b) The trend is for the average slope at x = 1 to decrease. The slope at x = 1 is 1.

Chapters 1 to 8 Course Review Chapters 1 to 8 Course Review Question 1 Page 509 a) i) ii) [2(16) 12 + 4][2 3+ 4] 4 1 [2(2.25) 4.5+ 4][2 3+ 4] 1.51 = 21 3 = 7 = 1 0.5 = 2 [2(1.21) 3.3+ 4][2 3+ 4] iii) = 0.12 1.11 0.1 = 1.2 b) The trend is for the average slope at x = 1 to decrease. The slope at x = 1 is 1. Chapters 1 to 8 Course Review Question 2 Page 509 a) Average rate, since the speed is over the period of time that Ali drove. b) Instantaneous rate, since the velocity was measured at a specific moment. c) Average rate, since the temperature dropped over the hours of the night. d) Instantaneous rate, since the leakage was measured at a specific moment. Chapters 1 to 8 Course Review Question 3 Page 509 a) i) 3(a + h) 2 3a 2 h = 3a2 + 6ah + 3h 2 3a 2 h 6ah + 3h2 = h = 6a + 3h 6(2) + 3(0.01) = 12.03 ii) ( a + h) a a + 3a h + 3ah + h a 3a h + 3ah + h = = h h h 2 2 = 3a + 3ah + h 3 3 3 2 2 3 3 2 2 3 3(2) 2 +3(2)(0.01) + (0.01) 2 = 12.0601 b) i) This is an approximation of the value of the slope of the tangent to f (x) = 3x 2 at x = 2. MHR Calculus and Vectors 12 Solutions 1020

ii) This is an approximation of the value of the slope of the tangent to f (x) = x 3 at x = 2. Chapters 1 to 8 Course Review Question 4 Page 509 a) 4.9[(2 + h)2 2 2 ]+ 6h h = 4.9(4h + h2 ) + 6h h = 19.6 4.9h + 6 = 13.6 4.9h The average rate of change is ( 13.6 4.9h) m/s. b) Choose the interval 1.9 t 2.1. c) 4.9(2.1 2 1.9 2 ) + 6(0.2) 0.2 4.9(0.2)(4) + 6(0.2) = 0.2 = 19.6 + 6 = 13.6 The instantaneous rate of change is 13.6 m/s. Alternatively, let h = 0 in the expression in part a). Chapters 1 to 8 Course Review Question 5 Page 509 a) No limit; the sequence does not converge. b) The limit is 5. c) No limit; the sequence does not converge. d) The limit is 0. MHR Calculus and Vectors 12 Solutions 1021

Chapters 1 to 8 Course Review Question 6 Page 509 a) lim x2 (3x 2 " 4x +1) = 5 5x + 40 b) lim x"8 x + 8 = lim 5(x + 8) x"8 x + 8 = 5 c) lim x6 + x " 6 = 0 d) The limit does not exist. The graph of the function has a vertical asymptote at x = 3. Chapters 1 to 8 Course Review Question 7 Page 509 a) b) i) lim x3 f (x) = 17 ii) The limit does not exist. The limits on the left and right sides are unequal. Chapters 1 to 8 Course Review Question 8 Page 509 a) dy dx = lim h0 f (x + h) " f (x) h ( ) " ( 4x 2 " 3) 4(x + h) 2 " 3 = lim h0 h 8xh + h 2 = lim h0 h = lim(8x + h) h0 = 8x MHR Calculus and Vectors 12 Solutions 1022

At x = 2, dy = 16 and y = 13. dx The equation of the tangent is: y 13 = 16( x 2) y = 16x 19 b) f (x) = lim h"0 f (x + h) # f (x) h ( ) # ( x 3 # 2x ) 2 (x + h) 3 # 2(x + h) 2 = lim h"0 h = lim h"0 3x 2 h + 3xh 2 + h 3 # 4xh # 2h 2 h = lim h"0 3x 2 + 3xh + h 2 # 4x # 2h = 3x 2 # 4x At x = 2, f (x) = 4 and f(x) = 0. The equation of the tangent is: y 0 = 4(x 2) y = 4x 8 c) g (x) = lim h"0 = lim h"0 = lim h"0 f (x + h) # f (x) h $ 3 x + h # 3 ' % & x ( ) h $ 3x # 3x # 3h' % & x(x + h) ( ) h #3h = lim h"0 xh(x + h) = lim h"0 #3 x(x + h) = # 3 x 2 At x = 2, g (x) = 3 4 and g(x) = 3 2. The equation of the tangent is: MHR Calculus and Vectors 12 Solutions 1023

y 3 2 = 3 (x 2) 4 y = 3 4 x + 3 d) h (x) = lim h"0 f (x + h) # f (x) h = lim h"0 (2 x + h # 2 x ) h = lim h"0 (2 x + h # 2 x )(2 x + h + 2 x ) h(2 x + h + 2 x ) = lim h"0 = (4(x + h) # 4x) h(2 x + h + 2 x ) = lim h"0 4h h(2 x + h + 2 x ) = lim h"0 4 (2 x + h + 2 x ) = 1 x 1 At x = 2, h (x) = and h(x) = 2 2. 2 The equation of the tangent is: y 2 2 = 1 (x 2) 2 y = 1 2 x + 2 Chapters 1 to 8 Course Review Question 9 Page 509 a) MHR Calculus and Vectors 12 Solutions 1024

b) Chapters 1 to 8 Course Review Question 10 Page 509 a) dy dx = 6x + 4 b) f (x) = "6x "2 +10x "3 c) f (x) = 2x " 1 2 = 2 x d) dy dx = " 1 (3x2 4x) $ # 1 2 x 2 % ' & + ( x 1 )(6x 4) e) dy dx = 3x(2x) (x2 + 4)(3) 9x 2 = 6x2 3x 2 12 9x 2 (x 2)(x + 2) = 3x 2 Chapters 1 to 8 Course Review Question 11 Page 510 Answers may vary. For example: a) Let f(x) = x 2 and g(x) = 2. ( f (x) g(x) )" = (x 2 2) = 4x L.S. R.S. f (x)" g (x) = 2x "0 = 0 Therefore, the statement is false. MHR Calculus and Vectors 12 Solutions 1025

b) Let f(x) = 2x 3 + x 2 and g(x) = x. f (x) $ ' " # g(x) % & = 2x3 + x 2 $ ' # " x & % ( ) ' = 2x 2 + x = 4x +1 L.S. R.S f (x) g (x) = 6x2 + 2x 1 = 6x 2 + 2x Therefore, the statement is false. Chapters 1 to 8 Course Review Question 12 Page 510 a) f (x) = 4x At x = 2, f (x) = 8 and f(x) = 7. Find b. y = mx + b b = 7 16 b = 9 The equation of the tangent is y = 8x 9. b) g (x) = 1 1 2 x" 2 At x = 4, g (x) = 1 4 Find b. y = mx + b b = 7 1 b = 6 and g(x) = 7. The equation of the tangent is y = 1 4 x 6. Chapters 1 to 8 Course Review Question 13 Page 510 a) v(t) = s (t) = 6t 2 "14t + 4 b) v(5) = 150 70 + 4 = 84 MHR Calculus and Vectors 12 Solutions 1026

The velocity is 84 m/s. Chapters 1 to 8 Course Review Question 14 Page 510 The slope is 0 when f (x) = 0. Set f(x) = 0. 3x 2 4x 4 = 0 x = 2 or x = 2 3 Evaluate for these values of x. f (2) = 8 8 8 + 4 = 4 " The points are (2, 4) and 2 3, 5 13 % # $ 27& '. " f 2 % # $ 3& ' = 8 27 8 9 + 8 3 + 4 = 40 27 + 4 = 5 13 27 Chapters 1 to 8 Course Review Question 15 Page 510 a) h(t) = 4.9t 2 + 28t + 2.5 b) v(t) = h (t) = "9.8t + 28 a(t) = v (t) = "9.8 c) h(2) = 4.9(4) + 28(2) + 2.5 = 38.9 v(2) = 19.6 + 28 = 8.4 a(2) = 9.8 Height is 38.9 m, velocity is 8.4 m/s, and acceleration is 9.8 m/s 2. MHR Calculus and Vectors 12 Solutions 1027

d) When t = 2, v = 8.4. Set v = 8.4 and solve for t. 9.8t + 28 = 8.4 t = 36.4 9.8 t = 3.7 After 3.7 s, the shell will have the same velocity and be falling downward. Chapters 1 to 8 Course Review Question 16 Page 510 Chapters 1 to 8 Course Review Question 17 Page 510 a) dy dx = 4(3x x1 )(3+ x 2 ) b) g (x) = 1 1 2 (2x + 2 5)" (2) = 1 2x + 5 c) dy dx = 1 3 (3x 2 1) (3) 2 3 = 2 3x 1 ( ) 3 d) y = 3 1 x 2 + 3x = (x 2 + 3x) 1 3 dy dx = 1 3 (x2 + 3x) 4 3 (2x + 3) (2x + 3) = 3 x 2 4 3 + 3x ( ) MHR Calculus and Vectors 12 Solutions 1028

Chapters 1 to 8 Course Review Question 18 Page 510 f (x) = x 2 (3(x 3 " 3x) 2 )(3x 2 " 3) + 2x(x 3 " 3x) 3 = (x 3 " 3x) 2 (3x 2 )(3x 2 " 3) + 2x(x 3 " 3x) At x = 1, f (x) = 16 and f(x) = 8. Find b. y = mx + b b = 8 16 b = 8 The equation of the tangent is y = 16x 8. Chapters 1 to 8 Course Review Question 19 Page 510 C(2000) = 15+ 0.1 2000 = 19.47 The cost is $19.47. " C (w) = 0.1 1 % # $ 2& 1 ' w( 2 " C (2000) = 0.1 1 % # $ 2& ' (2000)( = 0.0011 The rate of change of the cost is $0.0011/L. 1 2 Chapters 1 to 8 Course Review Question 20 Page 510 a) The demand function is the number of orders that can be sold. D(x) = 425 + 20x, where x is the number of $0.10 decreases and D(x) is the demand. b) Revenue = sales price of one order R(x) = (2.75 0.1x)(425+ 20x) c) R (x) = (2.75" 0.1x)(20) + ("0.1)(425+ 20x) = 55" 2x " 42.5" 2x = 12.5" 4x. MHR Calculus and Vectors 12 Solutions 1029

d) Set R (x) = 0 and solve for x. 0 = 12.5 4x 4x = 12.5 x = 3.125 Find the price when x = 1.125. 2.75 (0.1)(3.125) = 2.44 The price is $2.44. The total revenue will be maximized after 3.125 decreases of $0.10, and at a cost of $2.44 per order. Chapters 1 to 8 Course Review Question 21 Page 510 Let s = gt 2 + v 0 t + s 0, where t is time in seconds and g, v 0, and s 0 and constants. v(t) = 2gt + v 0 a(t) = 2g At t = 0, v 0 = 120 km/h or 100 3 m/s and s 0 = 0. a = 10 km/h/s or 25 9 m/s2 and g = 25 18. Find the value of t when v(t) = 0. 0 = 25 9 t + 100 3 t = 12 After 12 s, the car has stopped. s = 25 100 (144) + 18 3 (12) = 200 The car's stopping distance is 200 m. Chapters 1 to 8 Course Review Question 22 Page 512 a) A polynomial will have the limit ±. Test a large value. f (100) = 1 950 892. Therefore, lim x" (2x 3 # 5x 2 + 9x # 8) = ". MHR Calculus and Vectors 12 Solutions 1030

b) Use the laws of limits. x +1 1+ 1 lim x" x #1 = lim x x" 1# 1 x Divide the numerator and denominator by x. $ lim 1+ 1 ' x" % & x ( ) = $ lim 1+ 1 ' x" % & x ( ) = 1 1 = 1 c) Use the laws of limits. x 2 1" 3 " 3x +1 lim x"# x 2 + 4x + 8 = lim x + 1 x 2 x"# 1+ 4 x + 8 x 2 Divide the numerator and denominator by x 2. $ lim 1" 3 x"# x + 1 ' % & x 2 ( ) = $ lim 1+ 4 x"# x + 8 ' % & x 2 ( ) = 1 1 = 1 Chapters 1 to 8 Course Review Question 23 Page 512 Answers may vary. For example: The graph must have a local minimum at ( 2, 5), a point of inflection at ( 1, 1) and a local maximum at (0, 3). A possible graph is shown below. (x [ 4, 2], y [ 6, 6]) MHR Calculus and Vectors 12 Solutions 1031

Chapters 1 to 8 Course Review Question 24 Page 512 a) f (x) = x 3 + 2x 2 4x +1 f "(x) = 3x 2 + 4x 4 f ""(x) = 6x + 4 Find the critical numbers. 3x 2 + 4x 4 = 0 (3x 2)(x + 2) = 0 x = 2 or x = 2 3 Use the second derivative test to classify the critical points. f ("2) = "8 # 2& f $ % 3' ( = 8 Therefore, ( 2, 9) is a local maximum point and Find the possible points of inflection. 6x + 4 = 0 2 59 #, " $ % 3 27 & is a local minimum. x = 2 3 Have already tested the interval around this value. 2 115 Therefore, $ #, " % is a point of inflection. & 3 27 ' 2 2 Therefore, f is increasing for x < 2 and x > and decreasing for 2 < x <. 3 3 2 2 Also, f is concave down for x < and concave up for x >. 3 3 b) f (x) = 3x 4 2x 3 +15x 2 12x + 2 f "(x) = 12x 3 6x 2 + 30x 12 f ""(x) = 36x 2 12x + 30 Find the critical numbers. 12x 3 6x 2 + 30x 12 = 0 6(2x 3 + x 2 5x + 2) = 0 6(x 1)(2x 2 + 3x 2) = 0 6(x 1)(2x 1)(x + 2) = 0 MHR Calculus and Vectors 12 Solutions 1032

x = 1 or x = 2 or x = 1 2 Use the second derivative test to classify the critical points. f ("2) = "90 # 1& f $ % 2' ( = 15 f (1) = "18 Therefore, ( 2, 54) and (1, 0) are local maximum points and Find the possible points of inflection. 36x 2 12x + 30 = 0 6(6x 2 + 2x 5) = 0 x = x = 2 ± 124 12 1± 31 6 x 1.09 or x 0.76 1 " $, # 0.6875 % & 2 ' is a local minimum. Have already tested the interval around these values. Therefore, ( 1.09, 31.26) and (0.76, 0.33) are points of inflection. Therefore, f is increasing for x < 2 and 0.5 < x < 1 and decreasing for 2 < x < 0.5 and x > 1. Also, f is concave down for x < 1.09 and x > 0.76 and concave up for 1.09 < x < 0.76. 3 c) f (x) = x 2 +1 = 3(x 2 +1) 1 f (x) = "3(x 2 +1) "2 (2x) = "6x(x 2 +1) "2 f (x) = "6(x 2 +1) "2 + "6x( 2)(x 2 +1) "3 (2x) = ("6x 2 " 6 + 24x 2 )(x 2 +1) "3 = 6(3x 2 "1)(x 2 +1) "3 Find the critical numbers. 6x(x 2 +1) 2 = 0 x = 0 Use the second derivative test to classify the critical points. f (0) = "6 MHR Calculus and Vectors 12 Solutions 1033

Therefore, (0, 3) is a local maximum. Find the possible points of inflection. 6(3x 2 1)(x 2 +1) 3 = 0 x = ± 1 3 1 9 " 1 9 " Therefore, $ #, and, 3 4 % $ 3 4 % are points of inflection. & ' & ' Therefore, f is increasing for x < 0 and decreasing for x > 0. 1 1 1 1 Also, f is concave down for < x < and concave up for x < and x >. 3 3 3 3 Chapters 1 to 8 Course Review Question 25 Page 512 a) Follow the six step plan. f (x) = 3x 4 8x 3 + 6x 2 f "(x) = 12x 3 24x 2 +12x f ""(x) = 36x 2 48x +12 Step 1. Since this is a polynomial function, the domain is{x }. Step 2. f (0) = 0 The y-intercept is 0. For x-intercepts, let y = 0. 3x 4 8x 3 + 6x 2 = 0. x 2 (3x 2 8x + 6) = 0 x = 0 The second factor has not real roots. The only x-intercept is 0. Step 3. Find the critical numbers. 12x 3 24x 2 +12x = 0 12x(x 2 2x +1) = 0 12x(x 1)(x 1) = 0 x = 0, x =1 Use the second derivative test to classify the critical points. MHR Calculus and Vectors 12 Solutions 1034

f (0) = 12 f (1) = 0 Check further. f (1.1) = 2.76 f (0.9) = "2.04 Therefore, (0, 0) local minimum point and (1, 1) is a point of inflection. Step 4. Find all possible points of inflection. 36x 2 48x +12 = 0 12(3x 2 4x +1) = 0 12(3x 1)(x 1) = 0 x = 1 or x = 1 3 Test the intervals. Have already tested the intervals for these values. Therefore, (0.33, 0.41) is also a point of inflection. Step 5. From Step 3, f is increasing for 0 < x < 1 and x > 1 and decreasing for x < 0. From Step 4, f is concave down for 1 < x < 1 and concave up for 3 1 x < and x > 1. 3 Step 6. Sketch the graph. b) Follow the six step plan. 3 2 f x = x + x + 8x 3 ( ) ( ) ( ) f " x = x + x + 2 3 2 8 f "" x = 6x + 2 Step 1. Since this is a polynomial function, the domain is {x }. Step 2. f (0) = 3 The y-intercept is 3. For x-intercepts, let y = 0. x 3 + x 2 + 8x 3 = 0. This expression is not factorable. There may be up to three x-intercepts. MHR Calculus and Vectors 12 Solutions 1035

Step 3. Find the critical numbers. 3x 2 + 2x + 8 = 0 (3x 4)(x 2) = 0 x = 2 or x = 4 3 Use the second derivative test to classify the critical points. f (2) = "10 # f " 4 & $ % 3' ( = 10 Therefore, ( 1.33, 9.5) is a local minimum point and (2, 9) is a maximum. Step 4. Find possible points of inflection. 6x + 2 = 0 x = 1 3 Test the intervals. Have already tested the intervals for these values. Therefore (0.33, 0.26) is a point of inflection. 4 4 Step 4. From Step 3, f is increasing for < x < 2 and decreasing for x < and x > 2. 3 3 1 1 From Step 4, f is concave down for x > and concave up for x <. 3 3 Step 5. Sketch the graph. c) Follow the six step plan. x f (x) = x 2 +1 = x(x 2 +1) 1 (x [ 5, 10], Xscl = 2, y [ 6, 6], Yscl = 20) MHR Calculus and Vectors 12 Solutions 1036

f (x) = 1(x 2 +1) "1 + x( 1)(x 2 +1) "2 (2x) = (x 2 +1" 2x 2 )(x 2 +1) "2 = ("x 2 +1)(x 2 +1) "2 f (x) = "2x("x 2 +1) "2 + ("x 2 +1)( 2)(x 2 +1) "3 (2x) = ("x 3 " 2x + 4x 3 " 4x)(x 2 +1) "3 = (3x 3 " 6x)(x 2 +1) "3 Step 1. Since this is a rational function, look for asymptotes. The denominator cannot equal zero. There are no asymptotes. Therefore, the domain is {x }. Step 2. f (0) = 0 The y-intercept is 0. For x-intercepts, let y = 0. x x 2 +1 = 0 x = 0 The y-intercept is 0. Step 3. Find the critical numbers. (x 2 +1)(x 2 +1) 2 = 0 x 2 +1 = 0 x = ±1 Use the second derivative test to classify the critical points. f (1) = " 3 8 f ("1) = 3 8 Therefore, ( 0.375, 0.329) is a local minimum point and (0.375, 0.329) is a maximum. Step 4. Find possible points of inflection. (3x 3 6x)(x 2 +1) 3 = 0 3x 3 6x = 0 3x(x 2 2) = 0 x = 0 or x = 2 or x = 2 Test the intervals. f ("2) = 12 729 f (2) = " 12 729 MHR Calculus and Vectors 12 Solutions 1037

Therefore, (0, 0), ( 1.41, 0.47), and (1.41, 0.47) are points of inflection. Step 5. From Step 3, f is increasing for 0.375 < x < 0.375 and decreasing for x < 0.375 and x > 0.375. From Step 4, f is concave down for x < 2 and 0 < x < 2 and concave up for 2 < x < 0 and x > 2. Step 6. Sketch the graph. (x [ 5, 5], y [ 1, 1], Yscl = 0.25) Chapters 1 to 8 Course Review Question 26 Page 512 a) P(v) = 100v 3 16 v3 P "(v) = 100 9 16 v2 To find the maximum, let P (v) = 0. 100 9 16 v2 = 0 v 2 = 1600 9 v = ± 40 3 Since linear velocity cannot be negative, the maximum occurs when v = 13.3 m/s. b) P 40 $ " # 3 % & = 100 40 $ " # 3 % & ' 3 16 " # 888.9 40 3 The maximum power is about 889 A. $ % & 3 MHR Calculus and Vectors 12 Solutions 1038

Chapters 1 to 8 Course Review Question 27 Page 512 Draw a diagram. Let t be the time beginning when they first observe the second ship. The ship travelling north is (15 12t) km away from where they first observe the second ship. The second ship is 9t km away from this same location. The distances form a right triangle with d representing the distance between the two ships. 9t 15 12t d d(t) = (9t) 2 + (1512t) 2 1 = (225t 2 2 360t + 225) " d (t) = 1 % # $ 2& ' (225t 2 ( 360t + 225) ( 1 2 (450t ( 360) = (225t (180)(225t 2 ( 360t + 225) ( 1 2 For a minimum, the derivative must equal zero. (225t 180)(225t 2 360t + 225) 1 2 = 0 225t 180 = 0 t = 0.8 The ships are closest when t = 0.8 h. d(0.8) = (9(0.8)) 2 + (1512(0.8)) 2 = 9 The closest distance of approach of the two ships is 9 km. Chapters 1 to 8 Course Review Question 28 Page 512 C(x) = 0.000 25x2 + 8x +10 x = 0.000 25x + 8 +10x 1 C '(x) = 0.000 2510x 2 a) C(50) gives the cost per pizza at the 50 pizza per day production level. C(50) = 0.000 25(50)2 + 8(50) +10 50 = 8.2125 MHR Calculus and Vectors 12 Solutions 1039

For total cost: 50C(50) = 50(8.2125) = 410.63 The total cost of production is $410.63. b) For a minimum cost, let C (x) = 0. 0.000 2510x 2 = 0 0.000 25x 2 10 = 0 x 2 = 40 000 x = ±200 The negative answer has no meaning for this situation. The minimum average daily cost per pizza occurs at a production level of 200 pizzas per day. c) C(200) = 0.000 25(200)2 + 8(200) +10 200 = 8.1 The minimum average daily cost per pizza is $8.10. Chapters 1 to 8 Course Review Question 29 Page 511 a) dy dx = 3cos2 x( sin x) = 3(cos 2 x)(sin x) b) dy dx = 3x2 cos (x 3 ) c) f (x) = "5sin (5x " 3) # # x & # 1& & d) f (x) = sin 2 x % " sin $ % 2' ( $ % 2' ( ( $ ' + 2(sin x)(cos x) # cos # x & & % $ % 2' ( ( $ ' # # = 2(sin x)(cos x) cos x & & % $ % 2' ( $ ' ( " # 1 & # # x & & $ % 2 ' ( sin2 x % sin $ % 2' ( ( $ ' e) f (x) = 2(cos 4x 2 )(" sin (4x 2 )(8x)) = "16x(cos 4x 2 )(sin 4x 2 ) MHR Calculus and Vectors 12 Solutions 1040

(cos x " sin x)(" sin x) " cos x(" sin x " cos x) f) g (x) = (cos x " sin x) 2 = sin xcos x + sin2 x + cos x sin x + cos 2 x) (cos x " sin x) 2 1 = (cos x " sin x) 2 Chapters 1 to 8 Course Review Question 30 Page 511 y = 2 + cos 2x at x = 5 6. dy dx = 2sin 2x. Find dy dx when x = 5 6. dy dx = 2 " 3 % $ ' # 2 & = 3 When x = 5 6, dy dx = 3 and y = 5 2. Solve for b. y = mx + b b = 5 2 5 3 6 The equation of the tangent is y = 3x + 5 2 5 3" 6. Chapters 1 to 8 Course Review Question 31 Page 511 y = sec x = (cos x) 1 dy dx = (cos x)2 ( sin x) = sin x cos 2 x = sec x tan x MHR Calculus and Vectors 12 Solutions 1041

y = csc x = (sin x) 1 y = tan x = sin x cos x dy dx = (sin x)2 (cos x) = cos x sin 2 x = csc x cot x dy cos x(cos x) sin x( sin x) = dx cos 2 x = cos2 x + sin 2 x cos 2 x 1 = cos 2 x = sec 2 x Chapters 1 to 8 Course Review Question 32 Page 511 dy dx = 2sin x cos x 1 2 Set dy dx = 0. 2sin x cos x 1 2 = 0 2sin x cos x = 1 2 sin 2x = 1 2 2x = 2k +, for k 6 x = k + 12 and 2x = (2k "1) " 6 x = k " 7 12 for k d 2 y = 2cos2x is positive at x = 2 dx 12, 13 12, 25,..., k + 12 12 x = 5 12, 17 12, 29 12, 41 7,..., k 12 12. and negative at There are local minima at x = 12, 13 12, 25 12,..., k + 12. MHR Calculus and Vectors 12 Solutions 1042

There are local maxima at x = 5 12, 17 12, 29 12, 41 12,...k 7 12. The points of inflection are given by: d 2 y dx = 2cos2x 2 = 0 (2k +1) Points of inflection are 2x = or x = 2 (2k +1) 4 for k. Chapters 1 to 8 Course Review Question 33 Page 511 h (t) = 2 3 cos " 2 # $ 30 t % & ' At the maximum height, h (t) = 0. Solve for t. 0 = 2 3 cos " 2 # $ 30 t % & ' 2 30 t = 2 t = 7.5 h(7.5) = 10sin " 2 % " # $ 30 & ' # $ 30 4 % & ' +12 =10sin 2 +12 = 22 The maximum height is 22 m and it first occurs at 7.5 s. Chapters 1 to 8 Course Review Question 34 Page 511 a) d (t) = 6cos 6t + 24sin 6t d (1) = 6cos 6 + 24sin 6 = "0.945 The rate of change is 0.945 cm/s. MHR Calculus and Vectors 12 Solutions 1043

b) For the maximum and minimum, let d (t) = 0. 6cos6t + 24sin6t = 0 6cos6t = 24sin6t tan6t = 1 4 6t = 0.245 " 0.245, 2 " 0.245,... t = " 0.245 6 = 0.48, 2 " 0.245 6 = 1.01,... since t > 0. At t = 0.48, d (t) = "36sin6t +144cos6t is negative. So there is a maximum displacement at t = 0.48. d(0.48) = sin 6(0.48) 4cos 6(0.48) = 4.123 The maximum displacement is 4.123 cm at 0.48 s. At t = 1.01, d (t) = "36sin6t +144cos6t is positive. So there is a minimum displacement at t = 1.01. d(1.01) = sin6(1.01) 4cos6(1.01) = 4.12 The minimum displacement is 4.123 cm at 1.01 s. Chapters 1 to 8 Course Review Question 35 Page 511 a) b) y = e x increases faster as x increases. Both graphs have the same horizontal asymptote and pass through (0, 1). MHR Calculus and Vectors 12 Solutions 1044

The graphs are f (x) = e x and g (x) = ln 2(2 x ). f (x) increases faster as x increases and f (x) passes through (0, 1) while g (x) passes through (0, ln2). Both graphs have the same horizontal asymptote. c) y = e x increases faster as x increases. y = e x has a horizontal asymptote and passes through (0, 1), while the graph of y = 1nx has a vertical asymptote and passes through (1, 0). Chapters 1 to 8 Course Review Question 36 Page 511 a) ln 5 = 1.61 b) ln c) (ln 2 e = 2 2 e ) = 1 Chapters 1 to 8 Course Review Question 37 Page 511 x a) ln ( e ) = x b) ln x e = x c) e 2 ln x 2 = x Chapters 1 to 8 Course Review Question 38 Page 511 a) dy dx = 2ex b) g (x) = 5(ln 10)(10 x ) c) h(x) = e x sin (e x ) d) f (x) = "xe " x + e " x = e " x (1" x) MHR Calculus and Vectors 12 Solutions 1045

Chapters 1 to 8 Course Review Question 39 Page 512 f (x) = 1 2 ex+1 At x = 0, f (0) = e 2 and f (0) = e 2. The equation of the line perpendicular to f (x) = 1 2 ex+1 has a slope 2 e and passes through 0, e $ " # 2% & : y e 2 = 2 e x y = 2 e x + e 2 Chapters 1 to 8 Course Review Question 40 Page 512 The equation for exponential decay is N = N0e "t where N 0 = 20, is the initial amount, and the rate of decay. Since the half life is 1590: 10 = 20e "(1590) ln(0.5) = "(1590) " = ln(0.5) 1590 When N = 2, 2 = 20e "t ln(0.1) = "t t = ln(0.1)(1590) ln(0.5) = 5282 It will take 5282 years for 90% of the mass to decay. MHR Calculus and Vectors 12 Solutions 1046

Chapters 1 to 8 Course Review Question 41 Page 512 f (x) = x 2 e x + 2xe x Let f (x) = 0. x 2 e x + 2xe x = 0 e x (x 2 + 2x) = 0 x = 0 or x = 2 since e x > 0. Chapters 1 to 8 Course Review Question 42 Page 512 a) P (t) = 200("0.001)e "0.001t = "0.2e "0.001t b) P (200) = "0.2e "0.001(200) = "0.164 The rate of change of power is 0.164 Watts/day. Chapters 1 to 8 Course Review Question 43 Page 512 a) dy dx = 20.96(0.0329) " e 0.0329x e 0.0329x % $ # 2 ' & = 0.344 792(e 0.0329x e 0.0329x ) b) dy dx = 0.344 792(e0.0329(2) e 0.0329(2) ) = 0.0454 MHR Calculus and Vectors 12 Solutions 1047

c) The width can be found by letting y = 0. " 20.96 e0.0329x + e 0.0329x % $ 10.06 # 2 ' & = 0 e 0.0329x + e 0.0329x = 10.06 2 e 0.0329x + e 0.0329x = 20.12 Solve this equation using CAS: x = 91.1622 and x = 91.1622. Therefore, the width of the arch is about 182.3 m. The height can be found by letting dy dx = 0. 0.344792(e 0.0329x e 0.0329x ) = 0 e 0.0329x = e 0.0329x x = 0 Find the value of y when x = 0. " y = 20.96 e0.0329(0) + e 0.0329(0) % $ 10.06 # 2 ' & = 20.96(9.06) = 189.9 The height of the arch is 189.9 m. Chapters 1 to 8 Course Review Question 44 Page 512 MHR Calculus and Vectors 12 Solutions 1048

a) dy dx = e x cos x e x sin x Let dy dx = 0. cos x = sin x tan x = 1 x = 4, 5 4, 9 4,... d 2 y dx = 2 e x cos x e x sin x + e x sin x e x cos x = 2e x cos x d 2 y dx 2 is negative at x = 4 and positive at x = 5 4. There are maxima at x = 2k + 4, k and minima at x = 2k + 5 4, k. b) The maximum and minimum points rapidly get closer and closer to zero as x increases. When x = 4 : y = e " # 1 & 4 $ % 2 ' ( = 1.55 When x = 9 4 : y = e 9" # 1 & 4 $ % 2 ' ( = 0.0006. MHR Calculus and Vectors 12 Solutions 1049

Chapters 1 to 8 Course Review Question 45 Page 512 Answers may vary. For example: a) b) Chapters 1 to 8 Course Review Question 46 Page 512 Since BD AE and BD = AE, ABDE is a parallelogram and opposite sides are equal and parallel. MHR Calculus and Vectors 12 Solutions 1050

a) u b) v c) 1 5 u " d) AD " e) CD " = AE " " = u + v " = CB " + ED " + BD = 1 " "+ 5 u v " " " f) CE = CD + DE = 1 " " " 5 u + v + u = 6 " " 5 u + v ( ) Chapters 1 to 8 Course Review Question 47 Page 512 a) 3u + 5u b) 5 c 7u 6u " ( + d) 8 c = (3+ 5 7 6)u = 5u " ( d ) = 5c " 5d 8c " " + 8d " = 5c 8c 5d + 8d " = 13c + 3d Chapters 1 to 8 Course Review Question 48 Page 512 a) Draw a diagram. Use the Pythagorean theorem and trigonometry. MHR Calculus and Vectors 12 Solutions 1051

" R = 12 + 15 # 19.2 2 2 tan = 15 12 # 15& = tan "1 $ % 12' ( 51.3 o The resultant displacement is about 19.2 km in a direction N51.3ºE. b) Draw a diagram. Use the Pythagorean theorem and trigonometry. " R = 60 + 40 # 72.1 2 2 tan = 40 60 # 40& = tan "1 $ % 60' ( 33.7 o The resultant force is about 72.1 N in a direction 56.3 up from the horizontal. MHR Calculus and Vectors 12 Solutions 1052

Chapters 1 to 8 Course Review Question 49 Page 512 a) Draw a diagram. Use the Pythagorean theorem and trigonometry. " R = 450 + 15 # 450.2 2 2 It is not necessary to calculate angle θ since the ground velocity only involves the projection on the ground of this motion. The resultant ground velocity is about 450.2 km/h in the direction of the wind. Chapters 1 to 8 Course Review Question 50 Page 512 a) Draw a diagram of the situation. b) The angle between the forces in the diagram is 160º. Use the sine law to find the angle the second force makes with the resultant. sin sin 160o = 200 340 200sin 160o sin = 340 # 200sin 160 o & = sin "1 % $ 340 ( ' 11.6 o MHR Calculus and Vectors 12 Solutions 1053

To find the magnitude of the second force, use the sine law again. " F sin 8.4 = 340 o sin 160 o " 340sin 8.4o F = sin 160 o " F # 145.2 The second force has a magnitude of 145.2 N at an angle of 11.6 o from the resultant force. Chapters 1 to 8 Course Review Question 51 Page 512 Draw the vector diagram involving only two cables. They will support half of the scoreboard. 250(9.8) = 2450 The downward force will be 2450 N. It is useful to show the vectors forming an addition triangle since their sum must be the zero vector. " " Label the tension vectors as T1 and T2 as shown. To find T " 1, use the sine law. " T 1 sin 20 = 2450 o sin 140 o " 2450sin 20o T 1 = sin 140 o " # 1303.6 T 1 By symmetry the two tensions are identical. Each of the four tensions is approximately 1303.6 N at 70º to the horizontal. MHR Calculus and Vectors 12 Solutions 1054

Chapters 1 to 8 Course Review Question 52 Page 512 Draw a diagram. Use the Pythagorean theorem. " F v = 58 40 = 42 2 2 The vertical component of the force is 42 N. Chapters 1 to 8 Course Review Question 53 Page 512 The force makes a 40º with the east direction. 120cos 40 o 91.9 120sin 40 o 77.1 The rectangular components along the east-west and north-south lines are 91.9 km/h to the east and 77.1 km/h to the south. Chapters 1 to 8 Course Review Question 54 Page 512 Draw a diagram. " F ramp = 100cos 48 o # 66.9 " F perp. = 100sin 48 o # 74.3 The component parallel to the ramp (normal force by the surface) is 66.9 N. The component perpendicular to the ramp (friction) is 74.3 N. MHR Calculus and Vectors 12 Solutions 1055

Chapters 1 to 8 Course Review Question 55 Page 513 a) 5i + 6 j 4k b) 8 j + 7k Chapters 1 to 8 Course Review Question 56 Page 513 " a) PQ " = OQ " OP = "# 6, 3$ % "# 2, 4$ % = "# 8, 1$ % " PQ = (8) 2 + (1) 2 = 65 " b) PR " = OR " OP = "# 4, 10$ % "# 2, 4$ % = "# 2, 14$ % c) 3 "# 8, 1$ % 2 "# 2, 14$ % = "# 24, 3$ % "# 4, 28$ % = "# 28, 25$ % d) "# 8, 1$ % & "# 2, 14$ % = 8(2) + ( 1)( 14) = 16 +14 = 2 Chapters 1 to 8 Course Review Question 57 Page 513 a) Assume the axes to be the east-west and north-south axes. u = " 30cos 70 o, 30sin 70 o # $ " " 10.3, 28.2# $ b) Assume standard x- and y-axes. v = " 40cos 80 o, 40sin 80 o # $ " " 6.9, 39.4# $ MHR Calculus and Vectors 12 Solutions 1056

Chapters 1 to 8 Course Review Question 58 Page 513 The ship s vector is " # 20cos 56 o, 20sin 56 o $ %. The current s vector is " # 11cos 3 o, +11sin 3 o $ %. The resultant velocity is " # 20cos 56 o 11cos 3 o, 20sin 56 o +11sin 3 o " R = ( 22.2) 2 + ( 16) 2 # 27.4 # "16 & = tan "1 $ % "22.2' ( 35.8 o For the bearing, o o o 270 35.8 = 234.2. The resultant velocity is about 27 km on a bearing of 234.2 Chapters 1 to 8 Course Review Question 59 Page 513 Answers may vary. For example: a) Let u " = " 1, 1, 1# $, v = " 3, 2, 1# $, and w " LS = u + v ( ) + w ( ) + 0, %1, 4 = " 1, 1, 1# $ + " 3, 2, 1# $ = " 4, 3, 2# $ + " 0, %1, 4# $ = " 4, 2, 6# $ " # $ = " 0, %1, 4# $. L.S. = R.S. RS = u " ( + w) + v $ % " #22.2, 16.0$ %. ( ) = " 1, 1, 1# $ + " 3, 2, 1# $ + " 0, %1, 4# $ = " 1, 1, 1# $ + " 3, 1, 5# $ = " 4, 2, 6# $ MHR Calculus and Vectors 12 Solutions 1057

b) Let a = " 1, 1, 1# $, b = " 3, 2, 1# $, and k = %2. L.S. = k a b R.S. = a ( ) = ( 2)(1(3) +1(2) +1(1)) = ( 2)(6) = "12 = &12 L.S. = R.S. ( kb) = "# 1, 1, 1$ % "# &6, & 4, & 2$ % = 1( 6) +1( 4) +1( 2) Chapters 1 to 8 Course Review Question 60 Page 513 a) u v = u b) s t = s v cos " = (12)(21)cos 20 o " 236.8 t cos " = (115)(150)cos 42 o " 12 819.2 Chapters 1 to 8 Course Review Question 61 Page 513 " a) u ( v + w) = #$ 3, " 4% & #$ 6, 1% & + #$ "9, 6% & = #$ 3, " 4% & #$ "3, 7% & = 3( 3) + ( 4)(7) = "37 ( ) b) This is not possible because the result of the expression in brackets is a scalar. You cannot find the dot product of a vector and a scalar. " c) u( v w ) = #$ 3, " 4% & #$ 6, 1% & #$ "9, 6% & d) u ( v)" u ( ) = #$ 3, " 4% & (6( 9) +1(6)) = #$ 3, " 4% & ( 48) = "48#$ 3, " 4% & = #$ "144, 192% & ( + v ) = 3, 4 (#$ % & + #$ 6, 1% & )" #$ 3, 4% & #$ 6, 1% & = #$ 9, 3% & " #$ 3, 5% & = 9( 3) + ( 3)( 5) = 12 ( ) MHR Calculus and Vectors 12 Solutions 1058

Chapters 1 to 8 Course Review Question 62 Page 513 u and v are perpendicular if and only if u v = 0. "# 3, 7$ % & "# 6, k $ % = 0 18 + 7k = 0 k = 18 7 Chapters 1 to 8 Course Review Question 63 Page 513 " " AB = "# 4, 1$ % ; BC " = "# 7, 2$ % ;AC " " AB BC = #$ "4, "1% & #$ 7, " 2% & = "26 ' 0 " " ABAC = #$ "4, "1% & #$ 3, " 3% & = "9 ' 0 " " BCAC = #$ 7, " 2% & #$ 3, " 3% & = 27 ' 0 = "# 3, 3$ % None of the vectors are perpendicular. The triangle is not a right triangle. Chapters 1 to 8 Course Review Question 64 Page 513 cos = u "v u v cos = 76 cos = 120 51 cos " 0.9715 $% 4, 10, # 2& ' " $% 1, 7, #1& ' 4 2 +10 2 + ( 2) 2 1 2 + 7 2 + ( 1) 2 = cos #1 (0.9175) = 13.7 o The angle between the vectors is about 14º. MHR Calculus and Vectors 12 Solutions 1059

Chapters 1 to 8 Course Review Question 65 Page 513 " " a) W = F d = "# 1, 4$ % "# 6, 3$ % = 1(6) + 4(3) = 18 J " " b) W = F d = "# 320, 145$ % "# 32, 15$ % = 320(32) + 145(15) = 12 415 J Chapters 1 to 8 Course Review Question 66 Page 513 proj v u = u cos = 25cos 38 o " 19.7 The projection has magnitude 19.7 and has the same direction as v. Chapters 1 to 8 Course Review Question 67 Page 513 " " W = F s " " = F s cos " " 100 = F (3)cos 10 o " F " F 100 = 3cos10 o = 33.8 Roni applies a force with a magnitude of about 33.8 N. MHR Calculus and Vectors 12 Solutions 1060

Chapters 1 to 8 Course Review Question 68 Page 513 Answers may vary. For example: a) b) c) MHR Calculus and Vectors 12 Solutions 1061

Chapters 1 to 8 Course Review Question 69 Page 513 " " " AB = OB OA " " " OB = AB + OA = "# 6, 3, 2$ % + "# 1, 4, 5$ % = "# 7, 7, 3$ % The terminal point is B (7, 7, 3). Chapters 1 to 8 Course Review Question 70 Page 513 cos = a "b a b cos = #$ 7, 2, 4% & " #$ '6, 3, 0% & 7 2 + 2 2 + 4 2 ( 6) 2 + 3 2 + 0 2 '36 cos = 69 45 ( '36 + = cos '1 ) * 69 45, - " 130.2 o cos = a "c a c 7, 2, 4 cos = #$ % & " #$ 4, 8, 6% & 7 2 + 2 2 + 4 2 4 2 + 8 2 + 6 2 68 cos = 69 116 ( 68 + = cos '1 ) * 69 116, - " 40.5 o cos = b "c b c cos = $% #6, 3, 0& ' " $% 4, 8, 6& ' ( 6) 2 + 3 2 + 0 2 4 2 + 8 2 + 6 2 MHR Calculus and Vectors 12 Solutions 1062

cos = 0 45 116 ( ) = cos "1 0 = 90 o The angle between a and b is 130.2. The angle between a and c is 40.5. The angle between b and c is 90. Chapters 1 to 8 Course Review Question 71 Page 513 " L.S. = u ( v + w) ( ) = "# u 1, u 2, u 3 $ % "# v 1, v 2, v 3 $ % + "# w 1, w 2, w 3 $ % = "# u 1,u 2,u 3 $ % "# v 1 + w 1, v 2 + w 2, v 3 + w 3 $ % ( ) + u 2 ( v 2 + w 2 ) + u 3 ( v 3 + w 3 ) = u 1 v 1 + w 1 = u 1 v 1 + u 1 w 1 + u 2 v 2 + u 2 w 2 + u 3 v 3 + u 3 w 3 R.S. = u v " + u w = "# u 1, u 2, u 3 $ % "# v 1, v 2, v 3 $ % + "# u 1, u 2, u 3 $ % "# w 1, w 2, w 3 $ % = u 1 v 1 + u 2 v 2 + u 3 v 3 + u 1 w 1 + u 2 w 2 + u 3 w 3 = u 1 v 1 + u 1 w 1 + u 2 v 2 + u 2 w 2 + u 3 v 3 + u 3 w 3 Therefore, ( ) " " u v + w = u v + u w. L.S. = R.S. Chapters 1 to 8 Course Review Question 72 Page 513 Answers may vary. For example: Choose vectors that make a zero dot product with each of the vectors. a) [5, 1, 0] and [0, 4, 5] are two possibilities. b) [0, 1, 3] and [6, 5, 0] are two possibilities. Chapters 1 to 8 Course Review Question 73 Page 513 Ignoring wind the ground speed is 180cos 14 o 174.7 km/h. A top-view sketch shows the velocities. The vector for the plane is [ 174.7, 0 ]. MHR Calculus and Vectors 12 Solutions 1063

The wind vector is " 15cos 45 o, 15sin 45 o 185.31, 10.61. The resultant vector is [ ] " R = ( 185.31) 2 + ( 10.61) 2 # 185.6 # 10.61 & = tan "1 $ % 185.31' ( 3.3 o # $ " 10.61, 10.61# $. The ground velocity is about 185.6 km/h on a bearing of 86.7º. Chapters 1 to 8 Course Review Question 74 Page 513 a b = #$ 4, " 3, 5% & #$ 2, 7, 2% & = #$ "3(2) " 7(5), 5(2) " 2(4), 4(7) " 2( 3) % & = #$ "41, 2, 34% & b a = "# 2, 7, 2$ % "# 4, & 3, 5$ % = "# 7(5) & ( 3)(2), 2(4) & 5(2), 2( 3) & 4(7) $ % = "# 41, & 2, & 34$ % For the vectors to be orthogonal, the dot products must be zero. a ( a " b) = 4, # 3, 5 $% & ' $% #41, 2, 34& ' = 4( 41) + ( 3)(2) + 5(34) = #164 # 6 +170 = 0 b ( a " b) = #$ 2, 7, 2% & #$ '41, 2, 34% & = 2( 41) + 7(2) + 2(34) = '82 +14 + 68 You do not have to check b a since this vector is the opposite of a b and has the same direction. Chapters 1 to 8 Course Review Question 75 Page 513 The area of a triangle is given by A = 1 " " 2 PQ QR = 0 " PQ " = OQ " OP = "# 2, 5, 7$ % "# 3, 4, 8$ % = "# 5, 1, 1$ % " QR " = OR " OQ = "# 5, 1, 6$ % "# 2, 5, 7$ % = "# 3, 6, 1$ % MHR Calculus and Vectors 12 Solutions 1064

A = 1 2 = 1 2 " #5, 1, 1$ % & "# 3, 6, 1$ % 1( 1) ( 6)( 1), ( 1)( 3) ( 1)( 5), ( 5)( 6) ( 3) ( 1) = 1 2 "( # 7), 2, 33$ % = 1 2 ( 7)2 + ( 2) 2 + 33 2 16.9 The area of the triangle is approximately 16.9 units 2. Chapters 1 to 8 Course Review Question 76 Page 514 Answers may vary. For example: u = 1, 0, 1, v = 0, 1, 2, and k = 3. L.S. = ku Let [ ] [ ] ( ) v ( ) 0, &1, & 2 = 3"# 1, 0, 1$ % "# $ % = "# 3, 0, 3$ % "# 0, &1, & 2$ % = "# 0( 2) & ( 1)(3), 3(0) & ( 2)(3), 3( 1) & 0(0) $ % R.S. = u = "# 3, 6, & 3$ % ( kv) ( ) = "# 1, 0, 1$ % 3"# 0, &1, & 2$ % = "# 1, 0, 1$ % "# 0, & 3, & 6$ % = "# 0( 6) & ( 3)(1), 1(0) & ( 6)(1), 1( 3) & 0(0) $ % = "# 3, 6, & 3$ % L.S. = R.S. MHR Calculus and Vectors 12 Solutions 1065

Chapters 1 to 8 Course Review Question 77 Page 514 The cross product is orthogonal to both vectors. u v = #$ "1, 6, 5% & #$ 4, 9, 10% & = #$ 6(10) " 9(5), 5(4) "10( 1), "1(9) " 4(6) % & = #$ 15, 30, " 33% & or #$ 5, 10, 11% & Two orthogonal vectors are "# 5, 10, 11$ % and "# 5, 10, 11$ %. Note that any scalar multiple of these vectors will also answer the question. Chapters 1 to 8 Course Review Question 78 Page 514 " " " V = wu " v = #$ 5, 0, 1% & #$ '2, 3, 6% & " #$ 6, 7, ' 4% & = #$ 5, 0, 1% & #$ 3( 4) ' 7(6), 6(6) ' ( 4)( 2), ( 2)(7) ' 6(3) % & = #$ 5, 0, 1% & #$ '54, 28, ' 32% & = '302 = 302 The volume is 302 units 3. Chapters 1 to 8 Course Review Question 79 Page 514 = r " F sin " 26 = (0.35)95sin " 26 sin " = ( 0.35)95 " # 51.4 o The force is applied to the wrench at about a 51.4º angle. Chapters 1 to 8 Course Review Question 80 Page 514 Answers may vary. For example: " AB = " 7, 1, 17# $ A vector equation is " x, y, z# $ = " 5, 0, 10# $ + t " 7, 1, 17# $, t %. MHR Calculus and Vectors 12 Solutions 1066

Possible parametric equations are: x = 5+ 7t y = t z = 10 +17t, t Chapters 1 to 8 Course Review Question 81 Page 514 a) b) Chapters 1 to 8 Course Review Question 82 Page 514 a) The equation is of the form 4x + 8y + C = 0. Substitute the coordinates of P 0 to determine C. 4(3) + 8( 1) + C = 0 C = 4 The scalar equation is 4x + 8y 4 = 0 or x + 2y 1 = 0. b) The equation is of the form 6x 7y + C = 0. Substitute the coordinates of P 0 to determine C. 6( 5) 7(10) + C = 0 C = 40 The scalar equation is 6x 7y + 40 = 0 or 6x + 7y 40 = 0. Chapters 1 to 8 Course Review Question 83 Page 514 a) The y-axis has direction vector [0, 1]. A possible vector equation for the line is " x, y# $ = " 6, % 3# $ + t " 0, 1# $, t &. MHR Calculus and Vectors 12 Solutions 1067

b) The given line has normal vector [2, 7]. A line perpendicular to this line will have direction vector [2, 7] A possible vector equation is " x, y# $ = " 1, 6# $ + t " 2, 7# $, t %. Chapters 1 to 8 Course Review Question 84 Page 514 Answers may vary. For example: Choose arbitrary values for two of the coordinates and solve for the third. If x = 0 and y = 0, then z = 2. If x = 2 and y = 1, then z = 0. Two possible points are (0, 0, 2) and (2, 1, 0). Chapters 1 to 8 Course Review Question 85 Page 514 Parametric equations are: x = 4 + 2s + 6t y = 3+ s + 6t z = 5+ 4s 3t, t " For the " scalar " equation, need to find the normal. n = m1 m2 = "# 2, 1, 4$ % "# 6, 6, & 3$ % = "# 1( 3) & 6(4), 4(6) & ( 3)(2), 2(6) & 6(1) $ % = "# &27, 30, 6$ % or "# 9, &10, & 2$ % The scalar equation is of the form 9x 10y 2z + D = 0. Use the point (4, 3, 5) to determine C. 9(4) 10(3) 2( 5) + D = 0 D = 16 The scalar equation is 9x 10y 2z 16 = 0. Chapters 1 to 8 Course Review Question 86 Page 514 a) To find the x-intercept, let y = 0 and z = 0 and solve for s and t. x = 1+ s + 2t 0 = 8 12s + 4t " 0 = 6 12s 3t # MHR Calculus and Vectors 12 Solutions 1068

Solve and for s and t. 8 = 12s + 4t 6 = 12s 3t " 2 = 7t " t = 2 7 6 = 12s + 6 7 " s = 4 7 Now substitute t = 2 7 and s = 4 7 in equation. x = 1+ 4 7 + 2 " 2 % # $ 7& ' = 1 The x-intercept is 1. To find the y-intercept, let x = 0 and z = 0, and solve for s and t. 0 = 1+ s + 2t y = 8 12s + 4t " 0 = 6 12s 3t # Solve and for s and t. 1 = s + 2t 6 = 12s 3t " 18 = 21t 12+" t = 6 7 6 = 12s + 18 7 " s = 5 7 Now substitute t = 6 7 and s = 5 7 in equation. " y = 8 12 5 % # $ 7& ' + 4 " 6 % # $ 7 & ' = 4 MHR Calculus and Vectors 12 Solutions 1069

The y-intercept is 4. To find the z-intercept, let x = 0 and y = 0, and solve for s and t. 0 = 1+ s + 2t 0 = 8 12s + 4t " z = 6 12s 3t # Solve and for s and t. 1 = s + 2t 8 = 12s + 4t " 6 = 14s 2" s = 3 7 1 = 3 7 + 2t t = 5 7 Now substitute s = 3 7 and t = 5 7 in equation. " z = 6 12 3 % # $ 7& ' 3 " 5 % # $ 7 & ' = 3 The z-intercept is 3. b) To find the x-intercept, let y = 0 and z = 0, and solve for x. 2x 6(0) + 9(0) = 18 x = 9 To find the y-intercept, let x = 0 and z = 0, and solve for y. 2(0) 6y + 9(0) = 18 y = 3 To find the z-intercept, let x = 0 and y = 0, and solve for z. 2(0) 6(0) + 9z = 18 z = 2 The x-intercept is 9, the y-intercept is 3, and the z-intercept is 2. MHR Calculus and Vectors 12 Solutions 1070

Chapters 1 to 8 Course Review Question 87 Page 514 " a) AB = "# 9 5, 10 2, 3 8$ % n = "# 14, 8, 5$ % """ = AB """ BC " BC = "# 2 ( 9), ( 6) 10, 5 3$ % = "# 7, 16, 2$ % = #$ "14, 8, " 5% & #$ 7, "16, 2% & = #$ 8(2) " ( 16)( 5), ( 5)(7) " 2( 14), ( 14)( 16) " 7(8) % & = #$ "64, " 7, 168% & The scalar equation is of the form 64x + 7 y 168z + D = 0. Use the point A(5, 2, 8) to determine C. 64(5) + 7(2) 168(8) + D = 0 D = 1010 The scalar equation is 64x + 7 y 168z + 1010 = 0. b) Planes parallel to both the x- and y-axes have scalar equations of the form z = D. Since the plane passes through P( 4, 5, 6), the equation is z = 6. Chapters 1 to 8 Course Review Question 88 Page 514 Answers may vary. For example; If the line is perpendicular to the plane, its direction vector will be the same as the normal to the plane. n = [ 7, 8, 5] The vector equation of the line is " x, y, z# $ = " 6, 1, % 2# $ + t " 7, % 8, 5# $, t &. MHR Calculus and Vectors 12 Solutions 1071

Chapters 1 to 8 Course Review Question 89 Page 514 The angle between two planes is defined to be the angle between their normal vectors. cos = n 1 " n2 n1 n2 cos = cos = # $ 8, 10, 3% & " #$ 2, ' 4, 6% & #$ 8, 10, 3% & #$ 2, ' 4, 6% & 8(2) +10( 4) + 3(6) 8 2 +10 2 + 3 2 2 2 + ( 4) 2 + (6) 2 '6 cos = 98.4276 ( '6 + = cos '1 ) * 98.4276, - " 93.5 o The angle between the planes is about 93.5º. Chapters 1 to 8 Course Review Question 90 Page 514 Find the intersection points, if possible. Equate the expressions for like coordinates. 2 + 3s = 3 4t 3+ 4s = 4 + 5t 3s + 4t = 1 4s 5t = 7 2 10s = 2 + 3t 10s + 3t = 4 Solve equations and for s and t. 12s +16t = 4 4 12s 15t = 21 3" 31t = 17 43" t = 17 31 MHR Calculus and Vectors 12 Solutions 1072

Substitute t = 17 31 3s + 4 17 $ " # 31% & = 1 3s 68 31 = 1 3s = 99 31 s = 33 31 The lines intersect at into equation to solve for s. 33 17$, " # 31 31% &. Chapters 1 to 8 Course Review Question 91 Page 514 " " Let P 1 (2, 7, 3), P 2 (3, 4, 2), m1 = [ 3, 10, 1 ], and m2 = [ 1, 6, 1] " P 1 P 2 n " = m = "# 1, 11, 5$ % " 1 m = #$ "16, " 2, 28% & 2. d = " # 1, 11, 5$ % & "# 16, 2, 28$ % d = 178 1044 d 5.51 "# 16, 2, 28$ % Chapters 1 to 8 Course Review Question 92 Page 514 The distance between a point A and a plane is given by d = with normal n. Choose " Q(3, 0, 0) on the plane. AQ = " 3, 0, 0# $ % " 2, 5, % 7# $ = " 1, % 5, 7# $ n """ AQ where Q is any point on the plane n MHR Calculus and Vectors 12 Solutions 1073

d = " # 3, 5, 6$ % & "# 1, 5, 7$ % 3 2 + ( 5) 2 + 6 2 = 70 70 8.37 The distance is approximately 8.37 units. Chapters 1 to 8 Course Review Question 93 Page 514 The normal to the plane is: n = " 1, 3, 4# $ % " &5, 4, 7# $. = " 5, & 27, 19# $ n " m = #$ 5, " 27, 19% & #$ 1, 6, 5% & = "62 Therefore, the line and the plane are not parallel and they intersect at one point. Chapters 1 to 8 Course Review Question 94 Page 514 Answers may vary. For example: a) The normals must be parallel, but the equations of at least two planes cannot be multiples of each other. x + y + 2z = 3 2x + 2y + 4z = 17 x + y + 2z = 0 b) The planes must have identical equations (except for a scalar multiple). x + y + 2z = 3 2x + 2y + 4z = 6 x y 2z = 3 c) The normals must be coplanar (e.g., n1 = 5n2 + 3n3 ) and the constant terms must satisfy the same relationship as the normals (in this case, D1 = 5D2 + 3D3 ). 4x 6y + z + 2 = 0 3x 5y + 2z + 4 = 0 x y + z 2 = 0 MHR Calculus and Vectors 12 Solutions 1074

d) The normals must be non-coplanar. (i.e., n1 n2 " n3 # 0 ) Choose n1 = "# 1, 3, 2$ %, n2 = "# 1, 0, 0$ %, n3 = "# 0, 1, 0$ %. n2 n3 = "# 0, 0, 1$ % n1 n2 " n3 = $% 1, 3, # 2& ' $% 0, 0, 1& ' = #2 ( 0 x + 3y 2 = 8 x = 14 and y = 12 Chapters 1 to 8 Course Review Question 95 Page 514 a) First examine the normals. None are scalar multiples of each other. Check if the normals are coplanar. n1 n2 " n3 = 2, # 4, 5 3, 2, # 1 " 4, 3, # 3 [ ] [ ] [ ] [ 2, 4, 5] [ 3, 5, 3] = # # = # 11 The normals are not coplanar; there is an intersection point. Eliminate one of the variables from two pairs of equations. Eliminate z. 2x 4y + 5z = 4 15x +10y 5z = 5 5" 17x + 6y = 1 # +5" 9x + 6y 3z = 3 3 4x + 3y 3z = 1 " 5x + 3y = 4 # 3" Solve equations and for x and y. 17x + 6y = 1 10x + 6y = 8 2" 7x = 7 2" x = 1 Substitute x = 1 into equation. 5( 1) + 3y = 4 y = 3 MHR Calculus and Vectors 12 Solutions 1075

Substitute x = 1 and y = 3 into equation. 3(1) + 2(3) z = 1 z = 2 The three planes intersect at the point (1, 3, 2). b) First examine the normals. None are scalar multiples of each other. Check if the normals are coplanar. n1 n2 " n3 = 5, 4, 2 3, 1, # 3 " 7, 7, 7 [ ] [ ] [ ] [ 5, 4, 2] [ 28, 42, 14] = # = 0 Since the triple scalar product is zero, the normals are coplanar and the planes intersect either in a line or not at all. Eliminate one of the variables from two pairs of equations. Eliminate y. 5x + 4y + 2z = 7 12x + 4y 12z = 8 4" 7x + +14z = 1 # 4" 21x + 7 y 21z = 14 7" 7x + 7 y + 7z = 12 $ 14x 28z = 2 7"$ 7x +14z = 1 % Equations and are identical. Let z = t. 7x +14t = 1 x = 1 7 + 2t 1 Substitute x = + 2 t and z = t into equation. 7 3 1 " # 7 + 2t $ % & + y ' 3t = 2 y = 11 7 ' 3t The planes intersect in the line " x, y, z# $ = 1 7, 11 7, 0 # % & + t " 2, ' 3, 1# $, t (. " $ MHR Calculus and Vectors 12 Solutions 1076