Further applications of integration UNCORRECTED PAGE PROOFS

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1 . Kick off with CAS. Integrtion recognition. Solids of revolution. Volumes Further pplictions of integrtion. Arc length, numericl integrtion nd grphs of ntiderivtives.6 Wter flow.7 Review

2 . Kick off with CAS <To come> Plese refer to the Resources t in the Prelims section of our ebookplus for comprehensive step--step guide on how to use our CAS technolog.

3 . Integrtion recognition Deducing n ntiderivtive Becuse differentition nd integrtion re inverse processes, differentiting function f() with respect to will result in nother function g(). It follows tht n ntiderivtive of the function g() with respect to is the function f() with the ddition of n ritrr constnt c. In mthemticl nottion, if d ( f()) = g(), then d WorKeD exmple think g() d = f() + c. However, functions m not e ectl the derivtives of other functions, ut m differ constnt multiple. Let k e non-zero constnt. If d ( f()) = kg(), it follows tht d kg() d = k g() d = f(), since constnt multiples cn e tken outside integrl signs. Then, dividing oth sides the constnt multiple k, we otin g() d = f() + c, s we cn dd in the ritrr constnt t the end. This is k known s integrtion recognition. Differentite " + 6 nd hence find d. " + 6 WritE Write the eqution in inde form. Let = " + 6. = ( + 6) Epress in terms of u nd u in terms of. = u where u = + 6 Differentite with respect to u nd u with respect to. Find d d using the chin rule. d Sustitute ck for u nd cncel fctors. 6 Write the result s derivtive of one function. d du = u nd du d = = "u d = d du du d =!u d d = d d c" + 6 d = " + 6 " Use integrtion recognition. d = " + 6 " Mths Quest specilist MtheMtICs VCe units nd

4 8 Tke the constnt fctor outside the integrl sign. 9 Divide the constnt fctor, dd in the ritrr constnt +c, nd stte the finl result. d = " + 6 " + 6 " + 6 d = " c WorKeD exmple think Integrting other tpes of functions For the emple ove, integrtion techniques developed in erlier topics could hve een used to find the ntiderivtive. However, there re functions for which n ntiderivtive could not hve een otined using the rules developed so fr. Also, this method of deducing n ntiderivtive cn e used to evlute definite integrls or find res. To understnd the emples tht follow, it is necessr to review the differentition techniques developed in erlier topics. Recll the results for the derivtives of the inverse trigonometric functions. If is positive constnt, then: d d sin = d d cos = " for < " for < d d tn = for R. + Differentite cos! nd hence evlute d. "6 WritE Epress in terms of u nd u in terms of. Let = cos! Differentite with respect to u nd u with respect to. Find d d using the chin rule. d Sustitute ck for u. 8 = cos u where u = " = d du = nd du "6 u d = = " d = d du du d =! "6 u d d =!!6 d d = "6 since < < 6 topic Further PPLICtIons of IntegrtIon 97

5 Write the result s derivtive of one function. d d ccos! d = "6 6 Use integrtion recognition. d = cos! "6 7 Tke the constnt fctor outside the integrl sign. 8 Multipl the constnt fctor nd dd in the ritrr constnt +c. 9 However, in this cse we re required to evlute definite integrl. Sustitute in the upper nd lower terminls nd simplif. WorKeD exmple think Differentite cos() nd hence find sin() d. WritE Write the eqution. Let = cos() = u. v. Stte the functions u nd v. u = nd v = cos() Differentite u nd v with respect to. "6 d = cos! d = cos! "6 + c 8 d = c cos " 8 "6 d cos!8 ( cos ()) = cos! ( cos ()) Evlute. = π + π Stte the finl vlue of the definite integrl. d = π "6 8 using the product rule Some prolems require comintion of integrtion recognition nd the product rule. This is tpicl when integrting products of mied tpes of functions. Recll tht the product rule sttes tht if u = u() nd v = v() re two differentile functions of nd = u.v, then d d = u dv d + v du d. du dv = nd = sin() d d 98 Mths Quest specilist MtheMtICs VCe units nd

6 Find d using the product rule. d Sustitute for u, dv du, v nd d d. Write the result s derivtive of one function. d d = u dv d + v du d d = sin() + cos() d d [ cos()] = sin() + cos() d 6 Use integrtion recognition. ( sin() + cos()) d = cos() 7 Seprte the integrl on the left into two seprte integrls nd tke the constnt fctor outside the integrl sign. 8 Trnsfer one prt of the integrl which cn e performed on the right hnd side. 9 Perform the integrtion on the term on the right hnd side. Divide the constnt fctor, dd in the ritrr constnt +c, nd stte the finl result. WorKeD exmple think sin() d + cos() d = cos() sin()d = cos() cos() d sin() d = cos() sin() sin() d = cos() + sin() + c 6 Integrting inverse trigonometric functions The technique used in the lst emple cn lso e used to find ntiderivtives of inverse trigonometric functions. Find d d [ cos ()] nd hence find cos () d. WritE Write the eqution. Let = cos () = u. v. Stte the functions u nd v. u = nd v = cos () Differentite u nd v with respect to. Find d using the product rule. d Sustitute for u, dv du, v nd d d. du dv = nd d d = " d d = udv d + vdu d d d = + cos () " topic Further PPLICtIons of IntegrtIon 99

7 Write the result s derivtive of one function. d d cos () = + cos () " 6 Use integrtion recognition. q " + cos ()r d = cos () 7 Seprte the integrl on the left into two seprte integrls. 8 Trnsfer one prt of the integrl which cn e performed on the right-hnd side. 9 Consider now just the integrl on the right. Write the integrnd s power using inde lws. Use non-liner sustitution. Let t =. The integrl cnnot e done in this form, so differentite. Epress d in terms of dt inverting oth sides. Sustitute for d, noting tht the terms involving will cncel. Trnsfer the constnt fctors outside the front of the integrl sign. The integrl cn now e done. Antidifferentite using tn dt = tn+ n + with n =, so tht n + =. d + cos () d = cos () " d " = ( ) d = t d t = dt d = 8 d dt = 8 d = 8 dt = t 8 dt = t dt = t + c cos () d = cos () + d " 6 Write the epression with positive indices. =!t + c Mths Quest SPECIALIST MtheMtICS VCE Units nd

8 7 Sustitute ck for nd stte the result for this prt of the integrl. 8 Sustitute for the integrl, dd in the ritrr constnt +c, nd stte the finl result. d = " " cos () d = cos () " + c Eercise. Integrtion recognition PRctise Consolidte WE Differentite "(9 + 6) nd hence find "9 + 6 d. Determine d d [cos ()] nd hence find sin()cos () d. WE Differentite sin! nd hence evlute d. " If f() = rctn!, find f () nd hence evlute!( + 6) d. WE Find d d [ sin()] nd hence find cos() d. 6 Differentite e nd hence find e d. 7 WE Find d d [ sin ()] nd hence find sin () d. 8 Differentite tn () nd hence evlute rctn() d. 9 Differentite sin nd hence find cos d. Differentite cos() nd hence find sin() d. c Differentite e nd hence find e d. Determine d d c sin d nd hence find sin d. 8 6 Determine d d [ cos ()] nd hence find cos () d. c Determine d d c tn d nd hence find tn d. Topic Further pplictions of integrtion

9 If f() = sin(), find f ()nd hence evlute cos() d. If f() = cos(6), find f () nd hence evlute sin() cos()d. c If f() = e, find f (). Hence determine the shded 8 re ounded the grph of = e 6, the origin, the -is nd the line =, s shown. Determine d d ccos.... d nd hence evlute! d. " 6 Determine d d ctn! d nd hence evlute d. "(9 + ) c Determine d d csin d nd hence evlute d. " Differentite cos (). Hence determine the shded re ounded the grph of = cos (), the coordintes es nd =, s shown. Differentite sin () nd hence find the re ounded the curve = sin (), the -is, the origin nd the line =. c Differentite tn nd hence find the re ounded the grph of = tn, the -is, the origin nd the line =. Assuming tht is positive rel constnt: differentite e nd hence find e d π 8 π differentite sin() nd hence find cos() d c differentite cos() nd hence find sin() d. Mths Quest SPECIALIST MtheMtICS VCE Units nd

10 Use our results from question to evlute cos ()d. Use our results from question to evlute sin ()d. c Determine d d [( + ) log e ( + )] nd hence evlute 6 Assuming tht is positive rel constnt: differentite sin nd hence find sin d differentite cos nd hence find cos d c differentite tn nd hence find tn d. 7 Determine d d [ log e ()] nd hence find log e () d. Determine d d [ log e ()] nd hence find log e () d. c Determine d d [ log e ()] nd hence find log e () d. d Hence deduce n log e () d. Wht hppens if n =? π 8 π 9 log e ( + ) d. 8 If f() = e ( cos() + sin()), find f () nd hence find e cos() d. Determine d d [e ( cos() + sin())] nd hence find e sin() d. c Let f() = e ( cos() + sin()) find f () nd hence find e cos() d. Mster d Let g() = e ( sin() cos()) find g (). Hence find the re enclosed etween the curve = e sin(), the -is, the origin nd the first intercept it mkes with the positive -is. 9 Assuming tht is positive rel constnt: find the derivtive of tn nd hence find tn d Topic Further pplictions of integrtion

11 . i use the sustitution = sin(θ)to show tht d = " sin " ii find the derivtive of sin nd hence show tht sin d = sin + " + c. Find d d [ log e (tn() + sec())]. Hence, find the re enclosed etween the curve = sec(), the coordinte es nd the line = π. Given tht is positive rel constnt, find d d [ log e (cosec() + cot())]. Hence, find the re enclosed etween the curve = cosec(), the -is nd the lines = π 6 nd = π. sin() + cos() d c If = log e, show tht = sec(). Hence, find Å sin() cos() d the re enclosed etween the curve = sec(), the -is nd the lines = π 6 nd = π. Solids of revolution Rottions round the -is Suppose tht the curve = f() is continuous on the closed intervl. Consider the re ounded the curve, the -is nd the ordintes = nd =. If this re is rotted 6 out the -is, it forms solid of revolution nd encloses volume V. P(, ) = f() = f() Consider point on the curve with coordintes P(, ). When rotted it forms circulr disc with rdius of nd cross-sectionl re of A( ) = π. If the disc hs width of Δ, then the volume of the disc is the cross-sectionl re times the Mths Quest SPECIALIST MtheMtICS VCE Units nd

12 width, tht is A( )Δ. The totl volume, V, is found dding ll such discs etween = nd = nd tking the limit s Δ. This is given V = lim Δ = = π Δ = π d. We cn use the ove results to verif the volumes of some common geometricl shpes. Volume of clinder The volume of clinder of height h nd rdius r is given πr h. To verif this result, consider clinder ling on its side. r h r h To form the clinder, rotte the line = r 6 out the -is etween = h nd the origin. Since h is constnt nd r is constnt, the cn e tken outside the integrl sign. h V = π d h = π r d h = πr d = πr [] h = πr (h ) = πr h Volume of cone The volume of cone of height h nd rdius r is given πr h. To verif this result, it is esiest to consider the cone ling on its side. Consider line which psses through the origin, so its eqution is given = m. Now rotte the line 6 out the -is, etween = h nd the origin, so tht it forms cone. h r r h r h Topic Further pplictions of integrtion

13 The grdient of the line is m = r h, so the line hs the eqution = r. The volume is given h h V = π d = π h r h h d = πr h d = πr h c h d = πr h h = πr h Volume of sphere The volume of sphere of rdius r is given πr. To verif this result, consider the circle + = r with centre t the origin nd rdius r. If we rotte this circle 6 or rotte the top hlf of the circle = "r out the -is, etween = r nd = r, it forms sphere. r r r r V = π r r = π r r r d (r )d smmetr = π r (r )d = π cr d = π r r = πr r r r 6 Mths Quest SPECIALIST MATHEMATICS VCE Units nd

14 solid volumes formed rotting curves out the -is When rotting the curve = f() out the -is etween = nd =, the volume formed is given V = π d. The integrnd must e in terms of constnts nd -vlues onl. WorKeD exmple think The re ounded the curve =, the -is, the origin nd the line = is rotted out the -is to form solid of revolution. Find the volume formed. Sketch the grph nd identif the re to rotte. Write definite integrl which gives the volume. WritE/drW V = π d Antidifferentite using the rules. V = π c d =, =, = so = V = π d Evlute the definite integrl. V = π c d V = π Stte the volume. The volume formed is π cuic units. topic Further PPLICtIons of IntegrtIon 7

15 solid volumes formed rotting curves out the -is = f() = f() P (, ) WorKeD exmple 6 think When the curve = f( ) is rotted out the -is etween = nd =, the volume formed is given V = π d. The integrnd must e in terms of constnts nd -vlues onl. The re ounded the curve =, the -is, the origin nd the line = is rotted out the -is to form solid of revolution. Find the volume formed. Sketch the grph nd identif the re to rotte. Write definite integrl which gives the volume. WritE/drW 6 6 V = π d =, =, = V = π d Antidifferentite using the rules. V = π c d 8 Mths Quest specilist MtheMtICs VCe units nd

16 Evlute the definite integrl. V = π[8 ] V = 8π Stte the volume. The volume formed is 8π cuic units. WorKeD exmple 7 pplictions The volumes of some common geometricl ojects cn now e found using clculus, rotting lines or curves out the - or -es nd using the ove techniques. A drinking glss hs se dimeter of cm, top dimeter of 7 cm nd height of cm. Find the volume of the glss to the nerest ml. think Sketch the grph nd identif the re to rotte. Write the coordintes of the points A nd B. 7 WritE/drW Estlish the grdient of the line segment AB m = joining the points A, nd B 7,. 7 = Estlish the eqution of the line segment joining the points A nd B. Rotte this line out the -is to form the required volume. = A 7 (, ) B = m + c (, ) = The glss is formed when the line = for 7 is rotted out the -is. V = π d =, = topic Further PPLICtIons of IntegrtIon 9

17 Rerrnge to mke the suject. = = + = ( + ) = ( + ) 6 Write definite integrl which gives the volume. EErcisE. PrctisE Solids of revolution V = π 8 ( + ) d π 7 Antidifferentite using the rules. V = 8 c ( + ) d 8 Evlute the definite integrl. V = π 9 (77 ) 9 Since cm = ml, stte the volume of the glss. WE The re ounded the curve =!, the -is, the origin nd the line = is rotted out the -is to form solid of revolution. Find the volume formed. The re ounded the curve = nd the coordinte is is rotted out the -is to form solid of revolution. Find the volume formed. WE6 The re ounded the curve =!, the -is, the origin nd the line = is rotted out the -is to form solid of revolution. Find the volume formed. The re ounded the curve = nd the coordinte is is rotted out the -is to form solid of revolution. Find the volume formed. WE7 A plstic ucket hs se dimeter of cm, top dimeter of 6 cm nd height of cm. Find the volume of the ucket to the nerest litre. 6 V 99.9π cm The volume of the glss is ml. Mths Quest specilist MtheMtICs VCe units nd

18 6 A soup owl hs se rdius of 6 cm, top rdius of 8 cm, nd height of 7 cm. The edge of the owl is prol of the form = + c. Find the cpcit of the soup owl. = + c 8 Consolidte Find the volume of the solid of revolution formed when the re etween the line =, the -is, the origin nd = is rotted 6 out the -is. Find the volume of the solid of revolution formed when the re etween the line =, the -is, the origin nd = is rotted 6 out the -is. c A cone is formed rotting the line segment of + = 6 cut off the coordinte es out: i the -is ii the -is. Find the volume in ech cse. 8 If the region ounded the curve = sin(), the origin, the -is nd the first intercept it mkes with the -is is rotted 6 out the -is, find the volume formed. If the region ounded the curve = cos(), the coordinte es nd the first intercept it mkes with the -is is rotted 6 out the -is, find the volume formed. c If the region ounded the curve = sec(), the coordinte es nd = π is rotted 6 out the -is, find the volume formed. 8 d If the region ounded the curve = e, the coordinte es nd = is rotted 6 out the -is, find the volume formed. 9 i If the re etween the curve = +, the coordinte es nd the line = is rotted 6 out the -is, find the volume formed. ii If the re etween the curve = +, = nd = is rotted out the -is, find the volume formed. i For the curve = 9, find the re etween the curve nd the -is. ii If the re descried in i is rotted 6 out the -is, find the volume formed. iii If the re descried in i is rotted 6 out the -is, find the volume formed. Find the volume formed rotting the ellipse + 6 = out: i the -is ii the -is. i Determine the volume formed if the ellipse + =, hving semi-mjor nd semi-minor es nd respectivel, is rotted out the -is. Topic Further pplictions of integrtion

19 ii Determine the volume if the ellipse from prt i is rotted out the -is. iii If =, verif tht our results from i nd ii give the volume of sphere. c An egg cn e regrded s n ellipsoid. The egg hs totl length of 7 mm nd its dimeter t the centre is mm. Find its volume in cuic millimetres. The drive shft of n industril spinning mchine is metres long nd hs the form of the curve = e mesured in metres, rotted 6 out the -is etween the -is. i Find the volume in cuic metres. ii Wht is the length of similr shft if it encloses volume of.π m? A piece of plstic tuing hs its oundr in the form of the curve =. " + 9 When this curve is rotted 6 out the -is etween the -is nd the line =, find its volume. A wine glss is formed when the rc OB with the eqution = is rotted out the -is, s shown. The dimensions of the glss re given in cm. Find the volume of the glss in ml. B = The digrm shows wine rrel. The rrel hs totl length of 6 cm, totl height of 6 cm t the middle nd totl height of cm t the ends. i If the upper rc cn e represented prolic oundr, show tht its eqution is given = + for. 9, Mths Quest SPECIALIST MtheMtICS VCE Units nd

20 ii If the rc is rotted out the -is, find the cpcit of the wine rrel to the nerest litre. A vse hs se rdius of 6 cm, its top rdius is 8 cm nd its height is cm. Find the volume of the vse in cuic centimetres if the side of the vse is modelled : i stright line of the form = + c ii prol of the form = + c iii cuic = + c iv qurtic of the form = + c. 6 8 Another vse is modelled rotting the curve with the eqution ( ) = out the -is etween the -is nd =. Find the volume of wter needed to completel fill this vse. For the curve = e where > : i find the re A ounded the curve, the coordinte es nd = n ii if the region in i is rotted 6 out the -is, find the volume formed, V. iii Determine lim A nd lim V. n n The region ounded the rectngulr hperol =, the -is nd the lines = nd = hs n re of A nd volume V when rotted out the -is. i Find A nd V. ii Find lim A nd lim V if the eist. Topic Further pplictions of integrtion

21 Mster If the region ounded the curve = sin(n), the origin, the -is nd the first intercept it mkes with the -is is rotted out the -is, find the volume formed. If the region ounded the curve = cos(n), the coordinte es nd the first intercept it mkes with the -is is rotted out the -is, find the volume formed. 6 Prove tht the volume of right truncted cone of inner nd outer rdii r nd r respectivel nd height h is given πhr r (r r ). i A hemisphericl owl of rdius r contins wter to depth of h. Show tht the volume of wter is given πh (r h). ii If the owl is filled to depth of r, wht is the volume of wter in the owl? 7 The region ounded the curve + =, the -es nd = nd = is rotted out the -is. Find the volume formed. The region ounded the curve! +! = nd the coordinte es is rotted out the -is. Find the volume formed. 8 A fish owl consists of portion of sphere of rdius cm. The owl is filled with wter so tht the rdius of the wter t the top is 6 cm nd the se of the owl hs rdius of cm. If the totl height of the wter in the owl is 8 cm, find the volume of the wter in the owl.. Volumes Volumes of revolution Volumes round the -is Consider = g() nd = f() s two continuous non-intersecting curves on nd g() f(). If the re etween the curves is rotted 6 out the -is, it forms volume of revolution. = f() P(, ) = g() Mths Quest SPECIALIST MtheMtICS VCE Units nd

22 WorKeD exmple 8 think However, the volume is not solid nd hs hole in it. Consider cross-sectionl re shped like circulr wsher, with inner rdius r = = f() nd outer rdius r = = g(). The volume formed is V = π r r d. Note tht we must epress the inner nd outer rdii in terms of constnts nd -vlues onl. Find the volume formed when the re ounded the curve = nd the line = is rotted out: the -is Find the points of intersection etween the curve nd the line. Sketch the grph nd identif the re to rotte out the -is. Identif the inner nd outer rdii nd the terminls of integrtion. The volume formed hs hole in it. the -is. WritE/drW Let = nd =. = = = = ± = = = Write definite integrl for the volume. V = π V = π r r d =, =, r =, r = (( ) 9) d r r topic Further PPLICtIons of IntegrtIon

23 Epnd nd simplif the integrnd. V = π 6 Use smmetr to write the volume s definite integrl. V = π ( ) d (7 8 + ) d 7 Perform the integrtion. V = π c7 8 + d 8 Evlute the definite integrl. V = π c d 9 Stte the vlue of the volume. The volume is 6π cuic units. Sketch the grph nd identif the re = to rotte out the -is. This is solid volume of revolution. Write definite integrl for the volume. V = π d =, =, = V = π ( ) d Perform the integrtion. V = π c d Evlute the definite integrl. V = π c d Stte the vlue of the volume. The volume is π cuic units. Volumes round the -is In similr w, if = g( ) nd = f( ) re two continuous non-intersecting curves on where g() f() nd the re etween the curves is rotted 6 out the -is, it forms volume of revolution. However, the volume is not solid nd hs hole in it. Consider tpicl cross-sectionl re with inner rdius r = = f( ) nd outer rdius r = = g( ). 6 Mths Quest SPECIALIST MtheMtICS VCE Units nd

24 The volume formed is V = π r r d. Note tht we must epress the inner nd outer rdii in terms of constnts nd -vlues onl. WorKeD exmple 9 think Find the volume formed when the re ounded the curve =, the -is nd the line = is rotted out: the -is Find the points of intersection etween the curve nd the line. Sketch the grph nd identif the re to rotte out the -is. WritE/drW the -is. Let = nd = = 9 = 6 Identif the inner nd outer 6 rdii nd the terminls of integrtion. The volume formed hs hole in it. Write definite integrl for the volume. V = π r r r d =, =, r = =, r = V = π (9 ) d = π (9 ( + )) d topic Further PPLICtIons of IntegrtIon 7

25 Simplif the integrnd. V = π ( ) d 6 Perform the integrtion. V = π c d 7 Evlute the definite integrl. 8 Stte the vlue of the volume. Sketch the grphs nd identif the re to rotte out the -is. This is solid volume of revolution. Write definite integrl for the volume. V = π c d The volume is π cuic units. 6 = 6 V = π d Epnd the integrnd. V = π =, =, = V = π ( ) d ( 8 + 6) d Perform the integrtion. V = π c 8 + 6d Evlute the definite integrl. 6 Stte the vlue of the volume. = V = π c d The volume is π cuic units. Composite figures Sometimes we need to crefull consider the digrm nd identif the regions which re rotted out the es. 8 Mths Quest SPECIALIST MtheMtICS VCE Units nd

26 WorKeD exmple Find the volume formed when the re etween the curves = nd = 8 is rotted out: the -is the -is. think Find the points of intersection etween the two curves. Sketch the grphs nd identif the re to rotte out the -is. Identif the inner nd outer rdii nd the terminls of integrtion. Write definite integrl for the volume. Epnd nd simplif the integrnd. 6 Use smmetr to write the volume s definite integrl in simplest form. WritE/drW Let = nd = 8. = = 8 = 8 = = ± When = ±, = r V = π r r d = r = 8 =, =, r =, r = V = π V = π V = π ((8 ) ( ) ) d (6 6 + ) d (6 6 ) d topic Further PPLICtIons of IntegrtIon 9

27 7 Perform the integrtion. V = π c6 6 d 8 Evlute the definite integrl. V = π c 6 6 d 9 Stte the vlue of the volume. The volume is π Sketch the grph nd identif the re to rotte out the -is. The region comprises two sections. Write definite integrl for the totl volume V = π = π Perform the integrtion. V = π c d d + π d + π 8 cuic units. = = 8 8 d (8 ) d + π c8 d Evlute the definite integrl. V = π c d + π c d Stte the vlue of the volume. The volume is 6π cuic units. Eercise. PRctise Volumes WE8 Find the volume formed when the re ounded the curve = 9 nd the line = is rotted out: the -is the -is. Find the volume formed when the re ounded the curve =!, the -is, nd the line = is rotted out: the -is the -is. WE9 Find the volume formed when the re ounded the curve = 9, the -is nd the line = is rotted out: the -is the -is. 8 Mths Quest SPECIALIST MtheMtICS VCE Units nd

28 Consolidte Find the volume formed when the re ounded the curve =!, the -is nd the line = is rotted out: the -is the -is WE Find the volume formed when the re etween the curves = nd = 8 is rotted out: the -is the -is. 6 When the re etween the curves = + nd = + is rotted out the -is, it forms volume of revolution. Find the volume. 7 Consider the re ounded the curve = 6 nd the line =. If this re is rotted out the -is, it forms volume of revolution. Find the volume formed. If this re is rotted out the -is, it forms volume of revolution. Find the volume formed. 8 Consider the re ounded the curve = 6, the -is nd the line =. If this re is rotted out the -is, it forms volume of revolution. Find the volume formed. If this re is rotted out the -is, it forms volume of revolution. Find the volume formed. 9 Let R e the region for the re ounded the curve =, the -is nd the line =. If the region R is rotted out the -is, it forms volume of revolution. Find the volume formed. If the region R is rotted out the -is, it forms volume of revolution. Find the volume formed. The re etween the curve =! 8, the -is nd the line = 6 is rotted out: the -is the -is. Determine the volume of the resulting solid of revolution in ech cse. Find the volume formed when the re ounded the curve = cos nd the lines = nd = π is rotted out the -is. Find the volume formed when the re ounded the curve = sin, the -is nd the line = is rotted out the -is. Find the volume formed when the re etween the curves = nd = is rotted out: i the -is ii the -is. Find the volume formed when the re etween the curves = nd = is rotted out: i the -is ii the -is. If the re etween the two curves =! nd = is rotted out the -is, find the volume formed. If the re etween the two curves =! nd = where > is rotted out the -is, find the volume formed. Topic Further pplictions of integrtion

29 MstEr Find the volume formed when the re in the first qudrnt ounded the curve = sin π nd the line = is rotted out the -is. 6 Find the volume formed when the re in the first qudrnt ounded the curve = sin π nd the line = is rotted out the -is. 8 Find the volume formed when the re etween the curves = nd = is rotted out: i the -is ii the -is. Find the volume formed when the re in the first qudrnt ounded the curve = 8 sin π nd the prol = is rotted out the -is. 6 A clindricl hole of rdius cm is cut verticll through the centre of solid sphere of cheese of rdius cm. Find the volume of the cheese remining. A clindricl hole of rdius is drilled horizontll through the centre of sphere of rdius. Show tht the volume of the remining solid is given! π. 7 Find the volume of torus ( doughnutshped figure) formed when the re ounded the circle + ( 8) = 6 is rotted out the -is. Find the volume of torus ( doughnutshped figure) formed when the re ounded the circle ( ) + = r where > r > is rotted out the -is. 8 Find the volume formed when the re ounded the curve = nd the line =, where >, is rotted out: i the -is ii the -is. Find the volume formed when the re ounded the curve =, the -is nd the line =, where >, is rotted out: i the -is ii the -is. c Find the volume formed when the re etween the curves = nd =, where >, is rotted out: i the -is ii the -is.. Arc length, numericl integrtion nd grphs of ntiderivtives Length of curve In this section integrtion will e used to determine the rc length, s, of plne curve. Suppose tht the curve = f() is continuous curve on the closed intervl. The curve cn e thought of s eing mde up of infinitel mn short line segments s shown. Mths Quest specilist MtheMtICs VCe units nd

30 S = f() s WorKeD exmple think B Pthgors theorem, the length of tpicl smll segment Δs is equl to "(Δ) + (Δ). The totl length of the curve s from = to = is otined summing ll such segments nd tking the limit s Δ. s = lim s = Δ = = "(Δ) + (Δ) = lim Δ = Å + d d d = " + (f ()) d Find the length of the curve = + from = to =. 6 Find the grdient function d d differentiting nd epress ck with positive indices. Sustitute into the formul nd write definite integrl which gives the required length. WritE = + 6 = + 6 d d = 6 = 6 s = s = =Å + Δ Δ Δ Å + d d d with = nd = Å + d 6 Epnd using ( ) = +. s = Å + q r d Cncel terms nd simplif. s = Å d topic Further PPLICtIons of IntegrtIon

31 Simplif the integrnd. s = 6 Recognise the integrnd s perfect squre. 7 Epress the integrnd in form which cn e integrted. Å d s = Å + 6 d s = + 6 d s = + 6 d 8 Perform the integrtion. s = c 6 d numericl integrtion The worked emple ove is somewht contrived, ecuse for mn simple curves, the rc length formul results in definite integrl tht cnnot e evluted techniques of integrtion. In these situtions we must resort to numericl methods, such s using clcultors, which cn give numericl pproimtions to the definite integrls otined. = c 6 d 9 Evlute the definite integrl. s = 6 6 Stte the finl result. s = 9 units WorKeD exmple think Epress the length of the curve = from = to = s definite integrl nd hence find the length, giving our nswer correct to deciml plces. Find the grdient function d d differentiting. Sustitute into the formule nd write definite integrl which gives the required length. This definite integrl must e evluted using clcultor. Stte the finl result. WritE = d d = s = Å + d d d with = nd = s = " + 9 d s = 8.6 Mths Quest specilist MtheMtICs VCe units nd

32 WorKeD exmple think using numericl integrtion in differentil equtions For mn first-order differentil equtions, the integrl cnnot e found techniques of integrtion. In these situtions clcultors cn give numericl pproimtion. Consider the differentil eqution d d = f() nd ( ) =. We wnt to find the vlue of when =. We otin = f(t)d + c, where we hve ritrril used zero s the lower terminl nd t s dumm vrile. Note: is mintined s the independent vrile of the solution function. We use the given initil condition, = when =, then sustitute = f(t)dt + c to find tht the constnt of integrtion c = f(t) dt. Sustituting ck for c gives = f(t)dt + f(t)dt. = + B properties of the definite integrl, When =, Given tht d d =, () = : e f(t)dt + f(t)dt = + f(t)dt + = + = + f(t)dt f(t)dt. f(t)dt find definite integrl for in terms of determine the vlue of correct to deciml plces when =.. Antidifferentite the differentil eqution. WritE d d = e = e t dt + c topic Further PPLICtIons of IntegrtIon

33 Use the given initil conditions to find the vlue of the constnt of integrtion. Sustitute = when = : = e t dt + c c = e t dt Sustitute ck for the constnt nd simplif using the properties of definite integrls. Stte the solution for s definite integrl involving. = e t dt + e t dt = + e t dt + e t dt pproimting volumes For mn curves, the volume otined lso results in definite integrls tht cnnot e evluted n techniques of integrtion. Other volumes cn onl e evluted techniques tht tht we hve not s et covered. In either of these situtions we must gin resort to numericl methods, such s using clcultors, to find numericl pproimtions to the definite integrls otined. When question sks for n nswer correct to specified numer of deciml plces, clcultor cn e used to otin deciml pproimtion to the definite integrl. = + e t dt Find the vlue of t the required -vlue. Sustitute =.:. = + e t dt This definite integrl must e evluted using clcultor. Stte the finl result. WorKeD exmple =.6 Find the ect volume formed when the re ounded the curve = sin(), the -is, the origin nd the line = π is rotted out the -is. 6 Set up definite integrl for the volume formed when the re ounded the curve = sin(), the -is, the origin nd the line = π is rotted out the -is. Hence, find the volume correct to 6 deciml plces. 6 Mths Quest specilist MtheMtICs VCe units nd

34 THINK Sketch the grph nd identif the re to rotte out the -is. WritE/drW Write definite integrl for the volume. π π 6 π π 6 π π 6 V = π d π π π π 6 =, = π 6, = sin() π 6 V = π sin ()d Use the doule ngle V = π ( cos(6))d formul sin (A) = ( cos(a)). Perform the integrtion. V = π c 6 sin(6) d Evlute the definite integrl. V = π c π 6 6 sin(π) 6 sin() d π 6 π 6 6 Stte the vlue of the volume. V = π units Topic Further pplictions of integrtion 7

35 Sketch the grph nd identif the re to rotte out the -is. When = π 6, = sinπ =. r = π r = 6 Identif the inner nd outer rdii nd the terminls of integrtion. Write definite integrl for the volume. This definite integrl cnnot e evluted integrtion techniques. Find numericl vlue for the definite integrl using clcultor, nd stte the finl result. π π 6 6 π π V = π r r d =, =, r = = π 6, r = = sin() = sin() = sin = sin V = π q π V =.96 units 6 9 sin Grphs of ntiderivtives of functions Given function f(), we cn sketch the grph of the ntiderivtive F() = f()d noting ke fetures nd considering F () = f(). The tle elow shows the reltionships etween the grphs. Note tht the grph of the ntiderivtive cnnot e completel determined s it includes constnt of integrtion, which is verticl trnsltion of the grph of F() prllel to the -is. r d 8 Mths Quest SPECIALIST MtheMtICS VCE Units nd

36 WorKeD exmple Grph of function f negtive, f() < for (, ) positive, f() > for (, ) f() cuts the -is t = from negtive to positive f() cuts the -is t = from positive to negtive f() touches the -is t = f() hs turning point t = Grph of ntiderivtive F F() hs negtive grdient, or is decresing for (, ) F() hs positive grdient, or is incresing for (, ) F() hs locl minimum t = F() hs locl mimum t = F() hs sttionr point of infleion t = F() hs point of infleion t = (non-sttionr unless f() = ) In prticulr If f() is liner function, then the grph of the ntiderivtive F() will e qudrtic function. If f() is qudrtic function, then the grph of the ntiderivtive F() will e cuic function. If f() is cuic function, then the grph of the ntiderivtive F() will e qurtic function. Given the grph, sketch possile grph of the ntiderivtive. 6 The grph of the grdient function is shown. Sketch possile grph of the function. topic Further PPLICtIons of IntegrtIon 9

37 THINK The given grph crosses the -is t =, so the grph of the ntiderivtive hs sttionr point t =. At = the given grph chnges from positive to negtive s increses, so the sttionr point is mimum turning point. No further informtion is provided, so we cnnot determine the -vlue of the turning point or n vlues of the is intercepts. The grph of the ntiderivtive could e trnslted prllel to the -is. The given grph crosses the -is t = nd =, so the grph of the ntiderivtive hs sttionr point t these vlues. At = the grdient chnges from positive to negtive s increses, so the sttionr point is mimum turning point. At = the grdient chnges from negtive to positive s increses, so the sttionr point is minimum turning point. The grdient function hs turning point t = so the grph of the ntiderivtive hs point of infleion t =. No further informtion is provided. We cnnot determine the -vlues of the sttionr points or the point of infleion, or n vlues of the is intercepts. The grph could e trnslted prllel to the -is. Eercise. PRctise WritE/drW Arc length, numericl integrtion nd grphs of ntiderivtives WE Find the length of the curve = + from = to = 6. Find the length of the curve = + from = to = 8. WE Epress the length of the curve = from = to = s definite integrl nd hence find the length, giving our nswer correct to deciml plces. Mths Quest SPECIALIST MtheMtICS VCE Units nd

38 Epress the length of the curve = from = to = s definite integrl nd hence find the length, giving our nswer correct to deciml plces. WE Given tht d d = sin( ), (.) = : find definite integrl for in terms of determine the vlue of correct to deciml plces when =. 6 Given the differentil eqution d d = tn, π 8 =, find definite integrl for in terms of. 7 WE Find the ect volume formed when the re ounded the curve = tn(), the -is, the origin nd the line = π is rotted out the -is. 8 Set up definite integrl for the volume formed when the re ounded the curve = tn(), the -is, the origin nd the line = π is rotted out the 8 -is. Hence, find the volume correct to deciml plces. 8 Set up definite integrl for the volume formed when the re ounded the curve = sin (), the -is, the origin nd the line = is rotted out the -is. Hence, find the volume correct to deciml plces. Find the ect volume formed when the re ounded the curve = sin (), the -is, the origin nd the line = is rotted out the -is. 9 WE Given the grph, sketch possile grph of the ntiderivtive. The grph of the grdient function is shown. Sketch possile grph of the function Topic Further pplictions of integrtion

39 Consolidte The grph of = f () is shown. For the grph of = f(), stte the -vlues of the sttionr points nd their nture. Find the length of = + from = to = 6. Find the length of the curve = 6 + from = to =. c Find the length of the curve = + 8 from = to =. d Find the length of the curve = (e + e ) from = to =. Determine the length of the curve = " from = to =. For the curve 7 = ( ), find the length of the curve from = to = 8. c Find the length of the curve = "( ) from = to = 9. d Find the length of the curve = "( ) from = to = 9. Find the length of the curve = " from = to =. Find the length of the curve = "9 from = to =. Wht length does this represent? Set up definite integrls for the lengths of the following curves, nd hence determine the rc length in ech cse. Give our nswers correct to deciml plces. = + from = to = 6. = cos() from = to = π c = e from = to = d = log e ( + ) from = to = Set up definite integrl for the totl length of the ellipse 9 + =. Find this length, giving our nswer correct to deciml plces. Set up definite integrl for the totl length of the curve = cos. Find this length correct to deciml plces. 6 Find the ect volume formed when the re ounded the curve = cos(), the coordinte es nd = π is rotted out the -is. Mths Quest SPECIALIST MtheMtICS VCE Units nd

40 Set up definite integrl for the volume formed when the re ounded the curve = cos(), the coordinte es nd = π is rotted out the -is. Determine this volume correct to deciml plces 7 Given tht d d = e, () = : i find definite integrl for in terms of ii determine the vlue of when =. Given tht d d = sin ( ), (.) = : i find definite integrl for in terms of ii determine the vlue of when =.. c Given tht d d =, () = " + 8 : i find definite integrl for in terms of ii determine the vlue of when =. 8 The following grphs re of grdient functions. In ech cse sketch possile grph of the originl function. 6 6 c 6 8 d Let A e the re ounded the grph of = 6 nd the -is. Find the vlue of A. If the re A is rotted out the -is, find the volume formed. c Find the length of the curve = 6 from = to = 6, giving our nswer correct to deciml plces. d If the re A is rotted out the -is, find the volume formed. Topic Further pplictions of integrtion

41 Mster Let A e the re ounded the grph of = sin, the -is, the origin nd the line =. Let A e the re ounded the grph of = sin, the -is, the origin nd the line = π. Differentite sin nd hence find the vlue of A. Find the vlue of A. c If the re A is rotted out the -is, find the volume formed, giving our nswer correct to deciml plces. d If the re A is rotted out the -is, find the ect volume formed. e If the re A is rotted out the -is, find the ect volume formed. f If the re A is rotted out the -is, find the volume formed, giving our nswer correct to deciml plces. i Write definite integrl which gives the length of the curve = n from = to =. ii Hence, find correct to deciml plces the length of the curve =! from = to = 9. i Write definite integrl which gives the length of the curve = sin(k) from = to =. ii Hence, find correct to deciml plces the length of the curve = sin() from = to = π. c i Write definite integrl which gives the length of the curve = e k from = to =. ii Hence, find correct to deciml plces the length of the curve = e from = to =. For the line = m + c, verif tht the rc length formul gives the distnce long the line etween the points = nd =. If nd p re positive rel constnts, show tht the length of the curve = + from = = p is given (p )( p(p + p + ) + ). p c Prove tht the circumference of circle of rdius r is πr. d Show tht the totl length of the ellipse + = is given the definite integrl s = Å + ( ) d..6 Wter flow Torricelli s theorem Evngelist Torricelli (68 67) ws n Itlin scientist interested in mthemtics nd phsics. Mths Quest SPECIALIST MtheMtICS VCE Units nd

42 WorKeD exmple 6 He invented the rometer to mesure tmospheric pressure nd ws lso one of the first to correctl descrie wht cuses the wind. Hhe lso designed telescopes nd microscopes. Modern wether forecsting owes much to the work of Torricelli. His min chievement is the theorem nmed fter him, Torricelli s theorem, which descries the reltionship etween fluid leving continer through smll hole in the continer nd the height of the fluid in the continer. Bsicll, the theorem sttes tht the rte t which the volume of the fluid leves the continer is proportionl to the squre root of the height of the fluid in the tnk. This theorem pplies for ll tpes of continers. Prolem solving In solving prolems involving fluid flow, we need to use the techniques of finding volumes nd use relted rte prolems to set up nd solve differentil equtions. Sometimes we m need to use numericl methods to evlute definite integrl. A vse hs circulr se nd top with rdii of nd 9 cm respectivel, nd height of 6 cm. The origin, O, is t the centre of the se of the vse. The vse is formed when the curve =! + is rotted out the -is. Initill the vse is filled with wter, ut the wter leks out t rte equl to!h cm /min, where h cm is the height of the wter remining in the vse fter t minutes. Set up the differentil eqution for h nd t, nd determine how much time it tkes for the vse to ecome empt. think Set up simultneous equtions which cn e solved for nd. The height of the vse is 6 nd the height of the wter in the vse is h, so h 6. Note ll dimensions re in cm. WritE/drW 6 O The curve =! + psses through the points (, ) nd (9, 6). Sustituting: (, ) () = + (9, 6) () 6 = + Determine the vlues of nd. () () = 6, so =. The vse is formed when = 6! for 9 is rotted out the -is. Determine the volume of the vse. When curve is rotted out the -is, h the volume is V = π d. 9 h topic Further PPLICtIons of IntegrtIon

43 Rerrnge the eqution to mke the suject. = 6! Determine definite integrl for the volume of wter when the vse is filled to height of h cm. 6 Determine the given rtes in terms of time, t. The rte is negtive s it is decresing rte. 7 Note the result used is from the numericl techniques descried erlier. 8 Use relted rtes nd chin rule. 9 Set up the differentil eqution for the height h t time t. To solve this tpe of differentil eqution, invert oth sides. Set up definite integrl for the time for the vse to empt. Use clcultor to numericll evlute the definite integrl. 6! = + " = ( + ) 6 = ( + 6 ) = ( + ) 6 6 h ( + ) V = π 66 d Since the wter leks out t rte proportionl to the squre root of the remining height of the wter, dv dt =!h. h ( + ) V = π 66 d dv π(h + ) = dh 66 dh dt = dh dv dv dt Sustitute for the rtes nd use dh dv = /dv dh : dh 6 6!h = dt π(h + ) dt π(h + ) = dh 6!h π(h + ) t = dh 6 6"h Note the order of the terminls from h = 6 to h = t =.9; note tht the time is positive. Stte the result. The tnk is empt fter totl time of.9 minutes. 6 Mths Quest SPECIALIST MtheMtICS VCE Units nd

44 WorKeD exmple 7 think Another vse is formed when prt of the curve 6 = for 6,, is rotted out the -is to form volume of revolution. The - nd -coordintes re mesured in centimetres. The vse hs smll crck in the se, nd the wter leks out t rte proportionl to the squre root of the remining height of the wter. Initill the vse ws full, nd fter minutes the height of the wter in the vse is 6 cm. Find how much longer it will e efore the vse is empt. The vse is formed when the given curve is rotted out the -is. WritE/drW V = π d Determine the height of the vse. Find the vlue of when = 6. = 6 6 = = 6 = s > Sketch the region of the hperol which forms the vse. The height of the vse is cm nd the height of the wter in the vse is h, so h. 6 O Trnspose the eqution to mke the suject. Find definite integrl for the volume of wter when the vse is filled to height of h cm, where h. 6 = 6 = + = 6( + ) 6 Note the result used is from the numericl techniques descried erlier. = ( + ) h V = π ( + )d dv dh = π( + h ) 6 h 6 topic Further PPLICtIons of IntegrtIon 7

45 7 Determine the given rtes in terms of time t in minutes. 8 Use relted rtes nd chin rule. 9 Set up the differentil eqution for the height h t time t. Incorporte the constnts, into one constnt. To solve this tpe differentil eqution, invert oth sides. Since the wter leks out t rte proportionl to the squre root of the remining height of the wter, dv dt = k!h, where k is positive constnt. dh dt = dh dv dv dt Sustitute for the relted rtes, using dh dv = /dv dh dh dt = k!h π(h + ) dt dh = A(h + ) where A = π!h k dt dh = Ah + h Integrte with respect to h. () t = A h + h dh Perform the integrtion. t = A c h + h d + c, where c is the Two sets of conditions re required to find the vlues of the two unknowns. constnt of integrtion. Initill, when t =, h =, since the vse ws full. Sustitute t = nd h = : = A c () + d + c Simplif the reltionship. = A c + d + c c 6A = A = c 6 Find nother reltionship etween the unknowns. Since fter minutes the height of the wter in the vse is 8 cm, sustitute t = nd h = 6 into (): = A c (6) + 6 d + c 6 Simplif the reltionship nd solve for the constnt of integrtion. Sustitute for A: = c 6 c + d + c = c 6 c + d 6 6 c = c =.96 8 Mths Quest SPECIALIST MtheMtICS VCE Units nd

46 7 Determine when the vse will e empt. Since t = A c h + h d + c, the vse is empt when h =, tht is t time when t = c. 8 Stte the finl result. The vse is empt fter totl time of.96 minutes, so it tkes nother.96 minutes to empt. Eercise.6 PRctise Consolidte Wter flow WE6 A vse hs circulr se nd top with rdii of nd 9 cm respectivel, nd height of 6 cm. The origin, O, is t the centre of the se of the vse. The vse is formed when the line = + is rotted out the -is. Initill the vse is filled with wter, ut the wter leks out t rte equl to!h cm /min, where h cm is the height of the wter remining in the vse fter t minutes. Set up the differentil eqution for h nd t, nd determine how long it will e efore the vse is empt. A vse hs circulr se nd top with rdii of nd 9 cm respectivel nd height of 6 cm. The origin, O, is t the centre of the se of the vse. The vse is formed when the curve = + is rotted out the -is. Initill the vse is filled with wter, ut the wter leks out t rte equl to!h cm /min, where h cm is the height of the wter remining in the vse fter t minutes. Set up the differentil eqution for h nd t, nd determine how long efore the vse is empt. WE7 A vse is formed when prt of the curve 6 6 = for 9, 96, is rotted out the -is to form volume of revolution. The - nd -coordintes re mesured in centimetres. The vse hs smll crck in the se, nd the wter leks out t rte proportionl to the squre root of the remining height of the wter. Initill the vse ws full, nd fter minutes the height of the wter in the vse is 9 cm. Find how much time it will tke for the vse to ecome empt. A vse hs se rdius of cm, top rdius is 9 cm, nd height of 6 cm. The origin, O, is t the centre of the se of the vse. Initill the vse is filled with wter, ut the wter leks out t rte proportionl to the squre root of the remining height of the wter. The side of the vse is modelled qudrtic = +. Initill the vse ws full, nd fter minutes the height of the wter in the vse is 9 cm. Find how much time it will tke for the vse to ecome empt. A clindricl coffee pot hs se rdius of cm nd height of 9 cm nd is initill filled with hot coffee. Coffee is removed from the pot t rte equl to!h cm /sec, where h cm is the height of the coffee remining in the coffee pot fter t seconds. Set up the differentil eqution for h nd t, nd determine how long it will e efore the coffee pot is empt. A smll clindricl tepot with se rdius of cm nd height of 6 cm is initill filled with hot wter. The hot wter is removed from the tepot t rte proportionl to!h cm /min, where h cm is the height of the hot wter Topic Further pplictions of integrtion 9

47 remining in the tepot fter t minutes. Set up the differentil eqution for h nd t. If fter minutes the height of the hot wter is 9 cm, wht further time elpses efore the tepot is empt? 6 A rectngulr thtu hs length of. metres nd is.6 metres wide. It is filled with wter to height of metre. When the plug is pulled, the wter flows out of the th rte equl to!h m /min, where h is the height of the wter in metres in the thtu t time t minutes fter the plug is pulled. How long will it tke for the thtu to empt? A clindricl hot wter tnk with cpcit of 6 litres is 69 cm tll nd is filled with hot wter. Hot wter strts leking out through crck in the ottom of the tnk t rte equl to k!h cm /min, where h cm is the depth of wter remining in the tnk fter t minutes. If the tnk is empt fter 9 minutes, find the vlue of k. 7 The volume of hemisphericl owl is given πh ( h) cm, where h cm is the depth of the wter in the owl. h Initill the owl hs wter to depth of 9 cm. The wter strts leking out through smll hole in the owl t rte equl to k!h cm /min. After minutes the owl is empt. Find the vlue of k. A drinking trough hs length of m. Its cross-sectionl fce is in the shpe 9 of trpezium with height of cm nd with lengths nd 9 cm. Both h sloping edges re t n ngle of to the verticl, s shown in the digrm. The trough contins wter to height of h cm. The wter leks out through crck in the se of the trough t rte proportionl to!h cm /min. Initill the trough is full, nd fter minutes the height of the wter in the trough is 6 cm. How long will it e efore the trough is empt? 8 A plstic ucket hs se dimeter of cm, top dimeter of 6 cm nd height of cm. The side of the ucket is stright. Initill the ucket is filled with wter to height of 6 cm. However, there is smll hole in the ucket, nd the wter leks out of the ucket t rte proportionl to!h cm /min, where h cm is the height of the wter in the ucket. When the height of the wter in the ucket is 6 cm, the height is decresing t rte of. cm/min. Determine how long it will e efore the ucket is empt. A plstic coffee cup hs se dimeter of cm, top dimeter of 8 cm nd height of 9 cm. Initill the cup is filled with coffee. 8 9 Mths Quest SPECIALIST MtheMtICS VCE Units nd

48 However, the coffee leks out of the cup t rte equl to k!h cm /min, where h cm is the height of the coffee in the cup. If the cup is empt fter minutes, determine the vlue of k. 9 A conicl vessel with its verte downwrds hs height of cm nd rdius of cm. Initill it contins wter to depth of 6 cm. Wter strts flowing out through hole in the verte t rte proportionl to the squre root of the remining depth of the wter in the vessel. If fter minutes, the depth is 9 cm, how much longer will it e efore the vessel is empt? A conicl funnel with its verte downwrds hs height of cm nd rdius of cm. Initill the funnel is filled with oil. The oil flows out through hole in the verte t rte equl to k!h cm /s, where h cm is the height of the oil in the funnel. If the funnel is empt fter seconds, determine the vlue of k. When filled to depth of h metres, fountin contins V litres of wter, where V = h h. Initill the fountin is empt. Wter is pumped into the fountin t rte of litres per hour nd spills out t rte equl to!h litres per hour. Find the rte in metres per hour t which the wter level is rising when the depth is. metres. The fountin is considered full when the height of the wter in the fountin is metre. How long does it tke to fill the fountin? c When the fountin is full, wter is no longer pumped into the fountin, ut wter still spills out t the sme rte. How long will it e efore the fountin is empt gin? A wine glss is formed when the rc OB with the eqution = is rotted out B the -is. The dimensions of the glss re given in cm. = The wine leks out through crck in the se of the glss t rte proportionl to!h cm /min. Initill the glss is full, nd fter minutes the height of the wine in the glss is cm. Wht further time elpses efore the wine glss is empt? A lrge eer glss is formed when portion of the curve with the eqution = is rotted out the -is etween the origin nd = 6. The dimensions of this glss re given in cm. Beer leks out through crck in the se of this glss t rte proportionl to!h cm /min. Initill the glss is full, nd fter minutes the height of the eer in this glss is cm. How much longer will it e efore this glss is empt? Topic Further pplictions of integrtion

49 An ornmentl vse hs circulr top nd se, oth with rdii of cm, nd height of 6 cm. The origin, O, is t the centre of the se of the vse, nd the vse is formed when the hperol ( 8) = 6 is rotted out the -is, with dimensions in cm. Initill the vse is filled with wter, ut the wter leks out t rte equl to!h cm /s, where h cm is the height of the wter in the vse. Determine the time tken for the vse to empt nd the cpcit of the vse. A different ornmentl vse hs circulr top nd se, oth with rdii of.6 cm, nd height of 6 cm. The origin, O, is t the centre of the se of the vse, nd the vse is formed when the ellipse ( 8) + = 6 is rotted out the -is, with dimensions in cm. Initill this vse is filled with wter, ut the wter leks out t rte equl to!h cm /s, where h cm is the height of the wter in the vse. Determine the time tken for this vse to empt nd the cpcit of the vse. A clindricl vessel is initill full of wter. Wter strts flowing out through hole in the ottom of the vessel t rte proportionl to the squre root of the remining depth of the wter. After time of T, the depth of the wter is hlf its T initil height. Show tht the vessel is empt fter time of!. c A conicl tnk is initill full of wter. Wter strts flowing out through hole in the ottom of the tnk t rte proportionl to the squre root of the remining depth of the wter. After time of T, the depth of the wter is hlf its T initil height. Show tht the tnk is empt fter time of. " The digrm shows vse. The se nd the top re circulr with rdii of 9 nd cm respectivel, nd the height is 6 cm. The origin, O, is t the centre of the se of the vse, with the coordinte es s shown. Initill the vse is filled with wter, ut the wter leks out t rte equl to!h cm /s, where h cm is the height of the wter in the vse. Determine the time tken for the vse to empt nd the cpcit of the vse if the side of the vse is modelled : hperol, = + truncus, = +. h O 9 6 Mster A vse hs se rdius of cm, top rdius of 9 cm nd height of 6 cm. The origin, O, is t the centre of the se of the vse, nd the vse cn e represented curve rotted out the -is. Initill the vse is filled with wter, ut the wter leks out t rte equl to!h cm /s, where h cm is the height of the wter in the vse. Mths Quest SPECIALIST MtheMtICS VCE Units nd

50 9 6 7 O Determine the time tken for the vse to empt nd the cpcit of the vse if the side of the vse is modelled : cuic of the form = + qurtic of the form = +. 6 A hot wter tnk hs cpcit of 6 litres nd initill contins litres of wter. Wter flows into the tnk t rte of!t sin πt litres per hour over the time t T hours, where < T < 8. During the time intervl t, wter flows out of the tnk t rte of!t litres per hour. After hours, is the wter level in the tnk incresing or decresing? After hours, is the wter level in the tnk incresing or decresing? c At wht time is the wter level litres? d After hours, find the volume of wter in the tnk. e After hours, no wter flows out of the tnk. However, the inflow continues t the sme rte until the tnk is full. Find the time, T, required to fill the tnk. Topic Further pplictions of integrtion

51 ONLINE ONLY.7 Review the Mths Quest review is ville in customisle formt for ou to demonstrte our knowledge of this topic. the review contins: short-nswer questions providing ou with the opportunit to demonstrte the skills ou hve developed to efficientl nswer questions using the most pproprite methods Multiple-choice questions providing ou with the opportunit to prctise nswering questions using CAS technolog studon is n interctive nd highl visul online tool tht helps ou to clerl identif strengths nd weknesses prior to our ems. You cn then confidentl trget res of gretest need, enling ou to chieve our est results. Etended-response questions providing ou with the opportunit to prctise em-stle questions. summr of the ke points covered in this topic is lso ville s digitl document. REVIEW QUESTIONS Downlod the Review questions document from the links found in the Resources section of our ebookplus. Units & <Topic title to go here> Sit topic test Mths Quest specilist MtheMtICs VCe units nd