Question

1)Find the point of maximum curvature on the graph of y=e^(x)
2)Find the minimum and maximum curvatures of the ellipse r(t)=(acost,bsint),0<=t<=2pi, where a>b

Answer 1

Answer on Question #51516 – Math – Differential Geometry

Question

1) Find the point of maximum curvature on the graph of $y = e^x$

2) Find the minimum and maximum curvatures of ellipse $r(t) = (a*\cos t, b*\sin t)$, $0<=t<=2\pi$, where $a>b$.

Solution

1) Curvature is given by

$$k(x) = \frac{y''}{(1+(y')^2)^{3/2}} = \frac{e^x}{(1+e^{2x})^{3/2}}.$$

To find the point of maximum, we can start with the first derivative of function $k(x)$:

$$\begin{aligned} k'(x) &= \frac{e^x (1 + e^{2x})^{\frac{3}{2}} - \left(\frac{3}{2}\right) (1 + e^{2x})^{\frac{1}{2}} * 2 * e^{2x} e^x}{(1 + e^{2x})^3} = 0; \\ k'(x) &= \frac{e^x (1 + e^{2x})^{\frac{1}{2}}}{(1 + e^{2x})^3} (1 + e^{2x} - 3e^{2x}) = 0; \\ k'(x) &= \frac{e^x (1 + e^{2x})^{\frac{1}{2}}}{(1 + e^{2x})^3} (1 - 2e^{2x}) = 0; \\ e^{2x} &= \frac{1}{2}; \\ x_0 &= \frac{1}{2} \ln \frac{1}{2}. \end{aligned}$$

$k'(x)$ is going to be negative for $x$ greater than $x_0$ and positive for $x < x_0$. By the first sufficient condition of maximum, $x_0$ is the maximum value of curvature $k(x)$.

Answer: $x_0 = \frac{1}{2} \ln \frac{1}{2}$.

2) $x(t) = \cos t$; $y(t) = \sin t$, $0 \leq t \leq 2\pi$.

Curvature is given by

$$\begin{aligned} k(t) &= \frac{x' y'' - y' x''}{(x'^2 + y'^2)^{\frac{3}{2}}} = \frac{\text{ab } \sin(t) \sin(t) + \text{abcos}(t) \cos(t)}{\left((-\cos t)(2) + (\cos t)\right)^2)^{\frac{3}{2}}} = \frac{\text{ab } (\sin^2(t) + \cos^2(t))}{\left((\cos t)^2 + (\cos t)\right)^2)^{\frac{3}{2}}} = \\ &= \frac{\text{ab}}{((\cos t)^2 + (\cos t))^2)^{3/2}} = \frac{\text{ab}}{(b^2 + (a^2 - b^2) \sin^2(t))^{3/2}}. \end{aligned}$$

This quantity is minimum when denominator is maximum, i.e. for $\sin^2 t = 1$ where $a > b$. The minimum curvature points are at $t = t_{\min} = \pi / 2$; $3\pi / 2$, hence $x = 0$; $y = \pm b$ and the minimum curvature is $k_{\min} = \frac{ab}{a^3} = \frac{b}{a^2}$, where $a > b$.

This quantity is maximum when denominator is minimum, i.e. for $\sin^2 t = 0$ where $a > b$. The maximum curvature points are at $t = t_{\max} = 0$; $\pi$, hence $x = \pm a$; $y = 0$ and the maximum curvature is $k_{\max} = \frac{ab}{b^3} = \frac{a}{b^2}$, where $a > b$.

Note that the maximum curvature is $k_{\max} = \frac{b}{a^2}$ and the minimum curvature is $k_{\min} = \frac{a}{b^2}$, where $a < b$.

Answer: $k_{\min} = b / a^2$; $k_{\max} = a / b^2$, where $a > b$.

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