Tuesday, January 28 ∗ Solutions ∗ Parametric Curves Defined Using Vector Arithmetic1. (a) Plot of f (x) =x2+x −2<1,5>-4-3-2-1123-2246810(b) f0(x) =2x +1, so the equation for the tangent line to f (x) at x =2 is T (x) = f (2) + f0(2)(x −2) =4 +5( x −2) =5x −6.(c) A vector in the direction of the tangent line has a slope of 5, so the vector〈1,5〉is a goodchoice. It is shown on the graph above based at (2,4).2. (a) Plot of(x = ty = t2+t −2for 0 ≤ t < 4. This is different from the graph above because thedomain is restricted.123451015(b) The vectors based at (0,0) and ending at (x(t), y(t)) for t =0,1,2,3 are shown on the graphabove.(c)x0(2), y0(2)®=〈1,5〉. This represents velocity - this vector is shown on the curve in thegraph below 1.a.(d) The speed of the particle is the magnitude of the velocity, orp12+52=p26.3. • (a)-(d) shown below. The red arrows (from left to right) are the vectors〈−8,3〉,〈−5,2〉,〈−2,−1〉,and〈1,0〉. The black arrows show how these are obtained by adding the multiples −v, 0,v,and 2v of the vector v =〈3,−1〉to the vector〈−5,2〉.-8-6-4-20.51.01.52.02.53.0• (e) If we allow the scalar t to vary in the parametric equation〈−5,2〉+t〈3,−1〉we get a linethrough the point (−5,2) in the direction of the vector〈3,−1〉.4. (a) l(t ) =〈−5 +2 t , 2 +3t,1 −t〉=〈−5,2,1〉+t〈2,3,−1〉, so p =〈−5,2,1〉and v =〈2,3,−1〉.(b) Plot of the line from part (a)-20-10010x-20020y-50510z(c) v is called the direction vector because it points in the direction of the line.5. Let a =−p3,0,−1,0®and b =〈1,1,0,1〉be vectors in R4.(a) The distance between (−p3,0,−1,0) and (1,1,0,1) isq(1 +p3)2+12+12+12=p7 +2p3.(b) The angle between a and b is found by:arccosµa ·b|a||b|¶=arccosÃ−p32p3!=arccos(−1/2)
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