Question

+ Drag and drop your files or click to browse. Q4 (10 points) You are at a circular lake with radius 30 km, centered at the point(0,0). You want to travel from (30,0) to (-30,0). You will row from (30, 0) in a straight line to a point on the shore that is angle 8 radians around the lake from where you started, and then you will walk the rest of the way along the circular shore to (-30,0). Your rowing speed through the lake is 6 km/h, and your walking speed along the shore is 10 km/hr. Let T be the total time of the trip. Find the absolute minimum of the time T over domain 0 € 0, and explain how you know + Drag and drop your files or click to browse... Q5 (10 points) Time

Answer 1

Given; (30,0) is (60) IS A Rowing speed through the lake is 6km/ and walking 1-3004 speed along the shore is 10km/h T is the total time We have to find absolute minimum of the time 'T By the LC-90 properties of circe and triangle so; os (0) = 6/= di 30+30 or di = 60 cos(0) - (i) since, CCOB= 20 Now, 20= d2 - 30 <COB = LOAC + LALO - oto = 20 dz = 600 (11) since the - ut ti to distance speed is the time when walking along the shore. is the time when go through boat/ lake 10 similarly; ta = di = 60 cos(0) = 10 cos(0) let total tine-t So, t = t tt2 = 60+lows (0) f(0) = 6(0) + 10 wsco) to get absolute minimum differentiating with respect to d(+6) d160 +10 ws)) time 'T) o and make t(o)=0

since, a Coso) - Sino) t'(0)= 6 + Locsin (0)) 0 = 6 - 10 Sin (C) sin (O) = 4 = 's ori sin(o) = sin ( 36-8698°) o na t toº 36.869 0 = 1180° +61936.86° o= 180°-36.860 = 143. 14° or 2.49826 radian x) not possible no 36.8698 o or 0.6 4349 radian -

Since; to - 60 + 10 WS (0) o 20 radian t(e) - 6x0 +10 WS (0) = 0+10 = 10 hr o = 0.64 349 radian +(0.64349) = 6 x 0.64349 + 10 GS (0:64349) = 3.86 + 10X0.8 = 11.86 h. o = 1.57 radian +(1.57) = 64157 +10 Cs (157) = 9.42 to = 19.42 by] (so; at 0 = 1 radin Olandest = 1.57 radian i.e. along the shore t is minimum on T= 19.42 hr This is the required answer