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Unformatted text preview: Math 17C
Kouba
Discussion Sheet 3 1.) Find the Jacobi Matrix for each function. a.) rm) = (3””+4y) 2:1: — 5y
b.) may) = ( 1:2: in)
c) ﬂaw) = d) f (:6, y) = (COEESELQ
e) may) = 2.) Find the linearization L(w) of f (M) 2 (ﬂag?!) at (1:1). a.) Compute f(0.9, —1.2) and L(0.9, —1.2). What do you conclude ?
b.) Compute f(3, 2) and L(3, 2). What do you conclude ? d
3.) Assume that z = :vzy + 563/2, 3; = t3, and y = Compute 67: when t = 1 .
4.) Assume that z = ln(a;y) + 6296+”, a: = f(t), and y 2 g(t). If f(0) : —1, f’(0) = 2,
(1
9(0) 2 3, and g’(0) = 1 , then what is the value of 67: when t = O ? 5.) A Madagascar hissing cockroach is walking on the plane :v + 2y + 32 = 6 above the
circle $2 +312 2 1. Assume that the roach’s position (ac, y, 2) (distance measured in inches) at time t seconds is determined by a: 2 cost and y = sin 15. Determine the roach’s rate of dz
change of elevation ) when t = 0 seconds; t = 7r seconds. Y‘oqchls pa‘l‘h in
15+ oc‘l’ah't‘ 6.) Compute the derivative of f(;r, y) = ln(2;r +3y) at the point P = (2, 0) in the direction of vector A = . 7.) Consider the function f (513,14) = 333/3 and the point P = (2,1) . Determine all unit
vectors u so that D?f(2, 1) is a.) as large as possible (f increases most rapidly).
b.) as small as possible (f decreases most rapidly).
c.) equal to zero. (1.) equal to 1. 8.) Consider the surface given by x2 + y2 + 22 = 169 and the point P = (3, 4, 12) on the
surface. Find equations for a.) the plane tangent to the surface at point P.
b.) the line normal (perpendicular) to the surface at point P. 9.) Consider the surface (hyperbolic paraboloid or saddle) given by f (CL‘, y) 2: 33:2 — 23/2 +5
and the point P = (2,3, —1) on the surface. Find equations for a.) the plane tangent to the surface at point P.
b.) the line normal (perpendicular) to the surface at point P. 10.) Let f(a:, y) = 1:312 + 23 —— 3y . Use a linearization L(a:) at the point (0,0) to estimate
the value of f(0.2, —0.1). 11.) Find and classify critical points as determining relative maximums, relative mini
mums, or saddle points. a.)223m2—6$y+y2+12$~16y+1 b.)z=:c2y—a:2—2y2
c.) z :2 1:2 — 8ln(acy) + y2 d.) z = 3x23) — 6:122 + y3 — 6y2 12.) Find the point on the plane a: + 2y + 32 = 6 nearest the origin. 13.) Determine the dimensions and minimum surface area of a closed rectangular box with
volume 8 ft.3 14.) Determine the point on the sphere $2 + y2 + 22 = 4 which is a.) nearest the point
(1, ——1, 1). b.) farthest from the point (1, —1, 1). +++++++++++++++++++++++++++++++++ ” If you judge people, you have no time to love them.” — Mother Teresa ...
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 Summer '09
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