91影视

given,

$f\left(x\right)=x^2,(0,0)$

$f\left(x\right)=x^{2}at(0,0)$.

The objective is to find the equation of the osculating circle to the given scalar function $f\left(x\right)=x^2$at the specified point $(0,0)$.

Any function of the form $y=f\left(x\right)$has the parametrization $x=t,y=f\left(t\right)$.

So, $x\left(t\right)=t,y\left(t\right)=t^2$

Thus, the vector function $r\left(t\right)=<t,t^2>$.

First calculate the graph of a vector function $r\left(t\right)$, normal vector $N\left(t\right)$and the curvature $k$at $t=0$.

role="math"

Use the Quotient rule for derivatives,

Calculate unit normal vector $t=0$.

Since, $f\left(x\right)=x^2$

$\begin{array}{r}{f}^{\mathrm{鈥�}}\left(x\right)=2x\\ f^{鈥测赌�}\left(x\right)=2\end{array}$

The curvature of the graph of $f$is given by

$\begin{array}{r}k=\frac{\left|f^{鈥测赌�}\left(x\right)\right|}{{\left(1+{\left(f^{\mathrm{鈥�}}\left(x\right)\right)}^2\right)}^{\frac{3}{2}}}\\ k=\frac{2}{{\left(1+(2x{)}^2\right)}^{\frac{3}{2}}}\end{array}$

$\begin{array}{r}k=\frac{2}{{\left[1+(2鈰�0{)}^2\right]}^{\frac{3}{2}}}\\ =2\end{array}$

The radius of curvature $p$of $C$is given by $p=\frac{1}{k}$.

The radius of curvature is, $p=\frac{1}{2}$.

The osculating circle will have center is $r\left(0\right)+\frac{1}{k}N\left(0\right)$

The osculating circle will have center $(0,\frac{1}{2})$and radius $\frac{1}{2}$.

So the equation of the osculating circle is,

$\begin{array}{r}(x鈭�0{)}^2+{\left(y鈭�\frac{1}{2}\right)}^2={\left(\frac{1}{2}\right)}^2\\ x^2+{\left(y鈭�\frac{1}{2}\right)}^2=\frac{1}{4}\end{array}$

Therefore, the equation of the osculating circle to the given function at $t=0$is,

$f\left(x\right)=x^{2}at(0,0)$is localid="1649672543849" $x^2+{\left(y鈭�\frac{1}{2}\right)}^2=\frac{1}{4}$