91影视

We have to show that why the graph of the vector function r(t) = a + 伪 cos t, b + 伪 sin t is a circle C with center (a, b) and radius 伪.

Consider $r\left(t\right)=鉄�a+伪\mathrm{csot},b+伪\sin 鈦�t鉄�$where $伪>0$

Then $x=x\left(t\right)=a+伪\cos 鈦�t\text{and}y=y\left(t\right)=b+伪\sin 鈦�t$

Since, $x=a+伪\cos 鈦�t$

$x鈭�a=伪\cos 鈦�t$and

Since $y=b+伪\sin 鈦�t$

$y鈭�b=伪\sin 鈦�t$

$\begin{array}{r}y鈭�b=伪\sin 鈦�t\\ (x鈭�a{)}^2+(y鈭�b{)}^2={伪}^2{\cos }^2鈦�t+{伪}^2{\sin }^2鈦�t\\ (x鈭�a{)}^2+(y鈭�b{)}^2={伪}^2\left({\cos }^2鈦�t+{\sin }^2鈦�t\right)\end{array}$

$(x鈭�a{)}^2+(y鈭�b{)}^2={伪}^2$ which represents a circle with centre (a,b) and radius$伪$

$\begin{array}{r}鈭�r^{\mathrm{鈥�}}\left(t\right)脳r^{鈥测赌�}\left(t\right)鈭�=\sqrt{0^2+0^2+\left({伪}^2\right)}\\ \begin{array}{rl}鈭�r^{\mathrm{鈥�}}\left(t\right)鈭�& =\sqrt{{伪}^2{\sin }^2鈦�t+{伪}^2{\cos }^2鈦�t}\\ & =\sqrt{{伪}^2\left({\sin }^2鈦�t+{\cos }^2鈦�t\right)}\\ & =伪\end{array}\\ k=\frac{鈭�r^{\mathrm{鈥�}}\left(t\right)脳r^{鈥测赌�}\left(t\right)鈭�}{{鈭�r^{\mathrm{鈥�}}\left(t\right)鈭�}^3}=\frac{{伪}^2}{{伪}^3}\\ =\frac{1}{伪}\end{array}$

the radius is$伪$.

The radius of the curvature is $伪>0$, the osculating circle to the curve c at $r\left(t_0\right)$is the circle in the osculating plane.

Thus the osculating circle at every point on c is the circle c.

The answer to part (a) tally with the answers to parts (b) and (c).

These answers are consistent with the definition of the radius of curvature and osculating circle when k>0