Question

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

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

Short answer

Ans: Therefore, the equation of the osculating circle to $f\left(x\right)=e^{x}on(0,1)$is $(x+2{)}^2+(y-3{)}^2=8$.

Step-by-step solution

Step 1. Given information.

given,

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

Step 2. Consider the function,

$f\left(x\right)=e^x$

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

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

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

Thus,

Here, $t=0whenx=0$

First calculate the graph of a vector function $r\left(t\right)$, normal vector $N\left(t\right)$and the curvature $k$at role="math" localid="1649692759713" $t=0$

role="math" localid="1649692741000"

Step 3. Calculate unit tangent vector at t.

Use the Quotient rule for derivatives,

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

Calculate the unit normal vector at $t=0$.

The next step is to determine the curvature $k$.

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

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

Step 5. The curvature of the graph of f is given by 

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

The curvature of the graph of $f$at the point, $x=0$

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

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

Thus, $p=2\sqrt{2}$

Step 6. The radius of curvature is 1k=232.

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

The osculating circle will have a center, localid="1649693982688" $(-2,3)$and radius $2^{\frac{3}{2}}$.

The equation of the osculating circle is

localid="1649693860206" $\begin{array}{r}(x+2{)}^2+(y-3{)}^2={\left(2^{\frac{3}{2}}\right)}^2\\ (x+2{)}^2+(y-3{)}^2=2^3\\ (x+2{)}^2+(y-3{)}^2=8\end{array}$

Therefore, the equation of the osculating circle to localid="1649693924447" $f\left(x\right)=e^{x}on(0,1)$is $(x+2{)}^2+(y-3{)}^2=8$