Lesson 11 MA Nick Egbert

Transcription

1 Lesson MA 62 Nick Eert Overview In this lesson we return to stndrd Clculus II mteril with res etween curves. Recll rom irst semester clculus tht the deinite interl hd eometric menin, nmel the re under curve. The red reion in the rph elow is iven () d. Lesson In this lesson we re concerned with slihtl dierent prolem, nmel the re etween two curves. For emple, consider the rphs o nd elow, nd let s s we wnt to clculte the re ounded the two curves etween nd. I we clculte the re under ech curve seprtel, we ind the reen nd red res in the two rphs elow.

2 Lesson MA 62 Nick Eert Lin one on top o the other it s prett cler (i it wsn t lred) tht the re etween the curve nd the -is is counted twice. So i we tke the reen re nd sutrct the red re we precisel et the re etween the two curves. Usin interls to clculte, we hve Are () d (() ()) d. () d 2

3 Lesson MA 62 Nick Eert More enerll, i we wnt to ind the re etween two curves on n intervl, ], we do this computin Are Top Bottom (() ()) d Emple. Find the re o the reion ounded the curves () 2 nd () Solution. We wnt to interte the top minus the ottom. So it would e constructive (s with mn interl prolems) to rph nd. The i picture rph looks like this. But tht s hrd to see, so let s zoom in. 4/ Now we cn see tht strts out ove, then dips elow eore comin ck up in. So or the re etween the two, is the top unction. So we wnt ( ) d.

4 Lesson MA 62 Nick Eert But wht re nd? Points o intersection occur when the two unctions re equl. So settin () (), ( ) ( 4)( ), which hs solutions o nd 4. So there s our nd. Now onl to clculte 4/ ( ) d 92 4/ ln 2 7 ( ) ln. Intertin with respect to is nothin specil, it s just wht we usull do. Some prolems will lend themselves well to intertin with respect to insted, nd or others the onl w to solve the prolem will e intertin with respect to. How does this chne the prolem t hnd? The roles o nd switch. Given two curves nd which re unctions o, we wnt to interte the one with lrer vlues minus the one with the smller vlues. d c So we could mend our previous Top minus Bottom to Bier minus Smller. Speciicll or unctions o, we hve Are d c d c Riht d c Let (F () G()) d 4

5 Lesson MA 62 Nick Eert Emple 2. Find the re ounded the curves 2 nd +. Solution. The rphs in this emple re it esier to drw. I drwin the prol ives ou troule, tr rphin it s i it were 2 then lippin our pper lon the line To iure out where the two unctions intersect, we set them equl to ech other. We were iven the line s the eqution +, so irst we wnt to solve this or :. Now ( )( ). So the -intercepts re,. These re the ounds or our interl, nd now we use Riht minus let. ( 2 ) ( ) ] d ] d ] () + () 2 () 288 Emple. Find the re ounded the curves ] nd where. 5

6 Lesson MA 62 Nick Eert Solution. Let s cll the cuic one nd the qudrtic. To ind where nd intersect, we set them equl to ech other. ( + 2)( 9) So since we re stickin with, the points o intersection re, 9. Now rphin the two polnomils should e reeze. 9 Now to clculte the re etween nd, 9 ( ) d 9 9 ( ) ( 5) ] d ] d (9)4 + 7 (9) + 9(9) Emple 4. Find the eqution o the horizontl line tht divides the re o the reion ounded 8 2 nd in hl. Solution. We deinitel wnt to rph this one. We hve prol with verte t (, 8) nd the zeros re t ± 8 ± 2. Let s cll the horizontl line we re ter k. 9 6

7 Lesson MA 62 Nick Eert 8 k So we wnt the re in the ech o those two res to e equl to hl the re etween 8 2 nd. Tht mens we should irst iure out wht the totl re: 2 2 (8 2 ) d (8 2 ) d ( 2) Note. In the irst line we hve used tht 8 2 is smmetric out the -is. Tht mens tht whenever we plu in or we et the sme vlue. You could hve skipped tht step nd just evluted the interl normll. Thus we wnt the re etween 8 2 nd k to e equl to One w we could do this is to mke up n interl in terms o. I 8 2 then ± 8. So we could think o the prol s two unctions o put toether. So the re is iven Riht minus Let. The 7

8 Lesson MA 62 Nick Eert ounds re esil seen to e k nd 8. Now k 8 ] 8 ( 8 ) d k 8 k 8 k d u/2 u /2 du u /2 du 8 k u 8 du d 4 (8 k)/2 At this point we just need to do little ler to solve or k: (8 k)/ (8 k) / (8 k) /2 ( 27 2/ 2) 8 k ( k k ) 2/ Remrk. We could hve done this lst prt s n interl in terms o. In tht cse we would hve hd (8 2 ) k ] d, where nd re the vlues where the prol nd line intersect. How do we ind those points? Well in the worked solution we solved the prolic eqution or. We know tht the vlue is equl to k where the intersect, so the -vlues re ± 8 k. Then the interl we would set up would e 8 k 8 2 k ] d. 8 k I ou compute this, ou should still et 4 (8 k)/2, then the rest is the sme s eore. 8

9 Lesson MA 62 Nick Eert Remrk. Althouh we hven t covered n word prolems, homework hs ew, ut the re prett strihtorwrd pplictions. It s worth recllin ew deinitions thouh. In the contet o economics, revenue is the totl mone rouht in, cost hs the ovious deinition, nd proit is revenue minus cost. We should urther recll tht n time we re iven rte o somethin per somethin else, computin the deinite interl ives us n nswer in somethins. 9