Chapter 12 - Curves in the Plane

Introduction

Introduction

Objects of study in Vector Calc: Forces, Fields, Flows

What is Calc IV? Differential Geometry!

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Electricity and Magnetism

Electricity and Magnetism

Vector calculus was invented by O. Heaviside and J. W. Gibbs (1901) to simplify the presentation of electromagnetic theory that was originally written using quaternion fields by James Clerk Maxwell (1873). Since then it has been used extensively for fluid dynamics as well.

Maxwell:

Heaviside:
(also Telegrapher Equations)

Gibbs:
(also statistical mechanics)

Charged particles:

Current in solenoid:

Maxwell Equations:

EM Waves:

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Fluid Dynamics

Fluid Dynamics

SpaceX Starship reentry:

Rayleigh-Taylor instability:

Laminar vs. turbulent flow:

Conservation of Mass:

Conservation of Momentum:

Conservation of Energy:

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Plasma Physics: Fluids + EM Fields:

Plasma Physics - Fluids + EM Fields

Tokomaks:

Stellerators:

Hall Effect Thruster:

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Mathematics: Differential Geometry

Mathematics - Differential Geometry

Gravity:

Link: Gravitational lensing simulation

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Optimization

Optimization

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Parametric curves in 2D

Parametric curve

Parametric curve

A parametric curve or parameterized curve in the plane is a function $\gamma :[a,b]\to {\mathbb{R}}^2$ from an interval to the plane, given with coordinate functions by $(x(t),y(t))$.

The data of a parametric curve includes the parametrization, not just the set of points in the image.

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Example

Line in the plane from parametric equations

The parametric curve $(x(t),y(t))=(a+bt,c+dt)$ determines a line in ${\mathbb{R}}^2$.
Solve for $y$ in terms of $x$:

$$y=c+d\cdot \frac{1}{b}(x-a)=\frac{d}{b}x+(c-\frac{da}{b}).$$

The slope is $m=\frac{d}{b}$ and the $y$-intercept is $c-\frac{da}{b}$.

Observe that by setting $b=0,d\ne 0$, the image is a vertical line in ${\mathbb{R}}^2$ intersecting the $x$-axis at $x=a$.

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Warning!

Warning - same vertical line

The image of $(x,y)=(a,t^3)$ is the same vertical line!

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Warning!

Warning - Lissajous figure self-intersection

This “Lissajous figure” shows that parametric curves can intersect themselves!

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Example

Parametrization of a circle

Circle, radius $=r$, center $=(h,k)$
$(x(t),y(t))=(h+r\cos (t),k+r\sin (t)),t\in [0,2\pi ]$

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Example

Sine and cosine of 2t as a parametric curve

$(x,y)=(\sin (2t),\cos (2t)),t\in [0,2\pi ]$

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Example

Cubic curve with x equal to y cubed

$(x,y)=(-t^3,-t)$
The image curve satisfies $x=y^3$.

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Warning!

Warning - same image in the upper quadrant

In the upper quadrant, this curve has the same image:
$(x,y)=(e^{-3t},e^{-t})$
Check: $y^3=(e^{-t}{)}^3=e^{-3t}=x.$

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Exercise

01

Sketch the curve given by $(x,y)=(2t-4,3+t^2)$.

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Exercise

02

Describe the parametric curve $(x,y)=(2t-4,3+t^2)$ as the graph of a function.

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Exercise

03

Consider the curve $(50t,100t-5t^2)$. Find the point on this curve with largest $y$-value.

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Exercise

04

[Already done in class.]
Points satisfying ${\left(\frac{x}{2}\right)}^2+{\left(\frac{y}{5}\right)}^2=1$ lie on an ellipse. Find a parametric curve that covers this ellipse.

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Exercise

05

Find a parametric curve that is a line passing through $(3,-2)$ with slope $m=4$.

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Exercise

06

Eliminate $t$ to find an equation satisfied by $x,y$ alone for the following two parametric curves:
a. $(x,y)=(t^2-3,t+2)$
b. $(x,y)=(\csc (t),\cot (t))$

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Exercise

07

Describe the image of the curve parametrized by $(x,y)=(\sin (t),{\cos }^2(t))$.

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Exercise

08

One particle has position $(x,y)=(t+5,t^2+4t+6)$, a second particle has position $(2t+1,2t+6)$. Calculate the two points of intersection of the image curves determined by these formulas. Do the particles collide at those intersection points?

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Calculus with parametric curves in 2D

Traditional derivatives (unnatural)

Example

Slope of the tangent line to a parametric curve

Find a general formula for the slope of the tangent line to a parametric curve $\gamma (t)=(x(t),y(t))$ at the point determined by $t_0$.
Solution: Consider $t$ as a function of $x$. Remember the formula $\frac{dy}{dx}=\frac{1}{\frac{df}{dy}}$ when $x=f(y)$ so $y=f^{-1}(x)$. (Think about implicit differentiation.) Then using the Chain Rule and this formula, we have:

$$\frac{dy}{dx}=\frac{dy}{dt}\cdot \frac{dt}{dx}=\frac{dy}{dt}\cdot \frac{1}{\frac{dx}{dt}}=\frac{y^{\prime }(t)}{x^{\prime }(t)},$$

so the answer is $m(t_0)=\frac{y^{\prime }(t_0)}{x^{\prime }(t_0)}$.

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Warning!

Warning - traditional derivative is unnatural

This quantity is unnatural!

• From the point of view of graphs of functions, $\frac{dy}{dx}$ is natural and important. For a graph, space portrays the relation between input and output.
• From the point of view of parametric curves, it is unnatural. Here $x$ and $y$ should be treated on the same footing. Space portrays the output only.

Soon we consider more natural derivative operations in the parametric setting.

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Warning!

Warning - second derivative is not a ratio

$$\frac{d^{2}y}{d{x}^2}\ne \frac{\frac{d^{2}y}{d{t}^2}}{\frac{d^{2}x}{d{t}^2}}$$ Link to original

Example

Tangent line to a cycloid

A cycloid is given by the parametric curve with formula $(x,y)=(r(t-\sin (t)),r(1-\cos (t))$. Find the tangent line to this cycloid at $t=\pi /3$.
Solution: Using the formula for $\frac{dy}{dx}$, obtain:

$$\frac{dy}{dx}=\frac{r(0+\sin (t))}{r(1-\cos (t))}=\frac{\sin (t)}{1-\cos (t)}.$$

Then we have:

$$m={\frac{dy}{dx}|}_{t=\pi /3}=\sqrt{3}.$$

The image point when $t=\pi /3$ is $(x,y)=(r(\frac{\pi }{3}-\frac{\sqrt{3}}{2}),\frac{r}{2})$. After solving for the $y$-intercept we obtain the final equation:

$$y=\sqrt{3}x+r\left(\frac{\pi }{\sqrt{3}}-2\right).$$Link to original

Exercise

01

Find the equation of the tangent line to the parametric curve $(x,y)=(t^2+1,t^3-4t)$ at the point given by $t=3$. Where is the tangent line horizontal?

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Distance traveled

Distance traveled

The distance traveled by a particle whose position is given parametrically by $(x(t),y(t))$, as $t$ increases from $a$ to $b$, is computed using this integral:

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

The formula is quickly derived from the usual one for arc length of the graph of a function:

$$L={\int }_a^b\sqrt{1+{\left(\frac{dy}{dx}\right)}^2}dx$$

by using the formula for $\frac{dy}{dx}(t)$ together with the fact that $dx=\left(\frac{dx}{dt}\right)dt$.

It is better to see the formula as the summation of infinitesimal elements of distance $ds$, computing these with the Pythagorean Theorem:

$$ds=\sqrt{x^{\prime }(t{)}^2+y^{\prime }(t{)}^2}dt,$$

so the formula is really just:

$$L={\int }_a^{b}ds.$$ Link to original

Exercise

02

Find an integral formula for the length of the circumference of a general ellipse (centered at the origin). (Do not try to evaluate the integral!)

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Length function

Length function

Given a parametric curve $(x(t),y(t))$, the formula for distance traveled can be used to define a length function $s(t)$. It is simply the distance traveled from $t=t_0$ to the time $t$ itself. To write this we need a new dummy variable:

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

Speed

Speed

The speed is the rate of distance traveled per unit time: $v(t)=s^{\prime }(t)$. From the Fundamental Theorem of Calculus, we have:

$$v(t)=\sqrt{x^{\prime }(t{)}^2+y^{\prime }(t{)}^2}.$$ Link to original

Exercise

03

Suppose a particle traces the graph of a parabola with its position given parametrically by $(x,y)=(t,t^2)$. How fast is the particle moving at time $t=10$? What is its acceleration at $t=10$?

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Curvature

Curvature

The curvature $\kappa (t)$ of a parametric curve in the plane may be defined as the rate of change of the angle of inclination $\varphi$ per unit of arc length.

We can find a formula for this. Let the curve be given by $(x(t),y(t))$. Clearly $\tan (\varphi )=\frac{dy}{dx}$ so $\varphi ={\tan }^{-1}\left(\frac{y^{\prime }(t)}{x^{\prime }(t)}\right)$. Using the quotient rule we have:

$$\begin{array}{rl}\frac{d\varphi }{dt}& =\frac{1}{1+{\left(\frac{y^{\prime }(t)}{x^{\prime }(t)}\right)}^2}\cdot \frac{x^{\prime }(t)y^{\prime \prime }(t)-x^{\prime \prime }(t)y^{\prime }(t)}{x^{\prime }(t{)}^2}\\ & =\frac{x^{\prime }(t)y^{\prime \prime }(t)-x^{\prime \prime }(t)y^{\prime }(t)}{x^{\prime }(t{)}^2+y^{\prime }(t{)}^2}.\end{array}$$

Then

$$\frac{d\varphi }{ds}=\frac{d\varphi }{dt}\cdot \frac{dt}{ds}=\frac{d\varphi }{dt}\cdot \frac{1}{s^{\prime }(t)}=\frac{x^{\prime }(t)y^{\prime \prime }(t)-x^{\prime \prime }(t)y^{\prime }(t)}{{\left(x^{\prime }(t{)}^2+y^{\prime }(t{)}^2\right)}^{3/2}}.$$Link to original

Exercise

04

Compute the curvature function $\kappa (t)$ for the circle given parametrically by $(r\cos (t),r\sin (t))$. Is the answer what you should have expected?

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Polar coordinates

Introduction

The polar coordinate system uses parameters $(r,\theta )$ instead of $(x,y)$ to identify points in the plane. These coordinates are useful in problems having rotational symmetry. The two coordinate systems are related by these equations:

$$\begin{array}{rlrl}x& =r\cos (\theta )& r^2& =x^2+y^2\\ y& =r\sin (\theta )& \tan (\theta )& =y/x\end{array}$$ Link to original

Example

Circle with polar equation r = 2cos(theta)

The points whose polar coordinates satisfy $r=2\cos (\theta )$ define a circle:

We can verify this by converting to Cartesian coordinates:

$$x=r\cos (\theta )=\frac{1}{2}{r}^2=\frac{1}{2}(x^2+y^2),$$

so

$$(x-1{)}^2+y^2=1.$$ Link to original

Exercise

01

Find the polar form of the equation of a vertical line passing through $x=1$.

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Area under polar graphs

Area under polar graphs

There is a formula to compute the area of the “sector sum” formed by a polar graph.

$$A={\int }_a^b\frac{1}{2}r(\theta {)}^{2}d\theta$$

The area of a sector is $\frac{1}{2}{r}^2\theta$, so the area of an infinitesimal sector is $\frac{1}{2}{r}^{2}d\theta$, and the formula just sums these.

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Example

Area of one loop of a four-leafed rose

Find the area of one loop of the four-leafed rose $r=\cos (2\theta )$.

Solution: Plug into the formula, and use $a=-\frac{\pi }{4},b=\frac{\pi }{4}$, obtaining:

$$A={\int }_{-\frac{\pi }{4}}^{\frac{\pi }{4}}\frac{1}{4}(1+\cos (4\theta ))d\theta =\frac{\pi }{8}.$$Link to original

Exercise

02

Find the area inside the region bounded by $r=5+2\sin (\theta )$.

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