M408M Learning Module Pages
Main page ## Chapter 10: Parametric Equations and Polar Coordinates## Learning module LM 10.1: Parametrized Curves:## Learning module LM 10.2: Calculus with Parametrized Curves:Slope and areaArc length and surface area Summary and simplification Higher Derivatives ## Learning module LM 10.3: Polar Coordinates:## Learning module LM 10.4: Areas and Lengths of Polar Curves:## Learning module LM 10.5: Conic Sections:## Learning module LM 10.6: Conic Sections in Polar Coordinates:## Chapter 12: Vectors and the Geometry of Space## Chapter 13: Vector Functions## Chapter 14: Partial Derivatives## Chapter 15: Multiple Integrals |
## Slope and areaFinding the slope of the tangent line to a graph $y=f(x)$ is easy -- just compute $f'(x)$. Likewise, the area under the curve between $x=a$ and $x=b$ is just $\int_a^b f(x) dx$. But how do we compute slopes and areas with parametrized curves? For slopes, we are looking for $dy/dx$. This is a limit \begin{eqnarray*} \frac{dy}{dx} &=& \lim \frac{\Delta y}{\Delta x} \cr &=& \lim \frac{\Delta y/\Delta t}{\Delta x/\Delta t} \cr &=& \frac{\lim (\Delta y/\Delta t)}{\lim (\Delta x/\Delta t)} \cr &=& \frac {dy/dt}{dx/dt}, \end{eqnarray*}where the limits are as $\Delta t$ and $\Delta x$ and $\Delta y$ all go to zero. As long as we can take the derivatives of $x$ and $y$, we can compute $dy/dx$. For areas, we want $$\int_{x_1}^{x_2} y\; dx = \int_{t_1}^{t_2} y(t) \frac{dx(t)}{dt} dt.$$To find the area, we need to both compute a derivative and an integral. Here $x_1$ and $t_1$ are the starting values of $x$ and $t$, while $x_2$ and $t_2$ are the ending values. |