Budapest University of Technology and Economics



Calculus 2, test1.Solutions.1. Which statements are true and which are false? Give reasons for your answers.a) Let S is a nonempty subset of R2. If h is a boundary point of S, then h belongs to S. false: example: S=x,y:y>0 , (1,0) is a boundary point and is not in Sb) If the first order partial derivatives of f exist at (a,b), then f is continuous at (a,b). false: example: fx,y=xyx2+y2 , (x,y)≠(0,0)0 , x,y=(0,0) , the partial derivatives at (0,0) are 0, limit of f does not exist at (0,0) the function is not cont. at (0,0).c) The function fx,y=1x+1y+1 is differentiable at (3,6). yes: use the sufficient conditiond) If f is continuous on a closed bounded region A, then f takes on maximum and minimum values on A.yes: theorem ( Weierstrass theorem)c) in the second test2. Let fx,y=x+ysin1x+y if x+y≠0 and f(x,y)=0 if x+y=0.a) Is the function f continuous at (0,0)? yes: f(0,0)= 0 = lim(x,y)→(0,0)f(x,y)b) Does the partial derivative fy(0,0) exist?no: fy0,0=limk→0f0,k-f(0,0)k=limk→0sin1k does not exist3. Let fx,y=yex2-y2 .a) Show that 1xfx+1yfy=1y2f .e solution: fx=y2xex2-y2 , fy=ex2-y2-2yyex2-y2 find 1xfx+1yfy=1yex2-y2=1y2fb) Give the equation of the tangent plane to the surface z=f(x,y) at the point (1,1,1). fx1,1=2, fy1,1=-1 , n(2, -1,-1), equation: 2(x-1)-(y-1)-(z-1)=04. a) Find the local extrema and saddle points of fx,y=x3-3xy-3y2 . fx=3x2-3y=0, fy=-3x-6y=0 , the critical points (solutions of the system) are P(0,0) , Q(-1/2,1/4), D(x,y)= -36x-9D(P)= -9, P is saddle point, D(Q)=9>0, fxxQ=-3<0 , f has a max. at Q.b) Find the point(s) on the surface z2=xy+1 closest to the origin. d2=x2+y2+z2=x2+y2+xy+1=f(x,y) fx=2x+y=0, fy=2y+x=0 , the critical point is P(0,0)D(0,0)=4-1=3?0, fxx>0 , f has a min at P(0,0), the point on the surface is Q(0,0,1). 5. in the second testSeries1. Calculate the sum of the following series: n=2∞24n+2-3?(-4)n23n+1 .Sol.: The sum= n=2∞12n-32n=2∞-12n=1411-12-32?1411+12=12-14=1/42. Calculate the sum of n=1∞2(4n-1)(4n+3) .sol.: use the decomp. of the gen. term into the sum of partial fractions2(4n-1)(4n+3)=1214n-1-14n+3 , Sn=12(13-14n+3)the sum= 12lim13-14n+3=163. Are the following series conditionally convergent, absolutely convergent or divergent? If a series is conditionally convergent, then estimate the error if the sum is approximated by the sum of the first 1000 termsa) n=1∞1nn3+4 sol.: limit of the gen . term is 1(using limn→∞nn=1 ) , the series is div. b) n=1∞(-1)n3n-24n2+7n sol.: an=3n-24n2+7n→0 and decreasing , according to the Leibniz theorem the series is convergent, S-S1000<30014?10012+7007<40004?10002=10-3c) n=1∞(-1)nn2+2n5n6-6n3+3 sol.: it is abs. conv., (-1)nn2+2n5n6-6n3+3≈15n4 and n=1∞15n4 is conv. Using the limit comparison test the series is abs. conv. ( n=1∞1np num. series is conv. if p?1 )PAGE \* MERGEFORMAT2 ................
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