# Investigation Consider the helix represented by the vector-valued function r(t)= < 2 cos t, 2 sin t, t > (a) Write the length of the arc son the helix as a function of t by evaluating the integral s= int_{0}^{t} sqrt{[x'(u)]^{2} + [y'(u)]^{2} + [z'(u)]^{2} du}

Question
Investigation Consider the helix represented by the vector-valued function $$r(t)=\ <\ 2\ \cos\ t,\ 2\ \sin\ t,\ t\ >$$ (a) Write the length of the arc son the helix as a function of t by evaluating the integral $$s=\ \int_{0}^{t}\ \sqrt{[x'(u)]^{2}\ +\ [y'(u)]^{2}\ +\ [z'(u)]^{2}\ du}$$

2021-01-09
To calculate: The length of the arc s on the helix as a function of t The length of the curve is $$\underline{s=\ \sqrt{5t}}.$$ Used formula: $$s=\ \int_{0}^{t}\ \sqrt{[x'(t)]^{2}\ +\ [y'(t)]^{2}\ +\ [z'(t)]^{2}\ dt}$$ Calculation: The helix path is, $$r(t)=\ <\ 2\ \cos\ t,\ 2\ \sin\ t,\ t\ >$$ On differentiating this vector value function, $$r'(t)=\ <\ -2\ \sin\ t,\ 2\ \cos\ t,\ 1\ >$$ Calculate the length of the line segment for the given interval as $$s=\ \int_{0}^{t}\ \sqrt{[x'(t)]^{2}\ +\ [y'(t)]^{2}\ +\ [z'(t)]^{2}\ dt}$$
$$=\ \int_{0}^{t}\ \sqrt{(-2\ \sin\ t)^{2}\ +\ (2\ \cos\ t)^{2}\ +\ (1)^{2}\ dt}$$
$$=\ \int_{0}^{t}\ \sqrt{5dt}$$
$$= \sqrt{5t}$$ Thus, the arc length is $$s = \sqrt{5t}.$$

### Relevant Questions

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