Slope and Area

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

$dy/dx$ .

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.

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,

The slope of the tangent line at time $t$ is $\displaystyle{\frac{dx}{dt} = \frac{dy/dt}{dx/dt}}$ .
The area under the curve between $a=x(t_1)$ and $b=x(t_2)$ is $\displaystyle{\int_{t_1}^{t_2} y(t) \frac{dx}{dt} dt}$

In the following video, we derive these formulas, work out the tangent line to a cycloid, and compute the area under one span of the cycloid.

Integration by Parts

Integrals of Trig Functions

Trig Substitutions

Partial Fractions

Strategies of Integration

Improper Integrals

Modeling with Differential Equations

Euler's Method and Direction Fields

Separable Equations

Models of Growth

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Parametrized Curves

Calculus with Parametrized Curves

Polar Coordinates

Areas and Lengths of Polar Curves

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