{\prime }(\pi /4)}{x^{\prime }(\pi /4)}\gg \gg \frac{2\sqrt{2}}{4-2\sqrt{2}}\\ \\ \gg \gg \frac{2\sqrt{2}}{4-2\sqrt{2}}\cdot \frac{4+2\sqrt{2}}{4+2\sqrt{2}}\\ \\ \gg \gg \frac{8\sqrt{2}+8}{16-8}\gg \gg \sqrt{2}+1\end{array}$$

So:

$${\frac{dy}{dx}|}_{t=\pi /4}=\sqrt{2}+1$$

This is the slope $m$ for our line.

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(4) Need the point $P$ for our line. Find $(x,y)$ at $t=\pi /4$.

Plug $t=\pi /4$ into parametric formulas:

$$\begin{array}{c}(x(t),y(t)){|}_{t=\pi /4}\gg \gg \left(4\frac{\pi }{4}-4\sin (\pi /4),4-4\cos (\pi /4)\right)\\ \\ \gg \gg \left(\pi -2\sqrt{2},4-2\sqrt{2}\right)\end{array}$$

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(5) Point-slope formulation of tangent line:

$$x=a+t,y=b+mt$$

Inserting data:

$$x=(\pi -2\sqrt{2})+t,y=(4-2\sqrt{2})+(\sqrt{2}+1)t$$ Link to original

Vertical and horizontal tangents of the circle

Vertical and horizontal tangents of the circle

Consider the circle parametrized by $x=\cos t$ and $y=\sin t$. Find the points where the tangent lines are vertical or horizontal.

Solution

(1) For the points with vertical tangent line, we find where the moving point has $x^{\prime }(t)=0$ (purely vertical motion):

$$\begin{array}{c}{x}^{\prime }(t)=-\sin t,\\ \\ x^{\prime }(t)=0\gg \gg -\sin t=0\\ \\ \gg \gg t=0,\pi \end{array}$$

The moving point is at $(1,0)$ when $t=0$, and at $(-1,0)$ when $t=\pi$.

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(2) For the points with horizontal tangent line, we find where the moving point has $y^{\prime }(t)=0$ (purely horizontal motion):

$$\begin{array}{c}{y}^{\prime }(t)=\cos t,\\ \\ y^{\prime }(t)=0\gg \gg \cos t=0\\ \\ \gg \gg t=\frac{\pi }{2},\frac{3\pi }{2}\end{array}$$

The moving point is at $(0,1)$ when $t=\pi /2$, and at $(0,-1)$ when $t=3\pi /2$.

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Finding the point with specified slope

Finding the point with specified slope

Consider the parametric curve given by $(x,y)=(t^2,t^3)$. Find the point where the slope of the tangent line to this curve equals 5.

Solution

(1) Compute the derivatives:

$$x^{\prime }(t)=2t,y^{\prime }(t)=3t^2$$

Therefore the slope of the tangent line, in terms of $t$:

$$\begin{array}{c}m=\frac{dy}{dx}=\frac{y^{\prime }(t)}{x^{\prime }(t)}\\ \\ \gg \gg \frac{3t^2}{2t}\gg \gg \frac{3}{2}t\end{array}$$

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(2) Set up equation:

$$\begin{array}{c}m=5\\ \\ \frac{3}{2}t=5\\ \\ \gg \gg t=\frac{10}{3}\end{array}$$

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(3) Find the point:

$$(x,y){|}_{t=10/3}\gg \gg \left(\frac{100}{9},\frac{1000}{27}\right)$$ Link to original

Perimeter of a circle

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

Perimeter of an asteroid

Perimeter of an asteroid

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

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

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(2) Compute the infinitesimal arc element.

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

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 $\theta \in [0,2\pi ]$, 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:

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

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

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