# sol3 - MATH 237 Assignment 3 Winter 2007 Solutions...

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Unformatted text preview: MATH 237 Assignment 3 Winter 2007 Solutions 1.A2(iii) With 9(u, v, w) = ln(uv + 102), we ﬁnd the partial derivatives as ('9 (u v w) 1 d ( v+ 2) 1 v — J 3 = — u w = : any 7w + 1122 du m} + 1112?] M) + w2 (3 1 d 2 1 u — u,/u,w = — uv w = = 8vg( ) uv+w2 dv( + ) uv+w2u un+w2 8 1 d 2 1 2w (910901” 0’ w) — 1w + w2 Emu + w ) — U?) + 102 2w — at + 102' 1.A3 The tangent plane to the surface 2 = f(x, = 6 -— x2 —— y2 at (1, 2, 1) is given by (see page 36 of course notes) 0f z=1+5;(1,2)(\$—1)+8f EUJXy- 2). In this case 3% = —2x and %5 2 —2y so that gal, 2) = —2 and 655(1, 2) = —4. The tangent plane is thus z=1—2(a:—1)—4(y—2)=11—2x—4y. Substituting :1: = O and y = O we find the point at which this plane intersects the z-axis: (0, 0, 11). 1.A5(iii) a) From page 41 of the course notes, we have the general form of the linear approx— imation of a function f at a point a as LAX) = f(a) + Vf(a) ‘ (X - a)- b) With f(:c, y, z) = em+gy+32, we have )a:+2y+32 : 28x+2y+32 a_f : 3em+2y+3z g_ _ 833—6 ’83; ’32 To determine the linear approximation of f at the point a = (1,1,—1), we ﬁnd f (a) = €1+2(1)+3(“1) : e0 = 1, while = (61+2(1)+3(—1)’ 2el+2[1)+3(—1), 361+2(1)+3(—1)) : (1,2,3)- The linear approximation is then La(x,y,z)=1+(1,2,3)-(a:—1,y—1,z+1)=1+m—1+2(y—1)+3(2+1)=1+a3+2y+32. 1 1. B4 Referring to the sketch: and recalling that the volume of a sphere of radius 7‘ is §7TT3 While the volume of a right circular cylinder of radius 7“ and height h, is 7r'r2h,, we ﬁnd that the volume of the silo is given by 4 V(r, h) = 3777"3 + 7r7‘2h.. The linear approximation centred at a is Lab", h.) = l/(a) + VV(a) - ((7‘, h.) — a). We ﬁnd 2 VV(7‘, h) = (W, Vh) = (EWBT2 + 27r'rh,7rr2) = (27rr('r + h),7r7"2). At a = (5,25) we ﬁnd Vl/(5, 25) = (107r(30),7r52) = 257r(12,1). Now, if the radius is decreased by 0.05 and the height in increased by 0.01 (units of meters), the change in volume is approximated by La(4.95, 25.1) — V(5,25) VV(5, 25) . ((4.95, 25.1) — (5, 25)) 257r(12, 1) . (—0.05, 0.1) 2574—06 +0.1) = 323%, a decrease in volume. ADDITIONAL COMMENTS: in this simple case we can compare with the actual change in volume: (§w(4.95)3 + 71'(4.95)2(25.1))—(g7r52 + 7r(52)(25)) 3 7r(695.871 — 708.333) = —12.4627r V(4.95,10.1) — V(5, 25) 22 which is very close to our approximation of —12.57r. This might seems like a ridiculously large change (over 39 cubic meters) given the tiny adjustments made in the radius and height. However, considering the relative changes is more appropriate. The relative change in radius is -0505 = —0.01 = —1%. The relative change in height is % = 0.004 = 0.4%, and the resulting relative change in volume is 1133???" = ——0.017 = 17%. 2A3 We consider the function f deﬁned by r _ sin(93y) f(\$:y) — (18907400) f(0, 0) = 0. Evaluating the partial derivatives at the origin, we ﬁnd . h,0— 0,0 ham zkﬁn )hn > . 0—0 i113) h ’0 k—>O k 0 — 0 = HF?‘=0 H fy(0.~ 0) These partial derivatives exist. To verify that f is not continuous at (0,0), we consider the limit at this point along the line y = a: (we already know the limits along the axis are zero). We then ﬁnd . __ , sin(:52) 13513” f (as, y) — £135 _—“m(2x2 + 1)‘ This is an indeterminate form of the type %. Applying L’Hopital’s rule, we ﬁnd ‘ 2 2 2 2 2 2 hm_n§1_2mmi%\$2=hmeﬂQLaiﬂ=l_ cur—.0 ln(233 + l) m—>O m {Ii—>0 2 2 Since this limit is not equal to the function value at the origin, the function is not continuous. ADDITIONAL COMMENT: This example veriﬁes that existence of the partial derivatives is not sufficient to guarantee continuity. 2A5 We have f(x, = — 1). To verify differentiability at (0, 0) we must ﬁrst determine the linearization at (0,0). We ﬁnd f (0,0) = 0. The partial derivative with respect to x is 53—5 = |y| — 1, so %(0,0) = —1. To determine partial derivative with respect to y we must use the deﬁnition: NUS) —f(0,0) hmm =ta k _ 0—0 -ts77—0 3 The linearization of f at 0 = (0,0) is then 8 (9 L0(3:,y) = f(0,0) + (8—:(0,O), 8—£(0,0)) '(x — 0,3} — 0) = 0 — (1)56 + (0)y : —:I:. To verify differentiability, we compare this to the function values near (0, O) to ﬁnd lf(:v,:I/) — Lacey)! _ Wyl - 1) +\$l = {mllyl [Kmayl—(Oﬂlll _ ll(\$,'y)l| x/x2+y2' This tends to O as (:13, y) —> (O, O), as may be seen from the Squeeze Theorem : lmllyl lash/av2 + y2 0 S —— S —— S W /1-2 + y2 /m2 + y2 and —> O as (15,31) —» (O, 0) Hence, f is differentiable at (0,0). fOI‘ (any) as (0: 0), 2B6 We note that the surface 2 = f(:c,y) = Ix — yl can be written as {a3~y ifm—yZO z: . y~cc1fzr—y<0 Sketching these two planes intersecting at the line y = :r, we have: lb? x The surface does not have a tangent plan at any point along the line 9 = 9:. b) To verify non—differentiability, we establish that the partial derivatives at these points do not exist. At any point (11:, 51;), we ﬁnd [hi — O _. f(a:+h,:c)—f(a:,:r)_,_ fx(\$’x)‘l§li—_E_—“l.123 h - This limit is not deﬁned, and so there is no partial with respect to a: at this point. By symmetry, the same holds true for the partial derivative with respect to y. Either of these results tell us that f is not differentiable at such points (recall the definition of differentiability on page 45 of the course notes.) ...
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## This note was uploaded on 01/17/2011 for the course MATH 237 taught by Professor Wolczuk during the Spring '08 term at Waterloo.

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sol3 - MATH 237 Assignment 3 Winter 2007 Solutions...

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