A REVIEW OF CALCULUS CONCEPTS FOR JDEP 384H. Thomas Shores Department of Mathematics University of Nebraska Spring 2007

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1 A REVIEW OF CALCULUS CONCEPTS FOR JDEP 384H Thoms Shores Deprtment of Mthemtics University of Nebrsk Spring 2007 Contents Rtes of Chnge nd Derivtives 1 Dierentils 4 Are nd Integrls 5 Multivrite Clculus 6 Rtes of Chnge nd Derivtives Life would be simple if every function were constnt. It isn't nd they ren't. Clculus llows us to get hndle on how functions chnge with their rgument. Recll this denition. Denition. A function f is rule of correspondence tht ssigns to ech element x in set D, clled its domin, unique vlue f (x) in set R, clled its rnge (or trget). We write f : D R in this sitution Exmple. Let D = [0, 5], the intervl of rel numbers x R such tht π x π, nd dene function f : D R by the formul f (x) = x 2 + 2x. Certinly this formul yields one nd only one well-dened vlue f (x) for ech choice of x. (There is something to check: could the denomintor be zero for some x in D? Answer is no.) So we hve function. This function could be model of, for exmple, totl cost for certin product s function of output x. We re interested in how this function chnges with x. A trditionl denition of mrginl cost sys tht it is the dditionl costs incurred by the production of one 1

2 A REVIEW OF CALCULUS CONCEPTS FOR JDEP 384H 2 dditionl unit of the product. Thus the mrginl costs t production level x re f (x + 1) f (x) = f (x + 1) f (x) 1 = f (x + x) f (x), x = 1. x This is just step wy from the clculus denition of mrginl cost, nmely, the derivtive of f (x) t x dened by f (x) = df (x) = lim dx x 0 f (x + x) f (x). x Recll tht the derivtive hs nother interprettion, more geometricl in nture, nmely, f () is the slope of the tngent line y = f () (x ) + f() to the curve y = f (x) t the point (, f ()) on the curve. Observe tht this tngent line is relly limit of secnt lines of the form y = f ( + h) f () h (x ) + f () where we let h 0. See the gure below for comprison. y y = f ()(x ) + f() x df 01 f y = f(+ x) f() x (x ) + f() y = f(x) + x x As we know (nd won't give too much detil here) there re mny useful rules of dierentition, e.g., for given functions f (x), g (x) nd

3 constnts, b, n, A REVIEW OF CALCULUS CONCEPTS FOR JDEP 384H 3 (f (x) + bg (x)) = f (x) + bg (x) (f (x) g (x)) = f (x) g (x) + f (x) g (x) ( ) f (x) = g (x) f (x) f (x) g (x) g (x) g (x) 2 f (g (x)) = f (g(x)) g (x) (x n ) = nx n 1 (e x ) = e x (ln (x)) = 1 x (sin (x)) = cos (x) (cos (x)) = sin (x) (rctn (x)) = x 2 nd so forth. Recll tht F (x) is n ntiderivtive of f (x) if F (x) = f (x). Any two ntiderivtives of f (x) dier by constnt, so generl formul for the ntiderivtives of f (x) is given by f (x) dx = F (x) + C, where C is constnt of integrtion. Thus, ech derivtive formul gives rise to n ntiderivtive formul. For exmple, the lst derivtive formul bove implies tht 1 dx = rctn (x) + C. 1 + x2 Exmple. Find n eqution of the tngent line to the curve f (x) = x 2 + 2x t the point on the curve where x = 3. Solution. First use the derivtive properties to clculte ( ) 3 f (x) = + (2x) = 6x 1 + x 2 (1 + x 2 ) Evlute nd nd tht f (3) = 6.3, f (3) = 1.82, so tht the tngent line is given by y = 1.82 (x 3) = 1.82x

4 A REVIEW OF CALCULUS CONCEPTS FOR JDEP 384H 4 Differentils Tke closer look t the denition of derivtive. We cn think of the rte of chnge over the intervl [, + h] given by f f ( + x) f () = x x s n pproximtion to the derivtive f (). Or, if we re relly interested in subsequent vlue f ( + x) beyond given f (), we cn think of the derivtive s giving n pproximtion to f by wy of the formul f df f () dx where we tke x = dx. The quntity df dened in the bove equlity is clled the dierentil of f (x) t x =. In generl, we dene df = df (x, dx) f (x) dx. The dierentil is relly function of the independent vribles x nd dx.there is nice geometricl picture tht one cn drw tht shows tht we obtin the vlues df nd f from the tngent nd secnt curves t x. For smll vlues of dx the dierentil provides n excellent pproximtion to f nd conversely. Refer to the gure bove nd identify f nd df in the picture. Exmple. Use the clcultions of the previous exmple to pproximte f (2) nd f (4) using dierentils nd the vlues of f, f t x = 3. Solution. We obtin tht with dx = 1, so tht f = f (3 + dx) f (3) df (3, 1) = f (3) 1 = 1.82, f (3 + 1) f (3) = As mtter of fct, f (4) = For dx = 1, we obtin similrly tht so tht f = f (3 + dx) f (3) df (3, 1) ( 1) = 1.82, f (3 1) f (3) 1.82 = As mtter of fct, f (4) = nd f (2) = 4.2. Approximting vlues with dierentils mounts to using liner pproximtion to f (x) which is incresingly ccurte ner x =. The ide is tht for x ner, f (x) f () + f () (x ).

5 A REVIEW OF CALCULUS CONCEPTS FOR JDEP 384H 5 One cn pply this rgument to higher derivtives nd integrte (next section) to obtin the fmous Tylor formul f (x) f ()+f () (x )+f (x )2 () + +f (n) (x )n () P n (x). 2! n! In fct the error of pproximtion is well understood. One form of it is R n (x) f (x) P n (x) = f (n+1) (ξ) where ξ is some number between nd x. Are nd Integrls (x )n+1, (n + 1)! Let A (x) be the signed re between the curve y = f (x) nd the x-xis, with verticl line boundries t x = nd x. Thus A () = 0. We lso write A (x) = x f (x) dx. This is motivted by the pproximte equlity A (x + dx) A (x) f (x) dx whose ccurcy increses to equlity s dx 0. A grph of the re shows why this is so, so exmine the following gure. y 01 y = f(x) A(x) 1100 (x, f(x)) x x + x If we divide by dx nd pss to the limit, we see tht d A (x) = f (x). dx This is one form of the fundmentl theorem of clculus (FTOC). The other form follows from this rgument: As we sw erlier, ny two x

6 A REVIEW OF CALCULUS CONCEPTS FOR JDEP 384H 6 ntiderivtives of f (x) dier by constnt, so if F (x) = f (x), then F (x) + C = A (x), where C is some constnt. It follows tht b f (x) dx = A (b) = A (b) A () = (F (b) + C) (F () + C) = F (b) F (), which gives the second form of FTOC: If F (x) = f (x) is continuous on the intervl [, b], then b f (x) dx = F (b) F () F (x) b x= Exmple. Let f (x)be s in the rst exmple, nd clculte 4 f (x) dx. 0 Solution. Here we hve to nd n ntiderivtive f (x) which we write in the customry indenite integrl form F (x) = f (x) dx. We leve it to the reder to check tht ( ) x + 2x dx = 3 2 dx 1 + x 2 +2 x dx = 3 rctn (x)+2 x2 2 +C where C is n rbitrry constnt of integrtion. From this we deduce tht 4 ( ) x + 2x dx = 3 rctn (x) + x x=0 0 Perhps the rst form of the denite integrl tht you sw ws s limit of Riemnn sums, of which the following is specil cse: Let = x 0 < x 1 < x N = b with x j+1 x j = x b N b N 1 f (x) dx = lim f (x j ) x. x 0 Of course, this gin gives us the signed re between y = f (x) nd the x-xis. This is left Riemnn sum. The corresponding right Riemnn sum is N j=1 f (x j) x. A rther fortunte turn of events occurs when we verge the left nd right Riemnn sums for given N. j=0 Multivrite Clculus Life would be simpler if ll functions involved one vrible. They don't. For exmple, the volume of right circulr cylinder of rdius x nd height y is function of two vribles f (x, y) = πx 2 y. Here we understnd tht the domin of the function V consists of points (x, y) such tht x nd y re non-negtive. How do we mke sense of

7 A REVIEW OF CALCULUS CONCEPTS FOR JDEP 384H 7 rtes of chnge of such vribles when there re two (or more!) independent vribles. The nswer is tht we tke derivtives with respect to ech independent vrible seprtely, treting ll other vribles s constnt, nd use our single vrible rules. Such derivtives re prtil derivtives since they only tell us prt of the rte of chnge informtion bout f. Thus, in our exmple, f (x, y) x = 2πxy f y (x, y) = πx2. There is nice interprettion of these derivtives s simply ordinry derivtives of functions of one vrible obtined by intersection the surfce z = f (x, y) with verticl plnes prllel to the x- or y-xes. Similrly, there re higher nlogues of integrls nd dierentils. We won't go into detil here, but in nutshell, the double integrl over region R in the xy-plne of continuous function f (x, y) dened on tht region is number f (x, y) da R tht represents the signed volume between the grph of z = f (x, y), (x, y) R nd the xy-plne with verticl sides long the boundry of R. Finlly, there is the importnt ide of dierentils for functions of more thn one vrible. Just s dierentils represent tngent line pproximtions to curve y = f (x), dierentils for function of two vribles represent tngent plne pproximtions to surfce z = f (x, y). Here is the denition of dierentil for function f (x, y) of two vribles with continuous prtil derivtive: df = f f dx + x y dy. This denition is completely nlogous to dierentils in one vrible. It should be noted tht df is relly function of the four independent vribles x, y, dx nd dy. Just for the record, the denition bove gives rise to kind of chin rule for certin functions of two rguments. Suppose tht we know tht x = x (t) nd y = y (t) re both functions of t, so tht f = f (x (t), y (t))is relly function of the single independent vrible t. Then wht is df/dt? The nswer is chin rule for function of t tht hs two intermedite vribles x, y:

8 A REVIEW OF CALCULUS CONCEPTS FOR JDEP 384H 8 df dt = f dx x dt + f dy y dt. Exmple. Given tht f (x, y) = x 2 y + y 3 + x 2, nd formul for the dierentil df nd in prticulr, for the dierentil evluted t x = 2, y = 1. How does this help you describe the tngent plne pproximtion to z = f (x, y) for (x, y) ner the point (2, 1)? Solution. In this cse, f/ x = 2xy + 2x nd f/ y = x 2 + 3y 2. Therefore, we hve df = f f dx + x y dy = (2xy + 2x) dx + ( x 2 + 3y 2) dy so tht df (2, 3, dx, dy) = 8dx + 7dy. In prticulr, if we tke dx = x 2 nd dy = 1, then we obtin the expression df = 8 (x 2) + 7 (y 1) nd if we interpret df f for points (x, y) ner (2, 1), then we obtin the expression f (x, y) z, where z f (2, 1) = z 9 = 8 (x 2) + 7 (y 1), tht is, z = 8x + 7y 14, which is plne contining the point (2, 1, f (2, 1)) nd is in fct the eqution of the tngent plne to the surfce t (2, 1, 9).

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