Vector Calculus - Packet 02

Parametric curves in 2D

Parametric curve

A parametric curve or parameterized curve in the plane is a function $\gamma :[a,b]\to {ℝ}^2$ 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.

Example

The parametric curve $(x(t),y(t))=(a+bt,c+dt)$ determines a line in ${ℝ}^2$. 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 ${ℝ}^2$ intersecting the $x$-axis at $x=a$.

Warning!

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

Warning!

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

Example

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

Example

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

Example

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

Warning!

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

Exercise 02A-01

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

Exercise 02A-02

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

Exercise 02A-03

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

Exercise 02A-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.

Exercise 02A-05

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

Exercise 02A-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. $(x,y)=(\csc (t),\cot (t))$

Exercise 02A-07

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

Exercise 02A-08

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

Calculus with parametric curves in 2D

Traditional derivatives (unnatural)

Example

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 $m(t_0)=\frac{y^{\prime }(t_0)}{x^{\prime }(t_0)}$.

Warning!

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.

Warning!

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

Example

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:

$$y=\sqrt{3}x+r(\frac{\pi }{\sqrt{3}}-2).$$

Exercise 02B-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?

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:

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

Exercise 02B-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!)

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:

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

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:

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

Exercise 02B-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 $t=10$?

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:

$$\begin{array}{cccc}& \frac{d\varphi }{dt}& =\frac{1}{1+{\left(\frac{y^{\prime }(t)}{x^{\prime }(t)}\right)}^2}\cdot \frac{x^{\prime }(t)y^{\prime \prime }(t)-x^{\prime \prime }(t)y^{\prime }(t)}{x^{\prime }{(t)}^2}& \text{<span class="tml-eqn"></span>}\\ & & =\frac{x^{\prime }(t)y^{\prime \prime }(t)-x^{\prime \prime }(t)y^{\prime }(t)}{x^{\prime }{(t)}^2+y^{\prime }{(t)}^2}.& \text{<span class="tml-eqn"></span>}\end{array}$$

Then

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

Exercise 02B-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?

Polar coordinates

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:

$$\begin{array}{cccccc}& x& =r\cos (\theta )& r^2& =x^2+y^2& \text{<span class="tml-eqn"></span>}\\ & y& =r\sin (\theta )& \tan (\theta )& =y/x& \text{<span class="tml-eqn"></span>}\end{array}$$

Example

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

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

Exercise 02B-05

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

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.

Example

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:

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

Exercise 02B-06

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

Problems due 02 Sep 2023, 8:00pm

Problem 02-01

Find the self-intersection point of the curve parametrized by $(x,y)=(2t-t^3,t-t^2)$.

Problem 02-02

For a given parametric curve $\gamma (t)=(x(t),y(t))$, there is a formula for the second derivative:

$$\frac{d^{2}y}{d{x}^2}=\frac{x^{\prime }(t)y^{\prime \prime }(t)-y^{\prime }(t)x^{\prime \prime }(t)}{x^{\prime }{(t)}^3}.$$

Show how to derive this formula. Verify the formula for the parametric curve $(x,y)=(t,t^3)$ using the fact that this curve is the graph of a function. (You know how to find $\frac{d^{2}y}{d{x}^2}$ for graphs of functions.) Now compute $\frac{d^{2}y}{d{x}^2}$ for the parametric curve $(t^2,t^6)$, which has the same image.

Problem 02-03

Draw the graph of the limaçon given in polar coordinates by $r=1-2\sin (\theta )$. Find the area of the region lying between its two loops.

$★$ Problem 02-04

Do all 5 parts of the “Discovery Project” about Bézier Curves (p. 684 of 9e, 697-8 of 8e, 677-78 of 7e).