Question

Show that the curvature of the function of $y=\sqrt{1-x^2},x\in (-1,1)$, is constant, but its second derivative varies with x.

Short answer

The value of the second derivative depends on x.

The curvature of the function is 1 which is constant.

Step-by-step solution

Step 1. Given information.

We have to show that the curvature of the function of $y=\sqrt{1-x^2},x\in (-1,1)$, is constant, but its second derivative varies with x.

Step 2. To show that the second derivative varies with x 

if y=f(x) is a twice-differentiable function then the curvature of f is given by

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

Since,

$\begin{array}{r}y=f\left(x\right)\\ =\sqrt{1-x^2}\\ f^{\mathrm{\prime }}\left(x\right)=\frac{1}{2\sqrt{1-x^2}}(-2x)\\ \Rightarrow f^{\mathrm{\prime }}\left(x\right)=\frac{-x}{\sqrt{1-x^2}}\end{array}$

Use the quotient rule to find $f^{\prime \prime }\left(x\right)$:

$\begin{array}{r}{f}^{\prime \prime }\left(x\right)=\frac{\sqrt{1-x^2}(-1)+x\frac{1}{2\sqrt{1-x^2}}(-2x)}{1-x^2}\\ f^{\prime \prime }\left(x\right)=\frac{-\left(1-x^2\right)-x^2}{{\left(1-x^2\right)}^{\frac{3}{2}}}\\ f^{\prime \prime }\left(x\right)=\frac{-1}{{\left(1-x^2\right)}^{\frac{3}{2}}}\end{array}$

The value of the second derivative depends on x.

Step 3. To show that the curvature of the function is constant

Substituting $f^{\mathrm{\prime }}\text{and}{f}^{\prime \prime }$values in the formula for k we get :

$\begin{array}{r}k=\frac{\left|\frac{-1}{{\left(1-x^2\right)}^{\frac{3}{2}}}\right|}{{\left(1+{\left(\frac{-x}{\sqrt{1-x^2}}\right)}^2\right)}^{\frac{3}{2}}}\\ k=\frac{\left|{\left(1-x^2\right)}^{\frac{3}{2}}\right|}{{\left(1+\frac{x^2}{1-x^2}\right)}^{\frac{3}{2}}}\\ =\frac{\frac{-1}{{\left(1-x^2\right)}^{\frac{3}{2}}}}{{\left(\frac{1-x^2+x^2}{1-x^2}\right)}^{\frac{3}{2}}}\\ k=\frac{1}{{\left(1-x^2\right)}^{\frac{3}{2}}}\cdot \frac{{\left(1-x^2\right)}^{\frac{3}{2}}}{1}\\ k=1\end{array}$

The curvature of the function is 1 which is constant.