Space curves in Vector Analysis and proofs

Question

given the space curve $x=a\cos t$, $y=a\sin t$, $z=bt$ show that $k=\dfrac{a}{(a^2+b^2)}$

My solution :

Assuming $a,b$ and $c$ are positive: $k(a^2+b^2)=a$

$0\leq t\le2\pi$

$X^2+y^2=a^2$

Midpoints: $(x_0,y_0)$ With $x=x_0+a\cos t$ and $Y=y_0+b\sin t$

I haven't managed to get past here. Please assist

Answer 1

Assuming you meant $\;r(t)=(a\cos t,\,a\sin t,\,bt)\;,\;\;0\le t\le 2\pi\;,\;\;a,b>0\;$, we have:

$$r'(t)=(-a\sin t,\,a\cos t,\,b)\;,\;\;r''(t)=(-a\cos t,\,-a\sin t,\,0)\implies$$

$$r'\times r''=\begin{vmatrix}i&j&k\\-a\sin t&a\cos t&b\\-a\cos t&-a\sin t&0\end{vmatrix}=\left(ab\sin t,\,-ab\cos t,\,a^2\right)\implies$$

$$\kappa=\frac{\left\|r'\times r''\right\|}{\left\|r'\right\|^3}=\frac{\sqrt{a^2b^2+a^4}}{(a^2+b^2)^{3/2}}=\frac a{a^2+b^2}$$

If you do re-parametrization by arc length:

$$s=\int_0^t\left\|r'(x)\right\|dx=\int_0^t\sqrt{a^2+b^2}dx=\sqrt{a^2+b^2}t\implies t=\frac s{\sqrt{a^2+b^2}}$$

so our curve is now

$$r(s)=\left(a\cos\frac s{\sqrt{a^2+b^2}}\,,\,\,a\sin \frac s{\sqrt{a^2+b^2}}\,,\,\,\frac{bs}{\sqrt{a^2+b^2}}\right)\implies$$

$$T(s)=r'(x)=\left(-\frac a{\sqrt{a^2+b^2}}\sin\frac s{\sqrt{a^2+b^2}}\,,\,\,\frac a{\sqrt{a^2+b^2}}\cos\frac s{\sqrt{a^2+b^2}}\,,\,\,\frac b{\sqrt{a^2+b^2}}\right)\implies$$

$$T'(s):=\frac{dT(s)}{ds}=\left(-\frac a{a^2+b^2}\cos\frac s{\sqrt{a^2+b^2}}\,,\,\,-\frac a{a^2+b^2}\sin\frac s{\sqrt{a^2+b^2}}\,,\,\,0\right)\implies$$

$$\left\|T'(s)\right\|=\frac a{a^2+b^2}$$

exactly as before, of course.

Answer 2

\begin{align*} \boldsymbol{r} &= (a\cos t,a \sin t, bt) \\ \dot{\boldsymbol{r}} &= (-a\sin t,a \cos t, b) \\ |\dot{\boldsymbol{r}}| &= \sqrt{a^2+b^2} \\ \ddot{\boldsymbol{r}} &= (-a\cos t,-a\sin t,0) \\ \kappa &= \frac{|\dot{\boldsymbol{r}} \times \ddot{\boldsymbol{r}}|} {|\dot{\boldsymbol{r}}|^3} \\ &= \frac{|(ab\sin t,-ab\cos t,a^2)|} {\sqrt{(a^2+b^2)^{3}}} \\ &= \frac{a\sqrt{a^2+b^2}}{\sqrt{(a^2+b^2)^{3}}} \\ &= \frac{a}{a^2+b^2} \\ \dddot{\boldsymbol{r}} &= (a\sin t,-a\cos t,0) \\ \tau &= \frac{\dot{\boldsymbol{r}} \times \ddot{\boldsymbol{r}} \cdot \dddot{\boldsymbol{r}}} {|\dot{\boldsymbol{r}} \times \ddot{\boldsymbol{r}}|^2} \\ &= \frac{a^2b}{a^2(a^2+b^2)} \\ &= \frac{b}{a^2+b^2} \end{align*}