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Question: Find the quadratic polynomial that best approximates the unit circle centered at $(0,0)$ near the point, P $(0,1)$ (i.e. the osculating parabola).

Thoughts/Attempts:Right now, we are right now on the section on Taylor Series. So, I think this question is related to using Taylor Series of a higher order. Like, when we use the first derivative to approximate the tangent, now we use the the first and second derivative to approximate the quadratic polynomial (osculating parabola) (at least this what I think we do something like this).

In particular I think we do something like this:
enter image description here

Or maybe something like this: enter image description here which is the quadratic Taylor polynomial in two variables.

I'm honestly confused which one to use as the equation of a circle is $x^2+y^2=1$ . I'm also confused whether I should be taking derivatives or partial derivatives. Also, I'm somewhat confused whether $(x_0,y_0)$ is $(0,1)$? Or is it $(0,0)$? I'm pretty sure its $(0,1)$ as we are taking it near that point, but I just want to clarify. Sorry for what may seem like simple questions. Thank you.

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2 Answers 2

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HINT

The top half of the unit circle has the equation $y = \sqrt{1-x^2}$. Can you find the 1-variable quadratic Taylor polynomial in $x$ that best approximates $f(x) = \sqrt{1-x^2}$ near $x = 0$?

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use $f(x) = \sqrt {1-x^2}$

$f'(x) = \frac {x}{\sqrt {1-x^2}}\\ f''(x) = \frac {-1}{(1-x^2)^{\frac32}}$

Centered at $x = 0$

$F(x) = f(0) + f'(0) x + \frac 12 f''(0) x^2 = 1 - \frac 12 x^2$

Or, apply the binomial theorem to $(1-x^2)^\frac 12 = 1 - \frac 12 x^2$ (and stop afer you have found the $x^2$ term..)

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