Question

Q. Compute the curvature of the following curves
(i) γ(t)=4/5cos t, 1-sin⁡t,(-3)/5cos t
(ii) γ(t)=(t, cosht)
(iii) γ(t)=4/5 (〖cos〗^3t,〖sin〗^3t)
For the astroid in (iii), show that the curvature tends to ∞ as we approach one of the points (±1,0), (0,±1)

Answer 1

ANSWER on Question #71910 – Math – Differential Geometry | Topology

QUESTION

Compute the curvature of the following curves

1) $\gamma(t) = \left(\frac{4}{5} \cos t, 1 - \sin t, -\frac{3}{5} \cos t\right)$

2) $\gamma(t) = (t, \cosh t)$

3) $\gamma(t) = \frac{4}{5} (\cos^3 t, \sin^3 t)$

For the astroid in (3), show that the curvature tends to $\infty$ as we approach one of the points $(\pm 1,0)$, $(0,\pm 1)$

SOLUTION

By the definition, we want to briefly discuss the curvature of a smooth curve (recall that for a smooth curve we require $\overrightarrow{r'(t)}$ is continuous and $\overrightarrow{r'(t)} \neq 0$). The curvature measures how fast a curve is changing direction at a given point.

There are several formulas for determining the curvature for a curve. The formal definition of curvature is,

$$k = \left| \frac{d \vec{T}}{ds} \right|$$

Where $\vec{T}$ the unit tangent and $s$ is the arc length.

In general the formal definition of the curvature is not easy to use so there are two alternate formulas that we can use. Here they are.

$$k = \frac{\left| \overrightarrow{T'(t)} \right|}{\left| \overrightarrow{r'(t)} \right|}$$ $$k = \frac{\left| \overrightarrow{r'(t)} \times \overrightarrow{r''(t)} \right|}{\left| \overrightarrow{r'(t)} \right|^3}$$

In our case,

1)

$$\gamma(t) = \left(\frac{4}{5} \cos t, 1 - \sin t, -\frac{3}{5} \cos t\right)$$ $$\gamma'(t) = \left(-\frac{4}{5} \sin t, -\cos t, \frac{3}{5} \sin t\right)$$ $$\begin{aligned} |\gamma'(t)| &= \sqrt{\left(-\frac{4}{5} \sin t\right)^2 + (-\cos t)^2 + \left(\frac{3}{5} \sin t\right)^2} = \sqrt{\frac{15}{25} \sin^2 t + \cos^2 t + \frac{9}{25} \sin^2 t} = \\ &= \sqrt{\sin^2 t + \cos^2 t} = \sqrt{1} = 1 \rightarrow \boxed{|\gamma'(t)| = 1} \end{aligned}$$ $$T(t) = \frac{\gamma'(t)}{|\gamma'(t)|} = \frac{\left(-\frac{4}{5} \sin t, -\cos t, \frac{3}{5} \sin t\right)}{1} = \left(-\frac{4}{5} \sin t, -\cos t, \frac{3}{5} \sin t\right)$$ $$T'(t) = \left(-\frac{4}{5} \cos t, \sin t, \frac{3}{5} \cos t\right)$$ $$\begin{aligned} |T'(t)| &= \sqrt{\left(-\frac{4}{5} \cos t\right)^2 + (\sin t)^2 + \left(\frac{3}{5} \cos t\right)^2} = \sqrt{\frac{16}{25} \cos^2 t + \sin^2 t + \frac{9}{25} \cos^2 t} = \\ &= \sqrt{\cos^2 t + \sin^2 t} = \sqrt{1} = 1 \rightarrow \boxed{|T'(t)| = 1} \end{aligned}$$

Then,

$$k = \frac{|T'(t)|}{|\gamma'(t)|} = \frac{1}{1} = 1$$

Conclusion,

$$\gamma(t) = \left(\frac{4}{5} \cos t, 1 - \sin t, -\frac{3}{5} \cos t\right) \rightarrow k = 1$$

2)

$$\gamma(t) = (t, \cosh t)$$ $$\gamma'(t) = (1, \sinh t)$$ $$|\gamma'(t)| = \sqrt{(1)^2 + (\sinh t)^2} = \begin{bmatrix} \cosh^2 t - \sinh^2 t = 1 & \to \\ \cosh^2 t = 1 + \sinh^2 t & \end{bmatrix} = \sqrt{\cosh^2 t} = \cosh t$$ $$\boxed{|\gamma'(t)| = \cosh t}$$ $$T(t) = \frac{\gamma'(t)}{|\gamma'(t)|} = \frac{(1, \sinh t)}{\cosh t} = \left(\frac{1}{\cosh t}, \tanh t\right)$$ $$T'(t) = \left(- \frac{\sinh t}{\cosh^2 t}, \frac{1}{\cosh^2 t}\right)$$ $$|T'(t)| = \sqrt{\left(- \frac{\sinh t}{\cosh^2 t}\right)^2 + \left(\frac{1}{\cosh^2 t}\right)^2} = \sqrt{\frac{1 + \sinh^2 t}{\cosh^4 t}} = \begin{bmatrix} \cosh^2 t - \sinh^2 t = 1 & \to \\ \cosh^2 t = 1 + \sinh^2 t & \end{bmatrix} = \sqrt{\frac{\cosh^2 t}{\cosh^4 t}} = \sqrt{\frac{1}{\cosh^2 t}} = \frac{1}{\cosh t} \to \boxed{|T'(t)| = \frac{1}{\cosh t}}$$

Then,

$$k = \frac{|T'(t)|}{|\gamma'(t)|} = \frac{\frac{1}{\cosh t}}{\cosh t} = \frac{1}{\cosh^2 t}$$

Conclusion,

$$\gamma(t) = \gamma(t) = (t, \cosh t) \to k = \frac{1}{\cosh^2 t}$$

3)

$$\gamma(t) = \frac{4}{5} (\cos^3 t, \sin^3 t)$$ $$\gamma'(t) = \frac{4}{5} (3 \cos^2 t (-\sin t), 3 \sin^2 t \cos t)$$ $$\gamma'(t) = \frac{2 \cdot 3}{5} \left( (-\cos t)(2 \cos t \sin t), \sin t (2 \sin t \cos t) \right)$$ $$\gamma'(t) = \frac{6}{5} \left( (-\cos t) \sin 2t, \sin t \sin 2t \right)$$ $$|\gamma'(t)| = \frac{4}{5} \sqrt{ \left( 3 \cos^2 t (-\sin t) \right)^2 + \left( 3 \sin^2 t \cos t \right)^2 } =$$ $$= \frac{4}{5} \cdot 3 \sqrt{ \cos^4 t \sin^2 t + \sin^4 t \cos^2 t } = \frac{12}{5} \sqrt{ \cos^2 t \sin^2 t (\cos^2 t + \sin^2 t) } =$$ $$= \frac{12}{5} \sqrt{ \cos^2 t \sin^2 t } = \frac{12}{5} \cos t \sin t = \frac{6}{5} \cdot (2 \cos t \sin t) = \frac{6}{5} \cdot \sin 2t$$ $$\boxed{|\gamma'(t)| = \frac{6}{5} \cdot \sin 2t}$$ $$T(t) = \frac{ \gamma'(t) }{ | \gamma'(t) | } = \frac{ \frac{6}{5} \left( (-\cos t) \sin 2t, \sin t \sin 2t \right) }{ \frac{6}{5} \cdot \sin 2t } = (-\cos t, \sin t)$$ $$T'(t) = (\sin t, \cos t)$$ $$|T'(t)| = \sqrt{ (\sin t)^2 + (\cos t)^2 } = \sqrt{ \cos^2 t + \sin^2 t } = 1$$ $$\boxed{ |T'(t)| = 1 }$$

Then,

$$k = \frac{ |T'(t) | }{ | \gamma'(t) | } = \frac{1}{ \frac{6}{5} \cdot \sin 2t } = \frac{5}{6 \sin 2t}$$

Conclusion,

$$\gamma(t) = \gamma(t) = \frac{4}{5} (\cos^3 t, \sin^3 t) \rightarrow k = \frac{5}{6 \sin 2t}$$

For the astroid in (3), show that the curvature tends to $\infty$ as we approach one of the points $(\pm 1,0)$, $(0,\pm 1)$

If we approach the point $(1,0)$ - this means that $t = 0$

Then,

$$\lim_{t \to 0} k = \lim_{t \to 0} \frac{5}{6 \sin 2t} = \frac{5}{6 \cdot \sin (2 \cdot 0)} = \frac{5}{6 \cdot \sin 0} = \frac{5}{6 \cdot 0} = \infty$$

If we approach the point $(0,1)$ - this means that $t = \frac{\pi}{2}$

$$\lim_{t \to \frac{\pi}{2}} k = \lim_{t \to \frac{\pi}{2}} \frac{5}{6 \sin 2t} = \frac{5}{6 \cdot \sin \left(2 \cdot \frac{\pi}{2}\right)} = \frac{5}{6 \cdot \sin \pi} = \frac{5}{6 \cdot 0} = \infty$$

Answer provided by https://www.AssignmentExpert.com

.