1 2 x 2 y 2 2 25 xy at 1 2 ans by implicit di

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1.2(x2+y2)2= 25xyat(1,2).Ans:By implicit dierentiation4(x2+y2)·(2x+ 2yy0) = 25y+ 25xy0)y0=8x(x2+y2)-25y25x-8y(x2+y2)Thereforedydx|(1,2)=40-5025-80=1055=211The equation of the tangent at (1,2) isy-2 =211(x-1)-4-3-2-101234-3-2-11232.(x2+y2)3/2= 2xyat(1/p2,1/p2).Ans:By implicit dierentiation32(x2+y2)1/2(2x+ 2yy0) = 2y+ 2xy0)y0=3x(x2+y2)1/2-2y2x-3y(x2+y2)1/2thereforedydx|(1/p2,1/p2)=1p2-1p2=-1and the equation of the tangent at (1/p2,1/p2) isy-1p2=-(x-1p2)10
-2-1.6-1.2-0.8-0.400.40.81.21.622.4-1.2-0.8-0.40.40.81.211
Problem 16 (Linear Approximation)Use linearization process to approximatep1.43.And:We want to approximatep1.43 using the linearization of the functionf(x) =pxatx0= 1.44.f(1.44) = 1.2f0(x) =12pxf0(1.44) =12.4=512We use the formulaf(x)f(x0) +f0(x0)(x-x0)forxnearx0. We obtainp1.431210+512(1.43-1.44) =65-5121100=6·48-1240=287240= 1.1958¯3while 1.43 = 1.195826074....Problem 17 (Linear Approximation)Apply linear approximation to the functionf(x) =(1-x)21 + (1 +x)2atx= 0to approximate the number(0.99)21+(1.01)2.And:By linear approximation, ifxis near 0, we getf(x)f(x0) +f0(x0)(x-x0)In this case we havex0= 0andf(0) =12Thenf0(x) =-2(1-x)(1 + (1 +x)2)-2(1-x)2(1 +x)(1 + (1 +x)2)2andf0(0) =-32and we obtainf(0.01)⇠ -12-32·1100=97200= 0.485whilef(0.01) =(0.99)21 + (1.01)2= 0.485174Problem 18 (Linear Approximation)1. Use linearization process to approximatep35.9.Ans:Considerf(x) =px. We want to apply linearization atx0= 36. In this case we havef(36) = 6f0(x) =12pxf0(36) =112The equation of the tangent atx0isy= 6 +112(x-36)It follows thatf(35.9)'6 +112(35.9-36) = 6-112110=720-1120=719120= 5.991¯6On the other hand, using a calculatorp35.9 = 5.991660. . ..2. Consider the curve that is given byx3+xy3= 9(y-1). Notice that the point(1,2)lies on this curve.Approximate they–coordinate of the point of the curve withx–coordinate0.9.12
Ans:By implicit dierentiation,3x2+y3+ 3xy2y0= 9y0)y0=3x2+y39-3xy2Thereforedydx|(1,2)=-113and the equation of the tangent isy-2 =-113(x-1)-2-1.6-1.2-0.8-0.400.40.81.21.622.40.40.81.21.622.42.8Therefore, if we want to approximate they–coordinate corresponding tox= 0.9 it is enough to computethe corresponding value on the tangent liney|x=0.9'2-113(0.9-1) = 2 +1130= 2.3¯613
Problem 19 (Optimization)Find all pairs of positive integer numbersx, ysuch thatx+ 4y= 100and theirproduct is the largest possible.And:We have two variablesx, ysatisfyingx= 100-4y. We want to maximize the functionP(x, y) =xy)P(y) = (100-4y)y= 100y-4y2in the interval [0,25] (ify >25 thenx <0). We getP0(y) = 100-8y)P0(y) = 0,y=252We have a critical point aty= 12.5. In particular,y <12.5)P0(y)>0andy >12.5)P0(y)<0ThereforePattains the absolute maximum aty= 12.5 (moreover it’s obvious thatP(0) =P(25) = 0 at theendpoints). The maximum corresponds to the pair (x, y) = (50,12.5).Since we are interested in integer solutions, we should look at the closest integers and we obtain the pairs:(x, y) = (52,12))xy= 624and(x, y) = (48,13))xy= 624Problem 20 (Optimization)A metal storage tank with volumeVis to be constructed in the shape of a rightcircular cylinder surmounted by a hemisphere. What dimensions will require the least amount of metal?

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