04. 공간 운동Motion in Space

Solution. Since $\mathbf{r}(t)=(\cos t)\mathbf{i}+(\sin t)\mathbf{j}+t\mathbf{k}$, we have $\mathbf{r}'(t)=(-\sin t)\mathbf{i}+(\cos t)\mathbf{j}+\mathbf{k}$ and $|\mathbf{r}'(t)|=\sqrt{\sin^2t+\cos^2t+1}=\sqrt2$.

(1) $\displaystyle L=\int_0^{2\pi}|\mathbf{r}'(t)|\,dt=\int_0^{2\pi}\sqrt2\,dt=2\sqrt2\,\pi$.

(2) The arc length function with initial point $\mathbf{r}(0)$ is $\displaystyle s(t)=\int_0^t|\mathbf{r}'(u)|\,du=\int_0^t\sqrt2\,du=\sqrt2\,t$.

(3) From $s=\sqrt2\,t$ we get $t=s/\sqrt2$, so the arc length reparametrization is $$\mathbf{r}(s)=\Bigl(\cos\dfrac{s}{\sqrt2}\Bigr)\mathbf{i}+\Bigl(\sin\dfrac{s}{\sqrt2}\Bigr)\mathbf{j}+\dfrac{s}{\sqrt2}\mathbf{k}.$$

Solution. With $\mathbf{r}(t)=(\sqrt2\,t)\mathbf{i}+(\sqrt2\,t)\mathbf{j}+(1-t^2)\mathbf{k}$, we have $\mathbf{r}(0)=(0,0,1)$ and $\mathbf{r}(1)=(\sqrt2,\sqrt2,0)$, so $t$ ranges from $0$ to $1$.

Since $\mathbf{r}'(t)=\sqrt2\,\mathbf{i}+\sqrt2\,\mathbf{j}-2t\mathbf{k}$, $$|\mathbf{r}'(t)|=\sqrt{2+2+4t^2}=2\sqrt{1+t^2}.$$ Hence the length of the curve is $$L=\int_0^1 2\sqrt{1+t^2}\,dt=\Bigl[t\sqrt{1+t^2}+\ln\bigl(t+\sqrt{1+t^2}\bigr)\Bigr]_0^1=\sqrt2+\ln(1+\sqrt2).$$

Solution. Since $\mathbf{r}(t)=(1+2t)\mathbf{i}+(1+3t)\mathbf{j}+(6-6t)\mathbf{k}$, we have $\mathbf{r}'(t)=2\mathbf{i}+3\mathbf{j}-6\mathbf{k}$ and $$|\mathbf{r}'(t)|=\sqrt{4+9+36}=\sqrt{49}=7$$ (constant speed). The arc length is $$s=\int_0^t|\mathbf{r}'(u)|\,du=7t,$$ so $t=s/7$. Hence the arc length reparametrization is $$\mathbf{r}(s)=\Bigl(1+\dfrac{2s}{7}\Bigr)\mathbf{i}+\Bigl(1+\dfrac{3s}{7}\Bigr)\mathbf{j}+\Bigl(6-\dfrac{6s}{7}\Bigr)\mathbf{k}.$$

Solution. Since $\mathbf{r}(t)=(1+3\cos t)\mathbf{i}+(3\sin t)\mathbf{j}+t^2\mathbf{k}$, $$\mathbf{r}'(t)=(-3\sin t)\mathbf{i}+(3\cos t)\mathbf{j}+2t\mathbf{k}$$ and $$|\mathbf{r}'(t)|=\sqrt{9\sin^2t+9\cos^2t+4t^2}=\sqrt{9+4t^2}.$$ Hence the unit tangent vector is $$\mathbf{T}(t)=\dfrac{\mathbf{r}'(t)}{|\mathbf{r}'(t)|}=\dfrac{-3\sin t\,\mathbf{i}+3\cos t\,\mathbf{j}+2t\,\mathbf{k}}{\sqrt{9+4t^2}}.$$

Solution. Since $\mathbf{r}(t)=(5\sin t)\mathbf{i}+(5\cos t)\mathbf{j}+12t\mathbf{k}$, $$\mathbf{r}'(t)=(5\cos t)\mathbf{i}-(5\sin t)\mathbf{j}+12\mathbf{k},\qquad |\mathbf{r}'(t)|=\sqrt{25+144}=13$$ (constant speed). On $0\le t\le a$ the length is $13a$, and setting this equal to $26\pi$ gives $a=2\pi$.

Hence $a/2=\pi$, and $$\mathbf{r}'(\pi)=(5\cos\pi)\mathbf{i}-(5\sin\pi)\mathbf{j}+12\mathbf{k}=-5\mathbf{i}+12\mathbf{k},$$ so the unit tangent vector at $\mathbf{r}(a/2)$ is $$\mathbf{T}(\pi)=\dfrac{\mathbf{r}'(\pi)}{|\mathbf{r}'(\pi)|}=\dfrac{-5\mathbf{i}+12\mathbf{k}}{13}=-\dfrac{5}{13}\mathbf{i}+\dfrac{12}{13}\mathbf{k}.$$

Proof. (1) Since $\mathbf{r}(t)=\mathbf{C}+t\mathbf{v}$, we have $\mathbf{r}'(t)=\mathbf{v}$ and $\mathbf{r}''(t)=\mathbf{0}$. By the definition $\kappa=\dfrac{|\mathbf{r}'\times\mathbf{r}''|}{|\mathbf{r}'|^3}$, since $\mathbf{r}'\times\mathbf{r}''=\mathbf{v}\times\mathbf{0}=\mathbf{0}$, we get $\kappa=0$.

(2) Let $\mathbf{r}(t)=(a\cos t,a\sin t,0)$. Then $$\mathbf{r}'(t)=(-a\sin t,a\cos t,0),\qquad \mathbf{r}''(t)=(-a\cos t,-a\sin t,0).$$ Computing the cross product, $$\mathbf{r}'\times\mathbf{r}''=(0,0,a^2\sin^2t+a^2\cos^2t)=(0,0,a^2),$$ so $|\mathbf{r}'\times\mathbf{r}''|=a^2$. Also $|\mathbf{r}'(t)|=a$, so $|\mathbf{r}'|^3=a^3$. Therefore $$\kappa=\dfrac{a^2}{a^3}=\dfrac1a.$$

Solution. (1) Since $\mathbf{r}(t)=(a\cos t,a\sin t,bt)$, $$\mathbf{r}'(t)=(-a\sin t,a\cos t,b),\qquad \mathbf{r}''(t)=(-a\cos t,-a\sin t,0).$$ Computing the cross product, $$\mathbf{r}'\times\mathbf{r}''=(ab\sin t,\,-ab\cos t,\,a^2),$$ so $$|\mathbf{r}'\times\mathbf{r}''|=\sqrt{a^2b^2+a^4}=a\sqrt{a^2+b^2}.$$ Since $|\mathbf{r}'(t)|=\sqrt{a^2+b^2}$, $$\kappa=\dfrac{a\sqrt{a^2+b^2}}{(a^2+b^2)^{3/2}}=\dfrac{a}{a^2+b^2}.$$

(2) Since $\mathbf{r}(t)=(\cosh t,-\sinh t,t)$, $$\mathbf{r}'(t)=(\sinh t,-\cosh t,1),\qquad \mathbf{r}''(t)=(\cosh t,-\sinh t,0).$$ Computing the cross product, $$\mathbf{r}'\times\mathbf{r}''=(\sinh t,\cosh t,-\sinh^2t+\cosh^2t)=(\sinh t,\cosh t,1),$$ so, using $\cosh^2t-\sinh^2t=1$, $$|\mathbf{r}'\times\mathbf{r}''|=\sqrt{\sinh^2t+\cosh^2t+1}=\sqrt{2\cosh^2t}=\sqrt2\cosh t.$$ Similarly $|\mathbf{r}'(t)|^2=\sinh^2t+\cosh^2t+1=2\cosh^2t$, so $|\mathbf{r}'(t)|=\sqrt2\cosh t$. Therefore $$\kappa=\dfrac{\sqrt2\cosh t}{(\sqrt2\cosh t)^3}=\dfrac{1}{2\cosh^2t}.$$

Solution. (1) For $y=\ln(\cos x)$ $(-\pi/2 \lt x \lt \pi/2)$, $$y'=-\tan x,\qquad y''=-\sec^2x.$$ Substituting into the curvature formula for a plane curve $y=f(x)$, $\kappa=\dfrac{|f''|}{(1+f'^2)^{3/2}}$, and using $1+y'^2=1+\tan^2x=\sec^2x$, $$\kappa=\dfrac{\sec^2x}{(\sec^2x)^{3/2}}=\dfrac{\sec^2x}{\sec^3x}=\dfrac{1}{\sec x}=\cos x$$ (since $\cos x \gt 0$ on $-\pi/2 \lt x \lt \pi/2$, there is no sign ambiguity in $\sec^3x=|\sec x|^3$).

(2) Let $\mathbf{r}(t)=(\tan^{-1}(\sinh t),\ln(\cosh t))$. Then $$x'(t)=\dfrac{\cosh t}{1+\sinh^2t}=\dfrac{\cosh t}{\cosh^2t}=\operatorname{sech}t,\qquad y'(t)=\dfrac{\sinh t}{\cosh t}=\tanh t,$$ and $$x''(t)=-\operatorname{sech}t\tanh t,\qquad y''(t)=\operatorname{sech}^2t.$$ Using the plane-curve curvature formula $\kappa=\dfrac{|x'y''-y'x''|}{(x'^2+y'^2)^{3/2}}$, we get $$x'y''-y'x''=\operatorname{sech}t\cdot\operatorname{sech}^2t+\tanh t\cdot\operatorname{sech}t\tanh t=\operatorname{sech}t(\operatorname{sech}^2t+\tanh^2t)=\operatorname{sech}t$$ (since $\operatorname{sech}^2t+\tanh^2t=1$), and $$x'^2+y'^2=\operatorname{sech}^2t+\tanh^2t=1.$$ Hence $$\kappa=\dfrac{\operatorname{sech}t}{1^{3/2}}=\operatorname{sech}t=\dfrac{1}{\cosh t}.$$

Solution. (1) For $y=ax^2$ $(a\neq0)$, $y'=2ax$ and $y''=2a$, so $$\kappa(x)=\dfrac{|2a|}{(1+4a^2x^2)^{3/2}}.$$ The denominator $(1+4a^2x^2)^{3/2}$ attains its minimum value $1$ at $x=0$, so $\kappa(x)$ attains its maximum $$\kappa(0)=2|a|$$ at $x=0$.

(2) Parametrize the ellipse as $x=a\cos t,\ y=b\sin t$. Then $$x'=-a\sin t,\ y'=b\cos t,\qquad x''=-a\cos t,\ y''=-b\sin t,$$ and $$x'y''-y'x''=ab\sin^2t+ab\cos^2t=ab,$$ so $$\kappa(t)=\dfrac{ab}{(a^2\sin^2t+b^2\cos^2t)^{3/2}}.$$ Since $a \gt b \gt 0$, $$a^2\sin^2t+b^2\cos^2t=b^2+(a^2-b^2)\sin^2t$$ attains its minimum $b^2$ when $\sin t=0$ (i.e. $t=0,\pi$, the points $(\pm a,0)$), and its maximum $a^2$ when $\sin^2t=1$ (i.e. $t=\pi/2,3\pi/2$, the points $(0,\pm b)$). Since $\kappa$ is largest where the denominator is smallest, $$\kappa_{\max}=\dfrac{ab}{b^3}=\dfrac{a}{b^2}\ \ (\text{at }(\pm a,0)),\qquad \kappa_{\min}=\dfrac{ab}{a^3}=\dfrac{b}{a^2}\ \ (\text{at }(0,\pm b)).$$

Solution. (1) Since $\mathbf{r}(t)=(3\sin t)\mathbf{i}+(3\cos t)\mathbf{j}+4t\mathbf{k}$, $$\mathbf{r}'(t)=(3\cos t)\mathbf{i}-(3\sin t)\mathbf{j}+4\mathbf{k},\qquad |\mathbf{r}'(t)|=\sqrt{9+16}=5$$ (constant speed). Hence $$\mathbf{T}(t)=\dfrac{3\cos t}{5}\mathbf{i}-\dfrac{3\sin t}{5}\mathbf{j}+\dfrac{4}{5}\mathbf{k}.$$ Differentiating $\mathbf{T}$, $$\mathbf{T}'(t)=-\dfrac{3\sin t}{5}\mathbf{i}-\dfrac{3\cos t}{5}\mathbf{j},\qquad |\mathbf{T}'(t)|=\dfrac35,$$ so $$\mathbf{N}(t)=\dfrac{\mathbf{T}'(t)}{|\mathbf{T}'(t)|}=(-\sin t)\mathbf{i}-(\cos t)\mathbf{j}.$$ The unit binormal vector is $\mathbf{B}=\mathbf{T}\times\mathbf{N}$, so $$\mathbf{B}(t)=\begin{vmatrix}\mathbf{i}&\mathbf{j}&\mathbf{k}\\ 3\cos t/5 & -3\sin t/5 & 4/5\\ -\sin t & -\cos t & 0\end{vmatrix}=\dfrac{4\cos t}{5}\mathbf{i}-\dfrac{4\sin t}{5}\mathbf{j}-\dfrac35\mathbf{k}.$$

(2) From $\mathbf{T}'(t)$ above, $$\kappa=\dfrac{|\mathbf{T}'(t)|}{|\mathbf{r}'(t)|}=\dfrac{3/5}{5}=\dfrac{3}{25},$$ so the radius of curvature is $\rho=1/\kappa=25/3$. At $t=0$, $\mathbf{r}(0)=(0,3,0)$ and $\mathbf{N}(0)=(0,-1,0)$, so the center of the osculating circle (center of curvature) is $$\mathbf{r}(0)+\rho\,\mathbf{N}(0)=(0,3,0)+\dfrac{25}{3}(0,-1,0)=\Bigl(0,-\dfrac{16}{3},0\Bigr).$$ The osculating circle is centered at $\bigl(0,-\dfrac{16}{3},0\bigr)$ with radius $25/3$, lying in the plane spanned by $\mathbf{T}(0)$ and $\mathbf{N}(0)$ (the plane with normal $\mathbf{B}(0)=(4/5,0,-3/5)$, i.e. $4x-3z=0$).

Solution. Since $\mathbf{r}(t)=(\cos 2t)\mathbf{i}+(\sin 2t)\mathbf{j}$, $$\mathbf{r}'(t)=(-2\sin 2t)\mathbf{i}+(2\cos 2t)\mathbf{j},\qquad |\mathbf{r}'(t)|=2$$ (constant speed). Hence $$\mathbf{T}(t)=(-\sin 2t)\mathbf{i}+(\cos 2t)\mathbf{j}.$$ Differentiating $\mathbf{T}$, $$\mathbf{T}'(t)=(-2\cos 2t)\mathbf{i}-(2\sin 2t)\mathbf{j},\qquad |\mathbf{T}'(t)|=2,$$ so $$\mathbf{N}(t)=\dfrac{\mathbf{T}'(t)}{|\mathbf{T}'(t)|}=(-\cos 2t)\mathbf{i}-(\sin 2t)\mathbf{j}.$$

Solution. Since $\mathbf{r}(t)=(\cos t,\sin t,-t)$, $$\mathbf{r}'(t)=(-\sin t,\cos t,-1),\qquad |\mathbf{r}'(t)|=\sqrt2$$ (constant speed). At $t=\pi/4$, $$\mathbf{r}(\pi/4)=\Bigl(\dfrac{\sqrt2}{2},\dfrac{\sqrt2}{2},-\dfrac{\pi}{4}\Bigr)=P,\qquad \mathbf{T}(\pi/4)=\dfrac{\mathbf{r}'(\pi/4)}{\sqrt2}=\Bigl(-\dfrac12,\dfrac12,-\dfrac{\sqrt2}{2}\Bigr).$$

Since the speed is constant, $\mathbf{T}'(t)=\mathbf{r}''(t)/\sqrt2$; from $\mathbf{r}''(t)=(-\cos t,-\sin t,0)$ we get $\mathbf{r}''(\pi/4)=\bigl(-\sqrt2/2,-\sqrt2/2,0\bigr)$, so $$\mathbf{T}'(\pi/4)=\Bigl(-\dfrac12,-\dfrac12,0\Bigr),\qquad |\mathbf{T}'(\pi/4)|=\dfrac{1}{\sqrt2}.$$ Hence $$\mathbf{N}(\pi/4)=\dfrac{\mathbf{T}'(\pi/4)}{|\mathbf{T}'(\pi/4)|}=\Bigl(-\dfrac{\sqrt2}{2},-\dfrac{\sqrt2}{2},0\Bigr),$$ and $$\mathbf{B}(\pi/4)=\mathbf{T}(\pi/4)\times\mathbf{N}(\pi/4)=\Bigl(-\dfrac12,\dfrac12,\dfrac{\sqrt2}{2}\Bigr).$$

The osculating plane (normal $\mathbf{B}$), normal plane (normal $\mathbf{T}$), and rectifying plane (normal $\mathbf{N}$) are the planes satisfying $\mathbf{B}\cdot(X-P)=0$, $\mathbf{T}\cdot(X-P)=0$, $\mathbf{N}\cdot(X-P)=0$ respectively, which simplify to:

• Osculating plane: $-\dfrac12\bigl(x-\dfrac{\sqrt2}{2}\bigr)+\dfrac12\bigl(y-\dfrac{\sqrt2}{2}\bigr)+\dfrac{\sqrt2}{2}\bigl(z+\dfrac{\pi}{4}\bigr)=0$, i.e. $$x-y-\sqrt2\,z=\dfrac{\sqrt2\,\pi}{4}$$
• Normal plane: $-\dfrac12\bigl(x-\dfrac{\sqrt2}{2}\bigr)+\dfrac12\bigl(y-\dfrac{\sqrt2}{2}\bigr)-\dfrac{\sqrt2}{2}\bigl(z+\dfrac{\pi}{4}\bigr)=0$, i.e. $$x-y+\sqrt2\,z=-\dfrac{\sqrt2\,\pi}{4}$$
• Rectifying plane: $-\dfrac{\sqrt2}{2}\bigl(x-\dfrac{\sqrt2}{2}\bigr)-\dfrac{\sqrt2}{2}\bigl(y-\dfrac{\sqrt2}{2}\bigr)=0$, i.e. $$x+y=\sqrt2$$

Solution. Since $\mathbf{r}(t)=(\cos t+t\sin t)\mathbf{i}+(\sin t-t\cos t)\mathbf{j}$ $(t \gt 0)$, the product rule gives $-\sin t+\sin t+t\cos t=t\cos t$ and $\cos t-\cos t+t\sin t=t\sin t$, so $$\mathbf{r}'(t)=(t\cos t)\mathbf{i}+(t\sin t)\mathbf{j},$$ and $$|\mathbf{r}'(t)|=t\sqrt{\cos^2t+\sin^2t}=t\quad(t \gt 0).$$

The tangential component of acceleration is $$a_T=\dfrac{d}{dt}|\mathbf{r}'(t)|=\dfrac{d}{dt}(t)=1.$$ On the other hand, $$\mathbf{r}''(t)=(\cos t-t\sin t)\mathbf{i}+(\sin t+t\cos t)\mathbf{j},$$ so $$|\mathbf{r}''(t)|^2=(\cos t-t\sin t)^2+(\sin t+t\cos t)^2=\cos^2t+\sin^2t+t^2(\sin^2t+\cos^2t)=1+t^2.$$ Hence $$a_N=\sqrt{|\mathbf{r}''(t)|^2-a_T^2}=\sqrt{(1+t^2)-1}=\sqrt{t^2}=t\quad(t \gt 0).$$

Solution. Since $\mathbf{r}(t)=(e^t\cos t)\mathbf{i}+(e^t\sin t)\mathbf{j}+\sqrt2\,e^t\mathbf{k}$, $$\mathbf{r}'(t)=e^t(\cos t-\sin t)\mathbf{i}+e^t(\sin t+\cos t)\mathbf{j}+\sqrt2\,e^t\mathbf{k},$$ and since $$(\cos t-\sin t)^2+(\sin t+\cos t)^2+2=(1-\sin 2t)+(1+\sin 2t)+2=4,$$ we get $$|\mathbf{r}'(t)|=e^t\sqrt4=2e^t.$$ Hence $$a_T=\dfrac{d}{dt}(2e^t)=2e^t.$$

Differentiating $\mathbf{r}'(t)$ again, $$\mathbf{r}''(t)=e^t(-2\sin t)\mathbf{i}+e^t(2\cos t)\mathbf{j}+\sqrt2\,e^t\mathbf{k},$$ so $$|\mathbf{r}''(t)|^2=e^{2t}(4\sin^2t+4\cos^2t+2)=6e^{2t}.$$ Hence $$a_N=\sqrt{|\mathbf{r}''(t)|^2-a_T^2}=\sqrt{6e^{2t}-4e^{2t}}=\sqrt{2e^{2t}}=\sqrt2\,e^t.$$

Proof. Smoothness of the curve means $\mathbf{r}'(t)\neq\mathbf{0}$, so for $\mathbf{v}(t)=\mathbf{r}'(t)$ we have $|\mathbf{v}(t)|\neq0$. The acceleration decomposes as $$\mathbf{a}(t)=a_T\mathbf{T}(t)+a_N\mathbf{N}(t),\qquad a_N=|\mathbf{v}(t)|\,|\mathbf{T}'(t)|.$$ By assumption $a_N=0$ and $|\mathbf{v}(t)|\neq0$, so $|\mathbf{T}'(t)|=0$, i.e. $\mathbf{T}'(t)=\mathbf{0}$. Hence $\mathbf{T}(t)\equiv\mathbf{T}_0$ is a constant vector.

Since $\mathbf{v}(t)=|\mathbf{v}(t)|\mathbf{T}(t)=|\mathbf{v}(t)|\mathbf{T}_0$, for any $t_0$ we have $$\mathbf{r}(t)=\mathbf{r}(t_0)+\int_{t_0}^t\mathbf{v}(u)\,du=\mathbf{r}(t_0)+\Bigl(\int_{t_0}^t|\mathbf{v}(u)|\,du\Bigr)\mathbf{T}_0=\mathbf{r}(t_0)+s(t)\,\mathbf{T}_0,$$ where $s(t)=\int_{t_0}^t|\mathbf{v}(u)|\,du$ is a scalar function. Thus every point of the curve lies on the line through $\mathbf{r}(t_0)$ with direction $\mathbf{T}_0$, so $\mathbf{r}(t)$ is part of a line.

Solution. Since $\mathbf{r}(t)=(a\cos t,a\sin t,bt)$, $$\mathbf{r}'(t)=(-a\sin t,a\cos t,b),\quad \mathbf{r}''(t)=(-a\cos t,-a\sin t,0),\quad \mathbf{r}'''(t)=(a\sin t,-a\cos t,0).$$ From Exercise 03-1(1), $$\mathbf{r}'\times\mathbf{r}''=(ab\sin t,\,-ab\cos t,\,a^2),\qquad |\mathbf{r}'\times\mathbf{r}''|^2=a^2(a^2+b^2).$$ Now $$(\mathbf{r}'\times\mathbf{r}'')\cdot\mathbf{r}'''=ab\sin t\cdot a\sin t+(-ab\cos t)(-a\cos t)+a^2\cdot0=a^2b\sin^2t+a^2b\cos^2t=a^2b,$$ so, substituting into the torsion formula $\tau=\dfrac{(\mathbf{r}'\times\mathbf{r}'')\cdot\mathbf{r}'''}{|\mathbf{r}'\times\mathbf{r}''|^2}$, $$\tau=\dfrac{a^2b}{a^2(a^2+b^2)}=\dfrac{b}{a^2+b^2}.$$

Solution. The curve $\mathbf{r}(t)=(\cosh t)\mathbf{i}-(\sinh t)\mathbf{j}+t\mathbf{k}$ is the same as in Exercise 03-1(2), so we reuse that computation. $$\mathbf{r}'(t)=(\sinh t,-\cosh t,1),\quad \mathbf{r}''(t)=(\cosh t,-\sinh t,0),\quad \mathbf{r}'''(t)=(\sinh t,-\cosh t,0),$$ and from Exercise 03-1(2), $$\mathbf{r}'\times\mathbf{r}''=(\sinh t,\cosh t,1),\qquad |\mathbf{r}'\times\mathbf{r}''|=\sqrt2\cosh t.$$ Hence the unit binormal vector is $$\mathbf{B}(t)=\dfrac{\mathbf{r}'\times\mathbf{r}''}{|\mathbf{r}'\times\mathbf{r}''|}=\dfrac{(\sinh t,\cosh t,1)}{\sqrt2\cosh t}.$$

To find the torsion, compute $$(\mathbf{r}'\times\mathbf{r}'')\cdot\mathbf{r}'''=\sinh t\cdot\sinh t+\cosh t\cdot(-\cosh t)+1\cdot0=\sinh^2t-\cosh^2t=-1,$$ and since $|\mathbf{r}'\times\mathbf{r}''|^2=2\cosh^2t$, $$\tau=\dfrac{-1}{2\cosh^2t}.$$ (Note that $\kappa=|\mathbf{r}'\times\mathbf{r}''|/|\mathbf{r}'|^3=1/(2\cosh^2t)$, the same as in Exercise 03-1(2), and $|\tau|=\kappa$.)

Proof. ($\Rightarrow$) Suppose $\mathbf{r}(t)$ lies in a plane, i.e. there is a fixed unit normal $\mathbf{n}$ and a point $\mathbf{r}(t_0)$ with $(\mathbf{r}(t)-\mathbf{r}(t_0))\cdot\mathbf{n}=0$ for all $t$. Differentiating successively, $$\mathbf{r}'(t)\cdot\mathbf{n}=0,\quad \mathbf{r}''(t)\cdot\mathbf{n}=0,\quad \mathbf{r}'''(t)\cdot\mathbf{n}=0,$$ so $\mathbf{r}'(t),\mathbf{r}''(t),\mathbf{r}'''(t)$ all lie in the (two-dimensional) plane orthogonal to $\mathbf{n}$. Three vectors in a two-dimensional space are always linearly dependent, so the scalar triple product vanishes: $$(\mathbf{r}'\times\mathbf{r}'')\cdot\mathbf{r}'''=0.$$ Since $\kappa \gt 0$ gives $|\mathbf{r}'\times\mathbf{r}''|\neq0$, $$\tau=\dfrac{(\mathbf{r}'\times\mathbf{r}'')\cdot\mathbf{r}'''}{|\mathbf{r}'\times\mathbf{r}''|^2}=0.$$

($\Leftarrow$) Conversely, suppose $\tau=0$. By the Frenet–Serret formula $\mathbf{B}'(s)=-\tau(s)\mathbf{N}(s)$ (with $s$ the arc length parameter), $\tau\equiv0$ gives $\mathbf{B}'(s)=\mathbf{0}$, so $\mathbf{B}(s)\equiv\mathbf{B}_0$ is a constant vector. For any $s_0$, define $$f(s)=(\mathbf{r}(s)-\mathbf{r}(s_0))\cdot\mathbf{B}_0.$$ Then $$f'(s)=\mathbf{r}'(s)\cdot\mathbf{B}_0=\mathbf{T}(s)\cdot\mathbf{B}_0=0$$ since $\mathbf{T}\perp\mathbf{B}$. Hence $f(s)$ is constant, and since $f(s_0)=0$ we get $f(s)\equiv0$, i.e. $$(\mathbf{r}(s)-\mathbf{r}(s_0))\cdot\mathbf{B}_0=0$$ for all $s$. This means $\mathbf{r}(s)$ lies in the plane through $\mathbf{r}(s_0)$ with normal $\mathbf{B}_0$.

Therefore the torsion of $\mathbf{r}(t)$ is $0$ if and only if $\mathbf{r}(t)$ lies in a single plane.

Solution. In polar coordinates, the velocity and acceleration are given by $$\mathbf{v}=\dot r\,\mathbf{u}_r+r\dot\theta\,\mathbf{u}_\theta,\qquad \mathbf{a}=(\ddot r-r\dot\theta^2)\mathbf{u}_r+(r\ddot\theta+2\dot r\dot\theta)\mathbf{u}_\theta.$$

(1) Since $r=\theta$ and $\dot\theta=2$ (constant), $\dot r=\dot\theta=2$ and $\ddot r=\ddot\theta=0$. Hence $$\mathbf{v}=2\,\mathbf{u}_r+r\cdot2\,\mathbf{u}_\theta=2\,\mathbf{u}_r+2\theta\,\mathbf{u}_\theta,$$ and $$\mathbf{a}=(0-r\cdot4)\mathbf{u}_r+(0+2\cdot2\cdot2)\mathbf{u}_\theta=-4\theta\,\mathbf{u}_r+8\,\mathbf{u}_\theta.$$

(2) Since $r=a(1+\sin t)$ and $\theta=1-e^{-t}$, $$\dot r=a\cos t,\quad \ddot r=-a\sin t,\qquad \dot\theta=e^{-t},\quad \ddot\theta=-e^{-t}.$$ Hence $$\mathbf{v}=a\cos t\,\mathbf{u}_r+a(1+\sin t)e^{-t}\,\mathbf{u}_\theta,$$ and $$a_r=\ddot r-r\dot\theta^2=-a\sin t-a(1+\sin t)e^{-2t}=-a\bigl[\sin t+(1+\sin t)e^{-2t}\bigr],$$ $$a_\theta=r\ddot\theta+2\dot r\dot\theta=-a(1+\sin t)e^{-t}+2a\cos t\,e^{-t}=a(2\cos t-\sin t-1)e^{-t},$$ so $$\mathbf{a}=-a\bigl[\sin t+(1+\sin t)e^{-2t}\bigr]\mathbf{u}_r+a(2\cos t-\sin t-1)e^{-t}\,\mathbf{u}_\theta.$$

Solution. In cylindrical coordinates, the helix $\mathbf{r}(t)=(a\cos t,a\sin t,bt)$ has constant cylindrical radius $\rho=a$, with $\theta=t$ and $z=bt$. Hence $$\dot\rho=0,\ \ddot\rho=0,\qquad \dot\theta=1,\ \ddot\theta=0,\qquad \dot z=b,\ \ddot z=0.$$

The velocity is $$\mathbf{v}=\dot\rho\,\mathbf{u}_r+\rho\dot\theta\,\mathbf{u}_\theta+\dot z\,\mathbf{k}=0\cdot\mathbf{u}_r+a\cdot1\cdot\mathbf{u}_\theta+b\,\mathbf{k}=a\,\mathbf{u}_\theta+b\,\mathbf{k},$$ and the acceleration is $$\mathbf{a}=(\ddot\rho-\rho\dot\theta^2)\mathbf{u}_r+(\rho\ddot\theta+2\dot\rho\dot\theta)\mathbf{u}_\theta+\ddot z\,\mathbf{k}=(0-a\cdot1)\mathbf{u}_r+0\cdot\mathbf{u}_\theta+0\cdot\mathbf{k}=-a\,\mathbf{u}_r.$$