Unit 1: Coordinate Systems
Preface
This is the note of the MIT Course 18.01.2 Calculus. For more details, refer to MIT 18.x Catalog, and it can also be found at MIT Open Learning Library.
Parametric Curves
The parametric curve in the plane is defined by two equations:
$$
x\quad=\quad x(t)\\
y\quad=\quad y(t)
$$
where $t$ is called parameter. For each $t\in \R$, the point $(x(t), y(t))$ is a point on the curve.
Example 1:
$$ x\quad =\quad a\cos t\\ y\quad =\quad a\sin t $$
Since $$ x^2+y^2\ =\ a^2, $$ the graph of this parametric curve is a circle with radius $a$ and center at origin.
Tangent Line
The slope of a parametric curve $x=x(t)$, $y=y(t)$ is $$ \frac{dy}{dx}\quad=\quad \frac{dy/dt}{dx/dt}. $$
Therefore, we can use $\displaystyle\frac{y^{\prime}(t_0)}{x^{\prime}(t_0)}$ to compute the slope of the tangent line of the curve at $t=t_0$.
Arc Length
Consider a particle moving along a trajectory, which can be described by the parametric curve: $$ x\quad=\quad x(t)\\ y\quad=\quad y(t) $$
The speed of the particle is given by $$ \frac{ds}{dt}\quad=\quad\sqrt{\big(x^{\prime}(t)\big)^2+\big(y^{\prime}(t)\big)^2} $$
The differential arc length is given by $$ ds\quad=\quad \sqrt{\big(x^{\prime}(t)\big)^2+\big(y^{\prime}(t)\big)^2}\ dt $$
Let $s=s(t)$, where $s(t)$ is used to denote arc length. If $s_0=s(t_0)$ and $s_1=s(t_1)$, then the arc length from $t_0$ to $t_1$ can be calculated by $$ s_1-s_0=\int_{s_0}^{s_1}\ ds=\int_{t_0}^{t_1}\sqrt{\big(x^{\prime}(t)\big)^2+\big(y^{\prime}(t)\big)^2}\ dt $$
Surface Area
Consider the parametric curve: $$ x\quad=\quad x(t)\\ y\quad=\quad y(t) $$
Calculate the surface formed by rotating the curve about $y$-axis. $$ \begin{aligned} dA\quad=&\quad 2\pi x\ ds\\ =&\quad 2\pi x(t)\sqrt{\big(x^{\prime}(t)\big)^2+\big(y^{\prime}(t)\big)^2}\ dt \end{aligned} $$
Therefore, the surface area from time $t_0$ to time $t_1$ is given by the integral: $$ \int_{t_0}^{t_1} 2\pi x(t)\sqrt{\big(x^{\prime}(t)\big)^2+\big(y^{\prime}(t)\big)^2}\ dt $$
We can also calculate the surface formed by rotating the curve about $x$-axis. $$ \begin{aligned} dA\quad=&\quad 2\pi y\ ds\\ =&\quad 2\pi y(t)\sqrt{\big(x^{\prime}(t)\big)^2+\big(y^{\prime}(t)\big)^2}\ dt \end{aligned} $$
Therefore, the surface area from time $t_0$ to time $t_1$ is given by the integral: $$ \int_{t_0}^{t_1} 2\pi y(t)\sqrt{\big(x^{\prime}(t)\big)^2+\big(y^{\prime}(t)\big)^2}\ dt $$
Area
We have already known that the area under the a curve $y=f(x)$ between $x=a$ and $x=b$ is given by $\displaystyle \int_a^b f(x)\ dx$.
We can also use the parametric equation to express the curve: $$ x\quad=\quad x(t)\\ y\quad=\quad y(t) $$
Suppose that $x(t_0)=a$ and $x(t_1)=b$, then $$ \begin{aligned} \int_a^b f(x)\ dx\ =&\ \int_{t_0}^{t_1} f(x(t))\ x^{\prime}(t)\ dt\\ =&\ \int_{t_0}^{t_1} y(t)\ x^{\prime}(t)\ dt \end{aligned} $$