You don't have to know anything about parametric functions to work with polar functions, but if you do, keep in mind that [because more things are the same in mathematics than different] polar functions are parametric functions, parameterized by the angle, θ.
Just to remind you of the basics of polar coordinates, take a look at the figure on the right. Any point in the Cartesian (x, y) plane can be re-located by a pair of coordinates, (r, θ) on the polar plane, where r is a distance from the origin, and θ is the counterclockwise rotation from θ = 0, which is, by convention, along the +x axis.
By constructing a right triangle with the x-and y-coordinates and the r vector, we just get the transformations between polar and Cartesian coordinates by trigonometry.
x = r·cos(θ)
y = r·sin(θ)
Polar functions are just parametric functions parameterized by the angle, θ
In Cartesian coordinates, (x, y) ordered pairs are written (usually) with the independent variable first. Generally y = f(x).
In polar coordinates, θ is usually the independent variable. It's reversed. Generally, r = f(θ).
Who knows why this is. It's probably just an accident of history, and we're stuck with it. Don't let it throw you off.
Let's do an example that we already know from experience with implicit differentiation: Find the slope of the circle x2 + y2 = r2 at any point on the curve, then use it to find the horizontal and vertical tangents. Here is a picture of the circle.
The locations of the horizontal and vertical tangents are marked in the circle above; they're not too hard to find. But let's find them, first with implicit differentiation of the Cartesian equation, then using the polar form of the circle equation, r = R.
$$x^2 + y^2 = r^2$$
Taking the derivative of each term with respect to x (implicit differentiation), we get
$$2x + 2y \, \frac{dy}{dx} = 0$$
Solving for dy/dx gives the slope at any point:
$$\frac{dy}{dx} = \frac{-x}{y}$$
Now the zero(s) of the numerator (x = 0) give the locations of the horizontal tangent(s), and the zeros of the denominator are where the derivative is infinite, and thus the function has a vertical tangent. Because there are two locations for x = 0 and for y = 0, we have the tangent points:
In polar coordinates, the equation of a circle of radius R centered at the origin is simple:
$$r = R$$
Now the transformations between Cartesian and polar coordinates are
$x = r \cdot cos(\theta) y = r\cdot sin(\theta)$
where r is a function of θ:
$$r = f(\theta)$$
For x, we have
$$x = f(\theta) \cdot cos(\theta)$$
and its derivative with respect to θ (using the product rule) is
We can do the same procedure for the y-coordinate,
$$y = f(\theta) \cdot sin(\theta)$$
The derivative is
Now to find dy/dx, we take advantage of the fact that derivatives written this way are just ratios of differentials, and thus can be treated like variables in fractions:
$$\frac{\frac{dy}{d\theta}}{\frac{dx}{d\theta}} = \frac{dy}{d\theta} \frac{d\theta}{dx} = \frac{dy}{dx}$$
So the slope of our function is
$$\frac{dy}{dx} = \frac{\frac{dr}{d\theta} sin(\theta) + r \cdot cos(\theta)}{\frac{dr}{d\theta} cos(\theta) - r \cdot sin(\theta)}$$
That looks a bit complicated, but if you focus on remembering how it was derived, it's not that bad.
Because r = R is constant in our original polar function, we simplify dramatically:
$$r = R \, \longrightarrow \, \frac{dr}{d\theta} = 0$$
to get the final derivative:
$$\frac{dy}{dx} = \frac{-R \cdot cos(\theta)}{R \cdot sin(\theta)} = \frac{-cos(\theta)}{sin(\theta)}$$
Now for horizontal tangents, the numerator must be zero:
which gives
$$\theta = \frac{\pi}{2}, \: \frac{3\pi}{2}$$
Now we can convert those back to Cartesian coordinates:
$$ \begin{align} x &= r cos(\theta) = R cos \left( \frac{\pi}{2} \right) = 0 \\ \\ y &= r sin(\theta) = R sin \left( \frac{\pi}{2} \right) = R \end{align}$$
and
$$ \begin{align} x &= r cos(\theta) = R cos \left( \frac{3\pi}{2} \right) = 0 \\ \\ y &= r sin(\theta) = R sin \left( \frac{3\pi}{2} \right) = -R \end{align}$$
The vertical tangents are found by setting the denominator of the derivative to zero:
The solutions are:
$$\theta = 0, \: \pi$$
We can now convert to Cartesian coordinates:
$$ \begin{align} x &= r cos(\theta) = R cos(\pi) = -R \\ y &= r sin(\theta) = R sin(\pi) = 0 \end{align}$$
So we recover the same locations of horizontal and vertical tangents:
The derivative of a polar function specified by r = f(θ) is
Pro tip: Instead of memorizing this derivative formula, just remember how to find the derivative of a parametric function where x = f(t) and y = g(t), and derive this formula when you need it.
Here's what this figure, one of a class of curves called cardiods, looks like. The approximate locations of the horizontal and vertical tangents are shown.
Our derivative formula is:
$$\frac{dy}{dx} = \frac{\frac{dr}{d\theta} sin(\theta) + r \cdot cos(\theta)}{\frac{dr}{d\theta} cos(\theta) - r \cdot sin(\theta)}$$
Now r' = 3 sin(θ), so our derivative is
As the arrows above indicate, we can begin to simplify this derivative by distributing the (2 - 3 cos(θ)) term to get
$$= \frac{3 sin^2(\theta) - 3 cos^2(\theta) + 2 cos(\theta)}{6 sin(\theta) cos(\theta) - 2 sin(\theta)}$$
We'll eventually want to set the numerator equal to zero to find horizontal tangents, and the denominator equal to zero to find vertical tangents. Using the Pythagorean identity on the top and factoring 2 sin(θ) from the bottom gives:
$$= \frac{3(1 - cos^2(\theta)) - 3 cos^2{\theta) + 2 cos(\theta)}}{2 sin(\theta) (3 cos(\theta) - 1)}$$
Now it's clear why we used the trig identity; it creates a quadratic equation (in cos(θ)) in the numerator:
$$\frac{-6 cos^2(\theta) + 2 cos(\theta) + 3}{2 sin(\theta)(3 cos(\theta) - 1)}$$
Notice that getting the derivative was pretty simple. As with most problems in calculus, we make a calculus "move" or two and then the rest is just remembering our algebra skills.
Now let's just make sure that expressing this function as x = f(θ), y = f(θ) yields the same derivative.
Recall that for a polar function, x = r cos(θ) and y = r sin(θ). So x(θ) is:
$$ \begin{align} x &= (2 - 3 cos(\theta))\cdot cos(\theta) \\ &= 2 cos(\theta) - 3 cos^2 (\theta) \end{align}$$
and y(θ) is:
$$ \begin{align} y &= (2 - 3 cos(\theta))\cdot sin(\theta) \\ &= 2 sin(\theta) - 3 cos(\theta) sin(\theta) \end{align}$$
Now the derivatives of x and y can be easily found using the product rule:
$$\frac{dy}{d\theta} = 2 cos(\theta) - 3 cos^2(\theta) + 3 sin^2(\theta)$$
and
$$\frac{dx}{d\theta} = -2 sin(\theta) + 6 cos(\theta) sin(\theta)$$
These are precisely the numerator and denominator of our derivative above, so either way you choose to go with these problems is fine. Now let's find the horizontal tangents.
To find the horizontal tangents, the derivative is set equal to zero, which means that its numerator has to be zero:
$$-6 cos^2(\theta) + 2 cos(\theta) + 3 = 0$$
To solve this equation, we recognize that it looks a lot like a simple quadratic equation with x = cos(θ). We can solve the resulting simplified quadratic by completing the square (see quadratic functions):
$$ \begin{align} -6x^2 + 2x + 3 &= 0 \\ x^2 - \frac{1}{3} x &= \frac{1}{2} \\ x^2 - \frac{1}{3} x + \left( \frac{1}{6} \right)^2 &= \frac{1}{36} + \frac{1}{2} = \frac{19}{36} \\ \left( x = \frac{1}{6} \right)^2 &= \frac{19}{36} \\ x &= \frac{1 ± \sqrt{19}}{6} \end{align}$$
Now we re-substitute x = cos(θ), to find:
$$ \begin{align} cos(\theta) &= 0.88314 \: on \: [0, 2\pi] \\ \theta &= 0.4665 \, rad, \: 5.817 \, rad \\ \\ cos(\theta) &= -0.55982 \: on \: [0, 2\pi] \\ \theta &= 2.165 \, rad, \: 4.118 \, rad \end{align}$$
Here we've solved for cos(θ) for both the plus- and minus solutions for x, and accounted for the fact that we have to find all solutions on between 0 and 2π radians— thus θ and 2π-θ are solutions.
Here is a summary of the angles in radians and degrees.
It's easy to solve for the r that goes with each θ to find the points of horizontal tangency. Here they are on the graph of the function.
Vertical tangents exist at locations where the denominator of our derivative go to zero. That is:
$$2 sin(\theta)[3 cos(\theta) - 1] = 0$$
This equation is true when either sin(θ) = 0 or cos(θ) = 1/3. We need to take account of all solutions on the interval between 0 and 2π radians— thus θ and (in this case) π-θ are solutions. Here is a summary of those solutions in radians and degrees.
Finally, here are the (r, θ) coordinates of the vertical tangents plotted on the graph of our function.
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The function r(θ) = 1 + 2cos(θ) on the interval [0, 2π] is called a limaçon. (The little decoration on the c is called a cedilla; the word limaçon [lee' mə sahn] comes from the Latin word for snail.) Find an equation of the line tangent to this curve at θ = π/4. |
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