Finding 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?
The answer is to express everything in terms of the parameter $t$. By computing derivatives with respect to $t$ and integrals with respect to $t$, we can compute everything we want.
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 $dy/dx$.
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Example 1: Find the slope of the tangent to the unit circle
$x = \cos(t);$ $y=\sin(t)$ at time $t=\pi/6$, i.e. at
the point $(x,y) = \left ( \frac{\sqrt{3}}{2}, \frac12 \right )$.
Solution: $\displaystyle{\frac{dy}{dt} = \cos(t) = \frac{\sqrt{3}}{2}}$ and $\displaystyle{\frac{dx}{dt}=-\sin(t) = -\frac12}$, so $\displaystyle{\frac{dy}{dx} = \frac{\sqrt{3}/2}{-1/2} = - \sqrt{3}}$. |
To find the area under a curve between $x=a$ and $x=b$, we need to compute $$\int_a^b y\; dx = \int_{t_1}^{t_2} y(t) \frac{dx(t)}{dt} dt,$$ where $x(t_1)=a$ and $x(t_2)=b$. To find the area, we need to both compute a derivative as well as an integral.
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Example 2: Find the area under the top half of the unit circle between
$x=0$ and $x=1/2$.
Solution: Since $x=0$ when $t=\pi/2$ and $x=1/2$ when $t=\pi/3$, we are integrating from $t_1=\pi/2$ to $t_2 = \pi/3$. This may look strange, since $t_1 > t_2$, but it has to do with our tracing the upper semi-circle from right to left: $dx/dt=-\sin(t)$ is negative. Since $y(t)=\sin(t)$, our area is \begin{eqnarray*} \int_{t_1}^{t_2} y(t) \frac{dx}{dt} dt & = & \int_{\pi/2}^{\pi/3} - \sin^2(t) dt \cr & = & \int_{\pi/3}^{\pi/2} \sin^2(t) dt \cr &=& \int_{\pi/3}^{\pi/2} \frac12 (1-\cos(2t)) dt \cr &=& \frac{t}{2} - \frac{\sin(2t)}{4} \Big |_{\pi/3}^{\pi/2} = \frac{\pi}{12} + \frac{\sqrt{3}}{8} \end{eqnarray*} |
Summary: When working with a parametrized curve,
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