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202e1k04(Calc3)

# 202e1k04(Calc3) - Solutions for Vector Calculus First...

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Unformatted text preview: Solutions for Vector Calculus First Midterm 1 Calculate the length of the curve c : R ——> R3 deﬁned by the function C(t) = (t sint + cost, tcost — sin t, 2) as it varies between t = 0 and t = 2. solution: The velocity vector is c’(t) = (sint + tcost —— sint, cost — tsint — cost, 0) = t (cost, —— sint7 0), which has magnitude lc’(t)l = [t2 (00827: + sin2 t + 02]”2 = |t| (also known as the speed). The length of the arc from C(O) to C(2) is thus f2tdt—1t2‘=2—1[4 01—2 0 '_2 t=0_2 — ' 2 The curves a(t) = (t,t2,t3) and b(u) = (sin 2u,ucosu,u) interesect at the point (0,0,0). Find the cosine of the angle between the tangent lines to these curves at this intersection point. solution: The tangent vector to a(t) at t = O is given by the value a’(0) of the velocity vector there, and Similarly for the tangent vector of b(u) at u = 0; thus to solve the problem, all we need to do is calculate the ratio M = GOSH la’(0)l - lb’(0)| ’ where 0 is the angle between the vectors a’ (0) and b’ (0). But a’ (t) = (1, 2t, 3752) s—> (1,0,0) at t = 0, and b’(u) = (2 cosZu,cosu —usinu, 1) l—-> (2,1,1) at u = 0, so _ (1,0,0)-(2,1,1) _ 2 _ 1 COSQ‘ [(22 + 12 + 12) - (12 +02 +02)]1/2 — 76 ‘ §\/6. 3 Calculate the derivative of the function C(75) = WW): 2105), 205)) deﬁned by the path it n—> (m,y,z) = (sint, cost, 6‘) and the function 9(1', y, Z) =10g(962 + 212 + 22) ﬁrst directly, and then by using the chain rule. solution Substituting in the equations for the path, we get G“) = lowing t + +(et)2) = M1 + at) , and hence G’(t) —1o ’(1—l—e2t) (1 +e2t)’ — 262i _- g ‘ 1 + e” . On the other hand, the gradient of g is 2 Vg(\$,yvz) = m2+y2+22 (wnyZ) while (56’ (t),y’ (15)7 z’(t)) = (cost, — sin 15, ct). But the chain rule says 2 . . G’(t) = m (smt, cost, at) - (cost, —smt, ct) which equals 2 [sint cost + cos t( sin 15) + et at] — 26m 1 + e2t _ 1 + 621 7 in agreement with the previous calculation. 4 Find the points on the surface S = {(x,y72) I z = ﬁxay) = 337?; —\$3 —y3} where the tangent plane is parallel to the plane 3x+3y+z=1. [Hintz you might try a: = :121 . . .] solution S is the level surface 0=F(m,y,z)=z—f(\$,y) :Z-3\$y—\$3—y3 so the tangent plane at (20,11, 2) is perpendicular to the gradient VF(‘T7y7Z) = (—3y+3x2,—3x+ 3y271) F. To ﬁnd ponts where the tangent plane is parallel to ame as ﬁnding points where the gradient (3,3,1) of that surface; thus we need to of the deﬁning function the plane 39: + 3y + z = 1 is thus the 5 vector is parallel to the normal vector solve the equations ——3y+3x2=3, —3x+3y2=3. This is now a job for high—school algebra: we can rewrite the ﬁrst of these equations as y = 1'2 —— 1, and the second as y2 — a: = 1. Substituting the ﬁrst into the second gives (x2—1)2—x=x4—2x2+1—x=1, which simpliﬁes to m4—2x2—m=x(a:3—2x—1)=m(:z:+1)(a:2—x—1)=0. Thus a: = 0, a: = —17 and cc 2 %(1 :l: \/5) are solutions; the corresponding values of y are ~1, 0, and [combining the equations y = m2 — 1 with x2 = x + 1] 1 y = in i «5) . The corresponding values of z are 1, 1, and zz3xy—x3 —y3 =3x(a:2 —1)—av3 — (x2—1)(a:+1) (using the equations y = \$2 — 1 and y2 = a: + 1), which simpliﬁes to 1 2:1:3—x2—2x-1—1=\$(x+1)—-x2—2x+1=1—m=5(1q:\/5). ...
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