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University of Toronto at ScarboroughDepartment of Computer & Mathematical SciencesMAT B41H2007/2008Solutions #51. Marsden & Tromba, page 140, #10.f(x, y) =x2+y2andg(x, y) =-x2-y2+xy3. At (x, y) = (0,0), we havefx(0,0) =2xvextendsinglevextendsinglevextendsinglevextendsingle(0,0)= 0,fy(0,0) = 2yvextendsinglevextendsinglevextendsinglevextendsingle(0,0)= 0,gx(0,0) =-2x+y3vextendsinglevextendsinglevextendsinglevextendsingle(0,0)= 0 andgy(0,0) =-2y+ 3xy2vextendsinglevextendsinglevextendsinglevextendsingle(0,0)= 0.Hence the graph ofz=f(x, y) at (0,0,0) and the graph ofz=g(x, y) at (0,0,0) have the same tangent plane; thexy–plane (z= 0). Since theyhave the same tangent plane we can think of them as tangent.2. We know that the direction of maximum increase is the gradient direction; so the tem-perature will increase fastest in directionT=-80(1 +x2+ 2y2+ 3z2)2parenleftbigg2x,4y,6zparenrightbigg=-160(1 +x2+ 2y2+ 3z2)2parenleftbiggx,2y,3zparenrightbigg. At the point (1,1,-2), the temperature will in-crease fastest in directionT(1,1,-2) =-160256parenleftbigg1,2,-6parenrightbigg=-58parenleftbigg1,2,-6parenrightbigg.Themaximum rate of increase is the magnitude of the gradient,bardbl∇T(1,1,-2)bardbl=58vextenddoublevextenddoublevextenddoublevextenddoubleparenleftbigg1,2,-6parenrightbiggvextenddoublevextenddoublevextenddoublevextenddouble=5418.Themaximumrateofincreaseoftemperatureis5418C/m.3.(a) We regardzas a function ofxandyand differentiateF(x, y, z) = 0 w.r.t.xgiving∂F∂x∂x∂x+∂F∂y∂y∂x+∂F∂z∂z∂x= 0.Now∂x∂x= 1 and∂y∂x= 0, sinceyisindependent ofx. Hence we have∂F∂x+ 0 +∂F∂z∂z∂x= 0 =∂F∂z∂z∂x=-∂F∂x=∂z∂x=-∂F/∂x∂F/∂z,if∂F∂znegationslash= 0, as required.(b) IfwedifferentiateF(x, y, z)=0w.r.t.yinplaceofxwewillget0 +∂F∂y+∂F∂z∂z∂y= 0 or∂z∂y=-∂F/∂y∂F/∂z,if∂F∂znegationslash= 0.

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