Question

Find the equation of the osculating circle to the given scalar function at the specified point.

$f\left(x\right)=\cos x,(0,1)$

Short answer

Ans: The equation of the osculating circle to$f\left(x\right)=\cos xis{x}^2+y^2=1$.

Step-by-step solution

Step 1. Given information.

given,

$f\left(x\right)=\cos x,(0,1)$

Step 2. Consider the function 

$f\left(x\right)=\cos xat(0,1)$

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" $t=0$.

$\begin{array}{r}r\left(t\right)=⟨t,\cos t⟩\\ r\left(0\right)=⟨0,1⟩\\ r^{\mathrm{\prime }}\left(t\right)=⟨1,-\sin t⟩\\ ∥r^{\mathrm{\prime }}\left(t\right)∥=\sqrt{1+{\sin }^{2}t}\end{array}$

Step 3. Calculate unit target vector at t.

Use the Quotient rule for derivatives,

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

Step 4. Calculate the unit normal vector N(t).

Calculate unit normal vector at $t=0$

role="math"

The next step is to determine the curvature $k$

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

$\begin{array}{r}{f}^{\mathrm{\prime }}\left(x\right)=-\sin x\\ f^{\prime \prime }\left(x\right)=-\cos x\end{array}$

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

$\begin{array}{r}k=\frac{\left|f^{\prime \prime }\left(x\right)\right|}{{\left(1+f{\left(f^{\mathrm{\prime }}\left(x\right)\right)}^2\right)}^{\frac{3}{2}}}\\ =\frac{|-\cos x|}{{\left(1+(-\sin x{)}^2\right)}^{\frac{3}{2}}}\\ =\frac{\cos x}{{\left(1+{\sin }^{2}x\right)}^{\frac{3}{2}}}\end{array}$

Step 5. The curvature of the graph of f at the point x=0.

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

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

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

$\begin{array}{r}r\left(0\right)+\frac{1}{k}N\left(0\right)=⟨0,1⟩+1⟨0,-1⟩\\ =⟨0,0⟩\end{array}$

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

Thus, the equation of osculating circle is,

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

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