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I've been learning about derivatives and I've learnt that the derivative of a circunference for $x$ is given by $-\frac{x}{y}$ with the formula for the circunference being $y^2+x^2=1$

Knowing this, I have tried to calculate the slope of a line that goes through center of this circle and a point in the circunference. To make things simpler, I decided to use the trigonometic circle.

I did the following:

  • Let this line be $f(x)$ and the tangent on the point where $f(x) = 1$ be $g(x)$

  • I know that $f(x) = mx$ because it goes through through the point $(0,0)$ and so $b=0$

  • Because the derivative of the circunference is $-\frac{x}{y}$, I know that $g(x) = -\frac{x}{y}x+b = -\frac{x^2}{y}+b$

I think that the tangent is perpendicular to $g(x)$, which means that $g(x)$ is its normal line. I don't know this for sure nor do I know how to prove it, it just looks that way for me so I am going explore that possibility and see what happens.

Assuming that $g(x)$ is perpendicular to $f(x)$:

  • If the slope of $g(x)$ is $-\frac{x}{y}$, then the slope of $f(x)$ is $\frac{y}{x}$, and so $f(x) = \frac{y}{x}x = y$, which doesn't really bring me anywhere... I think.

Another way of calculating the slope of $f(x)$ would be the traditional way, replacing $x$ and $y$ with $(\cos(\theta);\sin(\theta))$ and so:

$$f(x) = mx \Leftrightarrow \sin(\theta) = m\cos(\theta) \Leftrightarrow m = \frac{\sin(\theta)}{\cos(\theta)} \Leftrightarrow m = \tan(\theta)$$

And so $f(x) = \tan(\theta)x$

Another thing that I tried to do was to calculate $b$ for $g(x)$, using the point $(\cos(\theta);\sin(\theta))$:

$$g(x) = -\frac{x^2}{y}+b \Leftrightarrow \sin(\theta) = -\frac{\cos^2(\theta)}{\sin(\theta)} \Leftrightarrow \sin(\theta)+\cos(\theta)\cot(\theta) = b \Leftrightarrow b = \csc(\theta)$$

And so $g(x) = -\frac{x^2}{y}+ \csc(\theta)$

EDIT: So, to recap:

$$f(x) = \tan(\theta)x \\ f'(x) = \tan(\theta) \\ g(x) = -\frac{x^2}{y}+ \csc(\theta) \\ g'(x) = -\frac{x}{y}$$

Now my questions:

  1. Did I do everything correctly?
  2. Are $f(x)$ and $g(x)$ perpendicular? Did I prove it above when I got $f(x) = y$? If not, could you show me how to prove whether or not if they are perpendicular?
  3. I could never really understand the purpose of $\tan$. Is it the slope of the line that contains the hypotenuse?

Feel free to expand on this subject if you want or point out any detail I might have missed.

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    $g(x)=-{x^2\over y}+b$ is not a line...you have to choose a point $(x_P,y_P)$ on the circle and calculate the derivative in $P$, so: $f(x)=-{x_P\over y _P}x+b$. – 2017-01-15
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    I meant $g(x)=-{x_P\over y_P}x+b$ in the previous comment...now it is easy to check that $f(x)$ and $g(x)$ are perpendicular at $P$ because the angular coefficient of $f(x)$ is given by: $m={y_P-0\over x_P-0}={y_P\over x_P}$ – 2017-01-15

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