Lines, Conics, Tangents, Limits and the Derivative

Transcription

1 Lines, Conics, Tangents, Limits and te Derivative Te Straigt Line An two points on te (,) plane wen joined form a line segment. If te line segment is etended beond te two points ten it is called a straigt line. Te points A (, ) and B (, ) represent te two general points on an line.. B (, ). A (, ) Te slope m of te line containing te points A and B is given b: m = Te lengt of te line segment AB or AB is given b: d = ( ) + ( ) Equations of Lines Parallel lines ave te same slope and perpendicular lines ave slopes tat are negative reciprocals of eac oter. ie. and. Lines tat are parallel to te ais ave a zero slope ( ie., k R ) and are given in te k form = b were b is te intercept. Lines tat are parallel to te ais ave a slope tat is undefined (ie. k, k R ) and are given in te form = a were a is te intercept. To find te equation of a particular line ou must be given or be able to find a point on tat line and its slope. Eac line is unique and tere will be onl one equation wit te given caracteristics. An of tree forms for te equation can be used. A) Slope - intercept = m + b m = slope of te line b = intercept B) Standard Form a + b + c = a, b, and c are integers and a > m = a b C) Slope Form ( ) = m or ( ) m = slope of te line ( - ) = m ( - ) (, ) is a point on te line. Eamples:

2 . Given te points A (-,4) and B (6, ) find te slope of AB and te lengt of AB. m AB = Use te formula for slope 4 = Substitute in te values for te and coordinates. 6 ( ) = Simplif. 9 AB = ( ) + ( ) Use te formula for lengt or distance. = ( 9) + ( ) Substitute in te values for te and coordinates. = 8 Simplif.. Find te equation of te line passing troug ( -, ) wit a slope of. = ( ) Use slope form for te equation of te line. ( - ) = ( + ) Multipl bot sides b te denominators. - = + Simplif = Rearrange in standard form.. Find te equations of te lines troug ( -, 5) and a) parallel to = - 7 b) perpendicular to = - 7 a) m = Te given line as slope as = - 7 Te required line is = + b Use slope intercept form 5 = (-) + b Substitute in te values of and. b = Solve for b. = + Substitute in te values for m and b. b) m = Use te negative reciprocal for perpendicular slope. = ( ) Substitute values in te slope form. ( - 5) = -( + ) Multipl bot sides b te denominators. - = - - Simplif = Write in standard form. 4. Find te equation of te line troug (, - 8) a) parallel to ais

3 b) perpendicular to ais a) = b Zero slope = - 8 Substitute in te value to find b. b) = a Tis is te equation for parallel to ais or perpendicular = to ais. Substitute in te value of te given point. X and Y intercepts Te - intercept ( i ) is te point were te line crosses te ais. It is found b letting = in te equation of te line and solving for. Te - intercept ( i ) is te point were te line crosses te ais. It is found b letting = and solving for. Points tat are on a line satisf te equation of te line. Eamples:. Determine weter te given points are on te line - + =. a) (, 5) - + Use te left side of te equation. = () - (5) + Substitute in te values for and = Evaluate. = Left side = Rigt side (, 5) is on te given line. b) (-, ) - + = = Left side Rigt side (-, ) is not on te line.. Find te and intercepts for te line - + = let = i - + = Solve for. = 6 let = i + = Solve for. = - 4

4 Conic Sections Te curves on te, plane are called conics. Teir equations sow te position and sape of te conic. Notice te differences and similarities of all te conic section equations. Wen solving a problem coose te most appropriate equation and find te values for te missing variables. Eac can be identified b teir equation. Look carefull at te plus and minus signs as well as were te terms are placed. Name Equation Description Sketc 4 Parabola = a - verte (,) - a > opens up - a < opens down - smmetric about ais a < a > Parabola = a - verte (, ) - inverse of = a - a > opens rigt - a < opens left - smmetric about ais a < a > Circle + = r - centre (, ) - radius = r r Ellipse a a > b + = b - centre (, ) -major ais (longest) on ais = a - minor ais on ais = b - vertices at ( ± a, ) (a, ) (, - b) Ellipse b a > b + = a - centre (, ) - major ais on ais = a - minor ais on ais = b - vertices at (, ± a) (, a) (b, )

5 5 Hperbola = a b no relation between a and b - centre (, ) - vertices on ais - transverse ais between vertices = a - conjugate ais = b - vertices ( ± a, ) - asmptotes = ± b a = b a (a,) Hperbola = b a no relation between a and b - centre (, ) - vertices on ais - transverse ais between vertices = a - conjugate ais = b - vertices (, ± a) - asmptotes = ± a b (, a) = a b Equilateral Hperbola or Rectangular Hperbola = a - centre (, ) - transverse ais on ais - vertices ( ± a, ) - asmptotes = ± = (a, ) Equilateral Hperbola or Rectangular Hperbola = a - centre (, ) - transverse ais on ais - vertices (, ± a) - asmptotes = ± (, a) = Eamples:. Find te equation of te following conics. a) circle centre (, ) wit radius 5. + = r Coose te appropriate equation. + = ( 5) + = 5 Substitute in te value of te radius. Simplif.

6 6 b) parabola smmetric about te ais troug (, 4) and verte (, ). = a Coose te correct parabola equation. = a (4) Substitute in values for and. a = Solve for a. = 8 8 Put a back in te original parabola equation. c) ellipse centre (, ) wit one verte at (- 5, ) and troug (- 4, ). a + = b ( 4) + = 5 b 4 6 b = b = 5 = b 9 + = = 5 or Coose te correct ellipse equation. Substitute in te values for, and a. Rearrange to solve for b. Substitute in te values for a and b. d) Hperbola centre (, ), one verte (, - 6) and one asmptote =. b = a = 4 6 Coose te correct perbola equation. a a = 6 from te verte and from =. b = 6 = b = b Substitute in te values for a and b. e) Rectangular perbola centre (, ) wit verte on te ais troug (-, ). = a Coose te correct rectangular perbola equation. (-) - = - a Substitute in te values for and. a = 5 Solve for a. = 5 Substitute back in te value for a.

7 7. Identif eac of te following b name, describe eac and sketc. a) + = (, 5) 9 5 Ellipse wit vertices on ais at (, ± 5), major ais = units (, ) minor ais = 6 units centre (, ) b) = 49 Rectangular perbola = vertices on te ais at ( ± 7, ) (7, ) asmptotes = ± centre (, ) c) = 64 Divide bot sides b 6. + = 4 Circle wit radius = (, ) centre (, ) d) 5 4 = Divide bot sides b 5 = = 4 5 Hperbola vertices on ais at ( ±, ) (, ) transverse ais = 4 conjugate ais = asmptotes = ± 5 centre (, )

8 Translation of Conics - Conics wit centre moved to (, k) Conics tat ave been translated ave teir centre moved from (, ) to anoter point on te, plane. Te retain teir original sape and all te properties of tose at (, ). Everting is translated eactl te same amount as te centre including te vertices and asmptotes. Te following are te equations in standard form for eac of te conics wit centre (, k). Circle + = r ( ) + ( k) = r 8 Parabola = a ( k) = a( ) or = a ( ) = a( k) Ellipse a b ( ) ( k) + = + = b a b or ( ) ( k) + = + = a b a Hperbola a b ( ) ( k) = = b a b or a ( ) ( k) = = b a or =± a ( ) ( k) =± a Eamples:. Name and describe eac of te following. ( ) ( + ) a) = 6 49 Hperbola centre (, -), vertices on = parallel to ais at (, 5) and (, -9), transverse ais = 4, conjugate ais = 8, asmptotes ( + ) = ± 7 ( - ). 4

9 9 b) ( + 5) + ( ) = 5 Circle wit centre (- 5, ) and radius = 5 = 5 = ( + ) + = ( + ) c) ( ) Parabola opening up wit verte (-, - ) Equations of Conics in Standard Form Wen te equations of conics wit centre (, k) ave been epanded te centre and te properties of te conic can not easil be seen. It is necessar to put te conic back into standard form using te metod of completing te square. Te metod retrieves te perfect square brackets tat sow te centre. See Formula for Success page and Unit in te section on Equations te won t factor for more information on completing te square. Eamples:. Find te centre of eac of te following. a) = + + 4= Group te terms togeter and te terms also. ( + 5 5) + ( ) = Add and subtract te square of alf te coefficient of te and terms. ( 5) 5+ ( + ) 4= Factor te first terms in eac bracket and remove last term in eac bracket. ( 5) + ( + ) = 9 Collect like terms. circle wit centre (5, -) and radius. b) = = 68 4( + ) 9( ) = 68 Factor out te coefficients of and ten add and subtract te square of alf te and coefficients 4( ) 4 9( ) + 6= 68. Factor te first terms in te brackets and multipl out te last term in eac bracket. (Watc te signs) 4( ) 9( ) = 6 Collect like terms. ( ) ( ) = Divide b perbola centre (, ) wit vertices on = parallel to te ais.

10 Solving Sstems of Equations Te solution to a sstem of two equations is te ordered pair(s) (, ) tat are te point or points of intersection. Tese points will satisf bot te equations. Tese ma be te points were cost and revenue are equal or suppl equals demand. Te metod of solving depends upon te tpes of equations in te particular question. Linear Sstems - intersection of two lines Two lines ma intersect in onl one point or ma not intersect at all if te lines are parallel. If te two equations represent te same line ten te points of intersection are all points on te line. line line line line and P(, ) line one point of intersection no points of intersection all points of intersection lines are parallel lines are same To solve tese equations two metods are generall used - elimination or substitution. Eiter metod can be used on most questions but one metod ma be sorter or easier on particular equation. Elimination - use addition or subtraction to eliminate one te variables Eamples:. 7 + = () Te coefficients of one of te variables must be te same 5-4 = 49 () to use elimination = 46 () = () Multipl () b so te coefficients are te same. 9 = 95 Add () and () to eliminate. = 5 Solve for. substitute = 5 into () Use te value in one of te original equations to find. 7(5) + = = - 5 = - = - 6 te point of intersection is (5, - 6).. - = 8 () - 8 = () - 8 = 48 () = () 6 Multipl () b 6. Bot and coefficients are te same. = 7 Subtract () - () Not true 7. no solution. Tese lines ave te same slope and are parallel. Tere is no point of intersection.

11 Substitution - substitute te epression for one variable into te oter equation. Eamples:. - = () + 4 = - 8 () = - () Rearrange () to solve for. + 4( - ) = - 8 Substitute tis epression into () for = - 8 Epand te bracket and solve for. = 44 = 4 Substitute = 4 into () = (4) - = - 5 te point of intersection is (4, - 5). Ceck our solutions b substituting te solution into te oter original equation. If te solution is correct it will satisf bot equations. Linear - Quadratic Sstems - intersection of a line and a conic Tese sstems are solved using te metod of substitution and ma ave,, or solutions. P (, ) P(, ) P (, ) Two points of contact. One point of contact. No points of contact. Line is a secant to conic. Line is tangent to conic. Eamples:. + = 5 () = () A line and a circle. + ( ) = 5 Substitute epression for into (). + 4 = 5 Simplif and solve for. 5 = 5 = = ± Substitute values for into () = () or = (-) = or = - points of intersection are (, ) and (-, - ).

12 . + 6 = 6 () - 4 = 4 () = () Rearrange () to solve for. ( 4+ 4) + 6 = 6 Substitute te epression for into () = 6 Epand te bracket. + = Collect like terms. ( + ) = Factor. (Common factor) = or = - Solve for. = 4 + 4() or = 4 + 4(- ) Solve for using (). = 4 or = te points of intersection are (4, ) and (, - ). Equations of Tangents to Curves - given te slope of te tangent or a point on te tangent Lines tat are tangent to one of te conics ave onl one point of intersection wit te curve. Tat means tat tere is onl one solution to te quadratic equation or tat b 4ac= wen using te quadratic formula to solve. Since a tangent is a line ou must know or be given eiter te slope of te line or a point on te line. Eamples:. Determine te equation(s) of te tangents to 6 + = 6 wit slope. = + b () General equation to line wit slope. 6 + = 6 () Equation of ellipse. 6 + ( + b) = b+ b = b+ b 6= b 4ac= ( b) 4( 7)( b 6) = 4b 68b + 88= 64b = 88 b = 7 b=± 7 Substitute te epression for into (). Square bracket. Collect like terms. Quadratic equation wit a = 7, b = b, and c = b 6. One solution for tangent. Substitute in values for a, b, and c. Simplif. Collect like terms. Solve for b. te equations wit slope tangent to 6 + = 6 are = ± 7.

13 . Find te equation of te tangents to + = wit intercept of -. = m - () General equation to line wit i = -. + = () Equation of circle. + ( m ) = Substitute epression for into (). + m 4m+ 4= Square bracket. ( + m ) 4m+ = Collect like terms. Quadratic wit a = +m and c =. b 4ac= Tangent as one solution. ( 4m) 4( + m )( ) = Substitute in values for a, b, and c. 6m m = 4m = m = m=± Epand te brackets. Collect like terms. Solve for m., b = - 4m Te equations of te tangents to + = wit intercept of - are = ± -. Limits Te limiting value of a function is te value for f() as approaces a given value. Eample: If f() = - find te limiting value of f() as approaces. B looking at te grap of f() around te value of we can see tat te values of or f() get closer and closer to. ie. lim ( ) = () - = As approaces f() approaces Limits of Indeterminate Form Wen finding te limit, if te grap is continuous (as no gaps or undefined areas) te limit can be found b substituting te value for into te function as above. If te answer for f() is indeterminate (ie. ) wen te value for is substituted into te question ten anoter approac must be taken. Tis ma take te form of factoring, using a substitute variable or rationalizing te numerator. Watever manipulation is used, te object is to remove te zero in te denominator wen te value of is substituted into f().

14 Eamples: 4. lim Wen - is substituted into te function te answer is. + = lim ( + )( ) Factor te numerator and divide out te common factor. ( + ) = lim ( ) Now - can be substituted into te epression. = lim Wen is substituted into te function te answer is. Let u = +4 Use substitution to remove te root. u = + 4 = u 4 Wen = for te limit u = 4 or u lim Substitute for and. u u 4 u = lim Factor and divide out. u ( u )( u+ ) = lim u u+ = Substitute in for u lim is te answer wen is substituted in for lim( )( ) = Multipl numerator and denominator b te conjugate of te numerator. Same roots but te sign between tem is opposite. + 4 ( + ) = lim Multipl out te numerator and leave te denominator ( ) factored. = lim Simplif te numerator and ten divide out te ( ) = lim Substitute in for =

15 Te Slope of te Tangent to a Curve - Given a point on te curve Te slope of a tangent line to a curve at a point P is te limiting slope of te secant line PQ as Q slides along te curve towards P. (Note: te points P and Q are on te curve f().) 5. Q [ +, f( + )] P [, f()] units orizontall Te orizontal distance between P and Q is units. As Q slides along te curve towards P te distance approaces zero. f ( + ) f ( ) slope of te secant PQ = ( + ) f ( + ) f ( ) = slope of te tangent at P f = lim ( + ) f ( ) Eamples:. Find te slope of te tangent to f() = at =. f te slope of te tangent at P = lim ( + ) f ( ) wen f() =, f( + ) = ( + ), =, and f() = 9 = lim ( + ) 9 = lim ( + )( + + ) Factor numerator (difference of squares) = lim ( 6+ ) Simplif numerator and divide out. = 6 Substitute in for. te slope of te tangent at = to f() = is 6.

16 6. Find te slope of te tangent to f() = wen = 4. f m of tangent = lim ( + ) f ( ) wen f() =, f(+) = +, = 4 and f(4) = 4+ = lim Let u = 4+, = u 4 u = lim u u 4 Wen =, u = 4 or. u = lim u ( u )( u+ ) Factor and divide out te common factor. = lim u u+ Substitute in for u. = 4 Te slope of te tangent to f() = at = 4 is. 4 b) Find te equation of te tangent to f() = wen = 4. m = and (4, ) is on te tangent 4 = m Use slope form of te equation of a line. Substitute in values for (, ) and m. 4 = 4 4( - ) = - 4 Multipl bot sides b te denominators. 4-8 = = Rearrange in standard form. Te equation of te tangent is =.. Find te slope of te tangent to f() = at te point P [, f()]. f m of tangent = lim ( + ) f ( ) wen f() = and f( + ) = ( + ) = lim ( + ) [ ][ ] = lim ( + ) ( + ) + ( + ) + Factor as a difference of cubes. [ ] = lim ( + ) + ( + ) + Simplif and divide out.

17 7 = lim( + ) + ( + ) + = ( + ) + ( + ) + Substitute in =. = Te slope of te tangent at all points on te curve f() = is. If te slope is needed at a particular value it can be substituted into. Te Derivative f Te limit lim ( + ) f ( ) is so important in te stud of calculus tat it is given its own name and notation. It is called te derivative and is denoted b f ( ) or. Tese are read as f prime of or prime. Te derivative and te slope of te tangent to a curve at te point P [, f()] are one and te same ting. Tus te slope of a tangent m to a curve is m = f () were f f () = lim ( + ) f ( ) Now tr te following questions for practice. Follow te steps as ou work our wa troug to ceck our answers. Practice. Given te points A (-, ) and B (4, - 5) find te following: a) te lengt of te line segment AB. AB = ( ) + ( ) ( ) ( ) Use te distance formula. = 4 ( ) + 5 Substitute in te values from points A and B. = 6 + ( 8) Simplif. = = Evaluate. b) te equation of te line troug AB m = Find te slope of te line first. Use information m = 8 4 = 6 = m from a) above. Use te slope form of equation.

18 8 = 4 Substitute in values for m and (, ). ( ) ( - ) = - 4( + ) Multipl b te denominators. - 9 = = Put equation in standard form. c) te equation of te line troug A parallel to te ais. = a Equation of all lines parallel to ais. = - Substitute in te coordinate of A.. Find te slope, and intercepts for te line =. = m + b = + 5 = + 5 Rearrange in slope intercept form. Coose te information from te equation. m = and b = 5 M is te slope and b is te intercept. Let = in te original equation. Find intercept b letting =. + 5 = = - 5 Solve for. Te slope is, te intercept is 5 and te intercept is Find te equation of te line troug (- 5, 7) tat is perpendicular to =. = + m = 7 m = 7 = ( 5) ( - 7) = - ( + 5) - 4 = = Rearrange equation in slope intercept form. Use te negative reciprocal of te slope for perpendicular slope. Use te point and slope for te new equation. Multipl troug b te denominators. Put in standard form. 4. Determine if te point (-, 4) is on, inside or outside te circle given b + = 6. L.S. = ( ) + 4 R.S. = 6 Substitute in te and values on te left side. = = 5 Evaluate and if te number is more tan, less tan 5 < 6 or equal to te rigt side ten te point is outside (-, 4) is inside te circle inside or on te circle.

19 5. Name and give ke features of te conics given b te following equations. a) ( ) ( + ) = 5 49 Look at te conic equations and find te appropriate perbola, centre (, - ) one. Negative sign in between is a ke feature. transverse ais parallel to ais Te positive indicates parallel to ais. transverse ais = conjugate ais = 4 b) = ( 7) + 4 Look at te equations to find te rigt matc. ( 4) = ( 7) parabola, opening rigt verte (4, 7) Move te 4 to be wit te. 6. Find te point(s) of intersection for eac of te following sstems of equations. a) = () = () = () = () Multipl te first equation b to matc s. 5-5 = () - () Subtract equation from equation. = Solve for. + 6() - 7 = Substitute in for to find. = 5 Solve for. Te point of intersection is (5, ). b) = () + = () = - () Rearrange equation to substitute into equation. ( ) = Square and collect like terms. = =± Divide bot sides b. Take square root of bot sides to solve for. Substitute in = and - Substitute in bot values for in () to find. = -() or = -(-) = - = te points of intersection are (-, ) and (, - ). 7. Determine te equation of te tangents wit slope to = 9. = + b () Equation of all lines wit slope. = 9 () Substitute () into (). ( + b) = 9 ( 4 + 4b+ b ) = 9 4b b 9= Square te bracket. Collect like terms and rearrange to use quadratic 9

20 b 4ac= ( 4b) 4( )( b 9) = 6b b 8= 4b = 8 b = 7 b=± 7 ± formula were a = -, b = - 4b and c = b 9 One solution as tangent touces one time. Substitute in all values for a, b, and c. Square and multipl out te bracket. Simplif. Solve for b. te equations of te tangents to = 9 wit slope are = ±. 8. Write in standard form to determine te coordinates of te centre of te conic and identif te conic given b te equation =. ( ) + 4 = ( ) 9+ 4 = ( ) + 4 = 8 Rearrange and ten add and subtract te square of alf te coefficient of te term. Factor te first tree terms in te bracket and bring out te subtracted term. ( ) + = Divide b 8 and look at te form to name te conic. 8 Te conic is an ellipse wit centre (, ). 9. Find te value of te following limits. + a) lim Substitute in - and te result is = lim Factor te denominator. ( + )( 5 ) = lim Divide out te common factor. ( 5 ) = Substitute in - for 6 9 b) lim Substitute in and te result is. 9 = lim ( )( + ) Factor numerator as a difference of squares. 9 = lim( + ) Divide out te common factor. 9 = 6 Substitute in 9 for.

21 9. Find te equation of te tangent to te curve = 4 at te point (-, ). f m of te tangent = lim ( + ) f ( ) were f() = 4, = - and f(-) = f = lim ( + ) f ( ) f(- + ) = 4 ( + ) = 4 - (4-4 + ) 4 = lim = 4 = lim ( 4 ) Factor and divide out. = 4 Substitute in for. te slope of te tangent to = 4 at = - is 4. Equation of line wit slope 4 troug (-, ) is = 4 Slope form of equation. ( ) = 4( + ) = Te equation of te tangent to te curve = 4 at (-, ) is = Review of Unit # 7. Given P (, - 4) and Q ( 5, 7) find te following: a) te slope of PQ. b) te lengt of PQ c) te equation of te line troug PQ. d) te equation of te line troug Q perpendicular to te ais.. Find te, intercepts and slope for =.. Find te equation of te line parallel to = troug (4, - ). 4. Name and give ke features for te following conics: a) = 6 b) = ( + ) 4 c) 6( + 5) + 9 = Find te equations for te following conic sections: a) perbola wit centre (, ), vertices (, ± 4), te equation of one asmptote =. b) ellipse wit centre ( - 5, ), major ais parallel to ais and lengt = 4, minor ais lengt = 8.

22 6. Find te points of intersection for te equations + = 7 and ( 4) + ( + ) =. 7. Determine te equation(s) of te tangent(s) to te curve + = 5 wit intercept = Cange to standard form to determine te coordinates of te centre and identif te conic wit te equation =. 9. Find te following limits: a) lim b) lim. a) Find te derivative of te curve = 4. b) Determine te slope of te tangent to te curve = 4 at =. Answers:. a) b) 5 c) = d) = 7. i = - i = m = = 4. a) rectangular perbola, centre (, ) vertices (, ± 6), asmptotes = ± b) Parabola opening up, verte (-, - 4) c) Ellipse centre (- 5, ) major ais parallel to ais, lengt = 8, minor ais = 6 vertices (- 5,4) and (- 5, - 4). 5. a) = b) (7, ) and (, ) =± + ( + 5) ( ) = 8. Hperbola wit centre ( -, ) transverse ais = 4, conjugate ais = 5 9. a) b). a) 4 b) m =

23