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Elliptical equation:

$(x+1)^2 + (\frac y 3)^2 = 1$

I need to find the unit tangent vector for it, so I need the derivative, but I have no idea how to differentiate this equation.

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  • $\begingroup$ Parametrize your curve as $(\cos(t) - 1, 3\sin (t))$ with $t \in [0, 2\pi)$ and differentiate. $\endgroup$ – MSDG Sep 10 '18 at 16:35
  • $\begingroup$ Can i derivate $y = 3 * sin t$ ? $\endgroup$ – tre038 Sep 10 '18 at 16:46
  • $\begingroup$ You should differentiate each component to get the tangent vector, i.e. if you parametrize your curve by $\gamma(t) = (x(t), y(t)) = (\cos(t)-1, 3\sin(t))$, then your tangent vector is $\gamma'(t) = \big((\cos(t)-1)', (3\sin(t))'\big).$ $\endgroup$ – MSDG Sep 10 '18 at 16:50
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By differentiation on $x$, and getting rid of the denominator

$$9(x+1)\,dx+y\,dy=0.$$

You can take the vector $(y,-9(x+1))$ as the direction of $(dx,dy)$.

After normalization,

$$\frac{(y,-9(x+1))}{\sqrt{y^2+81(x+1)^2}}.$$

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We can take the derivative of this equation "implicitly" as the following:

$$9x^2+18x+9+y^2 = 9\to9x^2+18x+y^2=0$$

So, now we implicitly differentiate$$18x\textrm{dx}+18\textrm{dx}+2y\textrm{dy}=0$$We now "divide" by $\textrm{dx}$ (Again, dividing by differentials isn't actually a permitted operation, but as you'll later learn in your mathematical journey, it happens to work nicely in cases like these).$$9(x+1)=-y\frac{dy}{dx}$$Hence,$$-\frac{9(x+1)}y=\frac{dy}{dx}$$

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Rephrasing:

Let $y \not =0$.

Differentiate the equation with respect to $x$:

$2(x+1)+ 2(y/3)(1/3)dy/dx=0$.

$9(x+1) +y dy/dx= 0.$

$dy/dx= -9(x+1)/y$.

Note :

$h(x):= f(g(x)).$

$h'(x)= f'(g(x))g'(x).$

In our equation :

$y=g(x)$, $f(y)=(y/3)^2$.

Hence :

$h'(x)= f'(y)y'= 2(y/3)(1/3)y'$.

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