Consider the helix represented by the vector-valued function r(t)=⟨2cos⁡t,2sin⁡t,t⟩.(a) Write the length of the arc s on the helix as a function of t by evaluating the integral(b) Solve for tin the relationship derived in part (a),and substitute the result into the original vector-valued function.This yields a parametrization of the curve interms of the arc length parameter s.(c) Find the coordinates of the point on the helix for arc lengths s=5 and s=4.(d) Verify that r′(s)=1.s=∫0t[x′(u)]2+[y′(u)]2+[z′(u)]2du.

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Anthony Caraballo · Accepted Answer

Step 1We have given that;A helix represented by the vector-valued function r(t)=(2cos⁡t,2sin⁡t,t)Step 2(a) From the given position vector, we have parametric equations;x(t)=2cos⁡y(t)=2sin⁡z(t)=twe differentiate these equations and get;x′(t)=−2sin⁡y′(t)=2cos⁡z′(t)=1We have given integrel;=∫0tx′(u)2+y′(u)2+z′(u)2du.=∫0t(−2sin⁡u)2+(2cos⁡u)2+12du.=∫0t4sin2u+4cos2u+1du=∫0t4+1du=∫0t5du=5tStep 3(b) From the part (a) we have;s=5tt=s5On plugging this value in the parametric equations,we get;x(s5)=2cos⁡s5y(s5)=2sin⁡s5z(s5)=s5So the function of given curve in the form of s will be;r(s)=⟨2cos⁡s5,2sin⁡s5,s5⟩Step 4{c) We have given the curve;r(s)=⟨2cos⁡s5,2sin⁡s5,s5⟩On plugging s=5 in the equation, we get;r(5)=⟨2cos⁡55,2sin⁡55,55⟩

Consider the helix represented by the vector-valued function \r(t)

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