Solutions to Homeork #05 MATH Kaai Section. (I) Exercise #. g x and g z : Product Rule: g x = x @ ln z + x + ln z + x @ [x] = x x z + x + ln z + x = x z + x + ln z + x x is eld constant. g z = x @ ln z + x z = x z + x = xz z + x (II) Exercise #0. f xx ; f xy ; f yx ; and f yy : Second order artials. f (x; y) = cos (xy) f x = sin (xy) @ [xy] = y sin (xy) From tis, e can derive f xy and f xx : f xx = @ [f x] = @ @ [ y sin (xy)] = y [sin (xy)] = y (cos (xy) y) = y cos (xy) f xy = @ @y [f x] = @ @y [ y sin (xy)] = y @ @y [sin (xy)] + sin (xy) @ @y [y] = (y (cos (xy) x) + sin (xy) ) = (xy cos (xy) + sin (xy)) No try te oter at: f y = sin (xy) @ [xy] = x sin (xy) @y From tis, e can derive f yx and f yy : f yx = @ [f y] = @ [ x sin (xy)] = x @ @ [sin (xy)] + sin (xy) [x] = (x (cos (xy) y) + sin (xy) ()) = (xy cos (xy) + sin (xy)) Tis veri es te equality of mixed artials (Clairaut s Tm.). f yy = @ @y [f y] = @ @y [ x sin (xy)] = x @ @y [sin (xy)] = x (cos (xy) x) = x cos (xy)
(III) Exercise #abcde. V (x; ) = x : (a) Te artial derivatives are: V x = @ x = x V = x @ @ [] = x : (b) For a box it = :5 meters, use linear aroximation to estimate V; if x: 0:50! 0:5: Tis ould make x = +0:0: V x ) V x = (x) (x) = (0:5) (:5) (0:0) = +0:05 cu. meters. (c) For a box it x = 0:5 meters, use linear arox. to estimate V; if : :50! :9: Tis ould make = 0:0: @ V ) V @ = x () = (0:5) ( 0:0) = 0:005 cu. meters. (d) For a xed eigt ; does a 0% cange in x alays roduce (arox.) a 0% cange in V? Tus, e ill ave x: x! :0x and x = 0:0x: V x = (x) (0:0x) = 0:0x = 0:0V: Tis is a 0% increased in volume, not 0%! Te artial derivative as tat extra factor of caused by te original x term. (e) For a xed base lengt x; does a 0% cange in alays roduce (arox.) a 0% cange in V? Tus, e ill ave :! :0 and = 0:0. @ V Yes, te cange is volume is exactly 0%. (IV) Exercise #5ab. Volume of serical ca: (a) Comute V and V r : ) V @ = x (0:0) = 0:0V: V (; r) = (r ) ; for 0 r: V = @ r = @ 6r = r V r = @ @r r = = :
(b) For a sere of ANY radius, is rate V r greater en = 0:r or en = 0:8r? We see tat tis artial deends solely on : V r (r; 0:r) = (0:r) = 0:0r V r (r; 0:8r) = (0:8r) = 0:6r > 0:0r : For a xed radius sere, te surface area above te lane ic slices te eigt of te ca is clearly larger en = 0:8r: Tis causes te increase in volume en r increases. (V) Exercise #58abce. Cobb-Douglas. Q (L; K) = L = K = : (a) Evaluate Q L and Q K : Q L = L = K = and Q K = L= K = : (b) If L = 0 and K: 0! 0:5; ten use linear arox. to estimate Q: Tis gives us K = +0:5: @Q @K Q K ) Q @Q @K K = 0 = 0 = (0:5) = = = = (c) If K = 0 and L: 0! 9:5; ten e ave L = Q @Q @L L = = = = 0 0 = : = 0:66 outut units. 0 = 0 = ( 0:5) = = 0:5 and by linear arox., e ave: 0 = 0 = : = 0:66 outut units. It turns out tat since a + b = ; e ave tis symmetry. (d) If e let L be our orizontal variable and K be our vertical variable, ten e ave te level curve: L = K = = ) x = y = = ) y = = x = ) y = = = x = = ) y = x = = : x K 5 Wen k = ; e ave x = y = = ) y = = x = ) y = = x = : It s te same curve as last time, but multilied by = = : 0 0 5 L Similarly, en k = ; e ave y = = x = and te original curve is multilied by :
(e) If you move along te vertical line L = in te ositive direction (u), o does Q cange? We are moving to te next iger level curve, so Q must increase. Is tis consistent it Q K = L= K =? YES. If L = ; ten Q K = = K = > 0 en K > 0: Section.5 Te instantaneous rate of cange is alays ositive as K increases. (VI) Exercise #6abc. Related rates! (a) We ave V (x; ) = x : If x = x (t) and = (t) ; ten ultimately, V ill be a function of t: Find V 0 (t) : @t = @t + @ @ @t = x x 0 (t) + x 0 (t) = V 0 (t) : (b) Suose tat x (t) = No evaluate V 0 (t) : We ave t 0 x 0 (t) = = t + t t + and (t) = ; for t 0: t + (t + ) and 0 (t) = (t + ) i 0 = (t + ) : Substitute! V 0 (t) = t t + t + = t t (t + ) : (t + ) + t t + (t + ) (c) Does te volume of te yramid in art (b) increase or decrease as t increases. Wen t = 0; V 0 (t) = 0: If e factor te numerator, e ave V 0 (t) = t ( t) (t + ) : Tis ill be ositive for 0 < t < ; ten V 0 (t) = 0 again for t = ; and ten V 0 (t) is negative (volume is decreasing) for t > :
(VII) Exercise #0. We ave By te Multivariable Cain Rule, e ave z (x; y) = xy x + y x (s) = cos (s) y (t) = sin (t) : @s = @s + @y @y @s : Evaluate te rst grou: Te second grou els us. z x = y ; z y = x + : x s = sin (s) ; y s = 0: Tus, e ave = (y ) ( sin (s)) + (x + ) (0) = ( y) sin (s) @s Our nal anser must be in terms of te loer variables, so e must substitute back in for y: Similarly, e ave @t = @t + @y @y @t (VIII) Exercise #. Imlicit di erentiation. Use te sortcut formula: = ( sin (t)) sin (s) : @s = (y ) (0) + (x + ) cos (t) = (x + ) cos (t) = (cos (s) + ) cos (t) : dy dx = Fx F y ; ere F (x; y) = c: We ave F (x; y) = y ln x + y + = ; so e evaluate te rst artials. F x = x y x + y + F y = y y x + y + ln x + y + + Te negative quotient is dy dx = 0 B @ xy x + y + y x + y + + ln (x + y + ) C A Multily to and bottom by x + y + : dy dx = xy y + (x + y + ) ln (x + y + ) : 5
(IX) Exercise #55a. Constant volume torus. V (R; r) = (R + r) (R r) ; R > r > 0: If R and r increase at te same rate, does te volume of te torus increase, decrease, or stay te same. We mean tis: Related rates. dv dt = dr @R dt + dr @r dt V R = V r = dr dt = dr dt = c > 0; a ositive constant. (R + r) ( (R r)) + (R r) = (R + r) ( (R r)) + (R r) = (R + r) (R r) + (R r) (R + r) (R r) + (R r) Since te oter to rates are equal to c; en e add tese roducts, e get: dv dt = c = c (R + r) (R r) + (R r) + ( ) (R + r) (R r) + (R r) (R r) = c (R r) > 0: Te derivative is ositive, so te volume must be increasing at tat oint. (X) Exercise #56abc. (a) Mosteller formula for surface area of uman body: S (; ) = 60 = 60 = = : Find S 0 (t) : Here te artials. S = 60 S = 0 ds dt = @S d @ dt + @S d @ dt! : = 0 Tis gives us ds dt = 0 (t) + 0 0! 0 (t) = 0 0 (t) +!! 0 (t) : 6
(b) If S remains constant, ten e must ave S 0 (t) = 0 and!! 0 (t) + 0 (t) = 0 ) 0 (t) + 0! 0 (t) = 0 Multily troug by : 0 (t) + 0 (t) = 0: X (c) So tat e must ave = C ; ere C is a ositive constant. Divide troug by : 0 (t) + 0 (t) = 0 ) Z 0 Z (t) 0 Z dt + (t) dt = 0 dt ln () + ln () = c ) ln () = c ) = e c : Let e c = C: Tis gives us = C ) = C : X 7