# Calculus

## Applications of the Derivative # Curvature and Radius of Curvature

Consider a plane curve defined by the equation y = f (x). Suppose that the tangent line is drawn to the curve at a point M(x, y). The tangent forms an angle α with the horizontal axis (Figure 1).

At the displacement $$\Delta s$$ along the arc of the curve, the point $$M$$ moves to the point $${M_1}.$$ The position of the tangent line also changes: the angle of inclination of the tangent to the positive $$x-\text{axis}$$ at the point $${M_1}$$ will be $$\alpha + \Delta\alpha.$$ Thus, as the point moves by the distance $$\Delta s,$$ the tangent rotates by the angle $$\Delta\alpha.$$ (The angle $$\alpha$$ is supposed to be increasing when rotating counterclockwise.)

The absolute value of the ratio $$\frac{{\Delta \alpha }}{{\Delta s}}$$ is called the mean curvature of the arc $$M{M_1}.$$ In the limit as $$\Delta s \to 0,$$ we obtain the curvature of the curve at the point $$M:$$

$K = \lim\limits_{\Delta s \to 0} \left| {\frac{{\Delta \alpha }}{{\Delta s}}} \right|.$

From this definition it follows that the curvature at a point of a curve characterizes the speed of rotation of the tangent of the curve at this point.

For a plane curve given by the equation $$y = f\left( x \right),$$ the curvature at a point $$M\left( {x,y} \right)$$ is expressed in terms of the first and second derivatives of the function $$f\left( x \right)$$ by the formula

$K = \frac{{\left| {y^{\prime\prime}\left( x \right)} \right|}}{{{{\left[ {1 + {{\left( {y'\left( x \right)} \right)}^2}} \right]}^{\frac{3}{2}}}}}.$

If a curve is defined in parametric form by the equations $$x = x\left( t \right),$$ $$y = y\left( t \right),$$ then its curvature at any point $$M\left( {x,y} \right)$$ is given by

$K = \frac{{\left| {x'y^{\prime\prime} - y'x^{\prime\prime}} \right|}}{{{{\left[ {{{\left( {x'} \right)}^2} + {{\left( {y'} \right)}^2}} \right]}^{\frac{3}{2}}}}}.$

If a curve is given by the polar equation $$r = r\left( \theta \right),$$ the curvature is calculated by the formula

$K = \frac{{\left| {{r^2} + 2{{\left( {r'} \right)}^2} - rr^{\prime\prime}} \right|}}{{{{\left[ {{r^2} + {{\left( {r'} \right)}^2}} \right]}^{\frac{3}{2}}}}}.$

The radius of curvature of a curve at a point $$M\left( {x,y} \right)$$ is called the inverse of the curvature $$K$$ of the curve at this point:

$R = \frac{1}{K}.$

Hence for plane curves given by the explicit equation $$y = f\left( x \right),$$ the radius of curvature at a point $$M\left( {x,y} \right)$$ is given by the following expression:

$R = \frac{{{{\left[ {1 + {{\left( {y'\left( x \right)} \right)}^2}} \right]}^{\frac{3}{2}}}}}{{\left| {y^{\prime\prime}\left( x \right)} \right|}}.$

## Solved Problems

### Example 1.

Calculate the curvature of the ellipse $\frac{{{x^2}}}{{{a^2}}} + \frac{{{y^2}}}{{{b^2}}} = 1$ at its vertices.

Solution.

Obviously, it suffices to find the curvature of the ellipse at points $$A\left( {a,0} \right)$$ and $$B\left( {0,b} \right)$$ (Figure $$2$$), because due to the symmetry of the curve, the curvature at the two opposite vertices of the ellipse will be the same.

To calculate the curvature, it is convenient to pass from the canonical equation of the ellipse to the equation in parametric form:

$x = a\cos t,\;\;\;y = b\sin t,$

where $$t$$ is a parameter. The parameter has the value $$t = 0$$ at the point $$A\left( {a,0} \right)$$ and is equal to $$t = \frac{\pi }{2}$$ at the point $$B\left( {0,b} \right).$$

Find the first and second derivatives:

$x' = {x'_t} = \left( {a\cos t} \right)^\prime = - a\sin t,\;\;\;x^{\prime\prime} = x^{\prime\prime}_{tt} = \left( { - a\sin t} \right)^\prime = - a\cos t;$
$y' = {y'_t} = \left( {b\sin t} \right)^\prime = b\cos t,\;\;\;x^{\prime\prime} = x^{\prime\prime}_{tt} = \left( { b\cos t} \right)^\prime = - b\sin t.$

The curvature of a parametrically defined curve is expressed by the formulas

$K = \frac{{\left| {x'y^{\prime\prime} - y'x^{\prime\prime}} \right|}}{{{{\left[ {{{\left( {x'} \right)}^2} + {{\left( {y'} \right)}^2}} \right]}^{\frac{3}{2}}}}}.$

Substituting the above derivatives, we get:

$K = \frac{{\left| {ab\,{{\sin }^2}t + ab\,{{\cos }^2}t} \right|}}{{{{\left( {{a^2}{{\sin }^2}t + {b^2}{{\cos }^2}t} \right)}^{\frac{3}{2}}}}} = \frac{{\left| {ab\left( {{{\sin }^2}t + {{\cos }^2}t} \right)} \right|}}{{{{\left( {{a^2}{{\sin }^2}t + {b^2}{{\cos }^2}t} \right)}^{\frac{3}{2}}}}} = \frac{{ab}}{{{{\left( {{a^2}{{\sin }^2}t + {b^2}{{\cos }^2}t} \right)}^{\frac{3}{2}}}}}.$

Now we calculate the values of the curvature at the vertices $$A\left( {a,0} \right)$$ and $$B\left( {0,b} \right):$$

$K\left( A \right) = K\left( {t = 0} \right) = \frac{{ab}}{{{{\left( {{a^2}{{\sin }^2}0 + {b^2}{{\cos }^2}0} \right)}^{\frac{3}{2}}}}} = \frac{{ab}}{{{{\left( {{b^2}} \right)}^{\frac{3}{2}}}}} = \frac{{ab}}{{{b^3}}} = \frac{a}{{{b^2}}};$
$K\left( B \right) = K\left( {t = \frac{\pi }{2}} \right) = \frac{{ab}}{{{{\left( {{a^2}{{\sin }^2}\frac{\pi }{2} + {b^2}{{\cos }^2}\frac{\pi }{2}} \right)}^{\frac{3}{2}}}}} = \frac{{ab}}{{{{\left( {{a^2}} \right)}^{\frac{3}{2}}}}} = \frac{{ab}}{{{a^3}}} = \frac{b}{{{a^2}}}.$

### Example 2.

Find the curvature and radius of curvature of the parabola $y = {x^2}$ at the origin.

Solution.

Write the derivatives of the quadratic function:

$y' = {\left( {{x^2}} \right)^\prime } = 2x;\;\;y^{\prime\prime} = \left( {2x} \right)^\prime = 2.$

Then the curvature of the parabola is defined by the following formula:

$K = \frac{{y^{\prime\prime}}}{{{{\left[ {1 + {{\left( {y'} \right)}^2}} \right]}^{\frac{3}{2}}}}} = \frac{2}{{{{\left[ {1 + {{\left( {2x} \right)}^2}} \right]}^{\frac{3}{2}}}}} = \frac{2}{{{{\left( {1 + 4{x^2}} \right)}^{\frac{3}{2}}}}}.$

At the origin (at $$x = 0$$), the curvature and radius of curvature, respectively, are

$K\left( {x = 0} \right) = \frac{2}{{{{\left( {1 + 4 \cdot {0^2}} \right)}^{\frac{3}{2}}}}} = 2,\;\;R = \frac{1}{K} = \frac{1}{2}.$

### Example 3.

Find the curvature and radius of curvature of the curve $y = \cos mx$ at a maximum point.

Solution.

This function reaches a maximum at the points $$x = {\frac{{2\pi n}}{m}},\;n \in Z.$$ By the periodicity, the curvature at all maximum points is the same, so it is sufficient to consider only the point $$x = 0.$$

Write the derivatives:

$y' = \left( {\cos mx} \right)^\prime = - m\sin mx,\;\;y^{\prime\prime} = \left( { - m\sin mx} \right)^\prime = - {m^2}\cos mx.$

The curvature of this curve is given by

$K = \frac{{\left| {y^{\prime\prime}} \right|}}{{{{\left[ {1 + {{\left( {y'} \right)}^2}} \right]}^{\frac{3}{2}}}}} = \frac{{\left| { - {m^2}\cos mx} \right|}}{{{{\left[ {1 + {{\left( { - m\sin mx} \right)}^2}} \right]}^{\frac{3}{2}}}}} = \frac{{\left| { - {m^2}\cos mx} \right|}}{{{{\left( {1 + {m^2}{{\sin }^2}mx} \right)}^{\frac{3}{2}}}}}.$

At the maximum point $$x = 0,$$ the curvature and radius of curvature, respectively, are equal to

$K\left( {x = 0} \right) = \frac{{{m^2}}}{{{{\left( {1 + {m^2}{{\sin }^2}0} \right)}^{\frac{3}{2}}}}} = {m^2},\;\;R = \frac{1}{K} = \frac{1}{{{m^2}}}.$

### Example 4.

Calculate the curvature and radius of curvature of the graph of the function $y = \sqrt x$ at $$x = 1.$$

Solution.

We write the derivatives of the square root:

$y' = \left( {\sqrt x } \right)^\prime = \frac{1}{{2\sqrt x }},\;\;y^{\prime\prime} = \left( {\frac{1}{{2\sqrt x }}} \right)^\prime = \left( {\frac{1}{2}{x^{ - \frac{1}{2}}}} \right)^\prime = - \frac{1}{4}{x^{ - \frac{3}{2}}} = - \frac{1}{{4\sqrt {{x^3}} }}.$

The curvature of the curve is defined by the formula

$K = \frac{{\left| {y^{\prime\prime}} \right|}}{{{{\left[ {1 + {{\left( {y'} \right)}^2}} \right]}^{\frac{3}{2}}}}} = \frac{{\left| { - \frac{1}{{4\sqrt {{x^3}} }}} \right|}}{{{{\left[ {1 + {{\left( {\frac{1}{{2\sqrt x }}} \right)}^2}} \right]}^{\frac{3}{2}}}}} = \frac{{\frac{1}{{4\sqrt {{x^3}} }}}}{{{{\left( {1 + \frac{1}{{4x}}} \right)}^{\frac{3}{2}}}}} = \frac{{\frac{1}{{4\sqrt {{x^3}} }}}}{{{{\left( {\frac{{4x + 1}}{{4x}}} \right)}^{\frac{3}{2}}}}} = \frac{{{4^{\frac{3}{2}}}\cancel{x^{\frac{3}{2}}}}}{{4\cancel{x^{\frac{3}{2}}}{{\left( {4x + 1} \right)}^{\frac{3}{2}}}}} = \frac{2}{{{{\left( {4x + 1} \right)}^{\frac{3}{2}}}}}.$

At the point $$x = 1,$$ we get the following values for the curvature and radius of curvature:

$K\left( {x = 1} \right) = \frac{2}{{{{\left( {4 \cdot 1 + 1} \right)}^{\frac{3}{2}}}}} = \frac{2}{{5\sqrt 5 }},\;\;\;R\left( {x = 1} \right) = \frac{1}{K} = \frac{{5\sqrt 5 }}{2}.$

See more problems on Page 2.