Fall 9 Instructor: Dr. Firoz Updated on October 4, 9 MAT 65 Basic Algebra and Trigonometr Review Algebra Factoring quadratic epressions Eamples Case : Leading coefficient. Factor 5 6 ( )( ), as -, - are factors of 6 and adds up to 5. Factor 5 6 ( 6)( ), as -6, are factors of -6 and adds up to 5 Case : Leading coefficient is not. Factor 5 ( ) ( )( ), as -, - are factors of 6 (= ) and adds up to 5. We insert leading coefficient in the boes.. Factor 6 7 ( ) (6 )( / ) ( )( ), as -, and -4 are factors of (= 6 ) and adds up to 7. We insert leading coefficient 6 in the boes and perform necessar algebraic works. 4 Case : Factors b grouping 4. Factor 4 ( ) ( ) ( )( ) Remember the formula Case 4: Using formula ( )( )( ) a ( a)( a a ) a a a ( )( ) 4. Factor ( )( ) ( )( )( ) Absolute value functions For an real number, Properties of,if,if. ab a b and, if b, a b a b
Fall 9 Instructor: Dr. Firoz Updated on October 4, 9. a b a b. a if and onl if a a 4. a if and onl if aor a Note: Properties and 4 are true for, as well. The function is alwas even and has -ais smmetr. To determine domain of a function: Practice following eamples on equations and inequalities. Solve the following equations:. 8 67 Answer: 8 4. Answer: 4. Solve ( 5) 4 ( ) Answer: 7 7 4. Solve Answer: 4 ( )( ) Solve the following Inequalities and write answers in interval(s), use real line test:. Solve the inequalit 9 4 and write our answer in interval notation.. Solve. Solve 6 4. Solve 7 5. Solve 5 6 6. Solve 5 6 7. Solve 6 8. Solve 6 9. Solve 5 6 5 6. Solve 5 6. Solve 5 6. Solve 5 6. Solve 4. Solve
Fall 9 Instructor: Dr. Firoz Updated on October 4, 9 Hint: Do not cross multipl to solve. Use etc. 5. Solve 6. Solve 7. Solve 5 6 5 6 8. Plot the rational function 9. Find the domain of f 9 ( ) 5 showing all asmptotes.. Find all zeros of the polnomial P ( ). Find the point(s) of intersection of the curves 5,. Find the slope of the line through the points (, 5) and (7, -9). Find the equation of a straight line through (, ) and parallel to the line through the points (5, 7) and (-, ) Eponential and Logarithmic Functions Remember the following formulas 6. Eponential function r e!! r! r r e ( )!! r! ( ln a) ( ln a) ( ln a) a ln a!! r! r. Hperbolic functions e e e e a) sinh b) cosh. Logarithmic functions:, r r ln( ) ( ) r r ln( ) r Properties: a) log : log, ln : log e b) log( ) log log, ln( ) ln ln
Fall 9 Instructor: Dr. Firoz Updated on October 4, 9 c) log( / ) log log, ln( / ) ln ln d) log, ln e p e) log p log f) log m, m m ; log, ln log ln g) log log ln 4. Solve the following equations for : 4 a) e 8 e b) e e c) ln 9ln d) ln( ) 8 4 e) e 5 ln f) 8e 5 5. Write the following result as a single logarithmic epression a) ln 5 ln(/5) b) ln(5/8) ln() ln5 6. Determine the following values a) log5 5 b) log 5(/ 5) c) log () 7. Round our answer to two decimal places if necessar a) Given f ( ) sinh( ), find f () b) Given f ( ) 9cosh( ), find f () b c) Given f ( ) ae, find when f (), f () 4, f ( ) Trigonometr h o The well known notations: soh, cah, to soh: s stands for sine, o stands for opposite and h stands for hpotenuse, sin cah: c stands for cosine, a stands for adjacent h stands for hpotenuse, cos a o h a h 4
Fall 9 Instructor: Dr. Firoz Updated on October 4, 9 toa: t stands for tan, o stands for opposite and a stands for adjacent, tan Where is the angle between the hpotenuse and the adjacent. Other three trigonometric functions have the following relations: h csc sin o, h sec cos a and a cot tan o o a. sin( n ) [?]sin, cos( n ) [?]cos, tan( n ) [?]tan, the sign? is for plus or minus depending on the position of the terminal side. One ma remember the four-quadrant rule: (All Students Take Calculus: A = all, S = sine, T = tan, C = cosine) sin all tan cosine Eample: Find the eact value of sin sin( 8 6 ) [ ]sin 6 = - quadrant four where sin is negative. Important values: sin. We ma write, in this case the terminal side is in sin 6 45 cos tan csc Undefined sec cot Undefined One ma create second row (for sin) as follows: 4 6 9 Undefined Undefined 5
Fall 9 Instructor: Dr. Firoz Updated on October 4, 9 sin 4 4 4 = 4 = If ou write these values from backward ou will get values for cos. 4 4 In the following diagram, each point on the unit circle is labeled first with its coordinates (eact values), then with the angle in degrees, then with the angle in radians. Points in the lower hemisphere have both positive and negative angles marked. The functions sin and cos are periodic of period, so that sin( ) sin, cos( ) cos On the other hand tan is periodic of period, thus tan( ) As a general form it is alwas true that sin( n ) sin, cos( n ) cos and tan( n ), for n is a positive integer. Also sin( / ) cos, cos( / ) sin 6
Fall 9 Instructor: Dr. Firoz Updated on October 4, 9 Trigonometric identities: cos ( ) sin ( ) tan ( ) sec ( ) cot ( ) csc ( ) cos( ) cos( )cos( ) sin( )sin( ) sin( ) sin( )cos( ) cos( )sin( ) tan( ) tan( ) tan( ) tan( ) tan( ) sin() sin( )cos( ) cos ( ) sin ( ) cos() cos ( ) sin ( ) tan() tan( ) tan ( ) c a b ab C cos( ) sin( A) sin( B) sin( C) a b c sin ( ) cos ( ) tan ( ) cos() cos() cos() cos() cos sin tan cos( ) cos( ) cos( ) cos( ) sin( )sin( ) cos( )cos( ) sin( )cos( ) cos( )sin( ) [cos [cos [sin [sin cos( cos( sin( sin( )] )] )] )] sin( ) sin( ) cos( ) cos( ) sin( ) sin( ) cos( ) cos( ) sin cos sin cos cos cos sin sin 7
Fall 9 Instructor: Dr. Firoz Updated on October 4, 9 Solution of trigonometric functions. Find all solutions of the equation sin, for an integer n. Solution: sin sin / sin( / ) sin( / ) Then n / or n /. Find all solutions of the equation cos 8cos, for an integer n. Solution: Factor the equation to (cos 5)(cos ) cos cos(), cos 5 Then n. Find all solutions of the equation sin sin, for an integer n. Solution: Factor the equation to sin (sin / ) sin sin() sin( ), sin / sin( / ) sin( / ) Then n or n / or n / Here we have two diagrams, one is given and another one is in Eample. and 4. Find all solutions of the equation sin, for an integer n. Trigonometric sum of angles For an real numbers and, the following identities hold:. sin( ) sin cos sin cos. cos( ) cos cos sin sin When = we have the following important identities. sin sin cos. cos cos sin sin and also sin / ( cos ), cos / ( cos ) cos 8
Fall 9 Instructor: Dr. Firoz Updated on October 4, 9 Sum to product form. sin sin sin cos 4. If ou replace b ou get sin sin sin cos 5. cos cos cos cos 6. cos cos sin sin Inverse Trigonometric functions (Remember the relations) Formula:. If If If. If If If. If If If sin ( t) sin ( t), tan ( t) tan ( t), cos ( t) cos ( t ) sin then sin sin(sin ) then [, ] and [ /, / ] sin (sin ) then [ /, / ] cos then cos and [, ] cos(cos ) then [, ] cos (cos ) then [, ] tan then tan and ( /, / ) tan(tan ) then (, ) tan (tan ) then ( /, / ) 4. If sec then sec and [, / ) ( /, ] 9
Fall 9 Instructor: Dr. Firoz Updated on October 4, 9 Chapter Functions and Limits Section. Functions and their representations A function f is a rule that assigns to each element in a set A eactl one element, called f(), in a set R. The set A is called the domain and the set R is called the range of the function. a) Following are the graphs of functions Fig. Fig. Fig. b) Following are not the graphs of functions Fig.4 Fig.5 Fig.6. Plot the function 5 Solution: Let us consider the following points Table - - f () - 5 8 Fig.
Fall 9 Instructor: Dr. Firoz Updated on October 4, 9. Find the domain: Eamples a) Find domain of f ( ) 5 6 Solution: The function is a polnomial; the domain is the entire real line, (, ) b) Find the domain of g ( ) 5 6 Solution: The function is a rational function, its denominator cannot be zero, 5 6,. The domain is (,) (,) (, ). c) Find the domain of g ( ) 5 6 Solution: The denominator must be strictl positive, 5 6. To solve this inequalit follow the steps: Step. Solve for 5 6 to get, Step. Use calculator (graph and trace at test points) to check sign for the function g ( ) 5 6 at each test point. Test points are circled..5 4 Sign of g ( ) + undefined + The domain is (,) (, ) d) Find the domain of f ( ) 5 6 Solution: For domain 5 6, (nonnegative) Step. Solve for 5 6, Step. Use calculator (graph and trace at test points) to check sign for the function f ( ) 5 6 at each test point. Test points are circled. For equal sign in the problem we consider solid circles to include solution points..5 4 Sign of g ( ) + undefined + The domain is (,] [, )
Fall 9 Instructor: Dr. Firoz Updated on October 4, 9 e) Find the domain of g ( ) 5 5 6 5 Solution: We must have with 5 6 5 6 To solve this inequalit follow the steps: Step. Solve for 5 6 to get, and also 5 5 Step. Use calculator (graph and trace at test points) to check sign for the function g ( ) 5 6 at each test point. Test points are circled. Eercise:.5 4 6 5 Sign of g ( ) undefined + undefined + The domain is (,) [5, ) a) g( ) b) g( ) c) g ( ) d) f( ) g) e) f( ) f ( ) 4 h) f( ) j) f ( ) ln( ) k) ( ) ln m) g ( ) 5 6 n) f) f( ) 7 i) f ( ) 4 f l) f ( ) ln( ) ln g( ) 6 o) g ( ) 5 6. Even and odd functions: A function f() is even if f ( ) f ( ). And a function f() is odd if f ( ) f ( ). The function f( ) is even and the function f( ) is odd. 4. Piecewise defined function: A function defined as follows is called piecewise defined function f( ) 8 5. Graph the function in eample 4.
Fall 9 Instructor: Dr. Firoz Updated on October 4, 9 Section. A Catalog of Essential Functions A function of the tpe f () a b is called a linear function. The domain is the entire real line. A function of the tpe f ( ) a b c is a quadratic function. The domain is the entire real line. g ( ) A function of the tpe f( ) h ( ) values where h ( ). h( ), g( ) are the polnomial functions. Transformation of functions: is a rational function. The domain is the set of. The function f() is verticall transferred b k units when represented as f () k. The function f() is horizontall transferred b h units when represented as f ( h ). Suppose c > cf (), Stretch (elongate) the graph of f() verticall b a factor of c f ( ), Compress (shrink) the graph of f() verticall b a factor c of c f ( c ), Compress (shrink) the graph of f() horizontall b a factor of c f, Stretch (elongate) the graph of f() horizontall b a c factor of c f () reflect the graph about ais f ( ) reflect the graph about ais Short cut form: The following diagram is useful to remember the stretch and shrink of a graph b a known factor c, with the following values. We consider the cases f ( c ) for horizontal stretch ( H ST ) or horizontal shrink ( H SH ) and cf () for vertical stretch ( V ST ) or vertical shrink ( V SH ).. c c H ST H SH V SH V ST
Fall 9 Instructor: Dr. Firoz Updated on October 4, 9 Rule on vertical stretch and shrink: The graph of g( ) cf ( ) is found b Shrinking verticall the graph of f () b a factor of c when c Stretching verticall the graph of f () b a factor of c when c Note: Remember that for g( ) cf ( ), when c is negative we have horizontal reflection or reflection along -ais. Rule on horizontal stretch and shrink: The graph of g( ) f ( c ) is found b Stretching horizontall the graph of f () b a factor of / c when c Shrinking horizontall the graph of f () b a factor of / c when c Note: Remember that for g( ) f ( c ), when c is negative we have vertical reflection or reflection along -ais. 4. Composition of functions For two given functions f ( ), g( ), the compositions are defined as f ( ) g( ) f g f ( g( )) and g( ) f ( ) g f g( f ( )) The domain of f gis defined as i) is in g ( ) and ii) g ( ) is in f(). For the functions and their domain 4 f ( ) and g( ) determine the following functions a) f ( ) g( ) b) g( ) f ( ) Solution: a) b) has domain all f ( ) g( ) f ( g( )) f ( ) g( ) f ( ) g( f ( )) g( ) 4 4 has domain all such that 4 which is true for all real values of. Thus the domain is the entire real line.. For the functions f( ) and g ( ) determine the following functions and their domain a) f ( ) g( ) b) g( ) f ( ) Solution: a) f ( ) g( ) f ( g( )) f 4
Fall 9 Instructor: Dr. Firoz Updated on October 4, 9 The domain is which is domain for g ( ) and also for g ( ) should be defined on f() when g( ). Thus the domain is all real values of, b) g( ) f ( ) g( f ( )) g The domain is which is domain for f( ) and also for f() should be defined on g ( ) when f ( ). Thus the domain is all real values of, Section. The limit of a function The graph of f ( ) 4 is not defined at = but it has limiting value 4 at that point. If ou graph and trace at = ou will find a hole. On the other hand the graph of f ( ) 4 has a vertical asmptote at =. This graph does not have a hole at = and limit at = also does not eist. We eamine the situation below:. Test the value of f ( ) 4 close to from left side.9.99.999.9999 f () -.9.99.999.9999 And the value of f ( ) 4 close to from right side.... f () - 4. 4. 4. 4. In both the cases the value approaches (converse to) 4. The first case is called the left hand limit and the second case is the right hand limit. 4 Now let us look at the second eample for f ( ) 4. Test the value of f ( ) close to from left side.9.99.999.9999 f () - -76. -796. -7996. -79996. 5
Fall 9 Instructor: Dr. Firoz Updated on October 4, 9 And the value of f ( ) 4 close to from right side.... f () - 84. 84. 84. 84. In the first case the value approaches to, on the other hand in second case the value approaches to. Thus the limit does not eist. Eercises. Test the limit of. Test the limit of f ( ) f ( ) sin for = cos for = Remember we have the following formulas sin tan cos lim, lim, lim, limcos Section.4 Calculating Limits Find the limits b direct method.. lim( ) lim cos ( ) cos ( / 4) /. / 4 [,] lim cos ( ) cos ( / 4).667 /4 is not a number, limit does not eist. Because Note: f ( ) arccos( ) cos ( ) is defined in the domain or [,] 4. lim 4 4 6. lim 4 e 8. lim e 4. lim.5 5. 7. 9. lim lim csc lim lim lim 9. lim sin sin 6
Fall 9 Instructor: Dr. Firoz Updated on October 4, 9. lim 5. 8. sin lim cos lim sin. lim 4 cos 6. lim 4. 7. e lim e lim cos 9. f( ) e find the following results. cos for 4 for is a piece-wise defined function. You need to a) lim f( ) b) lim f( ) c) lim f( ) for. f ( ) for for the following results. is a piece-wise defined function. You need to find a) lim f( ) b) lim f( ) c) lim f( ). lim 6. lim 4. lim 6 4 Squeeze Theorem Suppose that f ( ) h( ) g( ) for all ( c, d) I, ecept possibl at a I and that lim f ( ) lim g( ) l, for some real number l, then it a a follows that lim h ( ) l a 4. 5. sin lim cos lim 5 b squeeze theorem use squeeze theorem if applicable. 7
Fall 9 Instructor: Dr. Firoz Updated on October 4, 9 Section.5 Continuit If lim f ( ) lim f ( ) L f ( a ), we sa that f() is continuous at a. a a Otherwise the function is discontinuous or not continuous. The limit of the function eists if lim f ( ) lim f ( ) L, a finite value. a a Note; For a polnomial function f (), lim f ( ) lim f ( ) L f ( a ). The polnomial functions are alwas continuous. Eamples. for f ( ) 5 for for a a is a piece-wise defined function. You need to test the continuit. Solution: lim f ( ) 4, lim f ( ) 4, f ( ) 4, the function is continuous at. And lim f ( ), lim f ( ) 6, limit does not eist means the function is discontinuous at =.. Remove discontinuit if an of f( ) Solution: lim f ( ) lim f ( ) /, limit eists. For continuit we must have f () /.. Determine the value(s) of a, and b that makes the given function continuous. Eercise sin for f ( ) a for Solution: For continuit bcos for lim f ( ) b, lim f ( ), f () a lim f ( ) b lim f ( ) f () a a b 4. Determine the value(s) of a, and b that makes the given function continuous. 8
Fall 9 Instructor: Dr. Firoz Updated on October 4, 9 ae for f ( ) sin ( / ) for b for 5. Determine the value(s) of a, and b that makes the given function continuous. a (tan ) for b f ( ) e for ln( ) for Intermediate Value Theorem (IVT) Suppose that f is continuous on [a, b] and W is an number between f( a ) and f() b with f ( a) f ( b ), then there is a number c ( a, b ) for which f () c W Corollar: Suppose that f is continuous on [a, b] and f( a ) and f() b have opposite signs, then there is at least a number c ( a, b ) for which f( c ), the number c is called the zero of the function f(). Eamples Use intermediate value theorem to verif that f() has a zero in the interval. Then use bisection method to find an interval of length / that contains the zero.. f ( ) 7, [,] Solution: f() 4 7, f () 9 7, there is a zero in (, ) The bisection method: ( )/.5: f (.5).75, there is a zero in [.5, ] (.5 )/.75: f (.75).565, there is a zero in [.5,.75] (.5.75)/.65: f (.65).9, there is a zero in [.65,.75] (.65.75)/.6875: f (.6875)., there is a zero in [.65,.6875] (.65.6875)/.6565: f (.6565).55, there is a zero in [.65,.6565] having length.6565-.65 =.5 = / Eercise. f ( ) 4, [,] 9
Fall 9 Instructor: Dr. Firoz Updated on October 4, 9 Section.6 Limits Involving Infinit Verif the following limits:. lim (DNE). lim (DNE). lim 5 7 5 4. lim 4 4 is horizontal asmptote 5. 4 6 (DNE), has a slant asmptote K e 6. lim K K e K, a horizontal asmptote, K is positive constant. 7. lim ln( e ) e a horizontal asmptote 8. lim 9. lim ln( e ) Optional Epsilon-delta definition of limit: For a function f defined in some open interval containing a (but not necessaril at a), we sa that lim f ( ) L, if given an a number, there is another number, such that a guarantees that f ( ) L. Verif the limit lim( 9) Solution: using definition we have for there is a guarantee that 9 We consider 9 to approach to find relation with 9 ( ) since 7 as We can further consider limit. and we can conclude that 7 one ma consider that 7 and hence we prove the Determine relation between and to prove the following limits:. 4. lim( 4). 4 lim 6 lim( ). lim( )
Fall 9 Instructor: Dr. Firoz Updated on October 4, 9 Chapter Derivatives Section. Derivatives and Rates of Change You need to know the following results: Equation of a straight line through (, b) having slope m is b m Equation of a straight line through two given points (, )and (, ) is Rise m( ) where m = slope Run Now if P(, )and Q(, ) are two points on the graph of the function f (), then the line joining the given two points is called a secant line and has Rise the equation m( ) where m. On the other hand if Run the point Q approaches to P then the limiting position of the secant line is called the tangent line to the function at the point P. In this case the slope of the tangent line is given b mtan lim. The slope of the secant line is denoted b msec. As a general case we use the following notations and definitions Consider f () and two points on the function f ( a h) f ( a) P( a, f ( a))and Q( a h, f ( a h )) then mtan lim f ( a) h h f ( a h) f ( a) msec average value h and The equation of a tangent line to f () at = a is f ( a) f ( a)( a ) Eample. Find the slope of a secant line passing through the points (, 5) and (, 8) on 4. f ( a) f ( b) 8 5 Solution: slope msec =, average value a b. Find the slope and equation of the tangent line passing at the point (, 5) to 4. f ( h) f () ( h) 4 5 Solution: slope mtan f () lim lim h h h h The equation of the tangent line 5 ( )
Fall 9 Instructor: Dr. Firoz Updated on October 4, 9 Eercise. Find the slope of a secant line passing through the points (, ) and (, ) on.. Find the slope of a tangent line to. Find the equation of the tangent line to 4. Find the equation of the tangent line to Velocit at at at Suppose f () t gives the position of a particle at time t moving along a straight f ( b) f ( a) line. For t a and t b, the average velocit Vavg () t and the b a f ( a h) f ( a) velocit of the particle at t a is v( a) lim f ( a) which is h h also known as instantaneous velocit or rate of change of displacement at t a. 5. f ( t) 64 6 t, t.5, t, find average velocit 6. Find instantaneous velocit of f ( t) 64 6 t, t 7. Show that f () t t does not have a tangent line at t 8. Compute the slope of the secant line for f ( ) at i), ii).5, 9. Compute the slope of the secant line for f ( ) at i), ii).5,. Compute the slope of the secant line for f ( ) at i), ii).5,. Compute the slope of the secant line for f( ) at i), ii).5,. Compute the slope of the secant line for f( ) at i), ii).5,. Find the equation of the tangent line to f ( ) at i) ii) 4. Find the equation of the tangent line to ii) 5. Find the equation of the tangent line to f ( ) at i) f ( ) at i) ii)
Fall 9 Instructor: Dr. Firoz Updated on October 4, 9 6. Find the equation of the tangent line to f( ) 7. Find the equation of the tangent line to f( ) 8. For the given position functions calculate average velocit a) f ( t) 6t, t, t b) f ( t) t 8 t, t.9, t at i) ii) at i) ii) c) f ( t) 6t 5, t, t 9. For the given position functions calculate instantaneous velocit a) f ( t) 6t, t b) c) f ( t) t 8 t, t.9 f ( t) 6t 5, t Section. The Derivative as a Function Limit definition of derivative: Definition: The derivative of the function f () at a is defined as f ( a) f ( a h) f ( a) lim h h provided the limit eists. If this limit eists we sa that f() is differentiable at a. The general form derivative at ever point of, we write the limit definition of derivative f ( ) f ( h) f ( ) lim h h Other definitions: f ( a) f ( ) f ( a) lim. a a Eamples. Use limit definition to find the derivative of f ( ) Solution: We have the definition f ( ) f ( h) f ( ) lim h h Step. Find and simplif: f ( h) f ( ) ( h) h h Step. Evaluate f ( h) f ( ) h h f ( ) lim lim lim( h) h h h h h. Use limit definition to find the derivative of f ( ) f ( h) f ( ) Solution: We have the definition f ( ) lim h h Step. Find and simplif: ( h )( h ) h f ( h) f ( ) h h h f ( h) f ( ) h Step. Evaluate f ( ) lim lim h h h h( h )
Fall 9 Instructor: Dr. Firoz Updated on October 4, 9. Use limit definition to find the derivative of f( ) Solution: We have the definition f ( ) f ( h) f ( ) lim h h Step. Find and simplif: f ( h) f ( ) ( h) b common denominator ( ) (( h) ) h (( h) )( ) ( h )( ) h Step. Evaluate f ( ) f ( h) f ( ) ( h )( ) lim lim h h h h ( ) Eercise: Use limit definition of derivative. f ( ) 9, find f ( ), f (), f (). f ( ),, find f ( ), f (), f (). f ( ),, find f ( ), f (), f () 4. The graph of f () is given below, plot the graph of f () f () - -.5. 4 Note that f () is zero at,.5, and does not eist at. (because of the corner). Further f () is negative on (, ), (.5,), and (., ), positive everwhere else. Theorem: If f () is differentiable at a, then f() is continuous at a. The converse is false in general. The point(s) of nondifferentiabilit i) A hole or a jump ii) Vertical tangenc iii) A cusp or a corner 5. Find the point of nondifferentiabilit from eample 4 above. 4
Fall 9 Instructor: Dr. Firoz Updated on October 4, 9 6. sin,, show that the function is continuous and differentiable for all Section. The Basic Derivative Formulas You ma use the limit definition of derivative to verif the following results. For the d df f ( h) f ( ) function f () we use f ( ) lim d d h h Formulas dc (). d, the derivative of = c, a constant is zero d( ). d, the derivative of = is. n d ( ) n. n, for all real numbers. d d d 4. (sin ) cos 5. (cos ) d d sin If s( t), v( t ) are displacement and velocit functions respectivel then the acceleration dv d s function at (). dt dt Eamples: Evaluate the derivatives d ( ) 9. d d d. ( ) d d d( ) d / /. ( ) / d d d 4. ( ) 4 4 d 5. Find the points on the curve 4 8 5where the tangent line is horizontal. Solution: The tangent line is horizontal where the derivative is zero. We have the following: 4 4, 6 6 The tangent line is horizontal at the points (, 5) and (6, 47). Use calculator to check. 5
Fall 9 Instructor: Dr. Firoz Updated on October 4, 9 6. Application of rates of change The position of a particle is given b the equation s( t) t 9t 5t, where t is measured in seconds and s in meters a) Find the velocit at time t. b) What is the velocit after seconds? c) When is the particle at rest? d) When is the particle moving forward? e) Draw a diagram to represent the motion of the particle. f) Find the total distance traveled b the particle during the first five seconds. g) Find the acceleration at time t and after 4 seconds. Solution: a) The velocit at time t is v( t) s ( t) t 8t 5. b) Velocit after seconds v() s () () 8() 5 5 c) The particle is at rest when the velocit is zero: v( t) s ( t) t 8t 5 t, 5 d) We need to check the sign of the velocit to see when the particle is moving forward: Let us use a real line test. (Remember that the velocit is zero at and 5) Sign of 5 6 v( t) t 8t 5 + + at the test points The particle is moving forward when t [,) (5, ) e) t = 5 s = -5 t = t = s = v = v > s = 7 f) Total distance traveled in first 5 seconds = s() s() s(5) s() 7 9 meters g) Acceleration a( t) v ( t) 6t 8. Acceleration after 4 seconds a (4) 4 8 6 meter / sq. second Eercise: Find the derivative of the following functions with respect to (-9).. 5 4 4. 5. 7. 8. ( )( ) 9.. s( t) 4t 6t, find s(), v( t), v(), a( t), a (). 64 6 6. 5 4e 5 6
Fall 9 Instructor: Dr. Firoz Updated on October 4, 9. Find the equation of the tangent line to the curve 5 f ( ) 4 at = 5. Find the equation of the normal line to the curve f ( ) 4 at =. Suppose that. ut () measures the displacement (in inches) of a weight suspended from a spring t seconds after it is released and that u( t) 4cos t, find the velocit at an time t and determine the maimum velocit using calculator. 4. Draw a simple circuit with capacitance farad, the inductance Henr and impressed voltage t volts at time t. The model for a total charge Qt () in the circuit at time t is Q( t) sin t t 4 coulombs. The current is defined to be the rate of change of the charge with respect to time t and so is given b dq It () amperes. Compute the current at t = and t =. dt Section.4 The Product and Quotient Rules For two differentiable functions f : f ( ) and g : g( ) the product rule is d df dg ( fg) g f or in another form ( fg) f g fg and the quotient rule is d d d df dg g f d f d d f gf fg or in another form, g d g g g g Short cut form: Product ( fg) fg f g f f Derivative of Trigonometric Functions: d d. (sin ) cos. (cos ) d d sin d d 4. (sec ) sec tan 5. (csc ) d d csc cot Eamples g g Quotient f f g fg g g d. (tan ) d d 6. (cot ) d sec f f csc g g. ( )( ) 6 ( )(/( ) 6 ) ( ) 7
Fall 9 Instructor: Dr. Firoz Updated on October 4, 9. 6 ( ) ( )(/( ) 6 ) ( ) Eercise: Find the derivative of the following functions w. r. to (use product or quotient rules and also verif b direct method if possible).. 4. 5. ( )( ) Section.5 The Chain Rule For derivative of functions like epand the function to ( ) is not so difficult to perform as we can ( ) and then take the derivative. But functions like ( ) is ver time consuming to epand and then take the derivative term b term. So we need a convenient method to perform derivative of such functions. d We introduce the chain rule here ( f ( g ( ))) f ( g ( )) g ( ), if g is differentiable d at and f is differentiable at g(). Eamples. ( ) and ( ). But using chain rule ( d ) ( d ) ( ). ( 99 d 99 ), then ( ) ( ) ( ) d. ( ), then 99 d 99 ( ) ( ) ( ) ( d ) 8
Fall 9 Instructor: Dr. Firoz Updated on October 4, 9 Eercise Find derivative of the given functions w. r. to given variables. (4sin cos ). sin cos. 4. 7. 4sec 4sec 5. sec(cot ) 4 csc. sect cos t. Section.6 Implicit Differentiation 6. sin 4sec 8. sin tsect 9. sin cos sec. tan The functions so far we have used can be described b epressing one variable eplicitl in terms of another variable as f (). Some functions however are described implicitl b relations such as 6. Look at the graphs plotted below: Function: 6 Relation: Circle of radius 6: 6 Function: 6 We have some functions which are etremel difficult to epress as eplicit, such as the Folium of Descartes 6, where is a function of. Fortunatel we do not need to write eplicitl as a function of to find derivative. In this section we will discuss such derivatives known as the implicit derivative. The basic concept of chain rule will be applied considering as a function of. Suppose is a d d function of, then d d More eamples. Given, determine ( ), (), () Solution: ().44,,.44 (plug = and draw the graph of and find all zeros) Taking derivative on both sides w. r. to we have the following: and after simplification it is eas to find ( ) You ma now plug = and one value (from values of ) of to evaluate the derivative. 9
Fall 9 Instructor: Dr. Firoz Updated on October 4, 9. Find the equation of the tangent line to curve of at (, ) Solution: The equation of the tangent line to the graph at the given point (, ) is ()( ). We need to find (). Using implicit derivative we find that ( ) () 4 The equation of the tangent line is ()( ) 4( ) 4 9. Find the equation of the tangent line to curve of 4 4 at (, -) Solution: Tr ourself 4. Given e 6, find () when Solution: Using implicit derivative we have found the following ( ) e ( ) ( ( ) ) e Now plug = and = to get the following: ( ) e 4 ( ) ( ) ( ( ) ) 4 ( ) e Higher order derivatives (75) (5) ( n 5. Determine f ( ), f ( ) also f ) ( )(nth order derivative) a) f ( ) sin b) f ( ) cos Solution: a) f ( ) sin f ( ) cos sin Now f ( ) cos sin and also f ( ) cos sin ( n n We can conclude that f ) ( ) sin
Fall 9 Instructor: Dr. Firoz Updated on October 4, 9 And And (75) 75 f ( ) sin sin cos, use Precalculus concept. f ( ) sin sin 75 sin, use Precalculus concept. (5) 5 b) Tr ourself. Section.7 Related Rates Basic forms one needs to know: Circle: Consider the radius of the circle is r. Total arc or circumference = r, Area enclosed b the circle = r Bo: Consider l, w, h as length Circle Bo with a diagonal width, and height. The surface area = ( lw lh wh ) square units Length of a diagonal d l w h units Volume = lwh cubic units Clinder: Radius of the circular base is r and height is h. The volume = rh and total surface area (with two circular leads) = rh r Cone: Radius of the circular base is r and height is h. The volume = rh and total surface area (with one circular lead) = rh r Clinder Cone Sphere: Radius is r. The volume = 4 4 r r and total surface area = Trough: container for animal food or water: a long low narrow open container that holds feed or water for animals
Fall 9 Instructor: Dr. Firoz Updated on October 4, 9 Similar Triangle: Two triangles are said to be similar if the corresponding sides are proportional. We have the following results: A Eamples: AB AC BC DE DF EF D E F B C. An oil tanker has an accident and oil pours at the rate of 5 gallons per minute. Suppose that the oil spreads onto the water in a circle at a thickness of. Given that ft 7.5 gallons. Determine the rate at which the radius of the spill is increasing when the radius reaches 5 ft. Solution: The shape is like a clinder. We consider the volume equation dv dr V r h rh dt dt, where h inches ft dv 5 Also given that 5 gallons / m ft ft / m dt 7.5 dv dr dr dr Thus rh (5) dt dt dt dt 5 The rate at which diameter increases is.4 ft/m 5. A trough is formed b nailing together, edge to edge, two boards ft. in length, so that the right section is a right triangle. If 5 gallon of water is poured into the trough and if the trough is held level so that a right section of water is an isosceles right triangle, Find the cross sectional area of the water? (cu.in= gal.) Solution: The volume of water in the trough is 5 gallons or 5 = 465 cubic inches. The length of the trough is = inches and the volume of the water is the area of the triangular cross-section times the length so the area of the triangular crosssection. Thus the area of the triangular cross section is bd = 465/ = 6.5 square inches.
Fall 9 Instructor: Dr. Firoz Updated on October 4, 9. A car is traveling at 5 mph due south at a point / mile north of an intersection. A police car is traveling at 4 mph due west at a point /4 mile east of the same intersection. At that instant, the radar in the police car measurers the rate at which the distance between the cars is changing. What does the radar gun register? 4. A -foot ladder leans against the side of a building. If the top of the ladder begins to slide down the wall at the rate of ft/s, how fast is the bottom of the ladder sliding awa from the wall when the top of the ladder is 8 ft off the ground? 5. If a snowball melts so that its surface area decreases at a rate of sq. cm/min, find the rate at which the diameter decreases when the diameter is cm. Solution: The surface area of the snowball with radius r is S 4 r. Given that ds ds dr, and diameter r = cm. B differentiation we get 8 r. Using the dt dt dt dr given values we get rate of change of radius is. The rate of change of dt 4 d(r) diameter is. So the rate of decrease is cm / min. dt 6. Water is leaking out of an inverted canonical tank at a rate of, cubic meter per minute, at the same time water is being pumped into the tank at a constant rate. The tank has height 6 m and the diameter at the top is 4 m. If the water level is rising at a rate of cm/min when the height of the water is m, find the rate at which water is being pumped into the tank. dv Solution: Let us consider that the rate of water pump is c. Then c, where dt r v r h. Look at eample at page # 58 to understand that r h. h 6 dv dh dh Then we get v h h. Now for h m cm, cm / min 7 dt 9 dt dt we get c () c 89. 5 9 7. Gravel is being dumped from a conveor belt at a rate of cubic feet per minute, and its coarseness is such that it forms a pile in the shape of a cone whose base diameter and height are alwas equal. How fast is the height of the pile increasing when the pile is ft high? dv Solution: We are given, where v dt dh Then we get. 8. dt h r h h h h dv dt h 4 dh dt.
Fall 9 Instructor: Dr. Firoz Updated on October 4, 9 8. Two cars start moving from the same point. One travels south at 6 mi/h and the other travels west at 5 mi/h. At what rate is the distance between the cars increasing two hours later? d d Solution: We are given that z, 5, dt dt 6, 5 and 6. dz Find. dt Using Pthagorean Theorem we get z 5 z dz d d And b differentiation we get z. Using the given values we get dt dt dt dz dt 65 mi / h. 9. A particle moves along the curve. As it reaches the point (, ), the - coordinate is increasing at a rate of 4 cm/s. How fast is the -coordinate of the point changing at that instance? d d Solution: Given, 4, and. Find. dt dt d d We write. B differentiation we get. Using the given values dt dt d we get cm / sec. dt. Water is being poured into a inverted right circular conical tank as shown in the diagram below: (a) Epress the volume of the water in the tank at an given time as a function of the depth of the water, d, and the radius of the tank at the surface of the water, r. 4
Fall 9 Instructor: Dr. Firoz Updated on October 4, 9 (b) If the tank has a total depth of 5 m and a radius of 5 m, epress the volume of water in the tank as a function of d onl. (Hint: draw in lines for the total depth and radius, and use similar triangles to find a formula relating h and d). (c) If water flows into the tank at a rate of 5 m per minute, what is the depth of the water after (i) 5 min, (ii) min, (iii) hour Solution (a) Since the shape of the water is that of a right circular cone, its volume is: (b) Consider DABE and DACD, since the are similar Substituting this epression for r into the equation in part (a) we get: (c) Rearranging the formula in part (b) to solve for d we get: (i) Volume after 5 min = 5 min 5 m /min = 75 m Using the formula above for d, we get: 5
Fall 9 Instructor: Dr. Firoz Updated on October 4, 9 (ii) Volume after min = min 5 m /min = 5 m Using the formula above for d, we get: (iii) Volume after h = 6 min 5 m /min = m Using the formula above for d, we get: Formulas: Note: d d ( e ) e ( a ) a ln a d d d (ln p),, p is a constant d d (ln ), d Eamples: Find derivative of the following functions. f ( ) e e. / ( ) f e e. ( ) f e ln e 4. f ( ) ln ln e 5. f ( ) ln( e ) 6. ( ) f e ln 5 4 7. f ( ) 8. ( ) ( ) f 9. f( ) 9e. 4 f ( ) ln 4e. f ( ) ln e 4 e. f( ) e e e e d. Given sinh( ) and cosh( ) show that sinh( ) cosh( ) d d cosh( ) sinh( ) d e and Application: 4. If the value of a -dollar investment doubles ever ear, its value after t ears is given b ( t ). Find the rate of investment value, and the relative rate of change. 6
Fall 9 Instructor: Dr. Firoz Updated on October 4, 9 d t Solution: The rate of investment is ( )ln dt () t ln.69 t () and relative rate of change is 5. The velocit of a hanging weight is u() t where vertical displacement is given b ( ) t t u t Ae cos( t) Be sin( t ), AB,,, are constants. Find u() t t Solution: u ( t) A( cos( t) sin( t)) e B( sin( t) cos( t)) e t Section.8 Linear Approimations and Differentials We have the linear approimation f ( ) f ( ) f ( )( ) f ( a h) f ( a) We have used the rule of derivative f '( a) lim at a point (a, f (a)). We h h f ( ) f ( a) now use another modified rule f '( ) lim as a general rule of derivative at a h a a point (, f ()). From this general form we omit the limit and use the result as an f ( ) f ( a) approimation like f '( ). After simplification we find a f ( ) f ( a) f '( a)( a), which is called the linear approimation (also called tangent line approimation) of the function f about = a, ( is close to a). Eample. Prove that for close to, and illustrate this approimation b drawing the graphs of and on the same screen. We consider f ( ) f '( ) '(), as f a Now using linear approimation formula we find f ( ) f ( a) f '( a)( a) f ( ) f () f '()( ) 7
Fall 9 Instructor: Dr. Firoz Updated on October 4, 9 Rules for Differentials. The notation d( f ( )) is called the differential of the function f. We have the following rules. d( af ( ) bg( )) adf ( ) bdg( ) or simpl we write. d( af. d( fg). d( f / g) bg) gdf adf gdf g fdg fdg bdg Eample. Find the differentials (p, q, and r are constants): a. d ( ) b. Solution. a. d ( p q) d( ) 6d d r r b. d( p q) r( p q) ( p) d Eample. Prove that find approimation of Eamples r ( ) m m for close to and use this approimation to / ( 6.95).. Find the linear approimation of f ( ) at and thus approimate the values of 8. and 5. 5. Find a zero of f ( ) using Newton s Method with an initial guess Formulas: d. (sin ) d d. (tan ) d. 4. d d d d d d (cot ) (tan ) d d (cos ) (sin ) 8
Fall 9 Instructor: Dr. Firoz Updated on October 4, 9 5. d d (sec ) 6. d d (csc ) (sec ) d d Find the derivative of the following functions: 6. 7. 8. 9... f ( ) sin ( ) sin (/ ) cos ( ) f ( ) cos (ln ) csc (/ ) tan ( ) f ( ) sin (cos ) tan (/ ) f e ( ) sin ( ) f ( ) sec ( ) sin (cos ) cos (cos ) f ( ) tan, simplif the answer. Chapter Inverse Functions Section. Eponential Functions Section. inverse Functions and Logarithms Theorem: If f is differentiable at all and has an inverse function Then g ( ) f ( ), ( ( )) f ( g( )) f g Eample Given that f ( ), find does not eist. Solution: Set f ( ) and solve to get, now f (), where f ( ) f ( ) ( ) g( ) f ( ) f () g () and show that g () But for g () : set f ( ) and solve to get, then f () DNE. f () () iv) 5 Find derivative of f ( ) and compute g (7), where g( ) f ( ) v) Find derivative of vi) g( ) 9 Find derivative of 4 9
Fall 9 Instructor: Dr. Firoz Updated on October 4, 9 Section. Derivatives of Eponential and Logarithmic Functions Section.5 Inverse Trigonometric Functions Increasing and Decreasing Functions Under the following conditions the function will be called increasing ( ) or decreasing ( ) on some open interval. i) If f ( ) for all I ( a, b ), then f( ) on I. ii) If f ( ) for all I ( a, b ), then f( ) on I. Eamples: 4. Graph the function f ( ) 4 5 9showing all maimum, minimum values, open interval(s) where the function is increasing or decreasing. 4. Graph the function f ( ) 4.6. showing all maimum, minimum values, open interval(s) where the function is increasing or decreasing 5/ /. Graph the function f ( ) showing all maimum, minimum values, open interval(s) where the function is increasing or decreasing 4. Graph the function f ( ) 5 7 showing all maimum, minimum values, open interval(s) where the function is increasing or decreasing Section.5 Concavit and Second Derivative Test Under the following conditions the function will be called concave up ( down ( )on some open interval. )or concave i) If f ( ) for all I ( a, b ), then f ( ) then f is concave up ( ) on the open interval I. ii) If f ( ) for all I ( a, b ), then f ( ) then f is concave up ( ) on the open interval I. Eamples: 4. Graph the function f ( ) 6, showing all maimum, minimum values, open interval(s) where the function is increasing or decreasing and open interval(s) where the function is concave up or concave down. 4
Fall 9 Instructor: Dr. Firoz Updated on October 4, 9. Graph the function f ( ) 4, showing all maimum, minimum values, open interval(s) where the function is increasing or decreasing and open interval(s) where the function is concave up or concave down.. Graph the function f ( ), showing all maimum, minimum values, open interval(s) where the function is increasing or decreasing and open interval(s) where the function is concave up or concave down. 4/ / 4. Graph the function f ( ) 4, showing all maimum, minimum values, open interval(s) where the function is increasing or decreasing and open interval(s) where the function is concave up or concave down. 5. Graph the function f( ), showing all maimum, minimum 4 values, open interval(s) where the function is increasing or decreasing and open interval(s) where the function is concave up or concave down. 6. Sketch the graph of the function f () with following properties: f (), f ( ) for all and, f ( ) for, f ( ) for all,, and, f ( ) for all Section.7 Indeterminate Forms and L Hôpital s Rule We have noticed the following indeterminate forms,,,,,, f( ) L Hopital rule is applicable for lim a g if this limit has the form ( ) f ( ) f ( ) is as follows lim lim a g( ) a g ( ) indeterminate form appears.. The. The L Hopital rule. One ma appl the rule repeatedl as long as the Eamples with the above indeterminate forms:. lim lim. lim lim, does not eist (DNE). lim lim 4. e e lim lim, DNE as the result is not a finite number 4
Fall 9 Instructor: Dr. Firoz Updated on October 4, 9 5. lim lim lim e e e 6. lim lim e e 7. ln lim csc 8. lim ln( ) 9. lim ln. lim e. lim (sin ). lim( / ). lim 4 cos sin 4. lim sin sin 5. lim e 6. lim 7. lim sin sin(sin ) 8. lim sin 9. lim ln sin. lim DNE Section.9 Rates of Change in Economics and Sciences Eamples:. Cost function C( ). 4, where is the number of units produced. The marginal cost function is defined as MC( ) C ( ) and average 4
Fall 9 Instructor: Dr. Firoz Updated on October 4, 9 cost is defined as average cost. AC( ) C ( ). Find the etreme marginal cost and etreme. For the demand function f( p ), where p is the price, the elasticit is defined as p f ( p) E. For the demand function f ( p) 4( p ), where p is the price f( p) per unit. Find elasticit when p =, and find p for which E < -. Chapter 4 Applications of Differentiation Section 4. Maimum and Minimum Values For a function f () if f ( c ) or f () c is undefined for all c in the domain of f (), then c is called the critical number. If a function f () has either maimum or minimum value(s) at = c, then c must be a critical number. The converse is not true in general. Theorem: A continuous function f () defined on a closed, bounded interval [ ab, ] attains both an absolute maimum and absolute minimum. i) f() c is the absolute maimum of f in S if f ( c) f ( ), S ii) f() c is the absolute minimum of f in S if f ( c) f ( ), S Eamples:. Find all critical numbers of f( ). Find the absolute etrema of f ( ) 5 on [-, 4]. Find the absolute etrema of 4. Find the local etrema of 5/ 4 / 4 f ( ) 4 8 on [, 4] 4 f ( ) 5. Find the local etrema of f( ) Section 4. The Mean Value Theorem First of all we discuss the special case of Mean Value Theorem. The Rolle s Theorem: Suppose that f is continuous on the interval [ ab,, ] differentiable on ( ab, ) and f ( a) f ( b) then there is a number c ( a, b) such that f ( c ) 4 4
Fall 9 Instructor: Dr. Firoz Updated on October 4, 9 The Mean Value Theorem: Suppose that f is continuous on the interval [ ab,, ] differentiable on ( ab, ) then there is a number c ( a, b) such that f ( b) f ( a) f () c b a Constant Theorem: Suppose that f ( ) for all on an open interval I, then f must be a constant, that is f () c, a constant Corollar: Suppose that f ( ) g ( ) for all in some open interval I, then for some constant c, f ( ) g( ) c, I Section 4. Derivatives and the Shapes of graphs Section 4.4 Curve Sketching Section 4.5 Optimization Problems Eamples:. Find a positive number such that the sum of the number and its reciprocal is as small as possible.. A farmer with 75 feet fencing wants to enclose a rectangular area and divide it into four pens with fencing parallel to one side of the rectangle. What is the total largest area for the four pens?. A bo with an open top is to be constructed from a square cardboard feet wide b cutting out a square from each of the four corners and bending up the sides. Find the largest volume that such a bo can have. 4. If square cm of material is available to make a bo with square base and an open top, find the maimum volume of the bo. 5. Find the point(s) (, ) on the curve 4 4that is (are) farthest from (, ). Solution: Let (, ) be an point on the ellipse. Then distance of this point from (, ) is d ( ). Let us write D ( ) for simplicit. Now maimize D, use 4 4. D ( ) 4 4. Using derivative ou will find critical 4 number, then, and D d 6/, so d 4 /., which is maimum since D " 6. 44
Fall 9 Instructor: Dr. Firoz Updated on October 4, 9 6. A right circular clinder is inscribed in a sphere of radius r. To find the largest possible surface area of such a clinder. Solution: Let us consider the sphere with radius r. If is the radius of the base of the clinder and is the vertical distance of center from diameter of the base of the clinder then r The surface area of the clinder S () 4 r = Now find derivative of S, which is S ' (4 r 8 4 r ). r The critical numbers are 5 5 Observe that at r 5 5. 8565r, S is maimum. Then =.557r and ma S=.67 r. Section 4.7 Anti-derivatives A function F is called an antiderivative of f on an interval I if F ( ) f ( ) for all in I. An engineer who can measure the variable rate at which water is leaking from a tank wants to know the amount leaked over a certain time period. A biologist who knows the rate at which a bacteria population is increasing might want to deduce what the size of the population will be at some future time. A phsicist who knows the velocit of a particle might wish to know its position at a given time. In all these cases, the problem is to find a function F() whose derivative is a known function f(). When such a function eists, we sa that F() is the antiderivative of f(). If F() is the antiderivative of f() on an interval I, then the most general antiderivative of f() is the function F() + c, where c is an arbitrar constant. In this section we use the following formulas and smbols: The smbol is used for finding antiderivative. The antiderivative of the function f() with respect to is smbolized as f ( ) d F( ) Formulas: n n. d. e d e. n 5. cos d sin 6. sin d cos 7. a a d ln a sec d tan 4. d ln 45
Fall 9 Instructor: Dr. Firoz Updated on October 4, 9 8. d sin 9. d tan. kd k, k isa constant Eamples: Find the most general antiderivative of the following functions. 5 6. ( / e 5) d / 6 ln e 5 c. (e 7sec ) d e 7 tan c. 4. sin cos d tan sec d sec c 5. ( ( / ) ) d sin c Eercises 4 / 5/ ( ) / 9 / d d c.. 4 d. / 6 ( 6 ) d 4. 4 6 6 d ( 5( / ) ) d Find the function f: 5. Given 6. Given 7. Given f e f ( ) ( ), () f ( ) 8, f () 6 f ( ), f () f ( ) 8. Given f ( ) 4, f () 5, f () 9. Given f ( ) cos, f (), f ( / ). Given f ( ) e sin, f (), f (), f () Chapter 5 Integrals Section 5. Areas and Distances The area A of the region S that lies under the graph of the continuous function f is the limit of the sum of the areas of approimating rectangles: A lim R lim[ f ( ) f ( ) f ( )] lim f ( ) n n i n n n i n 46
Fall 9 Instructor: Dr. Firoz Updated on October 4, 9 Theorem: For an constant c and d, Formulas: n n n ( ca db ) c a d b i i i i i i i n( n ) n( n )(n ) n( n ) i, i, i 6 n n n i i i Eamples. Let A be the area of the region that lies under the graph of f ( ) e between = to =. a) Use right endpoints; find an epression for A as a limit. Do not evaluate the limit b) Use left endpoints; find an epression for A as a limit. Do not evaluate the limit c) Use middle points, find an epression for A as a limit. Do not evaluate the limit n in/ Solution: a) / n, / n, 4 / n,, i i / n, A lim e n n n. Find an epression for the area under the graph of f as a limit. Do not evaluate the ln limit. Given f( ),. Use all three rules as in eample. i. Compute the following sums i) 8 i (i ) ii) 8 i (i ) iii) (i ) i 6 Approimate the area under the curve rectangles and right endpoints. on the interval [, ] using Solution: A f ( ) [ f (.) f (.) f ()](.). i i Section 5. The Definite Integrals We have the following results to use in this section: b a n * ( ) lim ( i ) n i f d f Eamples and Eercises. Evaluate the Riemann sum for f ( ) 6 taking the sample points to be right endpoints and a, b, n 6 47
Fall 9 Instructor: Dr. Firoz Updated on October 4, 9 Solution:.5, 6. Evaluate ( 6 ) d 6 6 R f ( ).975 i i Solution: n i ( 6 ) d lim f 6.75 n i n n. Use the propert: If m f () M for a b, then b m( b a) f ( ) d M ( b a ) to estimate Solution: a ( e ) d b ( ) ( ) ( ) ( ) ( ) m b a f d M b a e e d a Where the absolute ma M f () and the absolute min function is decreasing on [, ]. 4. Prove that 5. Epress 6 b a d 5 d b a Section 5. Evaluating Definite Integrals m f () e as the as limit of Riemann sum. Do not evaluate the limit. Section 5.4 The Fundamental Theorem of Calculus We have the following results to use in this section: The fundamental theorem of calculus, Part : If f is continuous function on [a, b], then the function g defined b g( ) f ( t) dt, a b is continuous on [a, b] and differentiable on (a, b) and g ( ) f ( ) a Corollar: For h ( ) g( ) f ( t) dt, a b then g ( ) f ( h( )) h ( ) a 48