Question

Find the equation of the osculating circle to y=(1/2)x^2 at the point (1,1/2).

Answer 1

In class we found that y'=x and y''=1, curvature=\[\frac{ y'' }{ (1+y'^{2})^{\frac{ 3 }{ 2 }} }\]
so curvature=\[\frac{ 1 }{ 2^{\frac{ 3 }{ 2 }} }\] to find the center of the circle we did \[(1,\frac{ 1 }{ 2 })+2^{\frac{ 3 }{ 2 }}(-\frac{ 1 }{ \sqrt{2}}+\frac{ 1 }{ \sqrt{2} })\] simplifying: \[(-1,\frac{ 5 }{ 2 })\] so the equation of the circle is \[(x+1)^{2}+(y-\frac{ 5 }{ 2 })^{2}=8\] What I am confused about is where the \[(-\frac{ 1 }{ \sqrt{2} }+\frac{ 1 }{ \sqrt{2} })\] came from.

Answer 2

a

Answer 3

The correct answer is the 1st one.

Answer 4

@clara1223 the
\[(-\frac{1}{\sqrt{2}}, \frac{1}{\sqrt{2}})\]
comes from using polar form to obtain "x" distance and "y" distance from point to center of circle
Note from polar form that
\[x = r \cos \theta\]
\[y = r \sin \theta\]
where "r" is radius
Also
\[-\frac{1}{y'} = \tan \theta\]
Where tan is slope of normal line and y' is slope of tangent line

Using your example:

y' = x , at point (1,1/2) ----> y' = 1
\[\tan \theta = -1 \rightarrow \theta = \frac{3\pi}{4}\]
\[\cos \theta = -\frac{1}{\sqrt{2}}, \sin \theta = \frac{1}{\sqrt{2}}\]
\[r = 2^{3/2} \]
Therefore the center of circle is:
\[(x,y) + r(\cos \theta , \sin \theta)\]
\[(1, \frac{1}{2}) + 2^{3/2}(-\frac{1}{\sqrt{2}},\frac{1}{\sqrt{2}})\]

Hope that explains it for you.

Answer 5

@dallascowboys88 and @Vanesaretana no dirent answer plz

Answer 6

that is cheating