I have drawn a circle by doing this in Matlab: syms x;
ezplot(cos(x),sin(x))
I get the tangent point at which I want my tangent to be by taking $x = \cos(3.1415)$, $y = \sin(3.1415)$, but now I am confused because normally I would differentiate the function to get the slope.
Now I have two functions that created my curve, which one do I differentiate to get the slope?
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$\begingroup$Any point $P: (x,y)$ on the unit circle satisfies the relation $$x^2 + y^2 =1$$ Hence, we get that $$2x + 2y \dfrac{dy}{dx} = 0\implies \dfrac{dy}{dx} = - \dfrac{x}y$$ Since the usual parameterization of the circle is $x = \cos(\theta)$ and $y = \sin(\theta)$, the slope at a given $\theta$ is given by $$\text{Slope at }\theta = -\dfrac{\cos(\theta)}{\sin(\theta)} = - \cot(\theta)$$ For instance, if you are interested in the slope at $\theta = \pi/6$, then it is $-\cot(\pi/6) = - \sqrt{3}$.
$\endgroup$ $\begingroup$The prior answers have all used calculus. I'm going to post an answer using only trig.
The following diagram from Wikipedia's Trig Page is helpful.
However, that diagram also has a fault--the picture is very cluttered. :) Thus, I've redrawn it for you, labeling the components important for this problem:
Note that $\csc\theta$ returns the distance from the origin to the y-intercept of the tangent line, and $\sec\theta$ returns the distance from the origin to the x-intercept of the tangent line.
Let $m$ represent slope:$$m=\frac{\Delta y}{\Delta x}=\frac{\csc\theta}{\sec\theta}$$Simplifying, this gives:$$m=\cot\theta$$
Now, for your particular point, you will run into some difficulty:$$m=\cot\pi=\text{undefined}$$To interpret this, note that $\cot\pi$ is undefined because you are dividing by zero. This means that you had some change in the $y$ direction, but none in the $x$ direction. The only type of line like this is a vertical one. :)
Let me know if this needs more explaining--I'll edit the answer to make it clearer.
$\endgroup$ $\begingroup$If you seek $dy/dx$, note that$$\frac{{dy}}{{dt}} = \frac{{dy}}{{dx}}\frac{{dx}}{{dt}}$$by the chain rule, so$$\frac{{dy}}{{dx}} = \frac{{\frac{{dy}}{{dt}}}}{{\frac{{dx}}{{dt}}}}.$$Now let $x = \cos (t)$ and $x = \sin (t)$ and compute the derivatives. This is called parametric differentiation.
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