Investigation Consider the helix represented by the vector-valued functionr(t)= &lt; 2 cos⁡ t, 2 sin⁡ t, t &gt;(a) Write the length of the arc son the helix as a function of t by evaluating the integrals= ∫0t [x′(u)]2 + [y′(u)]2 + [z′(u)]2 du

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Investigation Consider the helix represented by the vector-valued functionr(t)= &amp;lt; 2 cos⁡ t, 2 sin⁡ t, t &amp;gt;(a) Write the length of the arc son the helix as a function of t by evaluating the integrals= ∫0t [x′(u)]2 + [y′(u)]2 + [z′(u)]2 du

crocolylec · Accepted Answer

To calculate: The length of the arc s on the helix as a function of tThe length of the curve is s= 5t―.Used formula: s= ∫0t [x′(t)]2 + [y′(t)]2 + [z′(t)]2 dtCalculation:The helix path is,r(t)= &amp;lt; 2 cos⁡ t, 2 sin⁡ t, t &amp;gt;On differentiating this vector value function,r′(t)= &amp;lt; −2 sin⁡ t, 2 cos⁡ t, 1 &amp;gt;Calculate the length of the line segment for the given interval ass= ∫0t [x′(t)]2 + [y′(t)]2 + [z′(t)]2 dt= ∫0t (−2 sin⁡ t)2 + (2 cos⁡ t)2 + (1)2 dt= ∫0t 5dt=5tThus, the arc length is s=5t.

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}

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