ANSWER The arc length is equal to k2∣k∣(ekb−1)
EXPLANATION. The legs LC of the arc of the curve specified parametrically C={(x(t),y(t)):a≤t≤b}
calculated using the formula LC=∫ab(x′(t))2+(y′(t))2dt . In the task , the point (1,0) corresponds to the
value of the parameter t=0. Hence , a=0.
x(t)=ektcost,x′(t)=kektcost−ektsint,y(t)=ektsint,y′(t)=kektsint+ektcost
(x′(t))2+(y′(t))2=(kektcost−ektsint)2+(kektsint+ektcost)2 =
=k2e2kt(2cos2t−2costsint+2costsint+2sin2t)= =2k2e2kt .
LC=∫0b2k2e2ktdt=2∣k∣∫0bektdt=k2∣k∣(ekb−1) .
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