91影视

given,

$f\left(x\right)=\cos 鈦�x,(0,1)$

$f\left(x\right)=\cos 鈦�xat(0,1)$

The objective is to find the equation of the osculating circle to the given scalar function role="math" localid="1649673072688" $f\left(x\right)=\cos x$at the specified point $(0,1)$.

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

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

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

First calculate the graph of a vector function $r\left(t\right)$, normal vector $N\left(t\right)$and the curvature role="math" localid="1649673442579" $k$at role="math" localid="1649673436031" $t=0$.

$\begin{array}{r}r\left(t\right)=鉄�t,\cos 鈦�t鉄�\\ r\left(0\right)=鉄�0,1鉄�\\ r^{\mathrm{鈥�}}\left(t\right)=鉄�1,鈭�\sin 鈦�t鉄�\\ 鈭�r^{\mathrm{鈥�}}\left(t\right)鈭�=\sqrt{1+{\sin }^2鈦�t}\end{array}$

Use the Quotient rule for derivatives,

$\begin{array}{r}=\sqrt{\frac{{\cos }^2鈦�t\left(1+{\sin }^2鈦�t\right)}{{\left(1+{\sin }^2鈦�t\right)}^3}}\\ =\sqrt{\frac{{\cos }^2鈦�t}{{\left(1+{\sin }^2鈦�t\right)}^2}}\\ =\frac{\cos 鈦�t}{1+{\sin }^2鈦�t}\end{array}$

Calculate unit normal vector at $t=0$

role="math"

The next step is to determine the curvature $k$

Since $f\left(x\right)=\cos x$

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

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

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

$\begin{array}{r}k=\frac{\cos 鈦�0}{{\left(1+{\sin }^2鈦�0\right)}^{\frac{3}{2}}}\\ k=1\end{array}$

The radius of curvature is $\frac{1}{k}=1$.

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

$\begin{array}{r}r\left(0\right)+\frac{1}{k}N\left(0\right)=鉄�0,1鉄�+1鉄�0,鈭�1鉄�\\ =鉄�0,0鉄�\end{array}$

The osculating circle will have a center $(0,0)andradius1$ .

Thus, the equation of osculating circle is,

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

Therefore, the equation of the osculating circle to$f\left(x\right)=\cos x$is$x^2+y^2=1$.