Solution to HW 4, Ma 1c Prac 2016
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1 Solution to HW 4 M c Prc 6 Remrk: every function ppering in this homework set is sufficiently nice t lest C following the jrgon from the textbook we cn pply ll kinds of theorems from the textbook without worrying too much bout the qulity of functions. Problem.4.4: Find the extrem of f(x y) = x y subject to the constrint g(x y) = x y =. Solution: Note tht g(x y) = (x y) which is nonsingulr everywhere wy from the origin so g(x y) = x y = is sufficiently well-behved levelcurve. Now we wish to find ll points (x y) C := {x y = } which re extrem for f nd to clssify those points. Setting f(x y) = λ g(x y) for λ R nd (x y) C we see tht s f(x y) = ( ) nd g(x y) = (x y) we obtin longside our constrint on the curve the system of equtions: λx = λy = x y =. Now noting tht the first two equtions bove imply λx = nd λy = one must hve λ. Solving for x nd y we obtin tht ny extremum of f(x y) = x y long x y = must hve x = λ = y nd hence be of the form (x y) = ( λ λ ). However ny such point must hve ( λ ) ( λ ) = nd consequently does not lie on our curve C. Thus the function f(x y) = x y does not hve ny extrem long the curve defined by x y =. This mkes sense since x y = defines hyperbol which is noncompct (it is certinly closed but is not bounded). Probldem.4.6: Find the extrem of f(x y z) = x + y + z subject to the constrints g (x y z) = x y = nd g (x y z) = x + z =. Solution: We wish to find the extrem of f long the curve S = {(x y z) R x y = nd x + z = } = {(x y z) R g (x y z) = nd g (x y z) = } = {(x y z) R x y = } {(x y z) R x + z = } (which is clerly curve s the intersection of two trnsverse surfces) for which we will use the method of Lgrnge multipliers on multiple constrints.
2 Thus we will ttempt to find point (x y z) S stisfying the eqution f(x y z) = λ g (x y z) + λ g (x y z) for some λ λ R. Noting tht f(x y z) = ( ) tht g (x y z) = (x y ) nd tht g (x y z) = ( ) we my rewrite the bove condition s ( ) = (λ x + λ λ y λ ). In totl we must solve the following system of five equtions: λ x + λ = λ y = λ = x y = x + z =. Since λ = the first eqution reduces to λ x+ = nd hence to λ x = so s the second eqution my be written s λ y = we in fct hve tht λ x = λ y. Furthermore the second eqution immeditely implies tht λ. Thus we my cncel λ from both sides to obtin tht x = y. However this immeditely implies tht ny extremum my not be on S s it implies tht x y = x x =. Thus f hs no extrem when subject to the constrints x y = nd x + z = (i.e. f hs no extrem when restricted to S). Problem.4.: Use the method of lgrnge multipliers to find the bsolute mximum nd minimum vlues of f(x y) = x + y x y + on the unit disk. Solution: Let D R denote the unit disk nd let D O denote its interior the open disk. To find ll the extreml vlues of f on D it suffices to check for extrem on D O nd then check for extrem on D = S = {(x y) x + y = } using the method of lgrnge multipliers. First note tht f(x y) = (x y ) so setting this equl to zero we see tht (x y ) = implies tht (x y) = ( ) is n extremum of f on D. Since there re no more possible zeros of the equtions x = nd y = this is the only extremum on D O. Now we turn our ttention to S. Noting tht S s we hve defined it is smooth level curve of the function g(x y) = x + y we my use the method of Lgrnge multipliers to find extrem. Note tht g(x y) = (x y). Any extrem (x y) S of f on S must stisfy f(x y) = λ g(x y). Thus rewriting this we obtin (x y ) = (λx λy).
3 Thus we obtin the following system of equtions: x = λx y = λy x + y =. Rerrnging the first two equtions we get tht ( λ)x = = ( λ)y. Thus one must hve tht λ so we my cncel ( λ) from both sides. From this we see tht we must hve x = y. Thus so x = y = ±. x + y = x = Thus S contributes two further possible extrem to f evluted on D nmely ±( ) D. Evluting f t ech of these points we get f( ) = ( ) + ( ) + = f( ) = ( ) + ( ) + = f( ) = ( ) + ( ) = + so ( ) is n bsolute minimum of f on D with vlue f( ) = nd ( ) is n bsolute mximum of f on D with vlue f( ) = +. Problem.4.: Find the bsolute mximum nd minimum vlues of f(x y z) = x + y z on the unit bll B = {(x y z) R x + y + z }. Solution: Tking B O to be the open unit bll we proceed nlogously to the previous exercise. Noting tht f(x y z) = ( ) we see tht the eqution f(x y z) = hs no solutions on B O. Thus we turn our ttention to B = S = {(x y z) R x + y + z = }. Clerly S s we hve defined it is smooth level set of g(x y z) = x + y + z nd so we my pply the method of Lgrnge multipliers. Any extremum (x y z) S for f must stisfy f(x y z) = λ g(x y z) for some λ R. Noting tht g(x y z) = (x y z) we my rewrite the bove condition s ( ) = (λx λy λz) which leves us with the following system of equtions: λx = λy = λz = x + y + z =.
4 Now the first three equtions clerly imply both tht λx = λy = λz = nd tht λ so cncelling by λ we get tht x = y = z. Thus one hs from the lst eqution tht x + y + z = x + x + ( x) = x = so one hs tht x = y = z = ±. Thus the bsolute extrem of f on D re ( ) nd ( ). Evluting f t ech of these points yields f( ) = + ( ) = = f( ) = ( ) + ( ) = =. Thus ( ) is n bsolute mximum of f on D with vlue f( ) = nd ( ) is n bsolute minimum of f on D with vlue f( ) =. Problem 4..: Find the rc length of the curve c(t) = (sin t cos t t / ) on the intervl t. Solution: Recll tht if we write the curve c(t) = (x(t) y(t) z(t)) the rc length of c long the intervl [ b] is given by L b (c) = c (t) dt = x (t) + y (t) + z (t) dt. Now given our prticulr choice of curve c(t) = (sin t cos t t / ) we see tht c (t) = ( cos t sin t t). Thus simplifying long the wy one my clculte the rc-length of c from to s L (c) = c (t) dt = ( cos t) + ( sin t) + ( t) dt = + tdt = [( + t) / ] = ( / ) = 4. Problem 4..: Compute the length of the curve c(t) = (log( t) t t ) for t. Solution: Recll tht if we write the curve c(t) = (x(t) y(t) z(t)) the length of c with t b is given by L b (c) = c (t) dt = 4 x (t) + y (t) + z (t) dt.
5 Now given our prticulr choice of curve c(t) = (log( t) t t ) we see tht c (t) = ( t t). Thus simplifying long the wy one my clculte the rc-length of c from to s L (c) = c (t) dt = 4t + + (6t + ) 9t dt = 4t dt = (6t + t )dt = [t + log t] = (9 + log ). Problem 4..5: Show tht the curve c(t) = (e t log t t ) t is flow-line of the vector field F (x y z) = (x z z ). Solution: To show tht given curve c : R R n is flowline to given vector field F : R n R n one need simply show tht c stisfies the functionl eqution c (t) = F (c(t)) for ll t where c is defined nd ll c(t) where F is defined. In our cse the curve we re given is c(t) = (e t log t t ) t nd the vector field we re given is F (x y z) = (x z z ). From here we my simply crry out the following string of computtions for ny t : c (t) = (e t t t ) = F (et log t t ) = F (c(t)). Thus one indeed hs tht c is flow-line of F. Problem 4..7: Show tht the curve c(t) = (sin t cos t e t ) is flow-line of the vector field F (x y z) = (y x z). Solution: To show tht given curve c : R R n is flowline to given vector field F : R n R n one need simply show tht c stisfies the functionl eqution c (t) = F (c(t)) for ll t where c is defined nd ll c(t) where F is defined. In our cse the curve we re given is c(t) = (sin t cos t e t ) nd the vector field we re given is F (x y z) = (y x z). From here we my simply crry out the following string of computtions for ny t R: c (t) = (cos t sin t e t ) = F (sin t cos t e t ) = F (c(t)). Thus one indeed hs tht c is flow-line of F. 5
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