University of. 2 nd. d Class

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1 Univesit of Technolog Electomechanical Depatment Eneg Banch Advanced Mathematics Patial Diffeentiation nd d Class st Lectue

2 Patial Deivatives Recall that given a function of one vaiable, f ( ), the deivative, f, epesents the ate of change of the function as changes. This is an impotant intepetation of deivatives and we ae not going to want to lose it with functions of moe than one vaiable. The poblem with functions of moe than one vaiable is that thee is moe than one vaiable. In othe wods, what do we do if we onl want one of the vaiables to change, o if we want moe than one of them to change? In fact, if we e going to allow moe than one of the vaiables to change thee ae then going to be an infinite amount of was fo them to change. Fo instance, one vaiable could be changing faste than the othe vaiable(s) in the function. Notice as well that it will be completel possible fo the function to be changing diffeentl depending on how we allow one o moe of the vaiables to change. f, moe commonl be witten as, f, = 4 and f, = 6 Let s stat with the function = the patial deivatives fom above will Since we can think of the two patial deivatives above as deivatives of single vaiable functions it shouldn t be too supising that the definition of each is ve simila to the definition of the deivative fo single vaiable functions. Hee ae the fomal definitions of the two patial deivatives we looked at above. f + h, - f, f, + h - f, f(, ) = lim f(, ) = lim hæ0 h hæ0 h Now let s take a quick look at some of the possible altenate notations fo patial deivatives. z = f, the following ae all equivalent notations, Given the function f f(, ) = f = = ( f (, )) = z = = Df f f(, ) = f = = ( f (, )) = z = = D f Fo the factional notation fo the patial deivative notice the diffeence between the patial deivative and the odina deivative fom single vaiable calculus. f fi df f = d f (, ) fi f f f(, ) = & f(, ) = Eample Find all of the fist ode patial deivatives fo the following functions. 4 f, = (a) (b) w= - 0z + 4-7tan( 4) h st, = t ln s + - s t Ê4 ˆ -5 f, = cosá Ë e (c) 7 (d)

3 4 f, = a) f (, ) = 4 f (, ) = (b) w= - 0z + 4-7tan( 4). w = h st, t ln s t s = - 0-8sec 4 w z (c) 7 ( ) 7 4 = + - = h st, = t ln s + 9t - s h Êsˆ 4 t 4 hs ( st, ) = = t Á - s = - s s Ë s 7 s 7 h 6-4 ht ( st, ) = = 7t ln( s ) -7t t Remembe how to diffeentiate natual logaithms. d g ( ln g ) = d g (d) f ( ) Ê4 ˆ -5, = cosá Ë e. Ê4ˆÊ 4 ˆ -5 Ê4ˆ -5 f, sin cos =- Á Á- e + Á e Ë Ë Ë 4 Ê4ˆ -5 Ê4ˆ -5 = sin cos Á e + Á e Ë Ë w =- 0z Also, don t foget how to diffeentiate eponential functions, d f f ( e ) = f e. d. Ê4 ˆ -5 f (, ) = ( -5 ) cosá Ë e Eample Find all of the fist ode patial deivatives fo the following functions. 9u (a) z = u + 5v sin ( ) (b) g( z,, ) = z (c) z = + ln( 5- ) 9u (a) z = u + 5v. 9( u + 5v ) -9u( u) - 9u + 45v zu = = u + 5v u + 5v z v ( 0)( u + 5v) -9u( 5) ( u + 5v) -45u ( u + 5v) = =

4 (b) (,, ) ( ) sin g z = z sin( ) cos g( z,, ) = g ( z,, ) = z z - g z,, = sin z z (,, ) sin - =- =- g z z (c) z = + ln( 5- ) sin z ( ) ( ( - ln 5 )) ln( 5 ) ( ) - =- + ln z = Ê 5 = + ln 5- Á+ Ë 5- Ê 5 ˆ = Á 5 ( - ) Ë ( ln( 5 )) - ( ( - ln 5 )) ln( 5 ) ( ln( 5 )) - Ê -6 ˆ = + - Á ( ) - ˆ z = Ë5- Eample Find d d fo = 5. The fist step is to diffeentiate both sides with espect to. d d + = 6 d 5-7 = d Eample 4 Find and fo each of the following functions. (a) z - 5 z = + 5 (b) sin ( - 5 z) = + cos ( 6 z) 5 (a) z - 5 z = + Let s stat with finding. We fist will diffeentiate both sides with espect to and emembe to add on a wheneve we diffeentiate a z. 5 5 z + z -5z- 5 = z = z, then an poduct of s and z s will be a Remembe that since we ae assuming poduct and so will need the poduct ule! Now, solve fo.

5 - 5 = = 5 z-5 z 5 z z 5 5 z z z w we ll do the same thing fo ecept this time we ll need to emembe to add on a heneve we diffeentiate a z. 4 5 z -5 z- 5 = - 5 = + 5 (b) sin ( - 5 z) = + cos ( 6 z) 5 4 z z + 5 = z-5 4 z z 5 We ll do the same thing fo this function as we did in the pevious pat. Fist let s find. Ê ˆ Ê ˆ sin( - 5z) + cos( -5z) Á- 5 =- sin( 6z) Á6z+ 6 Ë Ë Don t foget to do the chain ule on each of the tig functions and when we ae diffeentiating the inside function on the cosine we will need to also use the poduct ule. Now let s solve fo. sin( -5z) -5 cos( - 5z) =-6zsin( 6z) -6sin( 6z) sin( - 5z) + 6zsin( 6z) = ( 5 cos( -5z) -6sin( 6z) ) z sin( - 5z) + 6zsin( 6z = ) 5 cos -5z -6sin 6z Now let s take cae of. This one will be slightl easie than the fist one. Ê ˆ Ê ˆ cos( -5z) Á- 5 = cos( 6z) - sin( 6z) Á6 Ë Ë cos( -5z) -5 cos( - 5z) = cos( 6z) -6sin( 6z) ( 6sin( 6z) -5 cos( - 5z) ) = cos( 6z) - cos( -5z) ( z) - ( - z) z cos 6 cos 5 = 6sin 6z -5 cos -5z 4

6 Intepetations of Patial Deivatives This is a fail shot section and is hee so we can acknowledge that the two main intepetations of deivatives of functions of a single vaiable still hold fo patial deivatives, with small modifications of couse to account of the fact that we now have moe than one vaiable. The fist intepetation we ve alead seen and is the moe impotant of the two. As with functions of single vaiables patial deivatives epesent the ates of change of the functions as f, epesents the ate of change the vaiables change. As we saw in the pevious section, of the function f (, ) as we change and hold fied while f (, ) change of f (, ) as we change and hold fied. (a) if we allow to va and hold fied. (b) if we allow to va and hold fied. Eample Detemine if f (, ) = is inceasing o deceasing at (a) If we allow to va and hold fied. In this case we will fist need f (, ) and its value at the point. epesents the ate of,5, 4 f(, ) = fi f (,5) = > 0 5 So, the patial deivative with espect to is positive and so if we hold fied the function is,5 as we va. inceasing at (b) If we allow to va and hold fied. Fo this pat we will need f (, ) and its value at the point. 4 f, =- fi f,5 =- < 0 65 Hee the patial deivative with espect to is negative and so the function is deceasing at (,5 ) as we va and hold fied 5

7 Eample Find the slopes of the taces to z = at the point,. We sketched the taces fo the planes = and = in a pevious section and these ae the two taces fo this point. Fo efeence puposes hee ae the gaphs of the taces. Net we ll need the two patial deivatives so we can get the slopes. f, =- 8 f, =- slope of -4. To get the slopes all we need to do is evaluate the patial deivatives at the point in question. f, =- 8 f, =- 4 So, the tangent line at -8. Also the tangent line at, fo the tace to, fo the tace to z = fo the plane z = has a slope of = fo the plane = has a 6

8 Highe Ode Patial Deivatives Just as we had highe ode deivatives with functions of one vaiable we will also have highe ode deivatives of functions of moe than one vaiable. Howeve, this time we will have moe options since we do have moe than one vaiable.. Conside the case of a function of two vaiables, f (, ) since both of the fist ode patial deivatives ae also functions of and we could in tun diffeentiate each with espect to o. This means that fo the case of a function of two vaiables thee will be a total of fou possible second ode deivatives. Hee the ae and the notations that we ll use to denote them. ( f ) ( f ) ( f) ( f) Ê f ˆ f = f = Á = Ë Ê f ˆ f = f = Á = Ë Ê f ˆ f = f = Á = Ë Ê f ˆ f = f = Á = Ë Eample Find all the second ode deivatives fo We ll fist need the fist ode deivatives so hee the ae. f, =-sin -e Now, let s get the second ode deivatives. Claiaut s Theoem =- 5 e + f, 5 6 f f f f =-4cos -e =-0e =-0e e 6 =- + Suppose that f is defined on a disk D that contains the point (, ) ae continuous on this disk then, 5 f, = cos - e + 5 (, ) = (, ) f ab f ab Eample Veif Claiaut s Theoem fo (, ) 5 ab. If the functions f and f f = e -. We ll fist need the two fist ode deivatives. - - f, = e -e f, =- e - 7

9 Now, compute the two fied second ode patial deivatives f, =- e - 4e + 4 e =- 6e + 4e f, =- 6 e + 4e Sue enough the ae the same Ê f ˆ f f = ( f ) = Á = Ë Ê f ˆ f f = ( f ) = Á = Ë an etension to Claiaut s Theoem that sas if all thee of these ae continuous then the should all be equal, f = f = f Eample Find the indicated deivative fo each of the following functions. (a) Find f zz fo f ( z,, ) = z ln (b) Find (a) Find zz f fo f (, ) = e f fo f ( z,, ) = z ln In this case emembe that we diffeentiate fom left to ight. Hee ae the deivatives fo this pat. z f = z f =- z f =- (b) Find f f f z zz 6z =- z =- fo (, ) f = e Hee we diffeentiate fom ight to left. Hee ae the deivatives fo this function. f = e f = e f = e + e 8

10 Diffeentials This is a ve shot section and is hee simpl to acknowledge that just like we had diffeentials fo functions of one vaiable we also have them fo functions of moe than one vaiable. Also, as we ve alead seen in pevious sections, when we move up to moe than one vaiable things wok pett much the same, but thee ae some small diffeences. Given the function z f (, ) = the diffeential dz o df is given b, dz = f d+ f d o df = f d+ f d Thee is a natual etension to functions of thee o moe vaiables. Fo instance, given the w= g z,, the diffeential is given b, function Let s do a couple of quick eamples. dw= g d+ g d+ g dz z Eample Compute the diffeentials fo each of the following functions. + (a) z = e tan t (b) u = s 6 + (a) z = e tan Thee eall isn t a whole lot to these outside of some quick diffeentiation. Hee is the diffeential fo the function. t (b) u = s 6 ( ) tan sec e + e + e + tan( ) dz = + d+ d Hee is the diffeential fo this function t 6t t du = dt+ d- ds s s s 9

11 Chain Rule We ve been using the standad chain ule fo functions of one vaiable thoughout the last couple of sections. It s now time to etend the chain ule out to moe complicated situations. Befoe we actuall do that let s fist eview the notation fo the chain ule fo functions of one vaiable. The notation that s pobabl familia to most people is the following. F = f g F = f g g we ae going to be using in this section. Hee it is. ( ) d d d If = f and = g() t then = dt d dt Case : z = f (, ), = g( t), = ht and compute dz dt. This case is analogous to the standad chain ule fom Calculus I that we looked at above. In this case we ae going to compute an odina deivative since z eall would be a function of t onl if we wee to substitute in fo and. The chain ule fo this case is, dz f d f d = + dt dt dt Eample Compute dz fo each of the following. dt (a) z = e, = t, = t - (b) z = + cos, = ln ( t ), = sin( 4t) (a) z = e, = t, = t - Thee eall isn t all that much to do hee othe than using the fomula. dz f d f d = + dt dt dt ( e e )( t) e ( - t ) t( e e ) - t e = = + - So, technicall we ve computed the deivative. Howeve, we should pobabl go ahead and substitute in fo and as well at this point since we ve alead got t s in the deivative. Doing this gives, dz dt - 4 t t t t t = t e + te - t t e = te + t e 0

12 dz z = t e fi = te + t e dt t t t The same esult fo less wok. Note howeve, that often it will actuall be moe wok to do the substitution fist. (b) z = + cos, ln ( t ) =, = sin( 4t) Oka, in this case it would almost definitel be moe wok to do the substitution fist so we ll use the chain ule fist and then substitute. dz ʈ = ( - sin) Á + ( + cos) ( 4cos( 4t) ) dt Ë t ( t) t - ( t) ( t ) ( Î ) 4sin 4 ln sin 4 sin ln = + 4cos 4 sin 4 ln + cos ln t ( t) ( t) È t ( t ) Now, thee is a special case that we should take a quick look at befoe moving on to the net case. Let s suppose that we have the following situation, z = f, = g In this case the chain ule fo dz d becomes, dz f d f d f f d = + = + d d d d In the fist tem we ae using the fact that, d = d d d = Eample Compute dz z ln d fo = +, = cos( + ) We ll just plug into the fomula. dz Ê ˆ Ê ˆ = Áln + + Á + - sin + d Ë Ë Ê ˆ = ln cos sin + + cos + ( ( )) ( ) ( ) Á cos( ) Ë + ( ( )) ( ) ( ) ( ) = ln cos tan sin + cos + Now let s take a look at the second case. Case : z = f (, ), = g( st, ), h( st, ) = and compute s and t. In this case if we wee to substitute in fo and we would get that z is a function of s and t and so it makes sense that we would be computing patial deivatives hee and that thee would be two of them. Hee is the chain ule fo both of these cases. f f f f = + = + s s s t t t

13 Eample Find s and t fo z = e sin ( q ), = st- t, Hee is the chain ule fo s. z s = ( e sin( q) )( t) + ( e cos( q) ) s s + t q = s + t. ( st-t) ( + ) se cos s t ( st-t ) = t Ê sin( s t ˆ Á e + ) + Ë s + t Now the chain ule fo t. z t = ( e sin( q) )( s - t) + ( e cos( q) ) t s + t ( st-t ) ( + ) te cos s t ( st-t ) = ( s- t Ê ) sin( s t ˆ Á e + ) + Ë s + t Eample 4 Use a tee diagam to wite down the chain ule fo the given deivatives. (a) dw fo w= f ( z,, ), = g ( t), = g ( t), and z = g ( t) dt w (b) fo w= f ( z,, ), = g ( st,, ), g ( st,, ) z = g st,, (a) dw dt fo w= f ( z,, ), = g ( t), = g ( t), and z = g ( t) So, we ll fist need the tee diagam so let s get that. =, and dw f d f d f dz = + + dt dt dt dt which is eall just a natual etension to the two vaiable case that we saw above. (b) w fo w= f ( z,, ), = g ( st), = g ( st), and z = g ( st),,,,,, Hee is the tee diagam fo this situation. Fom this it looks like the deivative will be, w f f f = + +

14 f Eample 5 Compute fo f (, ) if = cosq and = sinq. q We will need the fist deivative befoe we can even think about finding the second deivative so let s get that. This situation falls into the second case that we looked at above so we don t need a new tee diagam. Hee is the fist deivative. f f f = + q q q f =- sin + ( q) cos( q) Oka, now we know that the second deivative is, f Ê f ˆ Ê sin ( q) f cos( q) f ˆ = Á = Á- + q q Ë q q Ë The issue hee is to coectl deal with this deivative. Since the two fist ode deivatives, f and, ae both functions of and which ae in tun functions of and q both of these tems ae poducts. So, the using the poduct ule gives the following, f cos f Ê sin f ˆ sin f Ê q q q cos( q) f ˆ =- - Á - + Á q q Ë q Ë Ê f ˆ We now need to detemine what Á q Ë and Ê f ˆ Á will be. These ae both chain ule q Ë poblems again since both of the deivatives ae functions of and and we want to take the deivative with espect to q. Befoe we do these let s ewite the fist chain ule that we did above a little. f q ( f ) =- sin( q) ( f ) + cos( q) ( f ) Note that all we ve done is change the notation fo the deivative a little. With the fist chain ule witten in this wa we can think of () as a fomula fo diffeentiating an function of and with espect to q povided we have = cosq and = sinq. This howeve is eactl what we need to do the two new deivatives we need above. Both of the f fist ode patial deivatives, and f, ae functions of and and = cosq and = sinq so we can use () to compute these deivatives. f () Ê f ˆ Hee is the use of () to compute Á q Ë. Ê f ˆ Ê f ˆ Ê f ˆ Á =- sin( q) Á + cos( q) Á q Ë Ë Ë f =- sin + ( q) cos( q) f Hee is the computation fo Ê f ˆ Á. q Ë

15 Ê f ˆ Ê f ˆ Ê f ˆ Á =- sin( q) Á + cos( q) Á q Ë Ë Ë f =- sin + ( q) cos( q) f The final step is to plug these back into the second deivative and do some simplifing. f f Ê f f ˆ =-cos ( q) -sin( q) Á- sin( q) + cos ( q) - q Ë Ê sin( q) + cos( q) Á- sin( q) + cos( q) Ë f f f =- cos( q) + sin ( q) - sin( q) cos( q) - f f f f f f sin( q) - sin( q) cos( q) + cos ( q) f =-cos( q) -sin implicit diffeentiation Eample 6 Find d f f + - ( q) sin ( q) f sin cos cos d ( q) ( q) + ( q) f F + F = 0 fi =- d d F cos + = -e. d fo 5 The fist step is to get a zeo on one side of the equal sign and that s eas enough to do. cos e = 0 5 Now, the function on the left is (, ) find the deivative. d F in ou fomula so all we need to do is use the fomula to 5 cos e = d d sin 5 e Eample 7 Find and fo sin ( - 5z ) = + cos ( 6z ). This was one of the functions that we used the old implicit diffeentiation on back in the Patial Deivatives section. You might want to go back and see the diffeence between the two. F ˆ Fist let s get evething on one side. sin - 5 z - - cos 6 z = 0 Now, the function on the left is (,, ) F z and so all that we need to do is use the fomulas developed above to find the deivatives. z sin( - 5z) + 6zsin( 6z =- ) -5 cos - 5z + 6sin 6z z =- ( - ) - cos 5 cos 6 z z -5 cos - 5z + 6sin 6z 4

16 Diectional Deivatives To this point we ve onl looked at the two patial deivatives f (, ) and f (, ). Recall that these deivatives epesent the ate of change of f as we va (holding fied) and as we va (holding fied) espectivel. We now need to discuss how to find the ate of change of f if we allow both and to change simultaneousl. The poblem hee is that thee ae man was to allow both and to change. Fo instance one could be changing faste than the othe and then thee is also the issue of whethe o not each is inceasing o deceasing. So, befoe we get into finding the ate of change we need to get a couple of pelimina ideas taken cae of fist. The main idea that we need to look at is just how ae we going to define the changing of and/o. Definition f in the diection of the unit vecto u = ab, The ate of change of (, ) diectional deivative and is denoted b D f (, ) deivative is, D f u ( ) u is called the. The definition of the diectional ( +, + ) - (, ) f ah bh f, = lim h Æ0 h If we now go back to allowing and to be an numbe we get the following fomula fo computing diectional deivatives. u u (, ) = (, ) + (, ) D f f a f b (,, ) = (,, ) + (,, ) + (,, ) D f z f z a f z b f z c z Eample Find each of the diectional deivatives. D f,0 f, = e + and u is the unit vecto in the diection D f (a) whee u p of q =. Du f z,, v = -,0,. (b) whee (,, ) (a) (,0) whee (, ) u p q =. f z z z z = + - in the diection of f = e + and u is the unit vecto in the diection of We ll fist find D f (, ) and then use this a fomula fo finding D f (,0) u giving the diection is, Êp ˆ Êp ˆ u = cos Á,sin Á = -, Ë Ë u. The unit vecto So, the diectional deivative is, Ê ˆ D u f Á Ë e e Á e Ë Ê ˆ (, ) ( = Á + ) 5

17 Now, plugging in the point in question gives, Ê ˆ Ê ˆ 5 - D u f (,0) = Á- () + ( 5) = Ë Á Ë (b) D f ( z,, ) whee (,, ) u f z = z+ z - z in the diection of v = -,0,. In this case let s fist check to see if the diection vecto is a unit vecto o not and if it isn t convet it into one. To do this all we need to do is compute its magnitude. v = = 0 So, it s not a unit vecto. Recall that we can convet an vecto into a unit vecto that points in the same diection b dividing the vecto b its magnitude. So, the unit vecto that we need is, u = -,0, = -,0, The diectional deivative is then, Ê ˆ Ê ˆ D u f z,, = Á- z- z + 0 z - z + Á + z- Ë 0 Ë 0 = ( + 6z-- z+ z) 0 u (,, ) = (,, ) + (,, ) + (,, ) D f z f z a f z b f z c = z f, f, f g abc,, z Now let s give a name and notation to the fist vecto in the dot poduct since this vecto will show up fail egulal thoughout this couse (and in othe couses). The gadient of f o gadient vecto of f is defined to be, f = f, f, f o f = f, f z O, if we want to use the standad basis vectos the gadient is, f = f i + f j + f k o f = f i + f j z The definition is onl shown fo functions of two o thee vaiables, howeve thee is a natual etension to functions of an numbe of vaiables that we d like. With the definition of the gadient we can now sa that the diectional deivative is given b, D u f = fg u whee we will no longe show the vaiable and use this fomula fo an numbe of vaiables. Note as well that we will sometimes use the following notation, D f = fg u u Eample Find each of the diectional deivative. (a) D u f fo f (, ) = cos( ) in the diection of v =,. (b) D u f fo f ( z,, ) = sin( z) + ln ( ) at (,,p ) in the diection of v =,, -. 6

18 D f (a) fo f (, ) cos( ) u = in the diection of v =,. Let s fist compute the gadient fo this function. f = cos,- sin Also, as we saw ealie in this section the unit vecto fo this diection is, u =, 5 5 The diectional deivative is then, D u f = cos ( ),-sin( ) g, 5 5 = ( cos ( ) - sin ( ) ) 5 (b) D f u fo f ( z,, ) sin( z) ln ( ) = + at,,p in the diection of v =,, -. In this case ae asking fo the diectional deivative at a paticula point. To do this we will fist compute the gadient, evaluate it at the point in question and then do the dot poduct. So, let s get the gadient. f ( z,, ) =, zcos ( z), cos( z) f (,, p) =, p cos ( p),cos( p) =, -p,- Net, we need the unit vecto fo the diection, v = u =,,- Finall, the diectional deivative at the point in question is, D (,, ),,,, ( ) u f p = -p - g - = - p + -p = Theoem D f (and hence then the maimum ate of change of the function The maimum value of u ( f ( ) ) is given b f ) and will occu in the diection given b f. Eample Suppose that the height of a hill above sea level is given b z ,00 in what diection is the elevation = - -. If ou ae at the point changing fastest? What is the maimum ate of change of the elevation at this point? Now on to the poblem. Thee ae a couple of questions to answe hee, but using the theoem makes answeing them ve simple. We ll fist need the gadient vecto. f = , 0.04 The maimum ate of change of the elevation will then occu in the diection of f ( 60,00) = -.,- 4 The maimum ate of change of the elevation at this point is, f 60,00 = = 7.44 =

19 Applications of Patial Deivatives 8

20 Tangent Planes and Linea Appoimations The equation of the tangent plane to the suface given b z = f (, ) at ( 0, 0) z- z = f (, )( - ) + f (, )( - ) is then, Also, if we use the fact that z0 f ( 0, 0) z- f ( 0, 0) = f( 0, 0)( - 0) + f( 0, 0)( - 0) z = f (, ) + f (, )( - ) + f (, )( - ) Eample Find the equation of the tangent plane to = we can ewite the equation of the tangent plane as, z = ln + at -,. f, = ln + z = f -, = ln = 0 f(, ) = f( -,) = + f(, ) = f( -,) = + The equation of the plane is then, z- 0= z = + - linea appoimation to be, 0 One nice use of tangent planes is the give us a wa to appoimate a suface nea a point. As, then the tangent plane should neal appoimate the long as we ae nea to the point 0 0 function at that point. Because of this we defi ne th e li nea app oim ati on to b e, (, ) = (, ) + (, )( - ) + (, )( - ) L f f f and as long as we ae nea (, ) then we should have that, 0 0 (, ) ª (, ) = (, ) + (, )( - ) + (, )( - ) f L f f f

21 Eample Find the linea appoimation to 6 9 z = + + at ( 4,) -. So, we e eall asking fo the tangent plane so let s find that. f (, ) = + + f (- 4,) = + + = f(, ) = f( - 4,) =- 8 f(, ) = f( - 4,) = 9 The tangent plane, o linea appoimation, is then, L, = Gadient Vecto, Tangent Planes and Nomal Lines The equation of the tangent plane is then, f, - + f, - - z- z = O, upon solving fo z, we get, z = f, + f, - + f, the equation of the nomal line is, t =,, z + t f,, z Eample Find the tangent plane and nomal line to Fo this case the function that we e going to be woking with is, F z,, = + + z + + z = 0 at the point (, -,5). and note that we don t have to have a zeo on one side of the equal sign. All that we need is a constant. To finish this poblem out we simpl need the gadient evaluated at the point. F =,,z The tang ent plane is then, F, -,5 =, -4,0 ( ) ( ) ( z ) = 0 The nomal line is, t = - + t - = + t - - t + t,,5, 4,0, 4,5 0 0

22 Relative Minimums and Maimums Definition. A function f (, ) has a elative minimum at the point ( ab, ) if f (, ) f ( ab, ) all points(, ) in some egion aound ( ab, ). fo. A function f (, ) has a elative maimum at the point ( ab, ) if f (, ) f ( ab, ) all points(, ) in some egion aound ( ab, ). Definition The point ( ab, ) is a citical point (o a stationa point) of f (, ) following is tue, f ab, 0 Fact fo povided one of the. = (this is equivalent to saing that f ( ab, ) = 0 and. f ( ab, ) and/o f ( ab, ) doesn t eist. If the point ( ab, ) is a elative etema of the function f (, ) then (, ) point of f (, ) and in fact we ll have f ( ab, ) = 0. Fact f ab, = 0), ab is also a citical Suppose that ( ab, ) is a citical point of f (, ) continuous in some egion that contains (, ) and that the second ode patial deivatives ae ab. Net define, (, ) (, ) (, ) È (, ) D D ab f ab f ab f ab = = - Î We then have the following classifications of the citical point.. If 0, 0 D > and f( ab ) > then thee is a elative minimum at (, ) D > and f( ab, ) < 0 then thee is a elative maimum at (, ) D < then the point ( ab, ) is a saddle point. D = then the point (, ). If 0. If 0 4. If 0 ab. ab. ab ma be a elative minimum, elative maimum o a saddle point. Othe techniques would need to be used to classif the citical point.

23 Eample Find and classif all the citical points of f, = We fist need all the fist ode (to find the citical points) and second ode (to classif the citical points) patial deivatives so let s get those. f = - f = - f = 6 f = 6 f =- Let s fist find the citical points. Citical points will be solutions to the sstem of equations, f = - = 0 f = - = 0 This is a non-linea sstem of equations and these can, on occasion, be difficult to solve. Howeve, in this case it s not too bad. We can solve the fist equation fo as follows, - = 0 fi = Plugging this into the second equation gives, ( ) ( ) - = - = 0 Fom this we can see that we must have = 0 o =. Now use the fact that citical points. = = = fi 0: 0 0 0,0 = = = fi :, = to get the So, we get two citical points. All we need to do now is classif them. To do this we will need D. Hee is the geneal fomula fo D. (, ) (, ) (, ) È (, ) = - Î D f f f ( 6)( 6) ( ) = - - = 6-9 To classif the citical points all that we need to do is plug in the citical points and use the fact above to classif them. 0,0 : So, fo (, ) : D= D 0,0 =- 9< 0 0,0 D is negative and so this must be a saddle point. D= D, = 6-9= 7> 0 f, = 6> 0 Fo (, ) D is positive and f is positive and so we must have a elative minimum. Fo the sake of completeness hee is a gaph of this function.

24 Notice that in ode to get a bette visual we used a somewhat nonstandad oientation. We can s 0,0 we do get a ee that thee is a elative minimum at (, ) and (hopefull) it s clea that at s addle point. f, = Eample Find and classif all the citical points fo As with the fist eample we will fist need to get all the fist and second ode deivatives. f = 6-6 f = + -6 f = 6-6 f = 6-6 f = 6 We ll fist need the citical points. The equations that we ll need to solve this time ae, 6-6= = 0 These equations ae a little tickie to solve than the fist set, but once ou see what to do the eall aen t teibl bad. Fist, let s notice that we can facto out a 6 fom the fist equation to get, 6 - = 0 So, we can see that the fist equation will be zeo if = 0 o =. Be caeful to not just cancel the fom both sides. If we had done that we would have missed = 0. To find the citical points we. can plug these (individuall) into the second equation and solve fo the emaining vaiable = 0 : = : - 6 = - = 0 fi = 0, = - = - = fi =- = 0, So, if = 0 we have the following citical points, 0,0 0, and if = the citical points ae, (,) (-,)

25 ( 0, ) : (, ) : (-,) : D= D 0,0 = 6> 0 f 0,0 =- 6< 0 D= D 0, = 6> 0 f 0, = 6> 0 D= D, =- 6< 0 D= D -, =- 6< 0 So, it looks like we have the following classification of each of these citical points. 0,0 : Relative Maimum 0, : Relative Minimum, : Saddle Point -, : Saddle Point Hee is a gaph of the suface fo the sake of completeness. Eample Detemine the point on the plane 4- + z = that is closest to the point (,,5) - -. Fist, let s suppose that ( z,, ) is an point on the plane. The distance between this point and the point in question, (-,-,5), is given b the fomula, equation of the plane to see that, ( ) ( 5) d = z- Plugging this into the distance fomula gives, z = - 4+ ( ) ( 4 5) d = ( ) ( ) ( 4 4 ) =

26 So, let s instead find the minimum value of that is closest to (-,-,5). (, ) = = ( + ) + ( + ) + ( ) f d We ll need the deivatives fist. f = = f = = f f f = 4 = 0 =-6 Now, befoe we get into finding the citical point let s compute D quickl. D = = 84> 0 So, in this case D will alwas be positive and also notice that f = 4> 0 is alwas positive and so an citical points that we get will be guaanteed to be elative minimums. Now let s find the citical point(s). This will mean solving the sstem = = 0 To do this we can solve the fist equation fo. = 6-6 = Now, plug this into the second equation and solve fo ( 8-8) + 0 = 0 fi = Back substituting this into the equation fo gives =-. So, it looks like we get a single citical point : ( 4, 5 ) - -. Ê 4ˆ Ê 5ˆ 07 z = -4Á- + Á- = Ë Ë So, the point on the plane that is closest to (-,-,5) is ( 4, 5, 07 )

27 Absolute Minimums and Maimums In this section we ae going to etend the wok fom the pevious section. In the pevious section we wee asked to find and classif all citical points as elative minimums, elative maimums and/o saddle points. In this section we want to optimize a function, that is identif the absolute minimum and/o the absolute maimum of the function, on a given egion in. Note that when we sa we ae going to be woking on a egion in we mean that we e going to be looking at some egion in the -plane. In ode to optimize a function in a egion we ae going to need to get a couple of definitions out of the wa and a fact. Let s fist get the definitions out of the wa. Definitions. A egion in is called closed if it includes its bounda. A egion is called open if it doesn t include an of its bounda points.. A egion in is called bounded if it can be completel contained in a disk. In othe wods, a egion will be bounded if it is finite. Let s think a little moe about the definition of closed. We said a egion is closed if it includes its bounda. Just what does this mean? Let s think of a ectangle. Below ae two definitions of a ectangle, one is closed and the othe is open. Open Closed - 5< < - 5 < < 6 6 In this fist case we don t allow the anges to include the endpoints (i.e. we aen t including the edges of the ectangle) and so we aen t allowing the egion to include an points on the edge of the ectangle. In othe wods, we aen t allowing the egion to include its bounda and so it s open. In the second case we ae allowing the egion to contain points on the edges and so will contain its entie bounda and hence will be closed. This is an impotant idea because of the following fact. Eteme Value Theoem f, is continuous in some closed, bounded set D in then thee ae points in D, If (, ) and (, ) so that f (, ) is the absolute maimum and (, ) minimum of the function in D. f is the absolute Note that this theoem does NOT tell us whee the absolute minimum o absolute maimum will occu. It onl tells us that the will eist. Note as well that the absolute minimum and/o absolute maimum ma occu in the inteio of the egion o it ma occu on the bounda of the egion. 6

28 Finding Absolute Etema. Find all the citical points of the function that lie in the egion D and detemine the function value at each of these points.. Find all etema of the function on the bounda.. The lagest and smallest values found in the fist two steps ae the absolute minimum and the absolute maimum of the function. The main diffeence between this pocess and the pocess that we used in Calculus I is that the bounda in Calculus I was just two points and so thee eall wasn t a lot to do in the second step. Fo these poblems the majoit of the wok is often in the second step as we will often end up doing a Calculus I absolute etema poblem one o moe times. Eample Find the absolute minimum and absolute maimum of f, = on the ectangle given b - and -. Let s fist get a quick pictue of the ectangle fo efeence puposes. The bounda of this ectangle is given b the following conditions ight side : =, - left side : =-, - uppe side : =, - lowe side : =-, - These will be impotant in the second step of ou pocess. We ll stat this off b finding all the citical points that lie inside the given ectangle. To do this we ll need the two fist ode deivatives. f = - 4 f = 8- Note that since we aen t going to be classifing the citical points we don t need the second ode deivatives. To find the citical points we will need to solve the sstem, - 4 = 0 8- = 0 We can solve the second equation fo to get, = 4 7

29 Plugging this into the fist equation gives us, Ê ˆ - 4Á = - = ( - ) = 0 Ë 4 This tells us that we must have = 0 o =± =± Now, ecall that we onl want citical points in the egion that we e given. That means that we onl want citical points fo which -. The onl value of that will satisf this is the fist one so we can ignoe the last two fo this poblem. Note howeve that a simple change to the bounda would include these two so don t foget to alwas check if the citical points ae in the egion (o on the bounda since that can also happen). Plugging = 0 into the equation fo gives us, 0 = = 0 4 The single citical point, in the egion (and again, that s impotant), is get the value of the function at the citical point. f 0,0 = 4 0,0. We now need to Eventuall we will compae this to values of the function found in the net step and take the lagest and smallest as the absolute etema of the function in the ectangle. Now we have eached the long pat of this poblem. We need to find the absolute etema of the function along the bounda of the ectangle. What this means is that we e going to need to look at what the function is doing along each of the sides of the ectangle listed above. Let s fist take a look at the ight side. As noted above the ight side is defined b =, - Notice that along the ight side we know that =. Let s take advantage of this b defining a new function as follows, g = f, = = Now, finding the absolute etema of f (, ) the absolute etema of Calculus I. We find the citical points of along the ight side will be equivalent to finding g in the ange -. Hopefull ou ecall how to do this fom g in the ange at the citical points and the end points of the ange of s. - and then evaluate g( ) Let s do that fo this poblem. g ( ) = 8- fi = 4 This is in the ange and so we will need the following function evaluations. g 9 (- ) = g () = 7 g Á Ê ˆ = = 4.75 Ë4 4 Notice that, using the definition of g( ) these ae also function values fo (, ) f. 8

30 g f g() = f = - =, - =, 7 ʈ Ê ˆ 9 gá = f Á, = = 4.75 Ë4 Ë 4 4 We can now do the left side of the ectangle which is defined b, =-, - Again, we ll define a new function as follows, g = f -, = = Notice howeve that, fo this bounda, this is the same function as we looked at fo the ight side. This will not alwas happen, but since it has let s take advantage of the fact that we ve alead done the wok fo this function. We know that the citical point is = 4 and we know that the function value at the citical point and the end points ae, g 9 (- ) = g () = 7 g Á Ê ˆ = = 4.75 Ë4 4 The onl eal diffeence hee is that these will coespond to values of f, at diffeent points than fo the ight side. In this case these will coespond to the following function values fo f,. g f g() = f (- ) = - = -,- =, 7 ʈ Ê ˆ 9 gá = f Á-, = = 4.75 Ë4 Ë 4 4 We can now look at the uppe side defined b, =, - We ll again define a new function ecept this time it will be a function of. h = f, = = 8- We need to find the absolute etema of point(s). h on the ange -. Fist find the citical 0 h =- fi = The value of this function at the citical point and the end points is, h - = 7 h = 7 h 0 = 8 and these in tun coespond to the following function values fo f (, ) h( - ) = f (-,) = 7 h() = f (,) = 7 h( 0) = f ( 0,) = 8 Note that thee ae seveal epeats hee. The fist two function values have alead been computed when we looked at the ight and left side. This will often happen. Finall, we need to take cae of the lowe side. This side is defined b, =-, - 9

31 The new function we ll define in this case is, h = f, - = = 8+ The citical point fo this function is, h = fi = 6 0 The function values at the citical point and the endpoint ae, h - = h = h 0 = 8 and the coesponding values fo f (, ) ae, h( - ) = f (-,- ) = h() = f (, - ) = h( 0) = f ( 0, - ) = 8 The final step to this (long ) pocess is to collect up all the function values fo f (, ) we ve computed in this poblem. Hee the ae, f 0,0 = 4 f, - = f, = 7 Ê f Á, ˆ= 4.75 Ë 4 f ( -,) = 7 f ( -, - ) = Ê ˆ f Á-, = 4.75 Ë 4 f ( 0,) = 8 f ( 0, - ) = 8 The absolute minimum is at maimum occus at (, - ) and (, ) that 0,0 since gives the smallest function value and the absolute - - since these two points give the lagest function value. Hee is a sketch of the function on the ectangle fo efeence puposes. 0

32 f, = Eample Find the absolute minimum and absolute maimum of on the disk of adius 4, + 6 Fist note that a disk of adius 4 is given b the inequalit in the poblem statement. The less than inequalit is included to get the inteio of the disk and the equal sig n is included to get the As this eample has shown these can be ve long poblems. Let s take a look at an easie poblem with a diffeent kind of bounda. bounda. Of couse, this also means that the bounda of the disk is a cicle of adius 4. Let s fist find the citical points of the function that lies inside the disk. This will equie the following two fist ode patial deivatives. f = 4 f =- + 6 To find the citical points we ll need to solve the following sstem. 4 = = 0 This is actuall a fail simple sstem to solve howeve. The fist equation tells us that = 0 and the second tells us that 0, and this is =. So the onl citical point fo this function is inside the disk of adius 4. The function value at this citical point is, f 0, = 9 Now we need to look at the bounda. This one will be somewhat diffeent fom the pevious eample. In this case we don t have fied values of and on the bounda. Instead we have, + = 6 We can solve this fo and plug this into the in f (, ) = g = = - + to get a function of as follows. We will need to find the absolute etema of this function on the ange (this is the ange of s fo the disk.). We ll fist need the citical points of this function. g =- + fi = 6 6

33 The value of this function at the citical point and the endpoints ae, g - 4 =- 40 g 4 = 8 g = 5 Unlike the fist eample we will still need to find the values of that coespond to these. We can do this b plugging the value of into ou equation fo the cicle and solving fo. =- 4: = 6-6= 0 fi = 0 = = - = fi = 4 : = = - = fi =± =± : The function values fo g( ) then coespond to the following function values fo (, ) g( - 4) =-40 fi f ( 0, - 4) =-40 g( 4) = 8 fi f ( 0,4) = 8 () g = 5 fi f - 5, = 5 and f 5, = 5 Note that the thid one actuall coesponded to two diffeent values fo f (, ) poduced two diffeent values of. f. since that also So, compaing these values to the value of the function at the citical point of f (, ) found ealie we can see that the absolute minimum occus at ( 0, - 4) while the absolute maimum occus twice at ( 5,) - and 5,. Hee is a sketch of the egion fo efeence puposes. that we In both of these eamples one of the absolute etema actuall occued at moe than one place. Sometimes this will happen and sometimes it won t so don t ead too much into the fact that it happened in both eamples given hee. Also note that, as we ve seen, absolute etema will often occu on the boundaies of these egions, although the don t have to occu at the boundaies. Had we given much moe complicated eamples with multiple citical points we would have seen eamples whee the absolute etema occued inteio to the egion and not on the bounda.

34 Small changes If is a function of, i.e. = f (), and the appoimate change in coesponding to a small change δ in is equied, then: δ δ d d and δ d d δ o δ f () δ Eample.Given = 4, detemine the appoimate change in if changes fom to.0. Since = 4, then d d = 8 Appoimate change in, δ d δ (8 )δ d When = and δ = 0.0, δ [8() ](0.0) 0.4 [Obviousl, in this case, the eact value of d ma be obtained b evaluating when =.0, i.e. = 4(.0).0 =.46 and then subtacting fom it the value of when =, i.e. = 4() =, giving δ =.46 = Using δ = d δ above gave 0.4, which shows d that the fomula gives the appoimate change in fo a small change in.] Eample.The time of swingt of a pendulum is given b T = k l, whee k is a constant. Detemine the pecentage change in the time of swing if the length of the pendulum l changes fom. cm to.0 cm. If T = k l = kl, then dt dl = k l = k l Appoimate change in T, δt dt k dl δl δl l k ( 0.) l

35 small changes Futhe Poblem. Detemine the change in if changes fom.50 to.5 when (a) = (b) = 5 [(a) 0.0 (b) 0.008]. The pessue p and volume v of a mass of gas ae elated b the equation pv = 50. If the pessue inceases fom 5.0 to 5.4, detemine the appoimate change in the volume of the gas. Find also the pecentage change in the volume of the gas. [ 0.0,.6%]. Detemine the appoimate incease in (a) the volume, and (b) the suface aea of a cube of side cm if inceases fom 0.0 cm to 0.05 cm. [(a) 60 cm (b) cm ] 4. The adius of a sphee deceases fom 6.0 cm to 5.96 cm. Detemine the appoimate change in (a) the suface aea, and (b) the volume. [(a) 6.0 cm (b) 8.0 cm ] 5. The ate of flow of a liquid though a tube is given b Poiseuilles s equation as: Q = pπ4 whee Q is the ate of flow, p 8ηL is the pessue diffeence between the ends of the tube, is the adius of the tube, L is the length of the tube and η is the coefficient of viscosit of the liquid. η is obtained b measuing Q, p, and L. If Q can be measued accuate to ±0.5%, p accuate to ±%, accuate to ±% and L accuate to ±%, calculate the maimum possible pecentage eo in the value of η. [.5%] ates of change. An altenating cuent, i ampees, is given b i = 0 sin πft, whee f is the fequenc in hetz and t the time in seconds. Detemine the ate of change of cuent when t = 0 ms, given that f = 50 Hz. [000π A/s]. The voltage acoss the plates of a capacito at an time t seconds is given b v = Ve t/cr, whee V, C and R ae constants. Given V = 00 volts, C = F and R = find (a) the initial ate of change of voltage, and (b) the ate of change of voltage afte 0.5 s. [(a) 65V/s (b) 0.5V/s] 4. The pessue p of the atmosphee at height h above gound level is given b p = p 0 e h/c, whee p 0 is the pessue at gound level and c is a constant. Detemine the ate of change of pessue with height when p 0 = pascals and c = at 450 metes. [.65 Pa/m] tangents and nomals Fo the cuves in poblems to 5, at the points given, find (a) the equation of the tangent, and (b) the equation of the nomal. [ ] (a) = 4. = at the point (, ) (b) 4 + = 9. = at the point (, 8) [ ] (a) = 0 (b) 0 + = 8 (. = at the point, ) [ (a) = + ] (b) = 0 4. = + at the point (, 5) [ ] (a) = (b) = 0 5. θ = ( t at the point, ) [ ] (a) 9θ + t = 6 (b) θ = 9t 6 o θ = 7t 80. The luminous intensit, I candelas, of a lamp is given b I = V, whee V is the voltage. Find (a) the ate of change of luminous intensit with voltage when V = 00 volts, and (b) the voltage at which the light is inceasing at a ate of 0. candelas pe volt. [(a) 0.4 cd/v (b) 50V] 4

36 maimum and minimum. = [ ( 4 + Minimum at, )]. = θ(6 θ) [Maimum at (, 9)]. = [ ] Minimum (, 88); Maimum(.5, 94.5) 4. = 5 ln [Minimum at (0.4000,.86)] 5. = e [Maimum at (0.69, 0.66)] 6. = t t t + 4 Minimum at (,.5); ( Maimum at ), = 8t + t [Minimum at (0.5, 6)] 8. Detemine the maimum and minimum values on the gaph = cos θ 5 sin θ in the ange θ = 0toθ = 60. Sketch the gaph ove one ccle [ showing elevant points. ] Maimum of at 7, Minimum of at Show that the cuve = (t ) + t(t ) has a maimum value of and a minimum value of. maimum and minimum. The speed, v, of a ca (in m/s) is elated to time tsb the equation v = + t t. Detemine the maimum speed of the ca in km/h. [54 km/h]. Detemine the maimum aea of a ectangula piece of land that can be enclosed b 00 m of fencing. [90000 m ]. A shell is fied veticall upwads and its vetical height, metes, is given b = 4t t, whee t is the time in seconds. Detemine the maimum height eached. [48 m] 9. Resistance to motion, F, of a moving vehicle, is given b F = Detemine the minimum value of esistance. [44.7] 0. An electical voltage E is given b E = (5 sin 50πt + 40 cos 50πt) volts, whee t is the time in seconds. Detemine the maimum value of voltage. [4.7 volts]. The fuel econom E of a ca, in miles pe gallon, is given b: E = v v 4 whee v is the speed of the ca in miles pe hou. Detemine, coect to significant figues, the most economical fuel consumption, and the speed at which it is achieved. [50.0 miles/gallon, 5.6 miles/hou] 4. A lidless bo with squae ends is to be made fom a thin sheet of metal. Detemine the least aea of the metal fo which the volume of the bo is.5 m. [.4 m ] 7. The powe developed in a esisto R b a batte of emf E and intenal esistance is given b P = E R. Diffeentiate P with (R + ) espect to R and show that the powe is a maimum when R =. 8. Find the height and adius of a closed clinde of volume 5 cm which has the least suface aea. [ ] height = 5.4 cm; adius =.7 cm 5. A closed clindical containe has a suface aea of 400 cm. Detemine the dimensions fo maimum volume. [ ] adius = cm; height = 9. cm 6. Calculate the height of a clinde of maimum volume which can be cut fom a cone of height 0 cm and base adius 80 cm. [6.67 cm] 5