Question at bottom of second picture Sketch the plane curve.r(t) = Pi + j, [0, 1]y.1.01.00.80.80.60.60.4F0.40.20.20.20.40.60.81.00.20.40.60.81.0y1.01.00.80.8 0.20.20.20.40.60.81.00.20.40.60.81.01.0-1.00.80.80.6-0.60.40.40.20.20.20.40.60.81.00.20.40.60.81.0Find its length over the given interval. (Round your answer to four decimal places.)

Name: Question at bottom of second picture Sketch the plane curve.r(t) = Pi
Uploaded: 2024-03-20T06:38:44.000Z
Channel: Bartleby
Description: Question at bottom of second picture Sketch the plane curve.r(t) = Pi + j, [0, 1]y.1.01.00.80.80.60.60.4F0.40.20.20.20.40.60.81.00.20.40.60.81.0y1.01.00.80.8 0.20.20.20.40.60.81.00.20.40.60.81.01.0-1.00.80.80.6-0.60.40.40.20.20.20.40.60.81.00.20.40.6

Length of a plane curve: The length of a plane curve or the arc length of a curve r(t)=f(t) i^ +g(t) j^ in the interval [a, b] is given by, L=∫ab(f'(t) )2+(g'(t))2dt Given, r(t)=t3 i^ + t2 j^ in the interval [0,1]Here f(t)=t3 and g(t)=t2 , a=0 and b=1 Therefore, f'(t)=3t2 and g'(t)=2t Substituting in the equation for arc length of the curve, L=∫ab[f'(t)]2+[g'(t)]2dt=∫01[3 t2]2+[2 t]2dt=∫019 t4+4 t2dt=∫01t2(9 t2+4)dt=∫01t(9 t2+4)dt Taking , 9 t2+4=u⇒18 t dt=du⇒t dt =118dut=0 ⇒u=4t=1⇒u=13 Therefore, substituting these values in L we have L=∫413118udu=118u12+112+1413=118u3232413=118×23133/2-43/2=127[46.8721668-8]=127×38.8721668=1.4397 The answer is 1.4397