Calculus  Activity 1 Rate of change of a function at a point.


 Darleen Parks
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1 Nme: Clss: p 77 Mths Helper Plus Resource Set. Copright 00 Bruce A. Vughn, Techers Choice Softwre Clculus  Activit Rte of chnge of function t point. ) Strt Mths Helper Plus, then lod the file: Clculus  Derivtives.mhp. The grph view displs grph of the function: = f() = ² + (See figure, below:) Figure Secnt line Q = f() P We wish to find the grdient of the curved function line t P where =. Another point Q lso lies on the curve. The two points define stright secnt line. If Q is ver close to P, then the grdient of the secnt will be good pproimtion of the grdient of the curve t P. The grdient of the secnt cn be found s follows: f(+h) Q = ( +h, f(+h) ) Thus the grdient of the secnt is given b: f(+h)  f() f ( + h) f ( ) P = (, f() ) h f() h +h ) Complete this tble. Use the P nd Q points in figure nd the digrm nd formul bove. f() h +h f(+h) f(+h)  f() grdient of secnt ) Use Mths Helper Plus to correct our nswers. Hold down Ctrl ke while ou press the T ke. The dt set titled: Grdient of secnt displs the clcultions. (Note tht A is used insted of h, nd X for the vlue.) Answers pper here, in the first row of numbers. ( Ignore the other rows. ) X f(x) A X+A f(x+a) (f(x+a)  f(x))/a $ $ $ $ $ $ To see the grph gin, use Ctrl G for just the grph, or Ctrl B for both tet nd grph together.
2 We wnt to estimte the grdient of the tngent line to the function = f() = ² + t point P = (,). To do this, we will move point Q long the curve towrds P. This will mke the secnt PQ shorter so tht its grdient will more closel mtch the grdient of the tngent line t P. ) Press F to displ the prmeters bo, then... ) Click here to select vrible A b) Click here to select the slider With the slider selected, ou cn use the up nd down rrow kes on the kebord to chnge A. If the rrow kes stop responding, click the slider button gin. ) Hold down Ctrl nd press B to view both tet nd grph together. Use the mouse to djust the splitter br tht divides tet nd grph views so tht ou cn see both the tet view clcultions nd the secnt line on the grph. 6) Use the rrow kes to chnge A s described in step, nd so complete the tble below. h = A grdient of secnt: f ( + h) f ( ) h = (f(x+a)  f(x))/a To set these vlues:  Click on the A edit bo  Tpe the vlue for A  Click the Updte button ) Wh cn t we clculte the grdient when h = 0? 8) As h gets closer to zero, the grdient of the secnt line pproches the ctul grdient of the curve t point P. From our results in the tble bove, estimte the grdient of the curve t P = (,): f ( + h) f ( ) The grdient of the curve t P is clled the limit of s h pproches zero. h f ( + h) f ( ) f ( + h) f ( ) This cn be written like this: lim, so in this cse, lim = h 0 h h 0 h The grdient of function t point is clled the derivtive of the function with respect to the vrible tht pproches limiting vlue. In this cse, vrible is pproching, so we hve found the derivtive of the function with respect to t =. d For the function: = f(), the derivtive of f() with respect to is written s: d So for the function: = ² +, we write: d = t =. d 9) Set prmeter A to, then slowl increse A towrds zero. Does (f(x+a)  f(x))/a pproch the sme limit s before?
3 Nme: Clss: p 79 Mths Helper Plus Resource Set. Copright 00 Bruce A. Vughn, Techers Choice Softwre Clculus  Activit Plotting the derived function. ) Strt Mths Helper Plus, then lod the file: Clculus  Derivtives.mhp. The grph view displs grph of the function: = f() = ² + (See below:) P d The derivtive of function = f() with respect to is defined s: = d The derivtive lso equls the grdient of the tngent to the function. f ( + h) f ( ) lim. h 0 h The tet view displs the pproimte derivtive of the function ner the point: (X, f(x)). ( NOTE: We re using BIG X to represent prticulr vlue. ) The dt set titled: Grdient of secnt displs the clcultions in the positions shown below: Note tht A is used insted of h, nd X for the vlue. Clcultions pper here: X f(x) A X+A f(x+a) (f(x+a)  f(x))/a $ $ $ $ $ $ This is the pproimte derivtive: ) From the pproimte derivtive shown on the tet view, estimte the true derivtive of = ² + t =. d At =, = d d NOTE: Sometimes, is written s: f '(), so we could lso write: f '() = d
4 ) Plot the point: (, f '()) on the grph on the other side of this sheet. ) Press F to displ the prmeters bo. (See below.) To chnge the vlue of X, ou... ) Click on the X edit bo b) Click the Updte button ) Chnge X to ech of the vlues in this tble, then complete using the tet view clcultions: ( Round the vlues to two deciml plces. ) 0 f '() Cop from question 6) Plot the points: (, f () ) from the tble in question on the grph pper on the front side of this sheet, then rule line through the points. 7) Wht is the eqution of the line through these points? = 8) The function ou obtined in question 7 is clled the derived function with respect to, or the derivtive with respect to. The vlue of this function t given vlue is the grdient of the function: = ² + t tht vlue, so d we normll write: = or f '() =. d The process of finding the derived function is clled differentiting. 9) Check our derivtive plot with Mths Helper Plus. The dt set clled Derived function clcultes the derivtive for mn points on long = f(). It then plots the points nd joins them with lines to mke smooth curve.  Double click on the tet view beside the dt set:  Select the options tb: Plot settings  Click on the check bo  Click OK Derived function = (f(+a)  f())/a The derived function will now displ s red line. Compre with our grph.
5 Nme: Clss: p 8 Mths Helper Plus Resource Set. Copright 00 Bruce A. Vughn, Techers Choice Softwre Clculus  Activit Investigting the derived function b mesuring grdients round function curve. ) Strt Mths Helper Plus, then lod the file: Clculus  Derivtives.mhp. The grph view displs grph of the function: = f() = ³ ² (See below:) At n given vlue, the derived function equls the grdient of the originl function. You will use the Mths Helper Plus line tool to mke quick estimtes of the grdient of the plotted function t vrious points. ) Select the line tool b clicking this toolbr button: The line tool dilog bo will be displed, s shown: If it covers n of the plotted grph, drg on the title br t the top of the line tool dilog bo to move it to more convenient loction. ) Move the mouse cursor over the grph view. The line tool cursor will displ on the grph view like this: To mke this line segment longer, hold down Ctrl ke while ou click nd drg the mouse to the right. Do this until the line tool cursor is two to three times its originl length: Ctrl ke + click nd drg right.
6 ) Click to select the Trce option on the line tool dilog bo. ) Click the + mouse cursor on n prt of the plotted grph to find the pproimte tngent t tht point. A tngent line will lock to the function plot. Move the tngent long the function b moving the mouse left nd right, or click on n other prt of the function grph to immeditel lock the tngent to tht point. HINT: Hold down the Shift ke while trcing for finer control. TRY IT: Move mouse Tngent line moves round curve + Click! 6) The tngent line eqution is displed t the top of the line tool dilog bo in the form: = m + c. Use the line tool to mesure the grdient ( m ) of the function grph close to the given vlues. Write the grdients in the second column of the tble below. Round the vlues to two deciml plces. d grdient, m = d grdient sign ( + or ) grdient chnge ( incresing or decresing ) ) On the grph on the front of this sheet, plot the points (, d/d) tht ou recorded in the tble. Join the points with smooth curve. This curve represents the derivtive of = ³  ² from = 0. to =. 8) In the third column of the tble, write + or for positive or negtive grdients. A positive grdient mens tht s increses, the function goes uphill. Inspect the grph nd verif tht the derivtive grph is positive when the function = ³ ² is going uphill s increses. Wht is hppening to the originl function curve when its derivtive grph is: () negtive (b) zero 9) In the lst column of the tble, write Incresing if tht grdient is greter thn the previous grdient (Above it in the tble), or Decresing if it is less. Use coloured pencil or mrker on the grph of = ³ ² on the front of this sheet to indicte where the grdient is decresing. Describe the behviour of the derivtive grph when the grdient of = ³ ² is: () decresing (b) incresing 0) The point on = ³ ² where the grdient chnges from decresing to incresing is clled point of inflection. How cn ou identif point of inflection from the derivtive grph? B inspection, estimte the pproimte point of inflection on = ³ ² nd mrk it with smll circle.
7 Nme: Clss: p 8 Mths Helper Plus Resource Set. Copright 00 Bruce A. Vughn, Techers Choice Softwre Clculus  Activit The definite integrl s limit. = f() Consider the problem of finding the re A between curved line: = f() nd the is, between = nd = b: Are = A We cn pproimte the re b slicing it into rectngles nd dding up the res of the rectngles. In the digrm below, there re three rectngles of equl width. The height of ech = = b rectngle equls the height of the function grph t its midpoint, nd the width of ech rectngle = (b )/: = f() A A A f( ) f( ) f( ) = = b = (b)/ The totl re under the curve from = to = b cn be pproimted b: A = A + A + A. Notice tht the re A is n overestimte of the re under the curve, A is bout right, nd A is too smll. Thus this method m be ccurte for some tpes of curves, but not for others. It depends on their shpe. One w to improve the ccurc for n shped curve is to increse the number of rectngles. If there re n rectngles, then s n increses, the width of ech rectngle decreses. For ver thin rectngles, res t the top tht don t fit the curve become insignificnt, then the sum of the rectngles becomes ver close to the re under the curved grph line. b Let F be the re under the grph: = f() from = to = b. We cn clculte the re b summing up the res of the rectngulr slices. If there re n slices, then the totl re is pproimted b: F b A + A + A + L The true re cn be written s the limit of this sum s n pproches infinit, like this: n b F = lim Ai n i= We will now use Mths Helper Plus to clculte the re under curved function grph b dding rectngles. + A n = n i= A i
8 ) Strt Mths Helper Plus nd lod the file: Clculus  Integrls.mhp The grph view will displ the curved function line s shown below: You will use the grph bove to estimte the re under the grph from = to = 9 using rectngles. ) Drw the rectngles on the grph bove with the top middle point of ech rectngle on the grph line, s shown on the front pge. ) Wht is the width, w, of ech rectngle? ) Clculte the re of ech rectngle, then dd the res: A = A = A = A = Totl re = ) Do ou think our totl is greter or less thn the true re under the curve? You will now find better estimte of the re under this curve b using much lrger numbers of rectngles. Mths Helper Plus cn do the clcultions for ou. 6) Move the tip of the mouse pointer onto the grph curve, then double click to displ the options dilog. Select the Integrls tb. Now set these options: In the Limits of integrtion edit bo, tpe: [,9] to specif = to = 9. In the Number of intervls edit bo, tpe severl vlues of n, like this:,,0,0,0,00,00,00 Select: Shde res on grph. In Methods of integrtion, select: Rectngulr (midpoint) Set Clcultion mode to re. Click OK to close the dilog bo. The tet view will displ the summed res for the different vlues of n. 7) Compre our nswer for question with the re clculted b Mths Helper Plus for n =. 8) Wht is the re under this function grph between = nd = 9 correct to four deciml plces? 9) About wht vlue of n is required to chieve this ccurc?
9 Nme: Clss: p 8 Mths Helper Plus Resource Set. Copright 00 Bruce A. Vughn, Techers Choice Softwre Clculus  Activit Primitives of polnomil functions b direct mesurement of res. The fundmentl theorem of the clculus sttes tht: f ( ) d = F( b) F( ) where F() is the primitive of the function f(). b nd b re two vlues. F(b) F(), clled the definite integrl, is number relted to the re between the curve nd the is, from = to =b. An prt of this re bove the is is positive, while prts of the re below the is re negtive. Adding these positive res nd negtive res gives the definite integrl. Thus the definite integrl cn be zero if equl mounts of re re bove nd below the is between nd b. Mths Helper Plus cn esil mesure the definite integrl of plotted function. You cn use this to find the primitive function: F(). The method is to choose some fied vlue for, then choose different vlues for = b. Since n vlues cn be chosen for b, we will replce b with : f ( ) d = F( ) F( ) Adding F() to both sides we hve: F( ) = f ( ) d + F( ) ) Strt Mths Helper Plus, then lod the file: Clculus  Integrls.mhp. The grph view will displ plot of the function: = ² (See below.) ) Select the Integrl tool b clicking this button: on the toolbr. The integrl tool dilog bo will be displed: Click here Relese here 0 D R A G Mke sure the Are option is NOT selected nd Snp to grid IS selected. Drg on the title br of this dilog to move it to convenient loction. ) Move the mouse cursor over the grph view. The integrl tool cursor will displ: ) Prctice finding definite integrl from =  to b =. To do this, move the integrl tool cursor so tht the grph curve is between the two rrows nd so tht the rrows line up with the strting vlue of . (See bove.) Now click the nd drg the mouse to the right until the grph is shded ll the w to =, then relese the mouse button. Red the Simpson s rule vlue for the definite integrl from the dilog bo. (It should be 6)
10 ) Now to record some vlues of the definite integrl: Let be , nd be integers from  to. f ( ) d (Click the Cler button on the integrl tool dilog bo to cler the lst mesurement red to strt nother.) To find definite integrl, click on the curve t =  s in () bove, nd drg the mouse right until ou rech the required vlue. Relese the mouse nd record the definite integrl vlue in the tble. When ou hve finished recording vlues, click the Cncel button on the integrl tool dilog bo. definite integrl from  to ) Plot our definite integrl vlues in the tble s set of (,) points, with the integrls s the coordintes. For emple, the integrl from =  to = is 6, so this would be the point (,6). To plot points ou: Click on the input bo. (On the tet view.) Tpe the points to plot, like this: (,0) (,... )... Click outside of the input bo. 7) With the mouse pointer ner the ect centre of one of the plotted points on the grph view, double click the mouse to displ the options dilog for the points. Select the Curve Fitting tb nd select the cubic option. Click the Plot button, then OK to close the dilog bo. 8) Cop the Best fit cubic eqution from the tet view: = 9) Round off ll numbers in this eqution to whole numbers or whole number frctions. = 0) All primitive functions include n unknown constnt term, often written s c. F( ) = f ( ) d + F( ) In our definition:, the unknown F() term is lso constnt vlue. ( Wh? ) Therefore, simpl replce the constnt term in (9) bove with letter c to represent n constnt. This is then the primitive function: F() = Etension Activit ) Repet to 0 bove, but with = . Mesure the definite integrl from  to = , 0,,,,,. ) Do ou end up with the sme primitive function: F()? Wh is this?
11 Nme: Clss: p 87 Mths Helper Plus Resource Set. Copright 00 Bruce A. Vughn, Techers Choice Softwre Clculus  Activit 6 Are under curve. The definite integrl cn be used to find the re between grph curve nd the is, between two given vlues. This re is clled the re under the curve regrdless of whether it is bove or below the is. When the curve is bove the is, the re is the sme s the definite integrl... Are = b f ( ). d = f() = = b but when the grph line is below the is, the definite integrl is negtive. The re is then given b: b Are = = f ( ). d = b = f() Sometimes prt of the grph is bove the is nd prt is below, then it is necessr to clculte severl integrls. When the re of ech prt is found, the totl re cn be found b dding the prts. f() = ² For emple, to find the re between the grph of: = ² nd the is, from = to =, we need to clculte three seprte integrls: The zeros of the function f() tht lie between nd form the boundries of the seprte re segments. In this cse there re zeros t = nd =, nd so three seprte res must be found: A, A nd A s follows: A = d A = A = ( ) ( ) ( ) d  d So the shded re, A = A + A + A. A   A A
12 Mths Helper Plus cn grph the function, locte the zeros nd clculte the definite integrls. Follow these steps to find the re under the curve s described in the emple on the front side of this sheet. ) Strt Mths Helper Plus, then lod the file: Clculus  Integrls.mhp. The grph view will displ plot of the function: = ². Do NOT use the integrl tool for this ctivit! ) We wnt to find the re under the curve between = nd =. The curve crosses the is between these vlues, so we use the intersection tool to locte the zeros, like this: Click the toolbr button to select the intersection tool. Now click the mouse cursor on the points where the grph cuts the is. In ech cse, red nd record the coordinte of the intersection point from the dilog bo. Zeros t: = nd t = CLICK! CLICK! Cncel the intersection tool b clicking the Cncel button in the dilog bo. ) Identif the seprte integrls tht need to be found. (There re three in this cse.) Write down the boundr vlues for ech of these integrls: Integrl : from to. Integrl : from to. Integrl : from to. ) Crefull point to the function curve with the mouse pointer then double click with the mouse to displ the options dilog for the function. (This works better on notsosteep prts of the grph.) Select the Integrls tb, then click on the Limits of integrtion edit bo. Tpe the integrl boundries ou require using squre brckets for ech seprte integrl, like this: [,] [,] [,] Click on the Number of intervls edit bo. Tpe : 00. (The lrger this number the greter the ccurc, but if it is too big the clcultions m be slow. This must be n even number to use Simpson s rule.) Select Shde res on grph. Also select Simpson s rule. Clcultion mode should be definite integrl. Click the OK button to close the dilog bo. The res will be shded on the grph, nd the integrls will be displed on the tet view. ) Add the bsolute vlue of the definite integrls to clculte the totl required re. Totl re = + + = 6) Use Mths Helper Plus to find the re under the functions below in the intervl shown. Use this tble to record our results. (The first row hs been prtl filled in s n emple.) function intervl zeros t = integrl boundries clculted res totl re = ² to, [, ] [,] [,] = ² + + to = ² to = ³ + ². to.
13 Nme: Clss: p 89 Mths Helper Plus Resource Set. Copright 00 Bruce A. Vughn, Techers Choice Softwre Clculus  Activit 7 Are between two curves with two intersection points b subtrction of res. To find the re between two intersecting curves tht onl intersect t two points, we first find the coordintes of the two intersection points: = nd = b. Definite integrls give us the re under ech curve from = to b, then we subtrct the two res. ) Strt Mths Helper Plus, then lod the file: Clculus  Integrls.mhp. The grph view will displ simultneous plot of the two functions: = ² + nd = + ) Click the toolbr button to select the intersection tool. Click the mouse cursor t the intersection points of the two curves. Record the coordintes of the intersection points: = = ² = + = b (left) = = (right) = b = Click Cncel on the intersection tool dilog to cncel the intersection tool. ) Crefull point to one of the function curves with the mouse pointer. Double click to displ the options dilog bo for tht function. (This m work better on prts of the grph tht re not so steep.)   o Select the Integrls tb, then click on the Limits of integrtion edit bo. Tpe the nd b vlues from () bove with squre brckets, like this: [,] o Click on the Number of intervls edit bo. Tpe : 00. (The lrger this number the greter the ccurc, but if it is too big the clcultions m be slow. This must be n even number to use Simpson s rule.) o Select Shde res on grph. Also select Simpson s rule. Clcultion mode should be definite integrl. Click the OK button to close the dilog bo. The re under this function will be shded on the grph, nd the definite integrl will be displed on the tet view. The bsolute (positive) vlue of the definite integrl equls the re of the first function, = ) Crefull point to the other function curve nd double click s before. o Select the Integrls tb nd set the sme options s for the first function. o Click the Shde colour... button, nd choose ver ple colour tht is different to the shding colour of the other function. Ple blue works well. Click the OK button to close the dilog bo. The bsolute vlue of the definite integrl equls the re of the second function, = 6) The re between the two grphs will now hve its own colour nd be es to identif. Clculte the re between the curves b subtrcting the smller re from the lrger re. Are between curves = =
14 7) For ech of the pirs of functions below, use Mths Helper Plus to help ou find the re between the plotted curves. In ech cse, write nswers nd clcultions in the spces provided. Also, sketch the grphs on the grph re provided, nd shde the re between the curves. In ech cse, strt b creting new document in Mths Helper Plus. ( new commnd in the File menu. ) ) Find the re between: = ² + + nd = + Intersection points: (left) = =. (right) = b =. First function: Definite integrl from = to = b =. Are under curve from = to = b =. Second function: Definite integrl from = to = b =. Are under curve from = to = b =. Are between curves = bigger re smller re = b) Find the re between: = ² nd = ² + + Intersection points: (left) = =. (right) = b =. First function: Definite integrl from = to = b =. Are under curve from = to = b =. Second function: Definite integrl from = to = b =. Are under curve from = to = b =. Are between curves = bigger re smller re = CHALLENGE QUESTION c) Find the re between: = ²( )( ) nd =
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