Calculus and parametric curves

$\newenvironment {prompt}{}{} \newcommand {\todo }[0]{} \newcommand {\oiint }[0]{{\large \bigcirc }\kern -1.56em\iint } \newcommand {\mooculus }[0]{\textsf {\textbf {MOOC}\textnormal {\textsf {ULUS}}}} \newcommand {\npnoround }[0]{\nprounddigits {-1}} \newcommand {\npnoroundexp }[0]{\nproundexpdigits {-1}} \newcommand {\npunitcommand }[1]{\ensuremath {\mathrm {#1}}} \newcommand {\RR }[0]{\mathbb R} \newcommand {\R }[0]{\mathbb R} \newcommand {\N }[0]{\mathbb N} \newcommand {\Z }[0]{\mathbb Z} \newcommand {\sagemath }[0]{\textsf {SageMath}} \newcommand {\d }[0]{\mathop {}\!d} \newcommand {\l }[0]{\ell } \newcommand {\ddx }[0]{\frac {d}{\d x}} \newcommand {\zeroOverZero }[0]{\ensuremath {\boldsymbol {\tfrac {0}{0}}}} \newcommand {\inftyOverInfty }[0]{\ensuremath {\boldsymbol {\tfrac {\infty }{\infty }}}} \newcommand {\zeroOverInfty }[0]{\ensuremath {\boldsymbol {\tfrac {0}{\infty }}}} \newcommand {\zeroTimesInfty }[0]{\ensuremath {\small \boldsymbol {0\cdot \infty }}} \newcommand {\inftyMinusInfty }[0]{\ensuremath {\small \boldsymbol {\infty -\infty }}} \newcommand {\oneToInfty }[0]{\ensuremath {\boldsymbol {1^\infty }}} \newcommand {\zeroToZero }[0]{\ensuremath {\boldsymbol {0^0}}} \newcommand {\inftyToZero }[0]{\ensuremath {\boldsymbol {\infty ^0}}} \newcommand {\numOverZero }[0]{\ensuremath {\boldsymbol {\tfrac {\#}{0}}}} \newcommand {\dfn }[0]{\textbf } \newcommand {\unit }[0]{\mathop {}\!\mathrm } \newcommand {\eval }[1]{\bigg [ #1 \bigg ]} \newcommand {\seq }[1]{\left ( #1 \right )} \newcommand {\epsilon }[0]{\varepsilon } \newcommand {\phi }[0]{\varphi } \newcommand {\iff }[0]{\Leftrightarrow } \DeclareMathOperator {\arccot }{arccot} \DeclareMathOperator {\arcsec }{arcsec} \DeclareMathOperator {\arccsc }{arccsc} \DeclareMathOperator {\si }{Si} \DeclareMathOperator {\scal }{scal} \DeclareMathOperator {\sign }{sign} \newcommand {\arrowvec }[1]{{\overset {\rightharpoonup }{#1}}} \newcommand {\vec }[1]{{\overset {\boldsymbol {\rightharpoonup }}{\mathbf {#1}}}} \DeclareMathOperator {\proj }{\vec {proj}} \newcommand {\veci }[0]{{\boldsymbol {\hat {\imath }}}} \newcommand {\vecj }[0]{{\boldsymbol {\hat {\jmath }}}} \newcommand {\veck }[0]{{\boldsymbol {\hat {k}}}} \newcommand {\vecl }[0]{\vec {\boldsymbol {\l }}} \newcommand {\uvec }[1]{\mathbf {\hat {#1}}} \newcommand {\utan }[0]{\mathbf {\hat {t}}} \newcommand {\unormal }[0]{\mathbf {\hat {n}}} \newcommand {\ubinormal }[0]{\mathbf {\hat {b}}} \newcommand {\dotp }[0]{\bullet } \newcommand {\cross }[0]{\boldsymbol \times } \newcommand {\grad }[0]{\boldsymbol \nabla } \newcommand {\divergence }[0]{\grad \dotp } \newcommand {\curl }[0]{\grad \cross } \newcommand {\lto }[0]{\mathop {\longrightarrow \,}\limits } \newcommand {\bar }[0]{\overline } \newcommand {\surfaceColor }[0]{violet} \newcommand {\surfaceColorTwo }[0]{redyellow} \newcommand {\sliceColor }[0]{greenyellow} \newcommand {\vector }[1]{\left \langle #1\right \rangle } \newcommand {\sectionOutcomes }[0]{} \newcommand {\HyperFirstAtBeginDocument }[0]{\AtBeginDocument } \newcommand {\epstopdfsetup }[0]{\setkeys {ETE}} changed the theorem header of amsthm failed\MessageBreak }}} \newcommand {\GetTitleStringSetup }[0]{\setkeys {gettitlestring}} \newcommand {\Sectionformat }[2]{#1}$

We discuss derivatives and integrals of parametric curves.

Derivatives

Let’s get right to it. Given a parametric function $(x(t),y(t))$ and recalling that

$\begin{align*} \d x &= x'(t) \d t\\ \d y &= y'(t) \d t, \end{align*}$ we can see how to compute the derivative of $y$ with respect to $x$ using differentials: $\frac {\d y}{\d x} = \frac {y'(t) \d t}{x'(t) \d t} = \frac {y'(t)}{x'(t)}$ provided that $x'(t) \ne 0$ .

We use this to define the tangent line.

Let a curve $C$ be parametrized by $x(t)$ and $y(t)$ , where $x$ and $y$ are differentiable functions on some interval $I$ containing $t=t_0$ . The tangent line to $C$ at $t=t_0$ is the line through $\big (x(t_0),y(t_0)\big )\text { with slope }m=\frac {y'(t_0)}{x'(t_0)},$ provided $x'(t_0)\neq 0$ .

The normal line to $C$ at $t=t_0$ is the line through $\big (x(t_0),y(t_0)\big )\text { with slope }m=\frac {-x'(t_0)}{y'(t_0)},$ provided $y'(t_0)\neq 0$ . The normal line is perpendicular to the tangent line.

The definition leaves two special cases to consider. When the tangent line is horizontal, the normal line is undefined by the above definition as $y'(t_0)=0$ . Likewise, when the normal line is horizontal, the tangent line is undefined. It seems reasonable that these lines be defined (one can draw a line tangent to the “right side” of a circle, for instance), so we add the following to the above definition.

If the tangent line at $t=t_0$ has a slope of $0$ , the normal line to $C$ at $t=t_0$ is the vertical line $x=x(t_0)$ .
If the normal line at $t=t_0$ has a slope of $0$ , the tangent line to $C$ at $t=t_0$ is the line $x=x(t_0)$ .

Time for some examples.

Let $x=5t^2-6t+4$ and $y=t^2+6t-1$ , and let $C$ be the curve defined by these equations. Find the equations of the tangent and normal lines to $C$ at $t=3$ .

We start by computing $x'(t) = \answer [given]{10t-6}$ and $y'(t) =\answer [given]{2t+6}.$ Thus $\dd [y]{x} = \answer [given]{\frac {2t+6}{10t-6}}.$ Make note of something that might seem unusual: $\dd [y]{x}$ is a function of $t$ , not $x$ . Just as points on the curve are found in terms of $t$ , so are the slopes of the tangent lines.

The point on $C$ at $t=3$ is $(31,26)$ . The slope of the tangent line is $m=1/2$ and the slope of the normal line is $m=-2$ . Thus,

the equation of the tangent line is $y=\answer [given]{\frac {(x-31)}{2}+26}$ , and
the equation of the normal line is $y=\answer [given]{-2(x-31)+26}$ .

Find where the circle, defined by $x=\cos t$ and $y=\sin t$ on $[0,2\pi ]$ , has vertical and horizontal tangent lines.

We compute the derivative $\dd [y]{x} = \frac {y'(t)}{x'(t)} = \answer [given]{-\frac {\cos t}{\sin t}}.$ The derivative is $0$ when $\cos t= \answer [given]{0}$ . This happens when $t=\pi /2$ , and $t= 3\pi /2$ . These are the points $(0,1)$ and $(0,-1)$ on the circle.

The normal line is horizontal (and hence, the tangent line is vertical) when $\sin t=\answer [given]{0}$ . This happens when $t= 0$ , $t=pi$ , $t=2\pi$ , corresponding to the points $(-1,0)$ and $(0,1)$ on the circle. These results should make intuitive sense.

Integrals

Assuming that the curve given by a parametric formula $(x(t),y(t))$ represents $y$ as a function of $x$ , that is traced out exactly once, we can integrate our parametric formula without too much additional trouble. Again, recall that $\d x = x'(t) \d t$ So we may write $\int _a^b y \d x = \int _\alpha ^\beta y(t) \cdot x'(t) \d t$ where $x(\alpha ) = a$ and $x(\beta ) = b$ .

We’ll be talking about this in more detail soon, so a simple example should suffice.

Let $x(t) = \cos (t)$ and $y(t)= \sin (t)$ as $t$ runs from $0$ to $\pi$ :

Compute the area between this region and the $x$ -axis on the interval $[-1,1]$ .

We would like to compute: $\int _{-1}^1 y \d x$ The parametric equations allow us to make a substitution: $\begin{align*} y(t) &=\answer[given]{\sin(t)}\\ \d x &=\answer[given]{-\sin(t)}\d t \end{align*}$ now then integral becomes $\int _{-1}^{1} y \d x= \int _\alpha ^\beta (\sin (t))(-\sin (t))\d t$ Now we need to find $\alpha$ and $\beta$ . Note that using $x(t)=\cos (t)$ When $x=-1$ , $t=\answer [given]{\pi }$ . Likewise, when $x=1$ , $t=\answer [given]{0}$ . Hence our integral becomes $\begin{align*} \int_{-1}^{1} y \d x&= \int_\pi^0 (\sin(t))(-\sin(t))\d t\\ &= \int_{\answer[given]{0}}^{\answer[given]{\pi}}\sin^2(t)\d t\\ &= \int_0^\pi\frac{1-\cos(2t)}{2}\d t\\ &= \int_0^\pi\frac{1}{2}-\frac{\cos(2t)}{2}\d t\\ &= \eval{\frac{t}{2}-\frac{\sin(2t)}{4}}_0^\pi\\ &= \answer[given]{\frac{\pi}{2}} \end{align*}$