I don't understand the concept of parametrization or how to go about solving this question. Any help would be appreciated!
Write the equation of the osculating circle of$ y=\sin(x)$ at $ (\pi/2,1)$
1
$\begingroup$
calculus
-
0Do you understand radians and trigonometry? If you understand radians, and the functions Sin(t) and Cos(t), then it should be fairly easy to solve this. – 2011-03-22
1 Answers
1
The osculating circle is one that passes through the point of interest with the same slope and curvature. You know the point of interest is $(\pi/2,1)$. The derivative at this point is $0$, so the tangent is horizontal. The center of the osculating circle is along the perpendicular to the tangent, which here is the line $x=\pi$, and on the concave side of the curve, below it in this case. So can you find the equation of a circle, which a) has center along $x=\pi$ and is below $(\pi/2,1)$, b)passes through $(\pi/2,1)$, and c) has the same second derivative value at $(\pi/2,1)$ as $\sin x$? a) and b) together take care of the tangent.