Differential Geometry Part 1

To review for my upcoming final in differential geometry, I will go over my last two midterms.

1. Parametrize the curve $y^2+y=x^2$ using as the parameter $t$ slope of the line from the origin to a point on it, i.e. $y=tx$.

The first thing we should do is substitute $tx$ for $y$ in the first equation to obtain $$(t^2-1)x^2+tx=0.$$ By the quadratic equation, we find that $$x=-\frac{t}{t^2-1}\text{or}x=0.$$ To find $y$ in terms of $t$, we substitute this $x$ into the second equation to obtain $$y=-\frac{t^2}{t^2-1}\text{or}y=0.$$ Therefore, the parametric equation of this curve is $$r\left(t\right)=\{\begin{array}{ll}-\frac{t}{t^2-1}{\partial }_x+-\frac{t^2}{t^2-1}{\partial }_y& \text{ if }t\in \mathbb{R}\setminus \{-1,1\},\\ 0{\partial }_x+0{\partial }_y& \text{ if }t\in \{-1,1\}.\end{array}$$ 2. The evolute of the parabola $y=x^2/2$ is $y=1+3/2x^{2/3}$. How many normals can be drawn to this parabola from $(1,-1)$? Explain.

Substituting this point into the evolute's equation yields $$y\left(1\right)=\frac{5}{2}>-1.$$ In other words, the point is below the evolute.

Some time ago, we concluded that under the evolute or on its cusps, only one normal can be drawn; above the evolute, only three normals can be drawn; and on the evolute, only two normals can be drawn.

Therefore, only one normal can be drawn.

3. Consider the curve $r\left(t\right)=e^t(\sin t-\cos t){\partial }_x+e^t(\sin t+\cos t){\partial }_y$ with $t\in \mathbb{R}$.

a. Find the speed.

The speed is defined as follows: $$v=\left|r^{\prime }\right|.$$ Therefore, we have that $$\begin{array}{rl}{r}^{\prime }\left(t\right)& =2e^t\sin t{\partial }_x+2e^t\cos t{\partial }_y,\\ \left|r^{\prime }\right|& =2e^t=v.\end{array}$$ b. Find a natural parameter.

The natural parameter is defined as follows: $$s=\int vdt.$$ Therefore, we have that $$s=2e^t+c.$$ c. Find involutes.

Involutes are defined as follows: $$h=r-T\int vdt.$$ Therefore, we have that $$\begin{array}{rl}T& =\frac{r^{\prime }}{v}=\sin t{\partial }_x+\cos t{\partial }_y,\\ h& =-[(e^t+c)\sin t+e^t\cos t]{\partial }_x+[e^t\sin t-(e^t+c)\cos t]{\partial }_y\end{array}$$ 4. Consider the curve $r\left(t\right)=t{\partial }_x+\ln \cos t{\partial }_y$ with $t\in (-\pi /2,\pi /2).$

a. Find the speed.
Hint: $1+{\tan }^{2}t={\sec }^{2}t$.

As before, we have that $$\begin{array}{rl}{r}^{\prime }\left(t\right)& ={\partial }_x-\frac{1}{\cos t}\sin t={\partial }_x-\tan t{\partial }_y,\\ v& =\sqrt{1+{\tan }^{2}t}=\sqrt{{\sec }^{2}t}=\sec t.\end{array}$$ b. Find the unit tangent.

As before, we have that $$T=\cos t{\partial }_x-\sin t{\partial }_y.$$ c. Find the right unit normal.

The right unit normal is defined as follows: $$\tilde{N}=-T_2{\partial }_x+T_1{\partial }_y.$$ Therefore, we have that $$\tilde{N}=\sin t{\partial }_x+\cos t{\partial }_y.$$ d. Find the signed curvature.

The signed curvature is defined as follows: $$\tilde{\varkappa }=\frac{r^{\prime }\times r^{″}}{v^3}$$ Therefore, we have that $$\begin{array}{rl}{r}^{″}\left(t\right)& =-{\sec }^{2}t{\partial }_y,\\ r^{\prime }\times r^{″}& =\left|\begin{array}{cc}1& -\tan t\\ 0& -{\sec }^{2}t\end{array}\right|=-{\sec }^{2}t,\\ \tilde{\varkappa }& =-\cos t.\end{array}$$ e. Find the principal unit normal.

The principal unit normal is defined as follows $$N=\text{sign}\left(\tilde{\varkappa }\right)\tilde{N}.$$ Therefore, we have that $$N=-\sin t{\partial }_x-\cos t{\partial }_y.$$ f. Find the evolute.

The evolute is defined as follows: $$c=r+\frac{1}{\tilde{\varkappa }}\tilde{N}.$$ Therefore, we have that $$c=(t-\tan t){\partial }_x+(\ln \cos t-1){\partial }_y.$$ 5. Fill in the blanks to form a true statement.

a. If the speed is $3$ and $T={\partial }_x$, then the velocity is "$3{\partial }_x$."

b. Natural parameter is unique up to "shift and sign change."

c. A plane has "two" unit normals.

d. Inflection point is where a curve "has $\varkappa =0$."

e. If the curvature is $\varkappa =e^{2t}+1$ and the natural parameter is $s=e^t$, then the natural equation of the curve is "$\varkappa \left(s\right)=s^2+1$."

f. Lines and circles are the only plane curves that "have constant curvature."

g. The unit tangent to a curve is drawn below, draw and label $N$ and $\tilde{N}$.

h. Tangents to a plane curve are normal to "its involutes."

i. Wavefronts of a plane curve have cusps on "its evolute."

j. Involute of the evolute is "a wavefront of the plane curve."

6. Prove that $\ddot{T}\cdot T=-{\varkappa }^2$, where $T$ is the unit tangent and $\varkappa$ is the curvature of a curve.
Hint: Differentiate $\dot{T}\cdot T$.

Proof. $\dot{T}=\varkappa N⟹\dot{T}\cdot T=\varkappa N\cdot T=0$. $$\begin{array}{rl}\frac{d}{dt}(\dot{T}\cdot T)& =\ddot{T}\cdot T+\dot{T}\cdot \dot{T}=\ddot{T}\cdot T+{\left|\dot{T}\right|}^2\\ & =\ddot{T}\cdot T+{\left|\varkappa N\right|}^2=\ddot{T}\cdot T+{\varkappa }^2=0\\ & ⟹\ddot{T}\cdot T=-{\varkappa }^2.◻\end{array}$$ 7. Let $r\left(t\right)$, $w\left(t\right)$ be two curves that are normal to the segment connecting them at each point, i.e. $r^{\prime }\perp (w-r)$ and $w^{\prime }\perp (w-r)$. Prove that they are equidistant, i.e. $|w-r|=\text{const}$.
Hint: Differentiate ${|w-r|}^2=(w-r)\cdot (w-r)$.

Proof. $$\begin{array}{rl}\frac{d}{dt}\left[{|w-r|}^2\right]& =\frac{d}{dt}[(w-r)\cdot (w-r)]\\ & =2(w^{\prime }-r^{\prime })\cdot (w-r)\\ & =2[w^{\prime }\cdot (w-r)-r^{\prime }\cdot (w-r)]=0.◻\end{array}$$