W12 - Notes

Parametric curves

Videos

Review Videos

Videos, Organic Chemistry Tutor

• Intro to parametric equations and graphing

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01 Theory

Theory 1

Parametric curves are curves traced by the path of a ‘moving’ point. An independent parameter, such as $t$ for ‘time’, controls both $x$ and $y$ values through Cartesian coordinate functions $x(t)$ and $y(t)$. The coordinates of the moving point are $(x(t),y(t))$.

Parametric curve

A parametric curve is a function from parameter space $ℝ$ to the plane ${ℝ}^2$ given in terms of coordinate functions:

$$t⟼(x(t),y(t))$$

Other notations

Be aware that sometimes the coordinate functions are written with $f$ and $g$ (or yet other letters) like this:

$$(x,y)=(f(t),g(t))$$

Or simply equating coordinate letters with functions: $x=f(t),y=g(t)$

Sometimes a different parameter is used, like $s$ or $u$.

For example, suppose:

$$x=t^2-2t,y=t+1$$

The curve traced out is a parabola that opens horizontally:

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Given a parametric curve, we can create an equation satisfied by $x$ and $y$ variables by solving for $t$ in either coordinate function (inverting either $f$ or $g$) and plugging the result into the other function.

In the example:

$$\begin{array}{c}y=t+1\gg \gg t=y-1\\ \\ \gg \gg x=t^2-2t\gg \gg x={(y-1)}^2-2(y-1)\\ \\ \gg \gg x=y^2-4y+3\gg \gg x={(y-2)}^2-1\end{array}$$

This is the equation of a parabola centered at $(-\mathrm{1,2})$ that opens to the right.

Image of a parametric curve

The image of a parametric curve is the set of output points $(x(t),y(t))$ that are traversed by the moving point.

A parametric curve has hidden information that isn’t contained in the image:

• The time values $t$ when the moving point is found in various locations.
• The speed at which the curve is traversed.
• The direction in which the curve is traversed.

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We can reparametrize a parametric curve to use a different parameter or different coordinate functions while leaving the image unchanged.

In the previous example, shift $t$ by $1$:

$$\begin{array}{c}x={(t+1)}^2-2(t+1),y=(t+1)+1\\ \\ \gg \gg x=t^2-1,y=t+2\end{array}$$

Since the parameter $t$ and the parameter $t+1$ both cover the same values for $t\in (-\infty ,\infty )$, the same curve is traversed. But the moving point in the second, shifted version reaches any given location one unit earlier in time. (When $t=-1$ in the second version, the input to $x(t)$ and $y(t)$ is the same as when $t=0$ in the first one.)

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02 Illustration

Example - Parametric circles

Parametric circles

The standard equation of a circle of radius $R$ centered at the point $(h,k)$:

$${(x-h)}^2+{(y-k)}^2=R^2$$

This equation says that the distance from a point $(x,y)$ on the circle to the center point $(h,k)$ equals $R$. This fact defines the circle.

Parametric coordinates for the circle:

$$x=h+R\cos t,y=k+R\sin t,t\in [\mathrm{0,2}\pi )$$

For example, the unit circle $x^2+y^2=1$ is parametrized by $x=\cos t$ and $y=\sin t$.

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Example - 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:

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

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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Example - 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:

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

Example - 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

Calculus with parametric curves

03 Theory - Slope, concavity

Theory 1

We can use $x(t)$ and $y(t)$ data to compute the slope of a parametric curve in terms of $t$.

Slope formula

Given a parametric curve $(x(t),y(t))$, its slope satisfies:

$$\frac{dy}{dx}=\frac{y^{\prime }(t)}{x^{\prime }(t)}(\text{where }{x}^{\prime }(t)\ne 0)$$

Concavity formula

Given a parametric curve $(x(t),y(t))$, its concavity satisfies the formula:

$\frac{d^{2}y}{d{x}^2}=\frac{d}{dt}\left(\frac{y^{\prime }(t)}{x^{\prime }(t)}\right)\cdot \frac{1}{x^{\prime }(t)}(\text{where }{x}^{\prime }(t)\ne 0)$

Extra - Derivation of slope and concavity formulas

For both derivations, it is necessary to view $t$ as a function of $x$ through the inverse parameter function. For example if $x=f(t)$ is the parametrization, then $t=f^{-1}(x)$ is the inverse parameter function.

We will need the derivative $\frac{dt}{dx}$ in terms of $t$. For this we use the formula for derivative of inverse functions:

$$\frac{dt}{dx}=\frac{1}{\frac{dx}{dt}}$$

Given all this, both formulas are simple applications of the chain rule.

For the slope:

$$\begin{array}{c}\frac{dy}{dx}=\frac{dy}{dt}\cdot \frac{dt}{dx}\gg \gg y^{\prime }(t)\cdot \frac{1}{dx/dt}\\ \\ \gg \gg \frac{y^{\prime }(t)}{x^{\prime }(t)}\end{array}$$

For the concavity:

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

(In the second step we inserted the formula for $\frac{dy}{dx}$ from the slope.)

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Pure vertical, Pure horizontal movement

In view of the formula $\frac{dy}{dx}=\frac{y^{\prime }(t)}{x^{\prime }(t)}$, we see:

• Pure vertical: when $x^{\prime }(t)=0$ and yet $y^{\prime }(t)\ne 0$
• Pure horizontal: when $y^{\prime }(t)=0$ and yet $x^{\prime }(t)\ne 0$

When $x^{\prime }(t_0)=y^{\prime }(t_0)=0$ for the same $t=t_0$, we have a stationary point, which might subsequently progress into pure vertical, pure horizontal, or neither.

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04 Illustration

Example - 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 $y^{\prime }(t)$:

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

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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: ${\displaystyle \frac{dy}{dx}=\frac{y^{\prime }}{x^{\prime }}}$:

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:

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

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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

Example - 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 $(\mathrm{1,0})$ when $t=0$, and at $(-\mathrm{1,0})$ 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 $(\mathrm{0,1})$ when $t=\pi /2$, and at $(0,-1)$ when $t=3\pi /2$.

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Example - 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:

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