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.

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

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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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 $\alpha$. $\tan \alpha =2$

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.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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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.

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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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?

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.

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?

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