# Calculus

Applications of the Derivative# Curvature and Radius of Curvature

Problem 1

Problems 2-10

Consider a plane curve defined by the equation \(y = f\left( x \right).\) Suppose that the tangent line is drawn to the curve at a point \(M\left( {x,y} \right).\) The tangent forms an angle \(\alpha\) with the horizontal axis (Figure \(1\text{).}\) 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 \(x\)-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.)

Figure 1.

The absolute value of the ratio \(\large\frac{{\Delta \alpha }}{{\Delta s}}\normalsize\) 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 \(M:\)

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

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

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

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:

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:

## Solved Problems

Click on problem description to see solution.

### ✓ Example 1

Calculate the curvature of the ellipse

at its vertices.

### ✓ Example 2

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

### ✓ Example 3

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

### ✓ Example 4

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

### ✓ Example 5

Consider the curve given by the equation \({y^2} + {x^3} = 0.\) Find its curvature at the point \(\left( { – 1,1} \right).\)

### ✓ Example 6

Find the curvature of the cardioid \(r = a\left( {1 + \cos \theta } \right)\) at \(\theta = 0.\)

### ✓ Example 7

Find the radius of curvature of the cycloid

{x = a\left( {t – \sin t} \right),}\;\;\;\kern-0.3pt

{y = a\left( {1 – \cos t} \right).}

\]

### ✓ Example 8

Determine the curvature of the curve \(y = \arctan x\) at \(x = 0\) and at infinity.

### ✓ Example 9

Determine the least radius of curvature of the exponential function \(y = {e^x}.\)

### ✓ Example 10

Find the least radius of curvature of the cubic parabola \(y = {x^3}.\)

### Example 1.

Calculate the curvature of the ellipse

at its vertices.

Figure 2.

*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:

where \(t\) is a parameter. The parameter has the value \(t = 0\) at the point \(A\left( {a,0} \right)\) and is equal to \(t = \large\frac{\pi }{2}\normalsize\) 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,}\;\;\;\kern-0.3pt

{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,}\;\;\;\kern-0.3pt

{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

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)}^{\large\frac{3}{2}\normalsize}}}} }

= {\frac{{\left| {ab\left( {{{\sin }^2}t + {{\cos }^2}t} \right)} \right|}}{{{{\left( {{a^2}{{\sin }^2}t + {b^2}{{\cos }^2}t} \right)}^{\large\frac{3}{2}\normalsize}}}} }

= {\frac{{ab}}{{{{\left( {{a^2}{{\sin }^2}t + {b^2}{{\cos }^2}t} \right)}^{\large\frac{3}{2}\normalsize}}}}.}

\]

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)}^{\large\frac{3}{2}\normalsize}}}} }

= {\frac{{ab}}{{{{\left( {{b^2}} \right)}^{\large\frac{3}{2}\normalsize}}}} }

= {\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)}^{\large\frac{3}{2}\normalsize}}}} }

= {\frac{{ab}}{{{{\left( {{a^2}} \right)}^{\large\frac{3}{2}\normalsize}}}} }

= {\frac{{ab}}{{{a^3}}} }

= {\frac{b}{{{a^2}}}.}

\]

Problem 1

Problems 2-10