Calculus

Solved Problems

Click or tap a problem to see the solution.

Example 5

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

Example 6

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

Example 7

Find the radius of curvature of the cycloid $x = a\left( {t - \sin t} \right), y = a\left( {1 - \cos t} \right).$

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 11

Find the curvature of a superellipse defined by the parametric equations $x\left( t \right) = a\, {\cos ^n}t,\; y\left( t \right) = b\,{\sin ^n}t,$ where $$a,b \text{ and } n$$ are positive real numbers.

Example 5.

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

Solution.

Calculate the derivative of this function using implicit differentiation:

${y^2} + {x^3} = 0,\;\; \Rightarrow \left( {{y^2} + {x^3}} \right)^\prime = 0,\;\; \Rightarrow 3{x^2} + 2yy' = 0,\;\; \Rightarrow y' = - \frac{{3{x^2}}}{{2y}}.$

Similarly, we find the second derivative:

$3{x^2} + 2yy' = 0,\;\; \Rightarrow \left( {3{x^2} + 2yy'} \right)^\prime = 0,\;\; \Rightarrow 6x + 2{\left( {y'} \right)^2} + 2yy^{\prime\prime} = 0,\;\; \Rightarrow yy^{\prime\prime} + {\left( {y'} \right)^2} + 3x = 0,\;\; \Rightarrow y^{\prime\prime} = - \frac{{{{\left( {y'} \right)}^2} + 3x}}{y}.$

Substitute the expression for the first derivative in the last formula:

$y^{\prime\prime} = - \frac{{{{\left( {y'} \right)}^2} + 3x}}{y} = - \frac{{{{\left( { - \frac{{3{x^2}}}{{2y}}} \right)}^2} + 3x}}{y} = - \frac{{\frac{{9{x^4}}}{{4{y^2}}} + 3x}}{y} = - \frac{{9{x^4} + 12x{y^2}}}{{4{y^3}}}.$

Calculate the values of the derivatives at the point $$\left( { - 1,1} \right):$$

$y'\left( { - 1,1} \right) = - \frac{{3 \cdot {{\left( { - 1} \right)}^2}}}{{2 \cdot 1}} = - \frac{3}{2},$
$y^{\prime\prime}\left( { - 1,1} \right) = - \frac{{9 \cdot {{\left( { - 1} \right)}^4} + 12 \cdot \left( { - 1} \right) \cdot {1^2}}}{{4 \cdot {1^3}}} = \frac{3}{4}.$

Then the curvature of the curve at the point $$\left( { - 1,1} \right)$$ is

$K = \frac{{\left| {y^{\prime\prime}} \right|}}{{{{\left[ {1 + {{\left( {y'} \right)}^2}} \right]}^{\frac{3}{2}}}}} = \frac{{\frac{3}{4}}}{{{{\left[ {1 + {{\left( { - \frac{3}{2}} \right)}^2}} \right]}^{\frac{3}{2}}}}} = \frac{{\frac{3}{4}}}{{{{\left( {1 + \frac{9}{4}} \right)}^{\frac{3}{2}}}}} = \frac{3}{4} \cdot \frac{{{4^{\frac{3}{2}}}}}{{{{13}^{\frac{3}{2}}}}} = \frac{6}{{{{13}^{\frac{3}{2}}}}} \approx 0,128.$

Example 6.

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

Solution.

To calculate the curvature of the curve we use 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 derivatives of the polar curve are given by

$r' = \left[ {a\left( {1 + \cos \theta } \right)} \right]^\prime = - a\sin \theta ,\;\;r^{\prime\prime} = \left( { - a\sin \theta } \right)^\prime = - a\cos \theta .$

Substituting this into the formula for the curvature, we get after simple transformations:

$K = \frac{3}{{{2^{\frac{3}{2}}}a{{\left( {1 + \cos \theta } \right)}^{\frac{1}{2}}}}}.$

Therefore when $$\theta = 0,$$ the curvature of the cardioid is

$K\left( {\theta = 0} \right) = \frac{3}{{{2^{\frac{3}{2}}}a{{\left( {1 + \cos \theta } \right)}^{\frac{1}{2}}}}} = \frac{3}{{{2^{\frac{3}{2}}}{2^{\frac{1}{2}}}a}} = \frac{3}{{4a}}.$

Example 7.

Find the radius of curvature of the cycloid $x = a\left( {t - \sin t} \right), y = a\left( {1 - \cos t} \right).$

Solution.

The radius of curvature of a parametric curve is expressed by the formula

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

Calculate the derivatives:

$x' = \left[ {a\left( {t - \sin t} \right)} \right]^\prime = a\left( {1 - \cos t} \right),\;\;\;x^{\prime\prime} = \left[ {a\left( {1 - \cos t} \right)} \right]^\prime = a\sin t,$
$y' = \left[ {a\left( {1 - \cos t} \right)} \right]^\prime = a\sin t,\;\;y^{\prime\prime} = \left( {a\sin t} \right)^\prime = a\cos t.$

Substituting the above expressions, we have

$R = \frac{{{{\left[ {2{a^2}\left( {1 - \cos t} \right)} \right]}^{\frac{3}{2}}}}}{{\left| {{a^2}\left( {\cos t - 1} \right)} \right|}} = \frac{{{2^{\frac{3}{2}}}{a^3}{{\left( {1 - \cos t} \right)}^{\frac{3}{2}}}}}{{{a^2}\left( {1 - \cos t} \right)}} = {2^{\frac{3}{2}}}a{\left( {1 - \cos t} \right)^{\frac{1}{2}}} = {2^{\frac{3}{2}}}a \cdot {\left( {2{{\sin }^2}\frac{t}{2}} \right)^{\frac{1}{2}}} = 4a{\left( {{{\sin }^2}\frac{t}{2}} \right)^{\frac{1}{2}}}.$

Confining ourselves to the first arc of the cycloid, i.e. the interval $$0 \le t \le 2\pi ,$$ we get the following expression for the radius of curvature of the cycloid:

$R = 4a\sin \frac{t}{2}.$

It follows that the radius of curvature has a maximum at $$t = \pi.$$ Its value at this point is $${R_{\max }} = 4a.$$

Example 8.

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

Solution.

We write the derivatives of the function $$y = \arctan x:$$

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

Then the curvature of the inverse tangent curve is given by

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

As it can be seen, at the point $$x = 0$$ the curvature is equal to zero: $$K\left( 0 \right) = {K_0} = 0.$$

In this case, the point $$x = 0$$ is a point of inflection of the function $$y = \arctan x.$$ Since the second derivative is zero at any inflection point, the curvature here must also be equal to zero, which coincides with the obtained solution.

Calculate the value of the curvature $${K_{\infty}}$$ in the limit as $$x \to \infty:$$

${K_\infty } = \lim\limits_{x \to \infty } K\left( x \right) = \lim\limits_{x \to \infty } \frac{{2x\left( {1 + {x^2}} \right)}}{{{{\left[ {{{\left( {1 + {x^2}} \right)}^2} + 1} \right]}^{\frac{3}{2}}}}} = \lim\limits_{x \to \infty } \frac{{2{x^3} + 2x}}{{{{\left( {{x^4} + 2{x^2} + 2} \right)}^{\frac{3}{2}}}}} = \lim\limits_{x \to \infty } \frac{{\frac{{2{x^3} + 2x}}{{{x^6}}}}}{{\frac{{{{\left( {{x^4} + 2{x^2} + 2} \right)}^{\frac{3}{2}}}}}{{{x^6}}}}} = \lim\limits_{x \to \infty } \frac{{\frac{2}{{{x^3}}} + \frac{2}{{{x^5}}}}}{{{{\left( {\frac{{{x^4} + 2{x^2} + 2}}{{{x^4}}}} \right)}^{\frac{3}{2}}}}} = \lim\limits_{x \to \infty } \frac{{\frac{2}{{{x^3}}} + \frac{2}{{{x^5}}}}}{{{{\left( {1 + \frac{2}{{{x^2}}} + \frac{2}{{{x^4}}}} \right)}^{\frac{3}{2}}}}} = \frac{0}{1} = 0.$

Thus, the curvature of the inverse tangent curve at infinity also approaches zero. This means that there is some intermediate value of $$x,$$ at which the curvature reaches a maximum.

Example 9.

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

Solution.

The exponential function $$y = {e^x}$$ is the only unique function whose derivatives of any order are equal to the function itself. Therefore, we can immediately write the following formula for the curvature of the given curve:

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

The absolute value sign in the numerator is omitted since the exponential function is always positive.

The radius of curvature, respectively, is

$R = \frac{1}{K} = \frac{{{{\left( {1 + {e^{2x}}} \right)}^{\frac{3}{2}}}}}{{{e^x}}}.$

The value of $$R$$ depends on the coordinate $$x.$$ Therefore considering $$R$$ as a function of $$x,$$ we can investigate it for extreme values. Calculate the derivative $$R'\left( x \right):$$

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

The function $$R\left( x \right)$$ has only one critical point:

$R'\left( x \right) = 0,\;\; \Rightarrow 2{e^{2x}} - 1 = 0,\;\; \Rightarrow {e^{2x}} = \frac{1}{2},\;\; \Rightarrow 2x = \ln \frac{1}{2} = - \ln 2,\;\; \Rightarrow x = - \frac{{\ln 2}}{2} \approx - 0,35.$

The derivative $$R^\prime\left( x \right)$$ is negative to the left of this value and is positive to the right. Therefore, this point is a point of minimum of the function $$R\left( x \right).$$ At this point, the exponential function has the least radius of curvature. Numerically it is equal to

${R_{\min }} = R\left( { - \frac{{\ln 2}}{2}} \right) = \frac{{{{\left[ {1 + {e^{2 \left( { - \frac{{\ln 2}}{2}} \right)}}} \right]}^{\frac{3}{2}}}}}{{{e^{ - \frac{{\ln 2}}{2}}}}} = \frac{{{{\left[ {1 + {e^{\ln \frac{1}{2}}}} \right]}^{\frac{3}{2}}}}}{{{e^{\ln \frac{1}{{\sqrt 2 }}}}}} = \frac{{{{\left( {1 + \frac{1}{2}} \right)}^{\frac{3}{2}}}}}{{\frac{1}{{\sqrt 2 }}}} = \sqrt 2 {\left( {\frac{3}{2}} \right)^{\frac{3}{2}}} \approx 2,60.$

Example 10.

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

Solution.

As

$y' = \left( {{x^3}} \right)^\prime = 3{x^2},\;\;y^{\prime\prime} = \left( {3{x^2}} \right)^\prime = 6x,$

the radius of curvature of the cubic function is determined by the following expression:

$R = \frac{{{{\left[ {1 + {{\left( {y'} \right)}^2}} \right]}^{\frac{3}{2}}}}}{{\left| {y^{\prime\prime}} \right|}} = \frac{{{{\left[ {1 + {{\left( {3{x^2}} \right)}^2}} \right]}^{\frac{3}{2}}}}}{{\left| {6x} \right|}} = \frac{{{{\left( {1 + 9{x^4}} \right)}^{\frac{3}{2}}}}}{{\left| {6x} \right|}}.$

Given that the cubic parabola is symmetric about the origin, we will consider only the portion of the curve for $$x \gt 0.$$ Omitting the absolute value sign, we write $$R$$ as a function of $$x:$$

$R\left( x \right) = \frac{{{{\left( {1 + 9{x^4}} \right)}^{\frac{3}{2}}}}}{{6x}}.$

Investigate extreme values. Find the derivative $$R'\left( x \right):$$

$R'\left( x \right) = \left[ {\frac{{{{\left( {1 + 9{x^4}} \right)}^{\frac{3}{2}}}}}{{6x}}} \right]^\prime = \frac{{{{\left( {{{\left( {1 + 9{x^4}} \right)}^{\frac{3}{2}}}} \right)}^\prime }6x - {{\left( {1 + 9{x^4}} \right)}^{\frac{3}{2}}}{{\left( {6x} \right)}^\prime }}}{{36{x^2}}} = \frac{{6{{\left( {1 + 9{x^4}} \right)}^{\frac{1}{2}}}\left[ {54{x^4} - \left( {1 + 9{x^4}} \right)} \right]}}{{36{x^2}}} = \frac{{{{\left( {1 + 9{x^4}} \right)}^{\frac{1}{2}}}\left( {45{x^4} - 1} \right)}}{{6{x^2}}}.$

At $$x \gt 0,$$ the function has only one critical point. The calculations lead to the following result:

$45{x^4} - 1 = 0,\;\; \Rightarrow {x^4} = \frac{1}{{45}},\;\; \Rightarrow x = \frac{1}{{\sqrt[4]{{45}}}} \approx 0,39.$

When passing through this point, the derivative $$R'\left( x \right)$$ changes sign from minus to plus. Therefore, this point corresponds to the least radius of curvature. We find its approximate value:

$y'\left( {\frac{1}{{\sqrt[4]{{45}}}}} \right) = 3 \cdot {\left( {\frac{1}{{\sqrt[4]{{45}}}}} \right)^2} = \frac{3}{{\sqrt {45} }},\;\;y^{\prime\prime}\left( {\frac{1}{{\sqrt[4]{{45}}}}} \right) = \frac{6}{{\sqrt[4]{{45}}}},$
$\Rightarrow R_{\min } = \frac{{{{\left[ {1 + {{\left( {y'} \right)}^2}} \right]}^{\frac{3}{2}}}}}{{\left| {y^{\prime\prime}} \right|}} = \frac{{{{\left[ {1 + {{\left( {\frac{3}{{\sqrt {45} }}} \right)}^2}} \right]}^{\frac{3}{2}}}}}{{\frac{6}{{\sqrt[4]{{45}}}}}} = \frac{{{{\left( {1 + \frac{9}{{45}}} \right)}^{\frac{3}{2}}}}}{{\frac{6}{{\sqrt[4]{{45}}}}}} = {\left( {\frac{{54}}{{45}}} \right)^{\frac{3}{2}}} \cdot \frac{{\sqrt[4]{{45}}}}{6} = \frac{{{2^{\frac{3}{2}}} \cdot {{\left( {{3^3}} \right)}^{\frac{3}{2}}} \cdot {5^{\frac{1}{4}}} \cdot {{\left( {{3^2}} \right)}^{\frac{1}{4}}}}}{{{{\left( {{3^3}} \right)}^{\frac{3}{2}}} \cdot {5^{\frac{3}{2}}} \cdot 2 \cdot 3}} = \frac{{{2^{\frac{1}{2}}} \cdot 3}}{{{5^{\frac{5}{4}}}}} = \frac{3}{5}\sqrt[4]{{\frac{4}{5}}} \approx 0,57.$

Example 11.

Find the curvature of a superellipse defined by the parametric equations $x\left( t \right) = a\, {\cos ^n}t,\; y\left( t \right) = b\,{\sin ^n}t,$ where $$a,b \text{ and } n$$ are positive real numbers.

Solution.

Superellipses are nice curves with the shape intermediate between circle and square. They have many applications in design and engineering.

The shape of a curve is determined by its curvature. In this problem, we derive the general formula for the curvature of a superellipse with an arbitrary positive exponent $$n.$$

Recall that the curvature of a parametric curve is expressed by the formula

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

Calculate the first derivatives:

$x^\prime = \left( {a\,{{\cos }^n}t} \right)^\prime = an\,{\cos ^{n - 1}}t \cdot \left( { - \sin t} \right) = - an\,{\cos ^{n - 1}}t\sin t,$
$y^\prime = \left( {b\,{{\sin }^n}t} \right)^\prime = bn\,{\sin ^{n - 1}}t \cdot \cos t = bn\,{\sin ^{n - 1}}t\cos t.$

The denominator includes the first derivatives squared which are given by

${\left( {x^\prime} \right)^2} = {\left( { - an\,{{\cos }^{n - 1}}t\sin t} \right)^2} = {a^2}{n^2}{\cos ^{2n - 2}}t\,{\sin ^2}t,$
${\left( {y^\prime} \right)^2} = {\left( {bn\,{{\sin }^{n - 1}}t\cos t} \right)^2} = {b^2}{n^2}{\sin ^{2n - 2}}t\,{\cos ^2}t.$

Hence, the denominator can be represented in the form

${\left[ {{{\left( {x^\prime} \right)}^2 + {\left( {y^\prime} \right)}^2}} \right]^{\frac{3}{2}}} = {\left[ {{a^2}{n^2}{{\cos }^{2n - 2}}t\,{{\sin }^2}t + {b^2}{n^2}{{\sin }^{2n - 2}}t\,{{\cos }^2}t} \right]^{\frac{3}{2}}}.$

Let's now consider the numerator. Compute the second derivative $$x^{\prime\prime}:$$

$x^{\prime\prime} = \left( { - an\,{{\cos }^{n - 1}}t\sin t} \right)^\prime = - an\left[ {\left( {n - 1} \right){{\cos }^{n - 2}}t \cdot \left( { - \sin t} \right) \cdot \sin t + {{\cos }^{n - 1}}t \cdot \cos t} \right] = an\,{\cos ^{n - 2}}t\left[ {\left( {n - 1} \right){{\sin }^2}t - {{\cos }^2}t} \right].$

This yields:

$y^\prime x^{\prime\prime} = bn\,{\sin ^{n - 1}}t\cos t \cdot an\,{\cos ^{n - 2}}t\left[ {\left( {n - 1} \right){{\sin }^2}t - {{\cos }^2}t} \right] = ab{n^2}{\sin ^{n - 1}}t\,{\cos ^{n - 1}}t \left[ {\left( {n - 1} \right){{\sin }^2}t - {{\cos }^2}t} \right].$

Similarly, we find $$y^{\prime\prime}$$ and $$x^\prime y^{\prime\prime}:$$

$y^{\prime\prime} = \left( {bn\,{{\sin }^{n - 1}}t\cos t} \right)^\prime = bn\left[ {\left( {n - 1} \right){{\sin }^{n - 2}}t \cdot \cos t \cdot \cos t + {{\sin }^{n - 1}}t \cdot \left( { - \sin t} \right)} \right] = bn\,{\sin ^{n - 2}}t\left[ {\left( {n - 1} \right){{\cos }^2}t - {{\sin }^2}t} \right],$
$x^\prime y^{\prime\prime} = \left( { - an\,{{\cos }^{n - 1}}t\sin t} \right) \cdot bn\,{\sin ^{n - 2}}t\left[ {\left( {n - 1} \right){{\cos }^2}t - {{\sin }^2}t} \right] = - ab{n^2}{\cos ^{n - 1}}t\,{\sin ^{n - 1}}t \left[ {\left( {n - 1} \right){{\cos }^2}t - {{\sin }^2}t} \right] = ab{n^2}{\cos ^{n - 1}}t\,{\sin ^{n - 1}}t \left[ {{{\sin }^2}t - \left( {n - 1} \right){{\cos }^2}t} \right].$

So, the expression in the numerator is given by

$x^\prime y^{\prime\prime} - y^\prime x^{\prime\prime} = ab{n^2}{\cos ^{n - 1}}t\,{\sin ^{n - 1}}t \left[ {{{\sin }^2}t - \left( {n - 1} \right){{\cos }^2}t} \right] + ab{n^2}{\sin ^{n - 1}}t\,{\cos ^{n - 1}}t \left[ {{{\cos }^2}t - \left( {n - 1} \right){{\sin }^2}t} \right] = ab{n^2}{\cos ^{n - 1}}t\,{\sin ^{n - 1}}t \left[ {\underbrace {{{\sin }^2}t + {{\cos }^2}t}_1 - \left( {n - 1} \right)\underbrace {\left( {{{\cos }^2}t + {{\sin }^2}t} \right)}_1} \right] = ab{n^2}{\cos ^{n - 1}}t\,{\sin ^{n - 1}}t\left[ {1 - \left( {n - 1} \right)} \right] = ab\left( {2 - n} \right){n^2}{\cos ^{n - 1}}t\,{\sin ^{n - 1}}t.$

Using the double angle formula, we can represent $${\cos ^{n - 1}}t\,{\sin ^{n - 1}}t$$ as follows:

${\cos ^{n - 1}}t\,{\sin ^{n - 1}}t = {\left( {\cos t\sin t} \right)^{n - 1}} = {\left( {\frac{{2\cos t\sin t}}{2}} \right)^{n - 1}} = {\left( {\frac{{\sin 2t}}{2}} \right)^{n - 1}} = \frac{{{{\sin }^{n - 1}}2t}}{{{2^{n - 1}}}}.$

The curvature of a curve is often considered as a measure of how sharply the curve bends. Therefore we take the absolute value of the expression in the numerator:

$\left| {x^\prime y^{\prime\prime} - y^\prime x^{\prime\prime}} \right| = \left| {ab\left( {2 - n} \right){n^2}\frac{{{{\sin }^{n - 1}}2t}}{{{2^{n - 1}}}}} \right| = ab{n^2}{2^{1 - n}}\left| {\left( {2 - n} \right){{\sin }^{n - 1}}2t} \right|.$

Thus, the curvature of a superellipse is given by the formula

$K \text = \frac{{ab{2^{1 - n}}\left| {\left( {2 - n} \right){{\sin }^{n - 1}}2t} \right|}}{{{{n\left[ {{a^2}{{\cos }^{2n - 2}}t\,{{\sin }^2}t + {b^2}{{\sin }^{2n - 2}}t\,{{\cos }^2}t} \right]}^{\frac{3}{2}}}}}.$

Let's consider some special cases. For a regular ellipse, we set $$n = 1$$. In this case,

${2^{1 - n}} = {2^0} = 1,\;\;{\sin ^{n - 1}}2t = {\left( {\sin 2t} \right)^0} \equiv 1,\;\;{\cos ^{2n - 2}}t = {\left( {\cos t} \right)^0} \equiv 1,\;\;{\sin ^{2n - 2}}t = {\left( {\sin t} \right)^0} \equiv 1,$

and the curvature is written in the form (see Example $$1$$):

${K_1} = \frac{{ab}}{{{{\left[ {{a^2}{{\sin }^2}t + {b^2}{{\cos }^2}t} \right]}^{\frac{3}{2}}}}}.$

Another interesting curve is the astroid which is defined by the parametric equations

$x = {\cos ^3}t,\;\;y = {\sin ^3}t.$

Substituting $$n = 3,$$ $$a = b = 1,$$ we can easily find the curvature of the astroid:

${K_3} = \frac{{{2^{ - 2}}\left| { - {{\sin }^2}2t} \right|}}{{3{{\left[ {{{\cos }^4}t\,{{\sin }^2}t + {{\sin }^4}t\,{{\cos }^2}t} \right]}^{\frac{3}{2}}}}} = \frac{{{{\sin }^2}2t}}{{12{{\left[ {{{\cos }^2}t\,{{\sin }^2}t\underbrace {\left( {{{\cos }^2}t + {{\sin }^2}t} \right)}_1} \right]}^{\frac{3}{2}}}}} = \frac{{{{\sin }^2}2t}}{{12{{\left[ {{{\left( {\cos t\sin t} \right)}^2}} \right]}^{\frac{3}{2}}}}} = \frac{{{{\sin }^2}2t}}{{12{{\left[ {{{\left( {\frac{{\sin 2t}}{2}} \right)}^2}} \right]}^{\frac{3}{2}}}}} = \frac{{{{\sin }^2}2t}}{{\frac{{12\,{{\sin }^3}2t}}{8}}} = \frac{2}{{3\sin 2t}}.$

Using the general expression for the curvature obtained above, we can derive particular formulas for a superellipse with a fractional value of $$n.$$ For example, if $$n = \frac{1}{2},$$ the curvature is given by

${K_{\frac{1}{2}}} = \frac{{\frac{{3\sqrt 2 ab}}{{\sqrt {\sin 2t} }}}}{{{{\left[ {{a^2}\frac{{{{\sin }^2}t}}{{\cos t}} + {b^2}\frac{{{{\cos }^2}t}}{{\sin t}}} \right]}^{\frac{3}{2}}}}}.$

Note that if the power $$n$$ is fractional, the bases in the parametric equations (the sine and cosine functions) should be positively defined. This puts limits on the parameter $$t,$$ which changes in the range $$0 \lt t \lt \frac{\pi}{2}.$$

The problem about superellipse has been suggested by John Erbes. We greatly appreciate for that!