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Chapter 10: Parametric Equations and Polar Coordinates

Learning module LM 10.2: Calculus with Parametrized Curves:

Slope and area
Arc length and surface area
Summary and simplification
Higher Derivatives

Summary and simplification

Summary and Simplification
Our first set of results applies to all parametrized curves:

 If we have a parametrized curve, where we know $x(t)$ and $y(t)$, then     The slope of the tangent line at time $t$ is $\displaystyle{\frac{dy/dt}{dx/dt}}$.     The area under the curve from time $t_1$ to $t_2$ is $\displaystyle{\int_{t_1}^{t_2} y(t) \frac{dx(t)}{dt} dt.}$     The distance traveled along the curve from time $t_1$ to time $t_2$ is $\displaystyle{\int_{t_1}^{t_2} \sqrt{ \left ( \frac{dx}{dt}\right )^2 + \left ( \frac{dy}{dt} \right )^2}\; dt}$     The area of the surface of revolution, obtained by rotating around the $x$-axis, is $$\displaystyle{\int_{t_1}^{t_2} 2 \pi y(t) \sqrt{ \left ( \frac{dx}{dt}\right )^2 + \left ( \frac{dy}{dt} \right )^2}\; dt}$$

If we happen to have a graph $y=f(x)$ then all of these formulas simplify, since we can take the parametrization $x(t)=t$, $y(t)=f(t)$. In that case $\frac{dx}{dt}=1$ and $\frac{dy}{dt} = f'(t)=f'(x)$, so

 The slope is $\displaystyle{\frac{dy/dt}{dx/dt} = \frac{f'(t)}{1} = f'(x)}$.     The area under the curve is $\displaystyle{\int_{t_1}^{t_2} f(t) dt = \int_{x_1}^{x_2} f(x) dx}$.     The arc length is $\displaystyle{\int_{t_1}^{t_2} \sqrt{1 + (f'(t))^2} dt =\int_{x_1}^{x_2} \sqrt{1 + (f'(x))^2} dx}.$     The surface area is $\displaystyle{\int_{t_1}^{t_2} 2 \pi f(t) \sqrt{1 + (f'(t))^2} dt =\int_{x_1}^{x_2} 2 \pi f(x) \sqrt{1 + (f'(x))^2} dx}.$

The first two are our usual formulas for slopes and areas. The third and fourth may (or may not) be familiar to you from applications of integration.