MAT183/ MAT421 : Calculus I 146 Example 4 : A company intends to produce boxes with cover. Each box has a volume of 8000 cm3 . The base of the box is a square with the length x cm and the height of the box is y cm. a. Show that the surface area of the box, A is given as x A x 32000 2 2 = + . b. Find the length of x and y that will minimize the area, A. Solution : a) 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 V = 8000 cm3 8000 2 x y = 2 8000 x y = ------------ x x y x x y V = 8000 cm3
Chapter 3 Applications of Differentiation 147 Step A3 : Write the other formula (base on the given shape) and name as equation Surface Area, A = top/bottom + right/left + front/back A = 2•x 2 + 2•xy + 2•xy A = 2x 2 + 4xy -------------- Step A4 : Substitute equation into equation (to write the equation in terms of x or r) By substituting into , A = 2x 2 + 4x 2 8000 x A = x x 32000 2 2 + # (shown) Part B : Application of Derivatives Step B1 : find dx dA (first derivative) = 2 2 + 32000 −1 = 4 − 32000 −2 = 4 − 32000 2 Step B2 : when dx dA = 0, find the value of x ℎ = 0 4 − 32000 2 = 0 4 = 32000 2 3 = 32000 4 3 = 8000 3 = 203 = 20 The formula to be differentiate
MAT183/ MAT421 : Calculus I 148 Step B3 : find 2 2 dx d A (second derivative) = 4 − 32000 −2 2 2 = 4 + 64000 −3 2 2 = 4 + 64000 3 = 20, 2 2 = 4 + 64000 203 = 4 + 64000 8000 = 4 + 8 = 12 (> 0) 2 2 > 0 ( ) = 20 ℎ ℎ = 20 , ( ①) = 8000 2 = 8000 202 = 8000 400 = 20 ℎ, = 20 = 20 ℎ Step B4 : To find the minimum surface area, substitute the value of x (obtained in B2) into Formula surface area (from step A4) , = 2 2 + 32000 = 2() 2 + 32000 = 800 + 1600 = 2,400 2
Chapter 3 Applications of Differentiation 149 Example 5 : Figure 3.31 shows an opened-top container full of water. The container is made of a hemispherical bowl of radius r, attached to the bottom of a cylinder with radius r and height h. Figure 3.31 a. If the surface area of the container is 288 cm2 , show that its volume is 3 3 1 V = 144 r − r . b. Determine r so that the volume is a maximum. Solution : a) Part A : Set up the formula (to be differentiate) Step A1 : draw an appropriate diagram and label the relevant quantity and r h V = 8000 cm3 r h opened-top no base r no cover
MAT183/ MAT421 : Calculus I 150 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 Surface area = 288 cm2 Surface area of curve side (rectangle) + surface area of hemisphere = 288 2rh + 2r 2 = 288 Dividing both sides by 2 rh + r 2 = 144 rh = 144 − r 2 h = r r 2 144 − ------------ Step A3 : Write the other formula (base on the given shape) and name as equation Volume, V = volume of cylinder + volume of hemisphere V = r 2h + 3 3 2 r -------------- Step A4 : Substitute equation into equation (to write the equation in terms of x or r) By substituting into , V = r 2 − r r 2 144 + 3 3 2 r V = r (144 − r 2 ) + 3 3 2 r V = 144r − r 3 + 3 3 2 r V = 144r − r 3 − 3 2 1 V = 144r − r 3 3 1 3 3 1 V = 144 r − r # (shown) The formula to be differentiate
Chapter 3 Applications of Differentiation 151 b) Part B : Application of Derivatives Step B1 : find dr dV (first derivative) = 144 − 1 3 3 = 144 − 3 2 Step B2 : when dr dV = 0, find the value of r ℎ = 0 144 − 3 2 = 0 3 2 = 144 2 = 144 3 = 48 = ξ48 = 6.93 Step B3 : find 2 2 dr d V (second derivative) = 144 − 3 2 2 2 = −6 = 6.93 , 2 2 = −6 = −6 (6.93) (< 0) 2 2 < 0 ( ) = 6.93 ℎ
MAT183/ MAT421 : Calculus I 152 Example 6 : A rectangular area of 3200 m2 is to be fenced off. Two opposite sides will use fencing costing RM 1 per meter and the remaining sides will use fencing costing RM 2 per meter. Find the dimension of the rectangle at the least cost. 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 Area, V = 3200 cm2 xy = 3200 y = x 3200 ------------ Step A3 : Write the other formula (base on the given shape) and name as equation Perimeter = 2x + 2y Cost, C = (RM 1)2x + (RM 2)2y C = 2x + 4y -------------- Step A4 : Substitute equation into equation (to write the equation in terms of x or r) By substituting into , C = 2x + 4 x 3200 C = 2x + x 12800 x Area = 3200 m y 2 The formula to be differentiate
Chapter 3 Applications of Differentiation 153 Part B : Application of Derivatives = 2 + 12800 = 2 + 12800 −1 Step B1 : find dx dC (first derivative) = 2 − 12800 −2 = 2 − 12800 2 Step B2 : when dx dC = 0, find the value of x ℎ = 0, 2 − 12800 2 = 0 2 = 12800 2 2 = 12800 2 = 6400 = 80 Step B3 : find 2 2 dx d C (second derivative) = 2 − 12800 −2 2 2 = 25600 −3 = 25600 3 = 80 , 2 2 = 25600 803 (> 0) 2 2 > 0 ( ) = 80 ℎ ℎ = 80 , ( ①) = 3200 = 3200 80 = 40 ℎ, = 80 = 40 ℎ
MAT183/ MAT421 : Calculus I 154 Example 7 A cylindrical can, open at the top, has a volume of 2000 cm3 . The material for the base will cost RM 4 per cm2 and the material for the side will cost RM 2 per cm2 . Find the minimum cost of constructing the cylindrical can. Ans : RM1757.51 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 V = 2000 cm3 r h 2 = 2000 h = 2 2000 r ------------ Step A3 : Write the other formula (base on the given shape) and name as equation Surface Area, A = base (circle) + curve side (rectangle) A = r 2rh 2 + Cost, C = RM4r RM2 2rh 2 + • C = 4r 4rh 2 + -------------- open top V = 2000 cm3 r h
Chapter 3 Applications of Differentiation 155 Step A4 : Substitute equation into equation (to write the equation in terms of x or r) By substituting into , = 4 2 + 4 ( 2000 2 ) = 4 2 + 8000 Part B : Application of Derivatives = 4 2 + 8000 = 4 2 + 8000 −1 Step B1 : find dr dC (first derivative) = 8 − 8000 −2 = 8 − 8000 2 Step B2 : when dr dC = 0, find the value of r ℎ = 0, 8 − 8000 2 = 0 8 = 8000 2 3 = 8000 8 = 1000 = √ 1000 3 = 6.83 Step B3 : find 2 2 dr d C (second derivative) = 8 − 8000 −2 2 2 = 8 + 16000 −3 = 8 + 16000 3 = 6.83 → 2 2 = 8 + 16000 (6.83) 3 (> 0) 2 2 > 0 ( ) = 6.83 ℎ The formula to be differentiate Step B4 : To find the minimum cost, substitute the value of r (obtained in B2) into Formula cost (from step A4) = 4 2 + 8000 = 4(. ) 2 + 8000 . = 1,757.51
MAT183/ MAT421 : Calculus I 156 3.5 Related Rates (Rate of Change) Procedure to solve Related Rates Step 1 : Draw an appropriate figure and label the quantities relevant to the problem. Step 2 : Put a suitable variable based on the given unit Step 3 : Write the question to mathematical expression Step 4 : Write an appropriate formula based on the given shape Step 5 : Differentiate both sides implicitly with respect to time (t) or use chain rule Step 6 : Write a suitable unit for the final answer Example 1/ Exercise 3.3/ Q1/ page 103 A soap bubble retains its spherical shape as it expands. If air is blown into it at the rate of 5 cm3 /s, how fast does its radius increase when the radius is 3 cm? Solution : Step 1 : Draw an appropriate figure and label the quantities relevant to the problem. Step 2 : Put a suitable variable based on the given unit dt dV = 5 cm3 /s Step 3 : Write the question to mathematical expression dt dr = ? when r = 3 cm Step 4 : Write an appropriate formula based on the given shape V = 3 3 4 r Step 5 : Differentiate both sides implicitly with respect to time (t) or use chain rule dt dV = dt dr r 2 3 4 3 • dt dV = dt dr r 2 4 5 = ( ) dt 2 dr 4 3 dt dr = 36 5 = 0.04 cm/s # r Step 6 : Write a suitable unit for the final answer
Chapter 3 Applications of Differentiation 157 Example 2 The width of a rectangle increases at a rate of 2 cm/s. The length is three times its width. Find the rate at which the area is increasing when the width is 4 cm. Solution : Step 1 : Draw an appropriate figure and label the quantities relevant to the problem. Step 2 : Put a suitable variable based on the given unit dt dx = 2 cm/s Step 3 : Write the question to mathematical expression dt dA = ? when x = 4 cm Step 4 : Write an appropriate formula based on the given shape Area, A = width × length A = x(3x) A = 3x 2 Step 5 : Differentiate both sides implicitly with respect to time (t) or use chain rule = 6 = 6(4)(2) = 48 2 / # 3x x put a suitable unit for the final answer
MAT183/ MAT421 : Calculus I 158 Example 3 Figure 3.5 shows an equilateral triangle with side p cm. Show that the area is 4 3 2 p A = . If the side changes at the rate of 0.05 cm/s, find the rate of change of the area when the side is 5 cm. Solution : Step 1 : Draw an appropriate figure and label the quantities relevant to the problem. Step 2 : Put a suitable variable based on the given unit dt dp = 0.05 cm/s Step 3 : Write the question to mathematical expression dt dA = ? when p = 5 cm p p p Figure 3.5 Height = h = ( ) 2 2 2 p p − = 4 2 2 p p − = 4 4 2 2 p −p = 4 3 2 p h = 2 3p Area, A = 2 1 • base • height = 2 1 • p • 2 3p A = 4 3 2 p # (shown) p p h
Chapter 3 Applications of Differentiation 159 Step 4 : Write an appropriate formula based on the given shape = ξ3 2 4 = ξ3 4 2 Step 5 : Differentiate both sides implicitly with respect to time (t) or use chain rule = ξ3 4 ∙ 2 = ξ3 2 = ξ3 2 ()(. ) = 0.2165 2 / # put a suitable unit for the final answer
MAT183/ MAT421 : Calculus I 160 Example 4 : A 10 m ladder is leaning against the wall of a house. Suppose x represents the distance between the foot of the ladder and the wall, while y represents the distance between the top of the ladder and the floor. If the top of the ladder slips down at the rate of 3 m/s, find the speed at which the foot of the ladder is moving away when it is 6 m from the wall. Solution : Step 1 : Draw an appropriate figure and label the quantities relevant to the problem. Step 2 : Put a suitable variable based on the given unit dt dy = −3 m/s Step 3 : Write the question to mathematical expression dt dx = ? when x = 6 m Step 4 : Write an appropriate formula based on the given shape 2 2 2 x + y = 10100 2 2 x + y = Step 5 : Differentiate both sides implicitly with respect to time (t) or use chain rule wall y x floor 10 m (ladder) 2 + 2 = 0 (÷ 2) → + = 0 () + ()(−) = 0 6 − 24 = 0 6 = 24 = 4 / # put a suitable unit for the final answer = 6 10 = ඥ102 − 6 2 = ξ64 = 8
Chapter 3 Applications of Differentiation 161 Example 5 : V is the volume of a right circular cone of base radius 5 m and height 15 m. If the volume decreases at a rate of 2 m3 /h, find the rate at which the radius of the cone is decreasing at the instant when the radius is 2 m. Solution : Step 1 : Draw an appropriate figure and label the quantities relevant to the problem. Step 2 : Put a suitable variable based on the given unit dt dV = −2 m3 /h (decrease → negative) Step 3 : Write the question to mathematical expression dt dr = ? when r = 2 m Step 4 : Write an appropriate formula based on the given shape V = r h 2 3 --------------- Substitute into : V = r 3r 3 2 • V = 3 r r = 5 m = = h = 3r ------------ h = 15 m put a suitable unit for the final answer Step 5 : Differentiate both sides implicitly with respect to time (t) or use chain rule = 3 = 3 2 − = 3() 2 − = 12 = − 2 12 = − 1 6 / #
MAT183/ MAT421 : Calculus I 162 Example 6 Sand is falling into a conical pile so that the radius of the base of the pile is always equal to half of its height. If the sand is falling at the rate of 10 cm3 /sec, how fast is the height of the pile increasing when the pile is 5 cm deep. Solution : Step 1 : Draw an appropriate figure and label the quantities relevant to the problem. Step 2 : Put a suitable variable based on the given unit dt dV = 10 cm3 /sec Step 3 : Write the question to mathematical expression dt dh = ? when h = 5 cm Step 4 : Write an appropriate formula based on the given shape V = r h 2 3 --------------- Substitute into : V = h h • 2 3 2 V = h h • • 3 4 2 V = 3 12 h r r = r = ------------ h r put a suitable unit for the final answer Step 5 : Differentiate both sides implicitly with respect to time (t) or use chain rule = 12 ℎ 3 = 12 ∙ 3ℎ 2 ℎ = 4 ℎ 2 ℎ = 4 () 2 ℎ = 25 4 ℎ ℎ = 40 25 = 8 5 / #
Chapter 3 Applications of Differentiation 163 Example 7 : Let p be the length of a diagonal of a rectangle whose sides have lengths x and y. Assume that x and y are vary with time. a. How are dt dp , dt dx and dt dy are related? b. If x increases at a constant rate of 0.5 m/min and y decreases at a constant rate of 0.25 m/min, how fast is the size of the diagonal changing when x = 3 m and y = 4 m? Solution : Step 1 : Draw an appropriate figure and label the quantities relevant to the problem. a) How are dt dp , dt dx and dt dy are related? Step 2 : Write an appropriate formula based on the given shape 2 2 2 p = x + y (Pythagoras Theorem) Step 3 : Differentiate both sides implicitly with respect to time (t) or use chain rule p x y 2 = 2 + 2 (÷ 2) → = +
MAT183/ MAT421 : Calculus I 164 = 0.5 / = −0.25 / = ? ℎ = 3 = 4 b) If x increases at a constant rate of 0.5 m/min and y decreases at a constant rate of 0.25 m/min, how fast is the size of the diagonal changing when x = 3 m and y = 4 m? Step 4 : Put a suitable variable based on the given unit Step 5 : Write the question to mathematical expression = + () = ()(. ) + ()(−. ) 5 = 1.5 − 1 5 = 0.5 = 0.5 5 = 0.1 / # = 3 = 4 = ඥ3 2 + 4 2 = ξ25 = 5 put a suitable unit for the final answer
Chapter 3 Applications of Differentiation 165 Example 8 A spherical balloon is being inflated at a rate of 8 cm3 /s. Find the rate of change of the surface area when the radius is 10 cm. Solution : Step 1 : Draw an appropriate figure and label the quantities relevant to the problem. Step 2 : Put a suitable variable based on the given unit dt dV = 8 cm3 /s Step 3 : Write the question to mathematical expression dt dA = ? when r = 10 cm Step 4 : Write an appropriate formula based on the given shape V = 3 3 4 r Step 5 : Differentiate both sides implicitly with respect to time (t) or use chain rule dt dV = dt dr r 2 3 4 3 • dt dV = dt dr r 2 4 8 = ( ) dt 2 dr 4 10 8 = dt dr 400 dt dr = 400 8 dt dr = 50 1 r put a suitable unit for the final answer To find dt dA , and use Step 4 & step 5 Step 4 : Write an appropriate formula based on the given shape ℎ, = 4 2 Step 5 : Differentiate both sides implicitly with respect to time (t) or use chain rule = 4 ∙ 2 = 8 = 8() ቀ ቁ = 1.6 2 / #
MAT183/ MAT421 : Calculus I 166 Example 9 : A cylindrical tank with radius 3 m is being filled with water at a rate of 5 m3 /min. How fast is the height of the water increasing? Hint : volume of cylinder = r h 2 Solution : Step 1 : Draw an appropriate figure and label the quantities relevant to the problem. Step 2 : Put a suitable variable based on the given unit dt dV = 5 m3 /min Step 3 : Write the question to mathematical expression dt dh = ? Step 4 : Write an appropriate formula based on the given shape V = r h 2 Substitute r = 3 m (because r is constant) V = ( ) h 2 3 V = 9 h r = 3 m h put a suitable unit for the final answer Step 5 : Differentiate both sides implicitly with respect to time (t) or use chain rule = 9ℎ = 12 ∙ 3ℎ 2 ℎ = 9 ℎ = 9 ℎ ℎ = 5 9 / #
Chapter 3 Applications of Differentiation 167 Example 10 : A ladder 8m long leans against a wall. The lower end of the ladder is pulled away from the wall at a rate of 2m/s. How fast is the angle between the top of the ladder and wall changing, dt d when the angle is 3 radian? Solution : Step 1 : Draw an appropriate figure and label the quantities relevant to the problem. Step 2 : Put a suitable variable based on the given unit dt dx = 2 m/s Step 3 : Write the question to mathematical expression dt d = ? when = 3 radian Step 4 : Write an appropriate formula based on the given shape wall 8 m (ladder) floor θ wall y x floor 8 m (ladder) θ Step 5 : Differentiate both sides implicitly with respect to time (t) or use chain rule = 8 = 8 ቀ ቁ = 8 ( ) = 4 = 0.5 / # = 8 = 8 put a suitable unit for the final answer
MAT183/ MAT421 : Calculus I 168 3.6 Mean-value Theorem of Differentiation The Mean Value Theorem for differentiation states that if f(x) is differentiable on (a, b) and continuous on [a, b], then there is at least one point c in (a, b) where ( ) ( ) ( ) b a f b f a f' c − − = Example 1/ OCT 2006/ MAT183/ Q3a (4 marks) The Mean Value Theorem for differentiation states that if f(x) is differentiable on (a, b) and continuous on [a, b], then there is at least one point c in (a, b) where ( ) ( ) ( ) b a f b f a f' c − − = Find the value of c if f(x) = 3 + x − 2 in the interval (2, 6). Solution : Step : Identify the values of a and b, f(a) and f(b) Step : Find f’(x) and f’(c) = 2 = 6 () = 3+ ξ − 2 () = (2) = 3+ ξ− 2 = 3+ 0 = 3 () = (6) = 3+ ξ− 2 = 3+ 2 = 5 () = 3+ ξ − 2 = 3+ ( − 2) 1 2 ′() = 0+ 1 2 ( − 2) − 1 2 = 1 2ξ − 2 ′() = 1 2ξ − 2 Step :Use ( ) ( ) ( ) b a f b f a f' c − − = to find the value of c. (note : the value of c should be between a and b) ′ () = ()− () − 1 2ξ − 2 = 5− 3 6− 2 1 2ξ − 2 = 2 4 1 2ξ − 2 = 1 2 2 = 2ξ − 2 ξ − 2 = 1 − 2 = 1 2 − 2 = 1 = 3 (2, 6)
Chapter 3 Applications of Differentiation 169 Example 2/ APR 2007/ MAT183/ Q4b (5 marks) The Mean Value Theorem for differentiation states that if f(x) is differentiable on (a, b) and continuous on [a, b], then there is at least one point c in (a, b) where ( ) ( ) ( ) b a f b f a f' c − − = If f(x) = 3x 2 + 2x + 1, find the value of c in the interval (1, 3) that satisfies the above theorem. Solution : Step : Identify the values of a and b, f(a) and f(b) Step : Find f’(x) and f’(c) = 1 = 3 () = 3 2 + 2 + 1 () = (1) = 3() 2 +2()+ 1 = 6 () = (3) = 3() 2 +2()+ 1 = 34 () = 3 2 + 2 + 1 ′ () = 6 + 2 ′ () = 6 + 2 Step :Use ( ) ( ) ( ) b a f b f a f' c − − = to find the value of c. (note : the value of c should be between a and b) ′ () = ()− () − 6 + 2 = 34− 6 3− 1 6 + 2 = 14 6 = 12 = 2 (1, 3)
MAT183/ MAT421 : Calculus I 170 Example 3/ OCT 2007/ MAT183/ Q4b (5 marks) The Mean Value Theorem for differentiation states that if f(x) is differentiable on (a, b) and continuous on [a, b], then there is at least one point c in (a, b) where ( ) ( ) ( ) b a f b f a f' c − − = Find the value of c in the interval (0, 4) that satisfies the above theorem if ( ) 2 4 2 + − = x x x f x . Solution : Step : Identify the values of a and b, f(a) and f(b) Step : Find f’(x) and f’(c) = 0 = 4 () = 2 − 4 + 2 () = (0) = () 2 −4() + 2 = 0 () = (4) = () 2 −4() + 2 = 0 () = 2 − 4 + 2 = ′ () = ′ −′ 2 = (2 −4)( + 2)− ( 2 − 4) ( + 2) 2 = 2 2 + 4 − 4 −8 − 2 + 4 ( + 2) 2 = 2 + 4 −8 ( + 2) 2 ′ () = 2 + 4 − 8 ( + 2) 2 Step :Use ( ) ( ) ( ) b a f b f a f' c − − = to find the value of c. (note : the value of c should be between a and b) ′ () = ()− () − 2 + 4 −8 ( + 2) 2 = 0 −0 4 −0 2 + 4 −8 ( + 2) 2 = 0 2 + 4 −8 = 0 = 1.46 (0, 4) = −5.46 (0, 4) = 2 −4 = + 2 ′ = 2 −4 ′ = 1
Chapter 3 Applications of Differentiation 171 Example 4/ APR 2008/ MAT183/ Q4b (5 marks) The Mean Value Theorem for differentiation states that if f(x) is differentiable on (a, b) and continuous on [a, b], then there is at least one point c in (a, b) where ( ) ( ) ( ) b a f b f a f' c − − = If f(x) = −2x 3 + 6x − 2, find the value of c in the interval (−2, 2) that satisfies the above theorem. Solution : Step : Identify the values of a and b, f(a) and f(b) Step : Find f’(x) and f’(c) = −2 = 2 () = −2 3 + 6 − 2 () = (−) = −2(−) 3 + 6(−)−2 = 16 − 12 −2 = 2 () = () = −2() 3 + 6() − 2 = −16+ 12 − 2 = −6 () = −2 3 + 6 − 2 ′ () = −6 2 + 6 ′ () = −6 2 + 6 Step :Use ( ) ( ) ( ) b a f b f a f' c − − = to find the value of c. (note : the value of c should be between a and b) ′ () = ()− () − −6 2 + 6 = −6− 2 2− (−2) = −8 4 = −2 −6 2 + 6 = −2 −6 2 = −8 2 = −8 −6 = 4 3 = ±√ 4 3 = 1.1547 (−2, 2) = −1.1547 (−2, 2)
MAT183/ MAT421 : Calculus I 172 Example 5// OCT 2010/ MAT183/ Q3e (5 marks) The Mean Value Theorem for differentiation states that if f(x) is differentiable on (a, b) and continuous on [a, b], then there is at least one point c in (a, b) where ( ) ( ) ( ) b a f b f a f' c − − = If f(x) = 9 − 3x + 3 , find the value of c in the interval (2, 11) that satisfies the above theorem. Solution : Step : Identify the values of a and b, f(a) and f(b) Step : Find f’(x) and f’(c) = 2 = 11 () = 9− ξ3 + 3 () = (2) = 9− ඥ3() + 3 = 9 − 3 = 6 () = (11) = 9 −ඥ3() + 3 = 9 − 6 = 3 () = 9− ξ3 + 3 = 9 −(3 + 3) 1 2 ′ () = 0 − 1 2 (3 +3) − 1 2 ∙ 3 = − 3 2ξ3 +3 ′ () = − 3 2ξ3 +3 Step :Use ( ) ( ) ( ) b a f b f a f' c − − = to find the value of c. (note : the value of c should be between a and b) ′ () = ()− () − − 3 2ξ3 + 3 = 3− 6 11− 2 −3 2ξ3 + 3 = −3 9 2ξ3 + 3 = 9 ξ3 + 3 = 9 2 3 + 3 = 81 4 = 20.25 3 = 17.25 = 5.75 (2, 11)
Chapter 3 Applications of Differentiation 173 Example 6/ APR 2011/ MAT183/ Q4c (5 marks) The Mean Value Theorem for differentiation states that if f(x) is differentiable on (a, b) and continuous on [a, b], then there is at least one point c in (a, b) where ( ) ( ) ( ) b a f b f a f' c − − = If f(x) = 5x 2 + 3x + 2, find the value of c in the interval (2, 4) that satisfies the above theorem. Solution : Step : Identify the values of a and b, f(a) and f(b) Step : Find f’(x) and f’(c) = 2 = 4 () = 5 2 + 3 + 2 () = (2) = 5() 2 +3()+ 2 = 28 () = (4) = 5() 2 +3()+ 2 = 94 () = 5 2 + 3 + 2 ′ () = 10 + 3 ′ () = 10 + 3 Step :Use ( ) ( ) ( ) b a f b f a f' c − − = to find the value of c. (note : the value of c should be between a and b) ′ () = ()− () − 10 + 3 = 98− 28 4− 2 10 + 3 = 35 10 = 32 = 3.2 (2, 4)
MAT183/ MAT421 : Calculus I 174 Example 7/ MAR 2013/ MAT183/ Q2c/ 5 marks The Mean Value Theorem for Differentiation states that if f(x) is differentiable on (a, b) and continuous on [a, b], then there is at least one point c in (a, b) where ( ) ( ) ( ) b a f b f a f' c − − = . If ( ) 1 3 2 + − = x x f x , find the value of c in the interval (1, 5) that satisfies the above theorem. Solution : Step : Identify the values of a and b, f(a) and f(b) Step : Find f’(x) and f’(c) = 1 = 5 () = 3− 2 + 1 () = (1) = 3− 2() +1 = 1 2 () = (5) = 3− 2() +1 = − 7 6 () = 3− 2 + 1 = ′ () = ′ −′ 2 = −2( + 1)− (3 − 2) ( +1) 2 = −2 −2 − 3 +2 ( + 1) 2 = − 5 ( + 1) 2 ′ () = − 5 ( + 1) 2 Step :Use ( ) ( ) ( ) b a f b f a f' c − − = to find the value of c. (note : the value of c should be between a and b) ′ () = ()− () − − 5 ( + 1) 2 = − 7 6− 1 2 5 − 1 − 5 ( + 1) 2 = − 5 12 ( + 1) 2 = 12 + 1 = ±ξ12 = ξ12− 1 = 2.46 (1, 5) = −ξ12− 1 = −4.46 (1, 5) = 3 − 2 = + 1 ′ = −2 ′ = 1
Chapter 3 Applications of Differentiation 175 Example 8/ OCT 2012/ MAT183/ Q3a/ 5 marks The Mean Value Theorem for Differentiation states that if f(x) is differentiable on (a, b) and continuous on [a, b], then there is at least one point c in (a, b) where ( ) ( ) ( ) b a f b f a f' c − − = . If f(x) = x + ln (x − 3) , find the value of c in the interval (4, 7) that satisfies the above theorem. Answer : c = 5.16 Solution : Step : Identify the values of a and b, f(a) and f(b) Step : Find f’(x) and f’(c) = 4 = 7 () = + ( − 3) () = (4) = + ( − 3) = 4 + (1) = 4 + 0 = 4 () = (7) = + ( − 3) = 7 + (4) = 8.3863 () = + ( − 3) ′ () = 1+ 1 −3 ′ () = 1+ 1 −3 Step :Use ( ) ( ) ( ) b a f b f a f' c − − = to find the value of c. (note : the value of c should be between a and b) ′ () = ()− () − 1 + 1 − 3 = 8.3863 − 4 7 − 4 1 + 1 − 3 = 1.4621 1 − 3 = 0.4621 1 0.4621 = − 3 = 3+ 1 0.4621 = 5.164 (4, 7)
MAT183/ MAT421 : Calculus I 176 Example 9/ MAR 2012/ MAT183/ Q4a/ 5 marks The Mean Value Theorem for Differentiation states that if f(x) is differentiable on (a, b) and continuous on [a, b], then there is at least one point c in (a, b) where ( ) ( ) ( ) b a f b f a f' c − − = . If f(x) = 4x + 1 , find the value of c in the interval (0, 6) that satisfies the above theorem. Answer : c = 2 Solution : Step : Identify the values of a and b, f(a) and f(b) Step : Find f’(x) and f’(c) = 0 = 6 () = ξ4 + 1 () = (0) = ඥ4() + 1 = 1 () = (6) = ඥ4() + 1 = 5 () = ξ4 + 1 = (4 +1) 1 2 ′ () = 1 2 (4 + 1) − 1 2 ∙ 4 = 2 ξ4 + 1 ′ () = 2 ξ4 + 1 Step :Use ( ) ( ) ( ) b a f b f a f' c − − = to find the value of c. (note : the value of c should be between a and b) ′ () = ()− () − 2 ξ4 + 1 = 5− 1 6− 0 2 ξ4 + 1 = 2 3 ξ4 + 1 = 3 4 + 1 = 9 4 = 8 = 2 (0, 6) END OF CHAPTER 3
Chapter 4 : Integration 177 CHAPTER 4 INTEGRATION List of topics : Indefinite & Definite Integrals (Definite integrals is used to find the volume of revolution, in chapter 5) Basic Rules of Integration : i) Constant Rule ii) Power Rule (for n −1) iii) Logarithmic Rule (for n = −1) iv) Sum and Difference Rules v) Exponential Rule vi) Trigonometric Rule Integration with u-substitution Powers of sine and cosine (even and odd power) Mean value theorem for Integrals Second Fundamental Theorem of Calculus
MAT183/ MAT421 : Calculus I 178 Basic Rules of Integration Differentiation Integration i) Constant Rule Example : If y = 2 dx dy = 0 In general, Constant Rule Example 1 : Indefinite integral : 2 dx = 2x + C definite integral : 2 2 2(1) 2(0) 2 1 0 1 0 = = − = dx x Example 2 : Indefinite integral : 5 dx =5x + C definite integral : 5(2) 5(1) 5 5 5 2 1 2 1 = = − = x dx In general, a dx = ax + C (a = constant) ii) Power Rule (for all values of n) Example : If y = x 3 2 3x dx dy = In general, If y = axn (a = constant) dx dy = na x n−1 Power Rule (for n −1) Example : Indefinite integral : C x x dx = + 4 4 3 definite integral : ( ) ( ) 4 15 4 1 4 2 4 4 4 2 1 2 4 1 3 = − = = x x dx In general, C n x ax dx a x dx a n n n + + = = + 1 1 = , (ℎ = ) = 0
Chapter 4 : Integration 179 Differentiation Integration iii) Logarithmic Rule (for n = −1) Wrong method : C x x dx = + − 0 0 1 = undefined Correct method : dx ln x C x x dx = = + −1 1 Why?? Because ( ) dx x d lnx 1 = iv) Sum and Difference Rules (differentiate one by one) Example : If y = x x x x 2 3 2 − + − y = 2 1 2 1 3 2 2 − x − x + x − x 2 3 2 1 2 1 2 3 − − = x − + x + x dx dy 2 3 2 1 2 3 − = − + + x x x dx dy Sum and Difference Rules (integrate one by one) Example : Indefinite integral : definite integral : ∫ ( 2 − 3 + √ − 2 √ ) = ∫ ( 2 − 3 + 1 2 − 2 − 1 2) = 3 3 − 3 2 2 + 3 2 ( 3 2 ) − 2 1 2 ( 1 2 ) + = 1 3 3 − 3 2 2 + 2 3 3 2 −2 ( 2 1 ) 1 2 + = 1 3 3 − 3 2 2 + 2 3 3 2 −4 √ + ∫ ( 2 − 3 + √ − 2 √ ) 1 0 = ∫ ( 2 − 3 + 1 2 − 2 − 1 2) 1 0 = [ 3 3 − 3 2 2 + 3 2 ( 3 2 ) − 2 1 2 ( 1 2 ) ] 0 1 = [ 1 3 3 − 3 2 2 + 2 3 3 2 − 2 ( 2 1 ) 1 2] 0 1 = [ 1 3 3 − 3 2 2 + 2 3 3 2 − 4 √] 0 1 = ( 1 3 − 3 2 + 2 3 − 4 )− 0 = − 9 2
MAT183/ MAT421 : Calculus I 180 Differentiation Integration v) Exponential Rule a) ( ) x x e dx d e = b) ( ) x x e dx d e 2 2 = 2 c) ( ) x x e dx d e 3 3 = 3 In general, ( ) ax ax ae dx d e = Exponential Rule a) e dx e C x x = + b) C e e dx x x = + 2 2 2 c) C e e dx x x = + 3 3 3 In general, C a e e dx ax ax = + vi) Trigonometric Rule • ( ) cos x dx d sin x = • ( ) sin x dx d cos x = − • ( ) sec x dx d tan x 2 = • ( ) sec x tan x dx d sec x = • ( ) cosecx cot x dx d cosecx = − • ( ) cosec x dx d cot x 2 = − Trigonometric Rule Note : never use product, quotient, generalized and chain rule • ∫ = + → ∫ = 1 + • ∫ = − + → ∫ = − 1 + • ∫ 2 = + → ∫ 2 = 1 + • ∫ = + → ∫ = 1 + • ∫ = − + → ∫ = − 1 + • ∫ 2 = − + → ∫ 2 = − 1 +
Chapter 4 : Integration 181 Examples for multiplication function (that possible to expand) a) ( x) dx x − + 2 1 2 3 2 1 7 = ( x x ) dx x − + + 2 1 2 9 12 4 1 7 (expand bracket square) = x dx x x x − + + − + 2 1 2 12 4 9 63 84 28 (expand) = dx x x x − + + 2 1 2 1 51 80 28 9 (simplify) = 2 1 2 3 9ln 3 28 2 80 51 − + + x x x x (integrate with respect to x) = − − + + − + + 9ln1 3 28 9ln 2 51 40 3 224 102 160 = 3 11 9ln 2 − = 2.5717 # b) (t )(t t ) dt − + − + 3 1 2 2 2 4 = (t t t t t ) dt − − + + − + 3 1 3 2 2 2 4 2 4 8 (expand) = (t ) dt − + 3 1 3 8 (simplify) = 3 1 4 8 4 − + t t (integrate with respect to t) = − − + 8 4 1 24 4 81 = 52 # Refer page 179 (chapter 4), dx ln x C x x dx = = + −1 1
MAT183/ MAT421 : Calculus I 182 c) t (t ) dt − 3 2 1 1 = t (t t )dt − 2 +1 3 2 1 (expand bracket square) = t t t dt − + 3 1 3 4 3 7 2 (expand and simplify using the properties of indices) = ( ) ( ) ( ) C t t t − • + + 3 4 3 7 3 10 3 4 3 7 3 10 2 (integrate with respect to t) = t − • t + t 3 + C 4 3 7 3 10 4 3 7 3 2 10 3 = t − t + t 3 + C 4 3 7 3 10 4 3 7 6 10 3 # Examples for division function with single denominator (that possible to separate) a) ( ) dx x x − 3 2 5 = dx x x x − + 3 10 25 2 1 (expand bracket square) = dx x x x x x − + 3 3 3 10 25 2 1 (separate) = x x x dx − + 1−3 −3 −3 10 25 2 1 (simplify using the properties of indices) = x x x dx − + −2 − −3 10 25 2 5 = ( ) C x x x + − + • − − • − − − − 2 10 25 1 2 2 3 1 2 3 (integrate with respect to x) = C x x x − − • − • − + − 2 2 25 3 2 10 1 2 3 = C x x x − + − + 2 2 25 3 1 20 2 3 #
Chapter 4 : Integration 183 b) ( ) dy y y + 2 2 2 1 = dy y y y + + 2 4 2 2 1 (expand bracket square) = dy y y y y y + + 2 2 2 2 4 2 1 (separate) = (y y ) dy − + + 2 2 2 (simplify using the properties of indices) = C y y y + − + + − 1 2 3 3 1 (integrate with respect to y) = C y y + y − + 1 2 3 1 3 # c) ( ) dx e e x x + 3 2 3 2 = dx e e e x x x + + 3 3 6 2 2 2 (expand bracket square) = dx e e e e e x x x x x + + 3 6 3 3 3 2 2 2 (separate) = ( e e ) dx x x + + −3 3 2 2 2 (simplify using the properties of indices) = C e x e x x + + + − • − 3 2 2 3 2 3 3 (integrate with respect to x) = e x e C x x − + + + −3 3 3 1 2 2 3 2 # ∫ = +
MAT183/ MAT421 : Calculus I 184 Integration with u-substitution ☺ Power Rule with u-substitution, for n ≠ −1 (u in bracket) From C n x ax dx a x dx a n n n + + = = + 1 1 ( ) ( ) C n u a u du a u du a n n n + + = = + 1 1 Example 1 : evaluate i) ( x) dx 5 2 − ii) ( x) dx − 2 1 5 2 Solution : i) ( x) dx 5 2 − = u • −du 5 = u du 5 − = C u − + 6 6 = − ( − x) + C 6 2 6 1 # ii) ( x) dx − 2 1 5 2 = ( x) dx x x = = − 2 1 5 2 = u du u u • − = = 0 1 5 = u du u u = = − 0 1 5 = 0 1 6 6 − u = − − 6 1 6 0 = 6 1 # Let u = 2 − x = −1 dx = −du Let u = 2 − x = −1 dx = −du from u = 2 − x, when x = 2, u = 0 when x = 1, u = 1 For definite integral with u-substitution
Chapter 4 : Integration 185 Example 2 : evaluate i) + dx x x 3 1 2 ii) + 1 0 2 3 1 dx x x Solution : i) + dx x x 3 1 2 = ( ) − x x + dx 2 1 3 1 2 = • • − x du x u 6 2 1 = • − 6 2 1 du u = − u du 2 1 6 1 = ( ) c u + 2 1 2 1 6 1 = u + c • 2 1 1 2 6 1 = u + c 2 1 3 1 = 3x +1 + c 3 1 2 # ii) + 1 0 2 3 1 dx x x = ( ) − + 1 0 2 2 1 x 3x 1 dx = = = − • • 1 0 6 2 1 x x x du x u = = = − • 4 1 6 2 1 u u du u = = = − 4 1 2 1 6 1 u u u du = ( ) 4 1 2 1 2 1 6 1 u = 4 1 2 1 1 2 6 1 • u = 4 1 3 1 u = 4 1 3 1 − = 2 1 3 1 − = 3 1 # Let u = 3x 2 +1 = 6x dx = from u = 3x 2 +1, when x = 1, u = 4 when x = 0, u = 1 Let u = 3x 2 +1 = 6x dx = For definite integral with u-substitution
MAT183/ MAT421 : Calculus I 186 Example 3 : evaluate i) x x +1 dx ii) − + 0 1 x x 1 dx Solution : i) x x +1 dx = ( ) x x + 2 dx 1 1 = ( ) u − •u du 2 1 1 = u − u du 2 1 2 3 = ( ) ( ) c u u − + 2 3 2 5 2 3 2 5 = u − u + c 2 3 2 5 3 2 5 2 = (x + ) − (x + ) 2 + c 3 2 5 1 3 2 1 5 2 # ii) − + 0 1 x x 1 dx = ( ) = =− + 0 1 2 1 1 x x x x dx = ( ) = = − • 1 0 2 1 1 u u u u du = − 1 0 2 1 2 3 u u du = ( ) ( ) 1 0 2 3 2 5 2 3 2 5 − u u = 1 0 2 3 2 5 3 2 5 2 u − u = 0 3 2 5 2 − − = 15 4 − # Let u = x +1 = 1 dx = dx from u = x +1 x = u − 1 when x = 0, u = 1 when x = −1, u = 0 Let u = x +1 = 1 dx = dx For definite integral with u-substitution
Chapter 4 : Integration 187 ☺ Logarithmic Rule with u-substitution, for n = −1 (u = denominator) From dx ln x C x = + 1 du ln u C u = + 1 Example 4 Evaluate the following integrals. a) dx x x − 2 3 4 = x du u x 2 4 − • = du u − 1 2 = − 2lnu + C = − − x + C 2 2ln 3 # b) x dx cot = dx x x sin cos = x du u x cos cos • = du u 1 = lnu + C = ln sin x + C # .c) dx x x tan sec 2 = x du u x 2 2 sec sec • = du u 1 = lnu + C = ln tan x + C #
MAT183/ MAT421 : Calculus I 188 d) dx x x 1+ 2cos sin , u = 1 + 2cosx dx x x 1+ 2cos sin = x du u x 2sin sin − • = du u − 1 2 1 = − lnu + C 2 1 = − ln1+ 2cosx + C 2 1 # e) dt e e t t + ln 2 0 2 2 2 , u = e 2t + 2 dt e e t t + ln 2 0 2 2 2 = t t e du u e 2 ln 2 0 2 2 • = du u 6 3 1 2 1 = 6 3 ln 2 1 u = ln6 ln3 2 1 − = 3 6 ln 2 1 = ln2 2 1 = 2 1 ln2 = ln 2 # When , = 1 + 2 = 3 When , = 4 + 2 = 6
Chapter 4 : Integration 189 ☺ Exponential Rule with u-substitution (u = exponent/power) From e dx e C x x = + e du e C u u = + Example 5 Evaluate the integrals. a) dx e e x x − − 4 − = dx e e e x x x − − − − 4 (single denominator : so, we separate the fraction) = ( e )dx x 4 − 1 = e x C x 4 − + # b) x e dx x 4 3 = 3 3 4x du x e u • • = e du u 4 1 = e C u + 4 1 = e C x + 4 4 1 # c) dx e e e x x x + 2 2 = dx e e e e x x x x + 2 2 (single denominator : so, we separate the fraction) = ( e )dx x x + − 2 1 2 = ( e )dx x 2 + 1 = e x C x 2 + + #
MAT183/ MAT421 : Calculus I 190 ☺ Trigonometric Rule with u-substitution (u = angle, if the trigonometric function exist in table of integral) Properties of the Definite Integral From sin x dx = − cos x + c sinu du = − cosu + c From cos x dx = sin x + c cos u du = sinu + c From sec x dx = tan x + c 2 sec u du = tanu + c 2 From cos ec x dx = − cot x + c 2 cos ec u du = − cot u + c 2 From sec x tan x dx = sec x + c sec u tanu du = sec u + c From cos ec x cot x dx = − cos ec x + c cos ec u cot u du = − cos ec u + c Extra Formula Derivative Integration ( ) ax ax e ae dx d = c a e e dx ax ax = + ( ) a ax dx d ax cos sin = c a cos ax sinax dx + − = ( ) a ax dx d ax sin cos = − c a sinax cos ax dx = + ( ) a ax dx d ax 2 sec tan = c a tan ax sec ax dx = + 2 ( ) a ec ax dx d ax 2 cos cot = − c a cot ax cos ec ax dx + − = 2 ( ) a ax ax dx d ax sec tan sec = c a sec ax sec ax tan ax dx = + ( ) a ecx ax dx d ecax cos cot cos = − c a cos ec ax cos ec ax cot ax dx + − = Important : this formula only suitable for the angle with x the power of positive one
Chapter 4 : Integration 191 Properties of the Definite Integral Sometimes we add or subtract definite integrals, multiply their integrand by constant and compare them with other definite integrals. This can be done by the following rules. Properties of the Definite Integral Multiply integrand by constant cf x dx b a ( ) = c f x dx b a ( ) Definite integral at a point ( ) = 0 f x dx a a Opposite of the definite integral f x dx b a ( ) = − f x dx a b ( ) Subdivision rule f x dx b a ( ) = f x dx c a ( ) + f x dx b c ( ) where a c b Example Given ( ) 3 8 2 = f x dx . Find, a. f x dx 8 2 3 ( ) b. 2 8 f(x)dx c. f x dx − 2 8 ( ) 3 d. f x dx 3 2 ( ) + f x dx 8 3 ( ) e. If f x kdx + 8 2 ( ) = 20, find k.
MAT183/ MAT421 : Calculus I 192 Solution a. f x dx 8 2 3 ( ) = 3 8 2 f(x) dx = 3(3) = 9 b. 2 8 f(x)dx = ( ) 3 8 2 − = − f x dx c. f x dx − 2 8 ( ) 3 = 2 8 f(x)dx − 2 8 3dx = − 8 2 f(x)dx − 2 8 3dx = −3 − 2 8 3x = −3 − (6 − 24)= 15 d. f x dx 3 2 ( ) + f x dx 8 3 ( ) = 8 2 f(x)dx = 3 e. f x kdx + 8 2 ( ) = 20 f x dx 8 2 ( ) + kdx 8 2 = 20 3 + 8 2 k x = 20 3 + 8k − 2k = 20 6k = 17 k = 6 17 #
Chapter 4 : Integration 193 Examples (from previous semester papers) Example 1/ APR 2011/ MAT183/ Q4 a) Evaluate (x 2)(x 5x) dx 3 2 1 − + (3 marks) b) Solve the following integrals using suitable substitutions. i) + dx x x 1 3 2 (3 marks) ii) x( x ) dx + 4 sin cos 1 (3 marks) Solution : a) (x 2)(x 5x) dx 3 2 1 − + = (x x x x)dx + − − 2 1 4 2 3 5 2 10 = 2 1 5 3 4 2 2 10 4 2 3 5 5 + − − x x x x = 2 1 2 5 3 4 5 3 2 5 5 + − − x x x x = − + − − + − − 5 2 1 3 5 5 1 8 20 3 40 5 32 = 30 139 − # b. i) + dx x x 1 3 2 = • x du u x 2 3 = du u 1 2 3 = lnu + C 2 3 = ln x + 1 + C 2 3 2 # b. ii) x( x ) dx + 4 sin cos 1 = x du x u sin sin 4 − • • = u du − 4 = C u − + 5 5 = − ( x + ) + C 5 cos 1 5 1 #
MAT183/ MAT421 : Calculus I 194 Example 2/ SEP 2011/ MAT183/ Q4 a) Evaluate x (x x 7) dx 3 2 1 0 5 1 + − (4 marks) b) Solve the following integrals using suitable substitutions. i) + dx x x tan2 3 sec 2 2 ii) y(y ) dy − 5 4 (7 marks) Solution : a) x (x x 7) dx 3 2 1 0 5 1 + − = + − 1 0 5 1 5 11 5 16 x x 7x dx (expand) = ( ) ( ) ( ) 1 0 5 6 5 16 5 21 5 6 5 16 5 21 7 + − x x x = 1 0 5 6 5 16 5 21 7 6 5 16 5 21 5 x + x − • x = 0 6 35 16 5 21 5 − + − = 336 1775 − # b. i) + dx x x tan2 3 sec 2 2 = • x du u x 2sec 2 sec 2 2 2 = du u 1 2 1 = lnu + C 2 1 = ln tan2x + 3 + C 2 1 # b. ii) y(y ) dy − 5 4 = y u du • 5 = (u )u du + 5 4 (expand) = (u u )du + 6 5 4 = C u u + + 6 4 7 7 6 = ( ) ( ) C y y + − + − 6 4 4 7 4 7 6 − (y − ) + (y − ) + C 7 6 4 3 2 4 7 1 # y = u + 4
Chapter 4 : Integration 195 Example 3/ MAR 2012/ MAT183/ Q4 c) Evaluate ( ) x x + 3x dx 2 (3 marks) d) Solve the following integrals using suitable substitutions. i) ( ) + dx x x 4 3 5 (5 marks) ii) ( ) tan 2 d (4 marks) Solution : c) ( ) x x + 3x dx 2 = ( ) x x + 3x dx 2 2 1 (expand) = x + x dx 2 3 2 5 3 = ( ) ( ) C x x + + 2 5 2 7 2 5 2 7 3 = x + • x 2 + C 5 2 7 3 5 2 7 2 = x + x 2 + C 5 2 7 5 6 7 2 # d. i) ( ) + dx x x 4 3 5 = du u x 4 5 = ( ) − du u u 4 3 5 = − du u u 4 3 5 = − du u u u 4 4 3 5 = ( ) − − u − u du 3 4 5 3 = C u u + − − − − − 3 3 2 5 2 3 = C u u − + + 2 3 5 2 5 = ( ) ( ) C x x + + + + − 2 3 3 5 2 3 5 # x = u − 3