Question

Reparametrize the curve r(t)=e^t i + e^t*sin(t) j +e^t*cos(t) k with respect to the arc length measured from the point (1,0,1) in the direction of increasing t and then show that r'(s)=T(s)

Answer 1

r(t) = <x(t), y(t), z(t)> = <e^t, e^t∙sin(t), e^t∙cos(t)>

r'(t) = <x'(t), y'(t), z'(t)> = <e^t, e^t∙[sin(t) + cos(t)], e^t∙[cos(t) - sin(t)]>

|r'(t)| = e^t √[1 + (sin(t) + cos(t))² + (cos(t) - sin(t))²]
...... = e^t √[1 + sin²(t) + 2sin(t)cos(t) + cos²(t) + cos²(t) - 2sin(t)cos(t) + sin²(t)]
...... = e^t √[1 + 2sin²(t) + 2cos²(t)]
...... = e^t √[1 + 2]
...... = e^t √3

L = ∫ |r'(t)| dt[a,b]
L = √3 ∫ e^t dt[1,t]
L = √3 [e^t - e]

Solve for t:
L/√3 = e^t - e
e^t = (L/√3) + e
t = ln[(L/√3) + e]

Rewrite the new vector:
r(s) = <x(s), y(s), z(s)> = <[s/√3 + e], [s/√3 + e]∙sin(ln[s/√3 + e]), [s/√3 + e]∙cos(ln[s/√3 + e])>

Answer 2

thanks.

Answer 3

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