math192_hw3sp06 - HW3 Solutions 13.3 Arc Length and the...

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HW3 Solutions13.3 Arc Length and the Unit Tangent Vector T8. T=v|v|= (cost)i-(sint)jand Length =R22tdt=ht22i2210.Let P(t0) denote the point. Thenv= (12 cost)i+ (12 sint)j+ 5kand-13π=Rt00p144 cos2t+ 144 sin2t+ 25t dt=Rt0013dt= 13t0t0=-π, and the point isP(-π) = (12 sin(-π),-12 cos(-π),-5π) = (0,12,-5π)14.r(1 + 2t)i+ (1 + 3t)j+ (6-6t)kv= 2i+ 3j-6k⇒ |v|=p22+ 33+ (-6)2= 7s(t) =R607= 7tLength =s(0)-s(-1) = 0-(-7) = 714.1 Functions of Several Variables8.(a)Domain: all (x, y) satisfyingx2+y29(b)Range: 0z3(c)level curves are circles centered at the origin with radiir3(d)boundary is the cirlex2+y2= 9(e)closed(f)bounded13.f14.e15.a16.c17.d18.b
20.-6-4-20246-505-40-30-20-10010xf(x,y) = 4-y2yFigure 1: (a)xyz = 4-y2-6-4-20246-6-4-20246Figure 2: (b)42.f(x, y, z) = ln(x2+y+z2) at (-1,2,1)w= ln(x2+y+z2); at (-1,2,1)w= ln(1+2+1) =ln 4ln 4 = ln(x2+y+z2)4 =x2+y+z246.Minimum offalong the line isf(1,0,9) = (1)(0)
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