Slope of the Unit Circle at Any Point

EvanJ

How do you calculate the slope of the unit circle at any point? Unlike curves that can be written in the form y = ax^2 + bx + c, which are easy to take the derivative of, writing the unit circle in y = form would have a positive or negative in front of a square root.

Greens

You can use Implicit Differentiation.

$\displaystyle x^2 + y^2 = 1$

Taking the derivative:

$\displaystyle 2x+2y\frac{dy}{dx} = 0$

$\displaystyle \frac{dy}{dx} = -\frac{x}{y}$

With the unit circle, $x=\cos{\theta}$, $y=\sin{\theta}$

So,

$\displaystyle \frac{dy}{dx} = -\cot{\theta}$

https://www.desmos.com/calculator/teoyll69g9

Here's a graph to mess around with.

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1 person

tahirimanov19

Think of it as a tangent of an angle formed by the intersection of the x-axis and tangent of the graph on a given point.

For example:
Draw graph of $f(x)=x^2$. Now, draw a tangent to the graph at point where x=2. Let the angle between tangent and x-axis be $\alpha$. $\tan \alpha =2$

EvanJ

You can use Implicit Differentiation.

$\displaystyle x^2 + y^2 = 1$

Taking the derivative:

$\displaystyle 2x+2y\frac{dy}{dx} = 0$

$\displaystyle \frac{dy}{dx} = -\frac{x}{y}$

With the unit circle, $x=\cos{\theta}$, $y=\sin{\theta}$

So,

$\displaystyle \frac{dy}{dx} = -\cot{\theta}$

https://www.desmos.com/calculator/teoyll69g9

Here's a graph to mess around with.
cot is adjacent/opposite. Take the point (0.8, 0.6) to make a right triangle of 0.6/0.8/1, which has the same ratios as 3/4/5. The adjacent is 0.6, and the opposite is 0.8, so is the slope -0.6/0.8 = -0.75? The website didn't have anything to calculate the slope at a point.

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SDK

What Greens has written is perfectly correct answer to your question. However, I think you are asking the wrong question. A better question is to find the tangent vector along a unit circle. I think is more useful since it is meaningful in any dimension, while "slope" implies you are choosing one special variable to measure with respect to a second and ignoring all others. In other words, even if it was defined everywhere (it isn't) there is no reason to prefer the slope of $y$ with respect to $x$ along a circle (or any other set).

So lets find the tangent vector. Write a vector $v \in (x,y)$ which we assume parameterizes the unit circle as a function of $t \in [0,1]$. Then by definition of a circle, $\left| \left| v(t) \right| \right|^2 = 1$ for all $0 \leq t \leq 1$. Rewriting this as a dot product we get $v^T v = 1$ and then we can take its derivative via the (vector) product rule
$\frac{d}{dt} v^T v = 2 v^T v' = 0$
which says that the tangent vector for $v$ is everywhere orthogonal to $v$. Thus, $v'$ is any normal vector to $v$ which is determined up to choosing the orientation of the parameterization. For example, if $v(t)$ has the usual orientation (counter-clockwise), then $v' = (-y, x)$.

Of course, if you are really after the slope of the tangent (say $y$ with respect to $x$), then you have it by dividing the components of $v'$ which gives you $\frac{dy}{dx} = \frac{-x}{y}$. However, notice that this slope is not defined everywhere e.g. $v = (1,0)$.On the other hand, the tangent vector we have computed is $v' = (0,1)$ which correctly identifies the fact that the tangent vector points straight in the $y$ direction.

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1 person

EvanJ

I wasn't trying to say that Greens gave a wrong answer. I was trying to say that the link didn't have what I wanted. I know that the slope is undefined at (-1, 0) where the tangent line is x = -1 and (1, 0) where the tangent line is x = 1. It's defined everywhere else, and I wasn't concerned with the tangent at those two points. I can do basic derivatives, but I can't work with vectors. Something that requires vectors is something that I just ask for the answer without trying to understand.

Take two unit circles with one centered at (0,0) and another centered at (a, 0) where 0 < a < 1 so that the circles overlap like a Venn Diagram. Given the value of a, how can I calculate the area of the overlapped area? Can the equation of the two circles be used to make an equation for the overlapped area and then can I integrate for the area under a curve to find the area with the area being double that because integrated finds the area above the x-axis which is half of the area of each circle?

Greens

Given $(x-a)^2+y^2=1$ and $x^2 + y^2 = 1$ you can show that the line goes through the overlap points is $x=\frac{a}{2}$

We can also solve for $y$ in $(x-a)^2+y^2=1$ to receive $y=\pm\sqrt{1-(x-a)^2}$

The overlap area is symmetric across $x=\frac{a}{2}$ so we can just double the area of the half covered by $(x-a)^2+y^2=1$ to receive the overlap area.

Since this circle is centered at $(a,0)$, its leftmost point is at $(a-1,0)$

Therefore, the area of this half is

$\displaystyle 2\int_{a-1}^{\frac{a}{2}}\sqrt{1-(x-a)^2}dx$

and thus the whole area is

$\displaystyle 4\int_{a-1}^{\frac{a}{2}}\sqrt{1-(x-a)^2}dx$

This integral can be done without too much trouble with trigonometric substitution.

PS. The website does calculate the slope, $c$ is the angle and $-\cot{c}$ is the slope, its on the list on the left side of the screen.

EvanJ

I looked at a textbook for integrals with radicals, and I found the integral of the square root of (1 - x^2) dx from 0 to 1 as pi/4, but not something I can generalize to an integral with anything inside the radical and any limits of integration.

I see the slope from that website, and I didn't know what it was yesterday.

Is it possible to take a function, the area under a curve I want, and one limit of integration, and solve for the other limit of integration? For example, if the function is y = x^2, the area is 10, and the lower limit of integration is x = 2, is it possible to solve for the upper limit of integration?