Question

Let

\(\mathbf{r}(t)=\left\langle R \cos \left(\frac{2 \pi N t}{h}\right), R \sin \left(\frac{2 \pi N t}{h}\right), t\right\rangle, \quad 0 \leq t \leq h\)

(a) Show that \(\mathbf{r}(t)\) parametrizes a helix of radius \(R\) and height \(h\) making \(N\) complete turns.

(b) Guess which of the two springs in Figure 5 uses more wire.

(c) Compute the lengths of the two springs and compare.

Answer 1

(a) In general the equation of a helix is \(x=R \cos t\)

\(y=R \sin t\)

\(z=c t\)

The vertical separation is the value when \(t=2 \pi .\) Then \(2 \pi=\frac{h}{N} \Rightarrow \frac{2 \pi N}{h}=1\)

Thus substituting \(\frac{2 \pi N}{h}\) in place of 1 in the above equations and taking \(c=1,\) we get \(x(t)=R \cos \left(\frac{2 \pi N t}{h}\right)\)

\(y(t)=R \sin \left(\frac{2 \pi N t}{h}\right)\)

\(z(t)=t\)

(b) I think the spring in \((A)\) of 3 turns and radius \(7 \mathrm{~cm}\) uses more wire.

\((\mathrm{c})\)

Now we will check which spring uses more wire i.e., which spring has more arc length. Arc length \(S\) \(=\int_{0}^{h} \sqrt{\left(x^{\prime}(t)\right)^{2}+\left(y^{\prime}(t)\right)^{2}+\left(z^{\prime}(t)\right)^{2}} d t\)

\(=\int_{0}^{h} \sqrt{\left(\frac{2 \pi N}{h}\right)^{2} R^{2}+1} d t\)

\(=h \sqrt{\left(\frac{2 \pi N}{h}\right)^{2} R^{2}+1}\)

\(=\sqrt{(2 \pi N R)^{2}+h^{2}}\) unit

Using this formula the arc length of the spring in figure \((A)\)

\(R=7 \mathrm{~cm}, \quad N=3, h=4 \mathrm{~cm}\)

Thus, \(S=\sqrt{(42 \pi)^{2}+16} \mathrm{~cm} \approx 132 \mathrm{~cm}\)

Using this formula the arc length of the spring in figure \((B)\)

Here \(R=4 \mathrm{~cm}, N=5, h=3 \mathrm{~cm}\)

Thus, \(S=\sqrt{(40 \pi)^{2}+9} \approx 125.7\)