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}

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 \(s=\ \int_{0}^{t}\ \sqrt{[x'(u)]^{2}\ +\ [y'(u)]^{2}\ +\ [z'(u)]^{2}\ du}\)

Answers (1)

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 \(s = \sqrt{5t}.\)
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