Question

Find the curvature of the ellipse\(x = 3\cos t,y = 4\sin t\)at the points\((3,0)\) and\((0,4)\).

Short answer

The curvature of the ellipse at\((3,0)\)is\(\frac{3}{{16}}\)and at\((0,4)\)is\(\frac{4}{9}\).

Step-by-step solution

Definition of the curve

A smooth drawn line or figure on a plane with turns or a bend is called a curve.

Evaluating the derivative.

Let \(C\)be a curve defined by the vector function\({\bf{r}}(t)\).

The formula\(k(t) = \frac{{\left| {{{\bf{r}}^\prime }(t) \times {{\bf{r}}^{\prime \prime }}(t)} \right|}}{{{{\left| {{{\bf{r}}^\prime }(t)} \right|}^3}}}\) describes the curvature of the curve\(C\).

Since the ellipse is given with the parametric equations\(x = 2\cos t,y = 4\sin t\), then the vector function of the following form is

\({\bf{r}}(t) = \langle 3\cos t,4\sin t\rangle \)

Let determine the curvature of the curve defined by the vector function\((2)\) .

To do this, first, find the first and second derivatives of\({\bf{r}}(t)\), then the cross product\({{\bf{r}}^\prime }(t) \times {{\bf{r}}^{\prime \prime }}(t)\), the length of the vector\({{\bf{r}}^\prime }(t)\), and replace the obtained results in equation (1).

Differentiate \({\bf{r}}(t)\)twice with respect to\(t\).

\(\begin{aligned}{l}{{\bf{r}}^\prime }(t) = \langle - 3\sin t,4\cos t\rangle {{\bf{r}}^{\prime \prime }}(t)\\ = \langle - 3\cos t, - 4\sin t\rangle \end{aligned}\)

Calculate the cross product of vectors\({{\bf{r}}^\prime }(t)\)and\({r^{\prime \prime }}(t)\).

\(\begin{aligned}{l}{{\bf{r}}^\prime }(t) \times {{\bf{r}}^{\prime \prime }}(t) = \left| {\begin{aligned}{*{20}{c}}{\bf{i}}&{\bf{j}}&{\bf{k}}\\{ - 3\sin t}&{4\cos t}&0\\{ - 3\cos t}&{ - 4\sin t}&0\end{aligned}} \right|\\ = {\bf{i}}\left| {\begin{aligned}{*{20}{c}}{4\cos t}&0\\{ - 4\sin t}&0\end{aligned}} \right| - {\bf{j}}\left| {\begin{aligned}{*{20}{c}}{ - 3\sin t}&0\\{ - 3\cos t}&0\end{aligned}} \right| + {\bf{k}}\left| {\begin{aligned}{*{20}{c}}{ - 3\sin t}&{4\cos t}\\{ - 3\cos t}&{ - 4\sin t}\end{aligned}} \right|\\ = {\bf{i}} \cdot 0 - {\bf{j}} \cdot 0 + {\bf{k}}\left( {12{{\sin }^2}t + 12{{\cos }^2}t} \right)\\ = {\bf{k}}\left( {12\left( {{{\sin }^2}t + {{\cos }^2}t} \right)} \right)\\ = 12{\bf{k}}\end{aligned}\)

The length of the vectors.

\({\bf{v}} = a{\bf{i}} + b{\bf{j}} + c{\bf{k}}\)is defined as:

\(|{\bf{v}}| = \sqrt {{a^2} + {b^2} + {c^2}} \)

Since, \({{\bf{r}}^\prime }(t) = \langle - 3\sin t,4\cos t\rangle \), it follows that

\(\begin{aligned}{l}\left| {{{\bf{r}}^\prime }(x)} \right| = \sqrt {{{( - 3\sin t)}^2} + {{(4\cos t)}^2} + {0^2}} \\ = \sqrt {9{{\sin }^2}t + 16{{\cos }^2}t} \end{aligned}\)

Substitute \({{\bf{r}}^\prime }(t) \times {{\bf{r}}^{\prime \prime }}(t)\) and \(\left| {{{\bf{r}}^\prime }(t)} \right|\)in equation (1)

\(\begin{aligned}{l}k(x) = \frac{{|12|}}{{{{\left( {\sqrt {9{{\sin }^2}t + 16{{\cos }^2}t} } \right)}^3}}}\\ = \frac{{12}}{{{{\left( {9{{\sin }^2}t + 16{{\cos }^2}t} \right)}^{\frac{3}{2}}}}}\end{aligned}\)

Finding the value of curvature

Finally, find the values of the curvature of the ellipse at point \((3,0)\)and\((0,4)\).

Note that point\((3,0)\)corresponds to the parameter\(t = 0\)and point\((0,4)\)to the parameter\(t = \frac{\pi }{2}\).

Substitute the values for\(t\)into equation\((3)\).

So, for\(t = 0\)we have that

\(\begin{aligned}{l}k(0) = \frac{{12}}{{{{\left( {9{{\sin }^2}0 + 16{{\cos }^2}0} \right)}^{\frac{3}{2}}}}}\\ = \frac{{12}}{{{{(9 \cdot 0 + 16 \cdot 1)}^{\frac{3}{2}}}}}\\ = \frac{{12}}{{16\frac{3}{2}}}\\ = \frac{{12}}{{64}}\\ = \frac{3}{{16}}\end{aligned}\)

and for \(t = \frac{\pi }{2}\)

\(\begin{aligned}{l}k\left( {\frac{\pi }{2}} \right) = \frac{{12}}{{{{\left( {9{{\sin }^2}\left( {\frac{\pi }{2}} \right) + 16{{\cos }^2}\left( {\frac{\pi }{2}} \right)} \right)}^{\frac{3}{2}}}}}\\ = \frac{{12}}{{{{(9 \cdot 1 + 16 \cdot 0)}^{\frac{3}{2}}}}}\\ = \frac{{12}}{{{9^{\frac{3}{2}}}}}\\ = \frac{{12}}{{27}}\\ = \frac{4}{9}\end{aligned}\)