Chapter 3 Applications of Differentiation 95 CHAPTER 3 APPLICATIONS OF DIFFERENTIATION List of Topics 3.1 dx dy as a slope/gradient of the tangent line to the curve y = f(x). 3.2 The graph of polynomial Functions • x-intercept and y-intercept • critical/stationary/turning points • the interval where f(x) is increasing and decreasing and the extremum point(s) • the interval where f(x) is concave up or concave down and the inflection points • based on the above answer sketch the graph of polynomial Functions 3.3 The graph of rational functions (linear/linear) • x-intercept and y-intercept • vertical and horizontal asymptotes • the interval where f(x) is increasing and decreasing and the extremum point(s) • the interval where f(x) is concave up or concave down and the inflection point(s) • based on the above answer sketch the graph of rational Functions 3.4 Maximum and Minimum Problems 3.5 Related Rates (Rate of Change) 3.6 Mean-value Theorem of Differentiation
MAT183/ MAT421 : Calculus I 96 3.1 dx dy as a slope/gradient of the tangent line to the curve y = f(x) Example 1/ OCT 2004/ MAT183/ Q2a (5 marks) Find the slope of the tangent to the curve (6 )( 3) 2 = + − x y x lncos x e at x = 0. Solution : = 6 + ( ) = 2− 3 ′ = 6 + 1 ∙ − ′ = 2 ∙ 2 − 0 ′ = 6 − ′ = 2 2 ′ = 6 − = (6 + )( 2− 3) = = ′ + ′ = (6 − )( 2− 3) + 2 2(6 + ( )) At = 0, = | =0 = (6 − 0)( 0 − 3) + 2 0(0 + ( 0)) = (6 − 0)(1 − 3) + 2(0 + (1)) = (6)(−2) + 2(0 + 0) = −12 #
Chapter 3 Applications of Differentiation 97 Example 2/ APR 2006/ MAT183/ Q2a (6 marks) Find the slope of the tangent to the curve ( 1) 8 3 + − = ln x sin x y at x = 0. Solution : = 3 ( + 1) − 8 = = ′ − ′ 2 = 3 3 (( + 1) − 8) − 3 + 1 (( + 1) − 8) 2 At = 0, = | =0 = 3 0 ((1) − 8) − 0 1 ((1) − 8) 2 = 3(1)(0 − 8) − 0 (0 − 8) 2 = 3(−8) (−8) 2 = −24 64 = − 3 8 # = 3 = ( + 1) − 8 ′ = 3 3 ′ = 1 + 1
MAT183/ MAT421 : Calculus I 98 Example 3/ OCT 2006/ MAT183/ Q1c (9 marks) Given that f(x) = 3x 2 − x i) Use the definition of derivative to find f ’(x). ii) Find the equation of the tangent line to the graph of f(x) at x = 1 Solution : ) () = 3 2 − ( + ℎ) = 3( + ℎ) 2 − ( + ℎ) = 3( 2 + 2ℎ + ℎ 2) − − ℎ = 3 2 + 6ℎ + 3ℎ 2 − − ℎ Using the first principle (or the definition of derivative) ′() = ℎ→0 ( + ℎ) − () ℎ = ℎ→0 (3 2 + 6ℎ + 3ℎ 2 − − ℎ) − (3 2 − ) ℎ = ℎ→0 3 2 + 6ℎ + 3ℎ 2 − − ℎ − 3 2 + ℎ = ℎ→0 6ℎ + 3ℎ 2 − ℎ ℎ = ℎ→0 ℎ(6 + 3ℎ − 1) ℎ = ℎ→0 (6 + 3ℎ − 1) = 6 + 0 − 1 = 6 − 1 # ii) to find coordinate y (when x = 1) → substitute x = 1 into f(x) = 3x 2 − x y = 3(1)2 − 1 = 2 to find the slope (when x = 1) → substitute x = 1 into f '(x) = 6x −1 slope, m = 6(1) −1 = 5 equation of tangent line : y = mx + C y = 5x + C 2 = 5(1) + C C = −3 Therefore the equation of tangent line at (1, 2) is given by : y = 5x − 3 # Q&A • What happen when we substitute the value of x into @ f ’(x)? Answer : we will get gradient/slope • What happen when we substitute the value of x into y @ f(x)? Answer : we will get coordinate y
Chapter 3 Applications of Differentiation 99 Example 4/ APR 2008/ MAT183/ Q1c (8 marks) Given that ( ) ( ) x e x f x 5 2 2 2 2 − = . i) Find f ’(x). ii) Find the equation of the tangent line at x = 0 Solution : ) () = (2 − 2) 2 2 5 ′() = ′() = ′ − ′ 2 ′() = −8 5(2 − 2) − 10 5(2 − 2) 2 (2 5) 2 = −2 5(2 − 2)(4 + 5(2 − 2)) 2 5 ∙ 2 5 = −(2 − 2)(4 + 5(2 − 2)) 2 5 = −(2 − 2)(4 + 10 − 5 2) 2 5 = ( 2 − 2)(4 + 10 − 5 2) 2 5 ii) to find coordinate y (when x = 0) → substitute x = 0 into () = (2− 2) 2 2 5 y = (2−0) 2 2 0 = (2) 2 2(1) = 2 to find the slope (when x = 0) → substitute x = 0 into ′() = ( 2−2)(4+10−5 2) 2 5 slope, m = (0−2)(0+10−0) 2 0 = (−2)(10) 2(1) = −10 equation of tangent line : = + = −10 + 2 = −10(0) + = 2 Therefore the equation of tangent line at (0, 2) is given by : = −10 + 2 # = (2 − 2) 2 = 2 5 ′ = 2(2 − 2) ∙ −2 ′ = 2 5 ∙ 5 ′ = −4(2 − 2) ′ = 10 5
MAT183/ MAT421 : Calculus I 100 Example 5/ OCT 2008/ MAT183/ Q2c (6 marks) Find the gradient of the tangent to the curve ( 2) 2 8 0 2 y x + + tan x − = at the point (0, 2). Solution : Given ( 2) 2 8 0 2 y x + + tan x − = Differentiate implicitly with respect to x : vu’ + uv’ + sec22x•2 − 0 = 0 2y(x +2) dx dy + y 2 + 2 sec22x = 0 2y(x +2) dx dy = −y 2 − 2 sec22x dx dy = 2 ( 2) 2sec 2 2 2 + − − y x y x Gradient at point (0, 2), dx (0,2) dy = ( ) ( ) 2(2)(0 2) 2 2sec 2 0 2 2 + − − dx (0,2) dy = 2(2)(2) 4 2sec 0 2 − − , where dx (0,2) dy = ( ) 8 − 4 − 2 1 dx (0,2) dy = 8 − 6 dx (0,2) dy = 4 3 − Therefore, the gradient at point (0, 2) is 4 3 − u = y 2 v = x + 2 u’ = 2y dx dy v’ = 1 sec20 = cos 0 1 2 = ( ) 2 cos0 1 = ( ) 2 1 1 = 1
Chapter 3 Applications of Differentiation 101 Example 6/ OCT 2010/ MAT183/ Q2c (6 marks) Find the equation of the tangent to the curve ( ) + + = x 1 e x f x ln x at the point x = 0. Solution : () = ( + + 1 ) () = ( + ) − ( + 1) ′() = 1 + ( + 1) − 1 + 1 (1) ′() = + 1 + − 1 + 1 at the point = 0 (into () ) → = ( 0 + 0 0 + 1 ) = ( 1 + 0 0 + 1 ) = (1) = 0 Slope at = 0 (into ′() ) = + 1 + − 1 + 1 = 0 + 1 0 + 0 − 1 0 + 1 = 1 + 1 1 − 1 1 = 2 − 1 = 1 Equation of tangent line at the point (0, 0) = + = (1) + = + 0 = (0) + = 0 Equation of tangent line at the point (0, 0) =
MAT183/ MAT421 : Calculus I 102 3.2 The graph of polynomial Functions (using First Derivative Test) i) x-intercept and y-intercept • to find x-intercept : when y = 0, find the value of x. Then write into the form of coordinate system. • to find y-intercept : when x = 0, find the value of y @ f(x). Then write into the form of coordinate system. ii) critical/ stationary/ turning points • find f ’(x) • when f ’(x) = 0, find the value(s) of x • to find the value(s) of y, substitute the value(s) of x (obtained above) into y @ f(x). Then write into the form of coordinate iii) the interval where f(x) is increasing and decreasing and the extremum points (using first derivative test) • use table of sign for f ’(x) • if f ’(x) 0 (positive value) : increasing • if f ’(x) 0 (negative value) : decreasing • if f ’(x) changes from positive to negative at x = a, then it has a maximum point at x = a • if f ’(x) changes from negative to positive at x = b, then it has a minimum point at x = b iv) the interval where f(x) is concave up or concave down and the inflection points (using second derivative test) • use table of sign for f ’’(x) • if f ’’(x) 0 (positive value) : concave up • if f ’’(x) 0 (negative value) : concave down • if the concavity changes at x = c (either from up to down @ from down to up) , then it has an inflection point at x = c • to find the coordinate of inflection point, substitute x = c into y @ f(x) then write into the form of coordinate. @ v) sketch the graph of polynomial Functions • use all the important point based on the above information : maximum/minimum point and inflection point
Chapter 3 Applications of Differentiation 103 Some examples for the Graph of Polynomial Function : Graph of Polynomial Function with maximum, minimum and Inflection points Graph of Polynomial Function with Inflection point and maximum points) Graph of Polynomial Function with Inflection point and minimum points) Graph of Polynomial Function with Inflection point only (no max/min point) Graph of Polynomial Function with maximum @ minimum point only inflection point inflection point inflection point maximum point inflection point maximum point minimum point inflection point maximum point minimum point inflection point maximum point inflection point minimum point inflection point minimum point maximum point minimum point
MAT183/ MAT421 : Calculus I 104 Example 1/ MAR 2004/ MAT183/ Q2b (15 marks) Consider the function f(x) = x 2 (x + 3). Find i) the x-intercept(s) and the y-intercept. ii) the interval where the function is increasing or decreasing. Hence determine the extremum point(s), if any. iii) the interval where the function is concave up or concave down. Hence determine the inflection point(s), if any. Using the above information, sketch the graph of f(x). Solution : i) -intercept (when = 0) ℎ = () = 0, 2( + 3) = 0 2 = 0, + 3 = 0 = 0, = −3 -intercepts : (0, 0) (−3, 0) -intercept (when = 0) ℎ = 0, = () = 2( + 3) = 0 2(0 + 3) = 0 -intercept : (0, 0) ii) increasing/decreasing and max/min points ′() () = 2( + 3) = 3 + 3 2 ′() = 3 2 + 6 ′() = 3 2 + 6 = 0 3( + 2) = 0 3 = 0, + 2 = 0 = 0, = −2 ℎ = 0, → = () = 2( + 3) = 0 2(0 + 3) = 0 ℎ = −2, → = () = 2( + 3) = (−) 2(− + 3) = 4 Critical points : (0, 0) (−2, 4) (−∞, −) (−, ) (, +∞) Test value −10 −1 1 Sign for ′() = 3 2 + 6 + − + Slope ′() Increasing : interval (−∞, −2) and interval (0, +∞) Decreasing : interval (−2, 0) Maximum point : (−2, 4) Minimum point : (0, 0)
Chapter 3 Applications of Differentiation 105 Graph () = 2( + 3) y min (0, 0) max (−2, 4) Inflection (−1, 2) x iii) concave up/down and inflection point ′′() ′() = 3 2 + 6 ′′() = 6 + 6 ′′() = 6 + 6 = 0 6 = −6 = −1 ℎ = −1, → = () = 2( + 3) = (−) 2(− + 3) = 2 (−∞, −) (−, +∞) Test value −10 0 Sign for ′′() = 6 + 6 − + concavity Concave down ′′() Concave up : interval (−1, +∞) Concave down : interval (−∞, −1) Inflection point : (−1, 2) Concave up −3
MAT183/ MAT421 : Calculus I 106 Example 2/ NOV 2005/ MAT183/ Q3b (15 marks) Given f(x) = x(x − 4)2 i) the x-intercept(s) and the y-intercept. ii) the interval where f(x) is increasing or decreasing. Hence determine the extremum point(s), if any. iii) the interval where f(x) is concave up or concave down. Hence determine the inflection point(s), if any. Sketch the graph of f(x) by using the above information. Solution : i) -intercept (when = 0) ℎ = () = 0, ( − 4) 2 = 0 = 0, ( − 4) 2 = 0 = 0, − 4 = ξ0 = 0 = 4 -intercepts : (0, 0) (4, 0) -intercept (when = 0) ℎ = 0, = () = ( − 4) 2 = 0(0 − 4) 2 = 0 -intercept : (0, 0) ii) increasing/decreasing and max/min points ′() () = ( − 4) 2 = ( 2 − 8 + 16) = 3 − 8 2 + 16 ′() = 3 2 − 16 + 16 ′() = 3 2 − 16 + 16 = 0 ( − 4)(3 − 4) = 0 − 4 = 0, 3 − 4 = 0 = 4, 3 = 4 = 4 3 ℎ = 4, → = () = ( − 4) 2 = 4(4 − 4) 2 = 0 ℎ = 4 3 , → = () = ( − 4) 2 = 4 3 ቀ 4 3 − 4ቁ 2 = 256 27 Critical points : (4, 0) ( 4 3 , 256 27 ) (−∞, ) ( , ) (, +∞) Test value 0 2 10 Sign for ′() = 3 2 − 16 + 16 + − + Slope ′() Increasing : interval (−∞, 4 3 ) and interval (4, +∞) Decreasing : interval ( 4 3 , 4) Maximum point : ቀ 4 3 , 256 27 ቁ Minimum point : (4, 0)
Chapter 3 Applications of Differentiation 107 Graph iii) concave up/down and inflection point ′′() ′() = 3 2 − 16 + 16 ′′() = 6 − 16 ′′() = 6 − 16 = 0 6 = 16 = 16 6 = 8 3 ℎ = 8 3 , → = () = ( − 4) 2 = 8 3 ቀ 8 3 − 4ቁ 2 = 128 27 (−∞, ) ( , +∞) Test value 0 10 Sign for ′′() = 6 − 16 − + concavity Concave down ′′() Concave up : interval ቀ 8 3 , +∞ቁ Concave down : interval ቀ 8 3 , −1ቁ Inflection point : ቀ 8 3 , 128 27 ቁ Concave up () = ( − 4) 2 y x min (4, 0) max ቀ 4 3 , 256 27 ቁ Inflection ቀ 8 3 , 128 27 ቁ 0
MAT183/ MAT421 : Calculus I 108 Example 3/ APR 2006/ MAT183/ Q3c (13 marks) Consider the function f(x) = 2x 3 − 3x2 − 12x + 8. i) Find the interval where the function is increasing or decreasing. Hence determine the extremum points. ii) Find the interval where the function is concave up or concave down. Hence determine the inflection point. iii) Sketch the graph of f(x) by using the above information. Solution : i) increasing/decreasing and max/min points ′() () = 2 3 − 3 2 − 12 + 8 ′() = 6 2 − 6 − 12 ′() = 6 2 − 6 − 12 = 0 6( 2 − − 2) = 0 6( + 1)( − 2) = 0 + 1 = 0, − 2 = 0 = −1, = 2 ℎ = −1, → = () = 2(−) 3 − 3(−) 2 − 12(−) + 8 = 15 ℎ = 2, → = () = 2() 3 − 3() 2 − 12() + 8 = −12 Critical points : (−1, 15) (2, −12) (−∞, −) (−, ) (, +∞) Test value −10 0 10 Sign for ′() = 6 2 − 6 − 12 + − + Slope ′() Increasing : interval (−∞, −1) and interval (2, +∞) Decreasing : interval (−1, 2) Maximum point : (−1, 15) Minimum point : (2, −12)
Chapter 3 Applications of Differentiation 109 iii) Graph () = 2 3 − 3 2 − 12 + 8 y min (2, −12) max (−1, 15) Inflection ቀ 1 2 , 3 2 ቁ x ii) concave up/down and inflection point ′′() ′() = 6 2 − 6 − 12 ′′() = 12 − 6 ′′() = 12 − 6 = 0 12 = 6 = 6 12 = 1 2 ℎ = 1 2 , → = () = 2 ቀ ቁ 3 − 3 ቀ ቁ 2 − 12 ቀ ቁ + 8 = 3 2 (−∞, ) ( , +∞) Test value 0 1 Sign for ′′() = 12 − 6 − + concavity Concave down ′′() Concave up : interval ቀ 1 2 , +∞ቁ Concave down : interval ቀ−∞, 1 2 ቁ Inflection point : ቀ 1 2 , 3 2 ቁ Concave up If x-intercept and y-intercept are not asked, we better look for them. This is to ensure our graph is in the right position.
MAT183/ MAT421 : Calculus I 110 Example 4/ APR 2007/ MAT183/ Q3b (13 marks) Consider the function f(x) = x 3 − 6x2 + 9x + 6. i) Find the interval where f(x) is increasing or decreasing. Hence determine the extremum point(s). ii) Find he interval where f(x) is concave up or concave down. Hence determine the inflection point(s). iii) Sketch the graph of f(x) by using the above information. Solution : i) increasing/decreasing and max/min points ′() () = 3 − 6 2 + 9 + 6 ′() = 3 2 − 12 + 9 ′() = 3 2 − 12 + 9 = 0 3( 2 − 4 + 3) = 0 3( − 1)( − 3) = 0 − 1 = 0, − 3 = 0 = 1, = 3 ℎ = 1, → = () = () 3 − 6() 2 + 9() + 6 = 10 ℎ = 3, → = () = () 3 − 6() 2 + 9() + 6 = 6 Critical points : (1, 10) (3, 6) Increasing : interval (−∞, 1) and interval (3, +∞) Decreasing : interval (1, 3) Maximum point : (1, 10) Minimum point : (3, 6) ′() (−∞, ) (, ) (, +∞) Test value 0 2 10 Sign for ′() = 3 2 − 12 + 9 + − + Slope
Chapter 3 Applications of Differentiation 111 iii) Graph () = 3 − 6 2 + 9 + 6 y min (3, 6) max (1, 10) Inflection (2, 8) x ii) concave up/down and inflection point ′′() ′() = 3 2 − 12 + 9 ′′() = 6 − 12 ′′() = 6 − 12 = 0 6 = 12 = 12 6 = 2 ℎ = 2, → = () = () 3 − 6() 2 + 9() + 6 = 8 (−∞, ) (, +∞) Test value 0 10 Sign for ′′() = 6 − 12 − + concavity Concave down ′′() Concave up : interval (2, +∞) Concave down : interval (−∞, 2) Inflection point : (2, 8) Concave up If x-intercept and y-intercept are not asked, we better look for them. This is to ensure our graph is in the right position. 6
MAT183/ MAT421 : Calculus I 112 Example 5/ OCT 2007/ MAT183/ Q3b (13 marks) Consider the function f(x) = 3 2 x 3 − 6x2 + 16x + 7. i) Find the interval(s) where f(x) is increasing or decreasing. Hence determine the extremum point(s). (7 marks) ii) Find the interval(s) where f(x) is concave up or concave down. Hence determine the inflection point(s). (4 marks) iii) Sketch the graph of f(x) by using the above information. (2 marks) Solution : i) increasing/decreasing and max/min points ′() () = 2 3 3 − 6 2 + 16 + 7 ′() = 2 2 − 12 + 16 ′() = 2 2 − 12 + 16 = 0 2( 2 − 6 + 8) = 0 2( − 2)( − 4) = 0 − 2 = 0, − 4 = 0 = 2, = 4 ℎ = 2, → = () = 2 3 () 3 − 6() 2 + 16() + 7 = 61 3 ℎ = 4, → = () = 2 3 () 3 − 6() 2 + 16() + 7 = 53 3 Critical points : ቀ2, 61 3 ቁ ቀ4, 53 3 ቁ Increasing : interval (−∞, 2) and interval (4, +∞) Decreasing : interval (2, 4) Maximum point : ቀ2, 61 3 ቁ Minimum point : ቀ4, 53 3 ቁ ′() (−∞, ) (, ) (, +∞) Test value 0 3 10 Sign for ′() = 2 2 − 12 + 16 + − + Slope
Chapter 3 Applications of Differentiation 113 iii) Graph () = 2 3 3 − 6 2 + 16 + 7 y min ቀ4, 53 3 ቁ max ቀ2, 61 3 ቁ Inflection (3, 19) x ii) concave up/down and inflection point ′′() ′() = 2 2 − 12 + 16 ′′() = 4 − 12 ′′() = 4 − 12 = 0 4 = 12 = 3 ℎ = 3, → = () = 2 3 () 3 − 6() 2 + 16() + 7 = 19 (−∞, ) (, +∞) Test value 0 10 Sign for ′′() = 4 − 12 − + concavity Concave down ′′() Concave up : interval (3, +∞) Concave down : interval (−∞, 3) Inflection point : (3, 19) Concave up If x-intercept and y-intercept are not asked, we better look for them. This is to ensure our graph is in the right position. 7
MAT183/ MAT421 : Calculus I 114 Example 6/ OCT 2008/ MAT183/ Q3a (12 marks) Given a function f(x) = x 3 + 6x2 + 12x + 7. i) Find the interval(s) where f(x) is increasing or decreasing. Hence determine the extremum point(s), if any. ii) Find he interval(s) where f(x) is concave upward or downward. iii) Determine the inflection point. iv) Find the y-intercept. v) Sketch the graph of f(x). (12 marks) Solution : i) increasing/decreasing and max/min points ′() () = 3 + 6 2 + 12 + 7 ′ () = 3 2 + 12 + 12 ′() = 3 2 + 12 + 12 = 0 3( 2 + 4 + 4) = 0 3( + 2)( + 2) = 0 + 2 = 0 = −2 ℎ = −2, → = () = (−) 3 + 6(−) 2 + 12(−) + 7 = −1 Critical points : (−2, −1) Increasing : interval (−∞, −2) and interval (−2, +∞) Decreasing : none Maximum point : Minimum point : ′() (−∞, −) (−, +∞) Test value −10 0 Sign for ′() = 3 2 + 12 + 12 + + Slope
Chapter 3 Applications of Differentiation 115 ii) concave up/down and inflection point ′′() ′() = 3 2 + 12 + 12 ′′() = 6 + 12 ′′() = 6 + 12 = 0 6 = −12 = − 12 6 = −2 ℎ = −2, → = () = (−) 3 + 6(−) 2 + 12(−) + 7 = −1 (−∞, −) (−, +∞) Test value −10 0 Sign for ′′() = 6 + 12 − + concavity Concave down ′′() Concave up : interval (−2, +∞) Concave down : interval (−∞, −2) iii) Inflection point : (−2, −1) Concave up −1 7 (−2, −1) inflection y x 0 () = 3 + 6 2 + 12 + 7 iv) -intercept (when = 0) ℎ = 0, = () = 3 + 6 2 + 12 + 7 = () 3 + 6() 2 + 12() + 7 = 7 -intercept : (0, 7) v) If x-intercept and y-intercept are not asked, we better look for them. This is to ensure our graph is in the right position.
MAT183/ MAT421 : Calculus I 116 Example 7/ OCT 2010/ MAT183/ Q3d (10 marks) Given f(x) = 16 − (x + 3)4 . Find i) the x-intercept(s) and the y-intercept. (2 marks) ii) the interval where f(x) is increasing or decreasing. Hence determine the extremum point(s), if any. (3½ marks) iii) the interval where f(x) is concave up or concave down. Hence determine the inflection point(s), if any. (2½ marks) Sketch the graph of f(x) by using the above information. (2 marks) Solution : ii) increasing/decreasing and max/min points ′() () = 16 − ( + 3) 4 ′() = −4( + 3) 3 ′() = −4( + 3) 3 = 0 ( + 3) 3 = 0 + 3 = ξ0 3 = 0 = −3 ℎ = −3, → = () = 16 − (− + 3) 4 = 16 − 0 = 16 Critical points : (−3, 16) Increasing : interval (−∞, −3) Decreasing : interval (−3, +∞) Maximum point : (−3, 16) Minimum point : ′() (−∞, −) (−, +∞) Test value −10 0 Sign for ′() = −4( + 3) 3 + − Slope i) -intercept (when = 0) ℎ = () = 0, 16 − ( + 3) 4 = 0 ( + 3) 4 = 16 ( + 3) 4 = (±2) 4 + 3 = 2, + 3 = −2 = −1, = −5 -intercepts : (−1, 0) (−5, 0) -intercept (when = 0) ℎ = 0, = () = 16 − ( + 3) 4 = 16 − (0 + 3) 4 = 16 − 81 = −65 -intercept : (0, −65)
Chapter 3 Applications of Differentiation 117 iii) Graph ii) concave up/down and inflection point ′′() ′() = −4( + 3) 3 ′′() = −12( + 3) 2 ′′() = −12( + 3) 2 = 0 ( + 3) 2 = 0 + 3 = ξ0 − 12( + 3) 2 = 0 = −3 + 3 = ξ0 ℎ = −3, → = () = 16 − (− + 3) 4 = 16 − 0 = 16 (−∞, −) (−, +∞) Test value −10 0 Sign for ′′() = −12( + 3) 2 − − concavity Concave down ′′() Concave up : none Concave down : interval (−1, +∞) ∪ (−∞, −1) Inflection point : ( ℎ ′ ℎ) Concave down (0, −65) (−5, 0) (−1, 0) maximum point (−3, 16) y x 0 () = 16 − ( + 3) 4
MAT183/ MAT421 : Calculus I 118 Example 8/ SEP 2011/ MAT183/ Q3a (16 marks) Given that the function f(x) = x 3 − 3x2 − 9x + 2, find (if any) i) the critical point(s). ii) Find the interval(s) where the function increases and/or decreases and hence determine the extremum point(s). iii) Find the interval(s) where the function concaves up and/or concaves down and hence determine the inflection point(s) and sketch the graph of f(x). (16 marks) Solution : i) Critical points ′() () = 3 − 3 2 − 9 + 2 ′() = 3 2 − 6 − 9 ′() = 3 2 − 6 − 9 = 0 3( 2 − 2 − 3) = 0 3( + 1)( − 3) = 0 + 1 = 0, − 3 = 0 = −1, = 3 ℎ = −1, → = () = (−) 3 − 3(−) 2 − 9(−) + 2 = 7 ℎ = 3, → = () = () 3 − 3() 2 − 9() + 2 = −25 Critical points : (−1, 7) (3, −25) (−∞, −) (−, ) (, +∞) Test value −10 0 10 Sign for ′() = 3 2 − 6 − 9 + − + Slope ′() Increasing : interval (−∞, −1) and interval (3, +∞) Decreasing : interval (−1, 3) Maximum point : (−1, 7) Minimum point : (3, −25) ii) increasing/decreasing and max/min points
Chapter 3 Applications of Differentiation 119 iii) Graph () = 3 − 3 y 2 − 9 + 2 min (3, −25) max (−1, 7) Inflection ቀ 1 2 , − 25 8 ቁ x ii) concave up/down and inflection point ′′() ′() = 3 2 − 6 − 9 ′′() = 12 − 6 ′′() = 12 − 6 = 0 12 = 6 = 6 12 = 1 2 ℎ = 1 2 , → = () = ቀ ቁ 3 − 3 ቀ ቁ 2 − 9 ቀ ቁ + 2 = − 25 8 (−∞, ) ( , +∞) Test value 0 1 Sign for ′′() = 12 − 6 − + concavity Concave down ′′() Concave up : interval ቀ 1 2 , +∞ቁ Concave down : interval ቀ−∞, 1 2 ቁ Inflection point : ቀ 1 2 , − 25 8 ቁ Concave up If x-intercept and y-intercept are not asked, we better look for them. This is to ensure our graph is in the right position. 2
MAT183/ MAT421 : Calculus I 120 Example 9/ SEP 2013/ MAT183/ Q3a (10 marks) Given ( ) ( 1) 8 3 f x = x + − , find i) the x-intercept and y-intercept ii) the interval(s) where f(x) is increasing or decreasing iii) the interval(s) where f(x) is concaves up or concaves down. Hence determine the inflection point(s) if any iv) sketch the graph of f(x) by using the above information (16 marks) Solution : ii) increasing/decreasing and max/min points ′() () = ( + 1) 3 − 8 ′ () = 3( + 1) 2 ′() = 3( + 1) 2 = 0 ( + 1) 2 = 0 + 1 = ξ0 = 0 = −1 ℎ = −1, → = () = (− + 1) 3 − 8 = 0 − 8 = −8 Critical points : (−1, −8) Increasing : interval (−∞, −2) and interval (−2, +∞) Decreasing : none Maximum point : Minimum point : ′() (−∞, −) (−, +∞) Test value −10 0 Sign for ′() = 3( + 1) 2 + + Slope i) -intercept (when = 0) ℎ = () = 0, ( + 1) 3 − 8 = 0 ( + 1) 3 = 8 ( + 1) 3 = 2 3 + 1 = 2 = 1 -intercepts : (1, 0) -intercept (when = 0) ℎ = 0, = () = ( + 1) 3 − 8 = (0 + 1) 3 − 8 = 1 − 8 = −7 -intercept : (0, −7)
Chapter 3 Applications of Differentiation 121 iii) Graph ii) concave up/down and inflection point ′′() ′() = 3( + 1) 2 ′′() = 6( + 1) ′′() = 6( + 1) = 0 + 1 = 0 = −1 ℎ = −1, → = () = (− + 1) 3 − 8 = 0 − 8 = −8 (−∞, −) (−, +∞) Test value −10 0 Sign for ′′() = 6( + 1) − + concavity Concave down ′′() Concave up : interval (−1, +∞) Concave down : interval (−∞, −1) Inflection point : (−1, −8) Concave up (0, −7) (1, 0) (−1, −8) inflection y x 0 x-intercept y-intercept f(x) = (x+1)3 − 8
MAT183/ MAT421 : Calculus I 122 3.3 The graph of rational functions (linear/linear) - (using First Derivative Test) i) x-intercept and y-intercept • to find x-intercept : when y = 0, find the value of x. Then write into the form of coordinate system. • to find y-intercept : when x = 0, find the value of y @ f(x). Then write into the form of coordinate system. ii) vertical and horizontal asymptotes • to find vertical asymptote : → denominator = 0, the find the value of x → vertical asymptote : line x = a • to find horizontal asymptote : → find lim f(x) x→ = b, → horizontal asymptote : line y = b iii) the interval where f(x) is increasing and decreasing and the extremum points (using first derivative test) • use table of sign for f ’(x) • when f ’(x) = 0, find the value(s) of x (which is no solution) → no extremum point • if f ’(x) 0 (positive value) : increasing • if f ’(x) 0 (negative value) : decreasing iv) the interval where f(x) is concave up or concave down and the inflection point (using second derivative test) • use table of sign for f ’’(x) • when f ’’(x) = 0, find the value(s) of x (which is no solution) ) → no inflection point • if f ’’(x) 0 (positive value) : concave up • if f ’’(x) 0 (negative value) : concave down v) sketch the graph of rational Functions • draw the vertical and horizontal asymptotes • point out the x-intercept and y-intercept • if the x-intercept and y-intercept in quadrant I, then the graph of rational function should be in quadrant I and III • if the x-intercept and y-intercept in quadrant II, then the graph of rational function should be in quadrant II and IV
Chapter 3 Applications of Differentiation 123 Possible shape for the Graph of Rational Function ( ) linear linear f x = : Graph of Rational Function ( ) linear linear f x = in Quadrant I and III Graph of Rational Function ( ) linear linear f x = in Quadrant II and IV y = b (horizontal asymptote) x = a (vertical asymptote) y = b (horizontal asymptote) x = a (vertical asymptote)
MAT183/ MAT421 : Calculus I 124 Example 1/ APR 2011/ MAT183/ Q3a/ 15 marks Consider the function 5 1 ( ) − + = x x f x i) Find the x-intercept(s) and y-intercept(s). ii) Find the vertical and horizontal asymptotes and its analysis iii) Find the interval where f(x) is increasing or decreasing iv) Find the interval where f(x) is concave up and concave down v) Sketch the graph of f(x) using the above information Solution : ii) vertical/horizontal asymptotes vertical asymptote () = − 5 = 0 = 5 →5− () = →5− + 1 − 5 = −∞ →5+ () = →5+ + 1 − 5 = +∞ vertical asymptote : = 5 i) -intercept (when = 0) ℎ = () = 0, + 1 − 5 = 0 + 1 = 0 = −1 -intercept : (−1, 0) -intercept (when = 0) ℎ = 0, = () = + 1 − 5 = 0+1 0−5 = − 1 5 -intercept : ቀ0, − 1 5 ቁ iii) increasing/decreasing ′() () = + 1 − 5 = ′ () = ′ − ′ 2 ′() = ( − 5) − ( + 1) ( − 5) 2 = − 5 − − 1 ( − 5) 2 = − 6 ( − 5) 2 ′() = − 6 ( − 5) 2 = 0 (no real solution & no extremum point) horizontal asymptote →±∞ () = →±∞ + 1 − 5 = →±∞ ( + 1 ) ( − 5 ) = →±∞ (1 + 1 ) (1 − 5 ) = 1 + 0 1 − 0 = 1 horizontal asymptote : = 1 = + 1 = − 5 ′ = 1 ′ = 1
Chapter 3 Applications of Differentiation 125 v) Graph iv) concave up/down ′′() ′() = − 6 ( − 5) 2 = −6( − 5) −2 ′′() = 12( − 5) −3 = 12 ( − 5) 3 ′′() = 12 ( − 5) 3 = 0 (no real solution & no inflection point) (−∞, ) (, +∞) Test value 0 10 Sign for ′′() = 12 ( − 5) 3 − + concavity Concave down ′′() Concave up : interval (5, +∞) Concave down : interval (−∞, 5) Inflection point : Concave up Increasing : none Decreasing : interval (−∞, 5) and interval (5, +∞) Maximum point : Minimum point : ′() (−∞, ) (, +∞) Test value 0 10 Sign for ′() = − 6 ( − 5) 2 − − Slope ′() = ( − 5) 2 = 0 − 5 = ξ0 = 0 = 5 ′′() = ( − 5) 3 = 0 − 5 = ξ0 3 = 0 = 5 (horizontal asymptote) y = 1 y x x = 5 (vertical asymptote) 0 −1 −
MAT183/ MAT421 : Calculus I 126 Example 2/ OCT 2012/ MAT183/ Q2b/ 14 marks Consider the function f(x) = 3 6 8 4 − + x x i) Find the x-intercept(s) and y-intercept(s). ii) Find the vertical and horizontal asymptotes and its analysis iii) Find the interval where f(x) is increasing or decreasing iv) Find the interval where f(x) is concave up and concave down v) Sketch the graph of f(x) using the above information Solution : ii) vertical/horizontal asymptotes vertical asymptote () = 3 − 6 = 0 3 = 6 = 2 →2− () = →2− 8 + 4 3 − 6 = −∞ →2+ () = →2+ 8 + 4 3 − 6 = +∞ vertical asymptote : = 2 i) -intercept (when = 0) ℎ = () = 0, 8 + 4 3 − 6 = 0 8 + 4 = 0 4 = −8 = −2 -intercept : (−2, 0) -intercept (when = 0) ℎ = 0, = () = 8 + 4 3 − 6 = 8+0 0−6 = 8 −6 = − 4 3 -intercept : ቀ0, − 4 3 ቁ iii) increasing/decreasing ′() () = 8 + 4 3 − 6 = ′ () = ′ − ′ 2 ′() = 4(3 − 6) − 3(8 + 4) (3 − 6) 2 = 12 − 24 − 24 − 12 (3 − 6) 2 = − 48 (3 − 6) 2 ′() = − 48 (3 − 6) 2 = 0 (no real solution & no extremum point) horizontal asymptote →±∞ () = →±∞ 8 + 4 3 − 6 = →±∞ ( 8 + 4 ) ( 3 − 6 ) = →±∞ ( 8 + 4) (3 − 6 ) = 0 + 4 3 − 0 = 4 3 horizontal asymptote : = 4 3 = 8 + 4 = 3 − 6 ′ = 4 ′ = 3
Chapter 3 Applications of Differentiation 127 v) Graph iv) concave up/down ′′() ′() = − 48 (3 − 6) 2 = −48(3 − 6) −2 ′′() = 96(3 − 6) −3 ∙ 3 = 288 (3 − 6) 3 ′′() = 288 (3 − 6) 3 = 0 (no real solution & no inflection point) (−∞, ) (, +∞) Test value 0 10 Sign for ′′() = 288 (3 − 6) 3 − + concavity Concave down ′′() Concave up : interval (2, +∞) Concave down : interval (−∞, 2) Inflection point : Concave up Increasing : none Decreasing : interval (−∞, 2) and interval (2, +∞) Maximum point : Minimum point : ′() (−∞, ) (, +∞) Test value 0 10 Sign for ′() = − 48 (3 − 6) 2 − − Slope ′() = (3 − 6) 2 = 0 3 − 6 = ξ0 = 0 3 = 6 = 2 ′′() = (3 − 6) 3 = 0 3 − 6 = ξ0 3 = 0 3 = 6 = 2 (horizontal asymptote) = 4 3 y x = 2 (vertical asymptote) 0 2 − 4 3
MAT183/ MAT421 : Calculus I 128 Example 3/ APR 2008/ MAT183/ Q3b/ 15 marks Consider the function f(x) = 3 2 1 − − x x , find (if any) i) the x-intercept(s) and y-intercept(s). ii) the vertical and horizontal asymptotes and its analysis iii) f ’(x) and f ’’(x) iv) the interval where f(x) is concave up and concave down v) the extremum point(s) and inflection point(s) vi) Sketch the graph of f(x) using the above information Solution : ii) vertical/horizontal asymptotes vertical asymptote () = − 3 = 0 = 3 →3− () = →3− 2 − 1 − 3 = −∞ →3+ () = →3+ 2 − 1 − 3 = +∞ vertical asymptote : = 3 i) -intercept (when = 0) ℎ = () = 0, 2 − 1 − 3 = 0 2 − 1 = 0 2 = 1 = 1 2 -intercept : ቀ 1 2 , 0ቁ -intercept (when = 0) ℎ = 0, = () = 2 − 1 − 3 = 0−1 0−3 = −1 −3 = 1 3 -intercept : ቀ0, 1 3 ቁ iii) increasing/decreasing ′() () = 2 − 1 − 3 = ′ () = ′ − ′ 2 ′() = 2( − 3) − (2 − 1) ( − 3) 2 = 2 − 6 − 2 + 1 ( − 3) 2 = − 5 ( − 3) 2 ′() = − 5 ( − 3) 2 = 0 (no real solution & no extremum point) horizontal asymptote →±∞ () = →±∞ 2 − 1 − 3 = →±∞ ( 2 − 1 ) ( − 3 ) = →±∞ (2 − 1 ) (1 − 3 ) = 2 − 0 1 − 0 = 2 horizontal asymptote : = 2 = 2 − 1 = − 3 ′ = 2 ′ = 1
Chapter 3 Applications of Differentiation 129 v) Graph iv) concave up/down ′′() ′() = − 5 ( − 3) 2 = −5( − 3) −2 ′′() = 10( − 3) −3 ∙ 1 = 10 ( − 3) 3 ′′() = 10 ( − 3) 3 = 0 (no real solution & no inflection point) (−∞, ) (, +∞) Test value 0 10 Sign for ′′() = 10 ( − 3) 3 − + concavity Concave down ′′() Concave up : interval (3, +∞) Concave down : interval (−∞, 3) Inflection point : Concave up Increasing : none Decreasing : interval (−∞, 3) and interval (3, +∞) Maximum point : Minimum point : ′() (−∞, ) (, +∞) Test value 0 10 Sign for ′() = − 5 ( − 3) 2 − − Slope ′() = ( − 3) 2 = 0 − 3 = ξ0 = 0 = 3 ′′() = ( − 3) 3 = 0 − 3 = ξ0 3 = 0 = 3 y = 2 (horizontal asymptote) y x x = 3 (vertical asymptote) 0 1 3 1 2
MAT183/ MAT421 : Calculus I 130 Example 4/ OCT 2006/ MAT183/ Q2b/ 13 marks Given the function f(x) = 3 3 3 + x − . i) Simplify f(x) in the form of ( ) q(x) p x , q(x) 0. ii) Find the x-intercept(s) and y-intercept(s). iii) Find the vertical and horizontal asymptotes and its analysis iv) Find the interval(s) where f(x) is decreasing v) the interval where f(x) is concave up and concave down vi) Sketch the graph of f(x) using the above information Solution : iii) vertical/horizontal asymptotes vertical asymptote () = − 3 = 0 = 3 →3− () = →3− 3 − 6 − 3 = −∞ →3+ () = →3+ 3 − 6 − 3 = +∞ vertical asymptote : = 3 ii) -intercept (when = 0) ℎ = () = 0, 3 − 6 − 3 = 0 3 − 6 = 0 3 = 6 = 2 -intercept : (2, 0) -intercept (when = 0) ℎ = 0, = () = 3 − 6 − 3 = 0−6 0−3 = −6 −3 = 2 -intercept : (0, 2) iv) increasing/decreasing ′() () = 3 − 3 + 3 = 3( − 3) −1 + 3 ′() = −3( − 3) −2 + 0 ′() = − 3 ( − 3) 2 ′() = − 3 ( − 3) 2 = 0 (no real solution & no extremum point) horizontal asymptote →±∞ () = →±∞ 3 − 6 − 3 = →±∞ ( 3 − 6 ) ( − 3 ) = →±∞ (3 − 6 ) (1 − 3 ) = 3 − 0 1 − 0 = 3 horizontal asymptote : = 3 ) () = 3 − 3 + 3 () = 3 + 3( − 3) − 3 () = 3 + 3 − 9 − 3 () = 3 − 6 − 3 = 3 − 6 = − 3 ′ = 3 ′ = 1 Or…( ′() ) () = 3 − 6 − 3 = ′ () = ′ − ′ 2 ′() = 3( − 3) − (3 − 6) ( − 3) 2 = 3 − 9 − 3 + 6 ( − 3) 2 = − 3 ( − 3) 2
Chapter 3 Applications of Differentiation 131 v) Graph iv) concave up/down ′′() ′() = − 3 ( − 3) 2 = −3( − 3) −2 ′′() = 6( − 3) −3 ∙ 1 = 6 ( − 3) 3 ′′() = 6 ( − 3) 3 = 0 (no real solution & no inflection point) (−∞, ) (, +∞) Test value 0 10 Sign for ′′() = 6 ( − 3) 3 − + concavity Concave down ′′() Concave up : interval (3, +∞) Concave down : interval (−∞, 3) Inflection point : Concave up Increasing : none Decreasing : interval (−∞, 3) and interval (3, +∞) Maximum point : Minimum point : ′() (−∞, ) (, +∞) Test value 0 10 Sign for ′() = − 3 ( − 3) 2 − − Slope ′() = ( − 3) 2 = 0 − 3 = ξ0 = 0 = 3 ′′() = ( − 3) 3 = 0 − 3 = ξ0 3 = 0 = 3 y = 3 (horizontal asymptote) y x x = 3 (vertical asymptote) 0 2 2
MAT183/ MAT421 : Calculus I 132 Example 5/ OCT 2004/ MAT183/ Q3b (13 marks) Consider the function f(x) = 1 2 + − − x x . Find i) the x and y-intercept(s). ii) the horizontal and vertical asymptotes (if any). Analyze the function near the vertical asymptote(s). iii) the interval where f(x) is increasing or decreasing. Hence determine the extremum point(s), if any. iv) the interval where f(x) the function is concave up or down. Hence determine the inflection point(s), if any. Use the above information to sketch the graph of f(x). Solution : ii) vertical/horizontal asymptotes vertical asymptote () = + 1 = 0 = −1 →−1− () = →−1− − 2 + 1 = −∞ →−1+ () = →−1+ − 2 + 1 = +∞ vertical asymptote : = −1 i) -intercept (when = 0) ℎ = () = 0, − 2 + 1 = 0 − 2 = 0 = 2 -intercept : (2, 0) -intercept (when = 0) ℎ = 0, = () = − 2 + 1 = 0−2 0+1 = −2 -intercept : (0, −2) iii) increasing/decreasing ′() () = − 2 + 1 = ′ () = ′ − ′ 2 ′() = ( + 1) − ( − 2) ( + 1) 2 = + 1 − + 2 ( + 1) 2 = 3 ( + 1) 2 ′() = 3 ( + 1) 2 = 0 (no real solution & no extremum point) horizontal asymptote →±∞ () = →±∞ − 2 + 1 = →±∞ ( − 2 ) ( + 1 ) = →±∞ (1 − 2 ) (1 + 1 ) = 1 − 0 1 + 0 = 1 horizontal asymptote : = 1 = − 2 = + 1 ′ = 1 ′ = 1 () = − 2 − + 1 () = − ( 2 − + 1 ) () = −1 1 ( 2 − + 1 ) () = − 2 + 1
Chapter 3 Applications of Differentiation 133 v) Graph iv) concave up/down ′′() ′() = 3 ( + 1) 2 = 3( + 1) −2 ′′() = −6( + 1) −3 = − 6 ( + 1) 3 ′′() = − 6 ( + 1) 3 = 0 (no real solution & no inflection point) (−∞, −) (−, +∞) Test value −10 0 Sign for ′′() = − 6 ( + 1) 3 + − concavity Concave up ′′() Concave up : interval (−∞, −1) Concave down : interval (−1, +∞) Inflection point : Concave down Increasing : interval (−∞, −1) and interval (−1, +∞) Decreasing : Maximum point : Minimum point : ′() (−∞, −) (−, +∞) Test value −10 0 Sign for ′() = 3 ( + 1) 2 + + Slope ′() = ( + 1) 2 = 0 + 1 = ξ0 = 0 = −1 ′′() = ( + 1) 3 = 0 + 1 = ξ0 3 = 0 = −1 (horizontal asymptote) = 1 y x = −1 (vertical asymptote) 0 2 −2
MAT183/ MAT421 : Calculus I 134 3.4 Maximum and Minimum Problems (using Second Derivative Test) Maximum/minimum Area (2D-diagram) Given A = f(x) Step 1 : Find dx dA Step 2 : When dx dA = 0, find the value(s) of x Step 3 : Find 2 2 dx d A . If 2 2 dx d A > 0 @ positive (min area) If 2 2 dx d A < 0 @ negative (max area) Note : If 2 2 dx d A is in terms of variable x, then substitute the value(s) of x into 2 2 dx d A to determine whether 2 2 dx d A is positive (min) or negative (max) value. Step 4 : To find the max/min area, substitute the value(s) of x into A = f(x) Maximum/minimum Volume (3D-diagram) Given V = f(x) Step 1 : Find dx dV Step 2 : When dx dV = 0, find the value(s) of x Step 3 : Find 2 2 dx d V . If 2 2 dx d V > 0 @ positive (min volume) If 2 2 dx d V < 0 @ negative (max volume) area Perimeter/ circumference Volume Surface area
Chapter 3 Applications of Differentiation 135 Note : If 2 2 dx d V is in terms of variable x, then substitute the value(s) of x into 2 2 dx d V to determine whether 2 2 dx d V is positive (min) or negative (max) value. Step 4 : To find the max/min volume, substitute the value(s) of x into V = f(x) Maximum/minimum Cost Given C = f(x) Step 1 : Find dx dC Step 2 : When dx dC = 0, find the value(s) of x Step 3 : Find 2 2 dx d C . If 2 2 dx d C > 0 @ positive (min cost) If 2 2 dx d C < 0 @ negative (max cost) Note : If 2 2 dx d C is in terms of variable x, then substitute the value(s) of x into 2 2 dx d C to determine whether 2 2 dx d C is positive (min) or negative (max) value. Step 4 : To find the max/min cost, substitute the value(s) of x into C = f(x) 2D : related with perimeter 3D : related with Surface area
MAT183/ MAT421 : Calculus I 136 2D-diagrams Square Rectangle Circle Semicircle Equilateral Triangle x x Area = x 2 Perimeter = 4x y x Area = xy Perimeter = 2x + 2y r Area = r 2 Perimeter = 2r Area = x• x • sin600 = Perimeter = 3x x Area = 2 1 r 2 Perimeter = r + 2r r
Chapter 3 Applications of Differentiation 137 3D-diagrams Sphere Hemispherical bowl Rectangular box with square base Cube r Volume = 3 4 r 3 Surface area = 4r 2 Volume = x 2y Surface area (open-top) = base + right/left + infront/back = x 2 + 2xy + 2xy Surface area (closed – with cover on top) = top/bottom + right/left + infront/back = 2x 2 + 2xy + 2xy Volume = x 3 Surface area (open-top) = base + right/left + infront/back = x 2 + 2x 2 + 2 x 2 Surface area (closed – with cover on top) = top/bottom + right/left + infront/back = 2x 2 + 2x 2 + 2x 2 x x y x x x Volume = r 3 Surface area = 2r 2 (open-top) Surface area = 2r 2 + r 2 (closed – with cover on top) r
MAT183/ MAT421 : Calculus I 138 Cylindrical can Inverted circular cone (Related Rates) prism Volume = r 2h Surface area (open-top) = base + curve side = area of circle + area of rectangle = r 2 + 2rh Surface area (closed – with cover on top) = top/bottom + curve side = area of two circle + area of rectangle = 2r 2 + 2rh Volume = 3 1 r 2h Surface area = 2 2 r r + h (without circle cover on top) Volume = 2 1 pqr Surface area = ps + pq + pr + qr r q s p r h r h How to set up the formula (to be differentiate) i. From the given value, write a suitable formula based on the given shape. Write y (or h) as a subject, name as equation ii. Write the other formula (base on the given shape) and name as equation iii. Substitute equation into equation (to write the equation in terms of x or r) Note : • x, y (the dimensions for rectangle, rectangular box) • r, h (te dimensions for cylinder, cone)
Chapter 3 Applications of Differentiation 139 Procedure to Solve Maximum and Minimum Problems Part A : Set up the formula (to be differentiate) Step A1 : draw an appropriate diagram and label the relevant quantity. Step A2 : From the given value, write a suitable formula based on the given shape. Write y (or h) as a subject, name as equation Step A3 : Write the other formula (base on the given shape) and name as equation Step A4 : Substitute equation into equation (to write the equation in terms of x or r) Part B : Application of Derivatives Step B1 : find dx dA (first derivative), (or find dx dV or dx dC or dr dA or dr dV or dr dC ) Step B2 : when dx dA = 0, find the value of x Step B3 : find 2 2 dx d A (second derivative) If 2 2 dx d A 0 (max) and If 2 2 dx d A 0 (min) Step B4 : x = (obtained in B2) will maximize/minimize the area/ volume/ cost Step B5 : To find the maximum area/ volume/ cost, substitute the value of x (obtained in B2) into Formula area/ volume/ cost (from step A4)
MAT183/ MAT421 : Calculus I 140 Example 1 : Find the maximum possible area of a rectangular of perimeter 200 m. Solution : Part A : Set up the formula (to be differentiate) Step A1 : draw an appropriate diagram and label the relevant quantity Step A2 : From the given value, write a suitable formula based on the given shape. Write y (or h) as a subject, name as equation perimeter = 200 2x + 2y = 200 x + y = 100 y = 100 − x ------------ Step A3 : Write the other formula (base on the given shape) and name as equation Area, A = xy -------------- Step A4 : Substitute equation into equation (to write the equation in terms of x or r) By substituting into , A = x(100 − x) A = 100x − x 2 x y The formula to be differentiate
Chapter 3 Applications of Differentiation 141 Part B : Application of Derivatives Step B1 : find dx dA (first derivative) dx dA = 100 − 2x Step B2 : when dx dA = 0, find the value of x 100 − 2x = 0 2x = 100 x = 50 m Step B3 : find 2 2 dx d A (second derivative) 2 2 dx d A = −2 ( 0) max x = 50 m will maximize the area. Step B3 : To find the maximum area, substitute the value of x (obtained in B2) into Formula area (from step A4) , From A = 100x − x 2 Maximum area = 100(50) − (50) 2 = 2,500 m2 #
MAT183/ MAT421 : Calculus I 142 Example 2 : A rectangular plot of farmland will be bounded on one side by a river and on the other three sides by a single-strand electric fence with 800 m of wire. What is the largest area can be enclosed? Solution : Part A : Set up the formula (to be differentiate) Step A1 : draw an appropriate diagram and label the relevant quantity Step A2 : From the given value, write a suitable formula based on the given shape. Write y (or h) as a subject, name as equation Fence = 800 2x + y = 800 y = 800 − 2x ------------ Step A3 : Write the other formula (base on the given shape) and name as equation Area, A = xy -------------- Step A4 : Substitute equation into equation (to write the equation in terms of x or r) By substituting into , A = x(800 − 2x) A = 800x − 2x2 farmland river x y x The formula to be differentiate
Chapter 3 Applications of Differentiation 143 Part B : Application of Derivatives Step B1 : find dx dA (first derivative) dx dA = 800 − 4x Step B2 : when dx dA = 0, find the value of x 800 − 4x = 0 4x = 800 x = 200 m Step B3 : find 2 2 dx d A (second derivative) 2 2 dx d A = −4 ( 0) max x = 200 m will maximize the area. Step B4 : To find the maximum area, substitute the value of x (obtained in B2) into Formula area (from step A4) , From A = 800x − 2x 2 Maximum area = 800(200) − 2(200) 2 = 80,000 m2 #
MAT183/ MAT421 : Calculus I 144 Example 3 : An empty container is filled with liquid. The height of the liquid in the container is x cm and its volume is V cm3 , where V = x(5 − x) 2 1 . Determine the value of x so that the volume of the container is maximum. Hence, calculate the maximum volume of the container. Solution : We skip Part A (Set up the formula to be differentiate) because the formula to be differentiate is given. We are now in Step A4. Given V = x(5 − x) 2 1 2 2 2 5 V x x = − Part B : Application of Derivatives Step B1 : find dx dV (first derivative) dx dV = x 2 2 2 5 − • dx dV = x − 2 5 Step B2 : when dx dV = 0, find the value of x x − 2 5 = 0 x = 2 5 2 5 x = 2 5 x = = 2.5 cm