b) The trend is for the average slope at x = 1 to decrease. The slope at x = 1 is 1.
|
|
- Kimberly Marshall
- 6 years ago
- Views:
Transcription
1 Chapters 1 to 8 Course Review Chapters 1 to 8 Course Review Question 1 Page 509 a) i) ii) [2(16) ][2 3+ 4] 4 1 [2(2.25) ][2 3+ 4] 1.51 = 21 3 = 7 = = 2 [2(1.21) ][2 3+ 4] iii) = = 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) = ii) ( a + h) a a + 3a h + 3ah + h a 3a h + 3ah + h = = h h h 2 2 = 3a + 3ah + h (2) 2 +3(2)(0.01) + (0.01) 2 = 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
2 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 = h + 6 = h The average rate of change is ( h) m/s. b) Choose the interval 1.9 t 2.1. c) 4.9( ) + 6(0.2) (0.2)(4) + 6(0.2) = 0.2 = = 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
3 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
4 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
5 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) = 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
6 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 x 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
7 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) = 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) = = 84 MHR Calculus and Vectors 12 Solutions 1026
8 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) = = 4 " The points are (2, 4) and 2 3, 5 13 % # $ 27& '. " f 2 % # $ 3& ' = = = Chapters 1 to 8 Course Review Question 15 Page 510 a) h(t) = 4.9t t b) v(t) = h (t) = "9.8t + 28 a(t) = v (t) = "9.8 c) h(2) = 4.9(4) + 28(2) = 38.9 v(2) = = 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
9 d) When t = 2, v = 8.4. Set v = 8.4 and solve for t. 9.8t + 28 = 8.4 t = 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) = (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 x ( ) MHR Calculus and Vectors 12 Solutions 1028
10 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) = = The cost is $ " C (w) = % # $ 2& 1 ' w( 2 " C (2000) = % # $ 2& ' (2000)( = 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) = x, where x is the number of $0.10 decreases and D(x) is the demand. b) Revenue = sales price of one order R(x) = ( x)( x) c) R (x) = (2.75" 0.1x)(20) + ("0.1)( x) = 55" 2x " 42.5" 2x = 12.5" 4x. MHR Calculus and Vectors 12 Solutions 1029
11 d) Set R (x) = 0 and solve for x. 0 = x 4x = 12.5 x = Find the price when x = (0.1)(3.125) = 2.44 The price is $2.44. The total revenue will be maximized after 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 m/s and s 0 = 0. a = 10 km/h/s or 25 9 m/s2 and g = Find the value of t when v(t) = 0. 0 = 25 9 t t = 12 After 12 s, the car has stopped. s = (144) (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) = Therefore, lim x" (2x 3 # 5x 2 + 9x # 8) = ". MHR Calculus and Vectors 12 Solutions 1030
12 b) Use the laws of limits. x 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
13 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 = #, " $ % 3 27 & is a local minimum. x = 2 3 Have already tested the interval around this value Therefore, $ #, " % is a point of inflection. & 3 27 ' 2 2 Therefore, f is increasing for x < 2 and x > and decreasing for 2 < x < 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 x 12 f ""(x) = 36x 2 12x + 30 Find the critical numbers. 12x 3 6x x 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
14 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 ± ± 31 6 x 1.09 or x " $, # % & 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 < 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 " x 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
15 Therefore, (0, 3) is a local maximum. Find the possible points of inflection. 6(3x 2 1)(x 2 +1) 3 = 0 x = ± " 1 9 " Therefore, $ #, and, 3 4 % $ 3 4 % are points of inflection. & ' & ' Therefore, f is increasing for x < 0 and decreasing for x > Also, f is concave down for < x < and concave up for x < and x > 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
16 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 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
17 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 > 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
18 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) = f (2) = " MHR Calculus and Vectors 12 Solutions 1037
19 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 < x < and decreasing for x < and x > 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) = v2 To find the maximum, let P (v) = v2 = 0 v 2 = v = ± 40 3 Since linear velocity cannot be negative, the maximum occurs when v = 13.3 m/s. b) P 40 $ " # 3 % & = $ " # 3 % & ' 3 16 " # The maximum power is about 889 A. $ % & 3 MHR Calculus and Vectors 12 Solutions 1038
20 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 t + 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) = x2 + 8x +10 x = x x 1 C '(x) = x 2 a) C(50) gives the cost per pizza at the 50 pizza per day production level. C(50) = (50)2 + 8(50) = MHR Calculus and Vectors 12 Solutions 1039
21 For total cost: 50C(50) = 50(8.2125) = The total cost of production is $ b) For a minimum cost, let C (x) = x 2 = x 2 10 = 0 x 2 = 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) = (200)2 + 8(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
22 (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 = The equation of the tangent is y = 3x " 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
23 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 x = 5 12, 17 12, 29 12, 41 7,..., k and negative at There are local minima at x = 12, 13 12, 25 12,..., k MHR Calculus and Vectors 12 Solutions 1042
24 There are local maxima at x = 5 12, 17 12, 29 12, 41 12,...k 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 = 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 sin 6 = "0.945 The rate of change is cm/s. MHR Calculus and Vectors 12 Solutions 1043
25 b) For the maximum and minimum, let d (t) = 0. 6cos6t + 24sin6t = 0 6cos6t = 24sin6t tan6t = 1 4 6t = " 0.245, 2 " 0.245,... t = " = 0.48, 2 " = 1.01,... since t > 0. At t = 0.48, d (t) = "36sin6t +144cos6t is negative. So there is a maximum displacement at t = d(0.48) = sin 6(0.48) 4cos 6(0.48) = The maximum displacement is cm at 0.48 s. At t = 1.01, d (t) = "36sin6t +144cos6t is positive. So there is a minimum displacement at t = d(1.01) = sin6(1.01) 4cos6(1.01) = 4.12 The minimum displacement is 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
26 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
27 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
28 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 Watts/day. Chapters 1 to 8 Course Review Question 43 Page 512 a) dy dx = 20.96(0.0329) " e x e x % $ # 2 ' & = (e x e x ) b) dy dx = (e0.0329(2) e (2) ) = MHR Calculus and Vectors 12 Solutions 1047
29 c) The width can be found by letting y = 0. " e0.0329x + e x % $ # 2 ' & = 0 e x + e x = e x + e x = Solve this equation using CAS: x = and x = Therefore, the width of the arch is about m. The height can be found by letting dy dx = (e x e x ) = 0 e x = e x x = 0 Find the value of y when x = 0. " y = e0.0329(0) + e (0) % $ # 2 ' & = 20.96(9.06) = The height of the arch is m. Chapters 1 to 8 Course Review Question 44 Page 512 MHR Calculus and Vectors 12 Solutions 1048
30 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 ' ( = MHR Calculus and Vectors 12 Solutions 1049
31 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
32 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 = ( )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
33 " R = # tan = # 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 = # tan = # 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
34 Chapters 1 to 8 Course Review Question 49 Page 512 a) Draw a diagram. Use the Pythagorean theorem and trigonometry. " R = # 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 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 = sin 160o sin = 340 # 200sin 160 o & = sin "1 % $ 340 ( ' 11.6 o MHR Calculus and Vectors 12 Solutions 1053
35 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 # The second force has a magnitude of 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 " # T 1 By symmetry the two tensions are identical. Each of the four tensions is approximately N at 70º to the horizontal. MHR Calculus and Vectors 12 Solutions 1054
36 Chapters 1 to 8 Course Review Question 52 Page 512 Draw a diagram. Use the Pythagorean theorem. " F v = = 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 sin 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
37 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) = = 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
38 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 = The resultant velocity is about 27 km on a bearing of 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
39 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 " t cos " = (115)(150)cos 42 o " 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
40 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 $ % = k = 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 = cos " $% 4, 10, # 2& ' " $% 1, 7, #1& ' ( 2) ( 1) 2 = cos #1 (0.9175) = 13.7 o The angle between the vectors is about 14º. MHR Calculus and Vectors 12 Solutions 1059
41 Chapters 1 to 8 Course Review Question 65 Page 513 " " a) W = F d = "# 1, 4$ % "# 6, 3$ % = 1(6) + 4(3) = 18 J " " b) W = F d = "# 320, 145$ % "# 32, 15$ % = 320(32) + 145(15) = 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
42 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
43 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% & ( 6) '36 cos = ( '36 + = cos '1 ) * 69 45, - " o cos = a "c a c 7, 2, 4 cos = #$ % & " #$ 4, 8, 6% & cos = ( 68 + = cos '1 ) * , - " 40.5 o cos = b "c b c cos = $% #6, 3, 0& ' " $% 4, 8, 6& ' ( 6) MHR Calculus and Vectors 12 Solutions 1062
44 cos = ( ) = cos "1 0 = 90 o The angle between a and b is The angle between a and c is 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 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
45 The wind vector is " 15cos 45 o, 15sin 45 o , The resultant vector is [ ] " R = ( ) 2 + ( 10.61) 2 # # & = tan "1 $ % ' ( 3.3 o # $ " 10.61, 10.61# $. The ground velocity is about km/h on a bearing of 86.7º. Chapters 1 to 8 Course Review Question 74 Page 513 a b = #$ 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 $% & ' $% #41, 2, 34& ' = 4( 41) + ( 3)(2) + 5(34) = #164 # = 0 b ( a " b) = #$ 2, 7, 2% & #$ '41, 2, 34% & = 2( 41) + 7(2) + 2(34) = ' 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
46 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) 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
47 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
48 Possible parametric equations are: x = 5+ 7t y = t z = t, 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
49 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
50 Solve and for s and t. 8 = 12s + 4t 6 = 12s 3t " 2 = 7t " t = = 12s " s = 4 7 Now substitute t = 2 7 and s = 4 7 in equation. x = " 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 = = 12s " s = 5 7 Now substitute t = 6 7 and s = 5 7 in equation. " y = % # $ 7& ' + 4 " 6 % # $ 7 & ' = 4 MHR Calculus and Vectors 12 Solutions 1069
51 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 = = t t = 5 7 Now substitute s = 3 7 and t = 5 7 in equation. " z = % # $ 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
52 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 = 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
53 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) ( 4) 2 + (6) 2 '6 cos = ( '6 + = cos '1 ) * , - " 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 s = 3 4t 3+ 4s = 4 + 5t 3s + 4t = 1 4s 5t = s = 2 + 3t 10s + 3t = 4 Solve equations and for s and t. 12s +16t = s 15t = 21 3" 31t = 17 43" t = MHR Calculus and Vectors 12 Solutions 1072
54 Substitute t = s $ " # 31% & = 1 3s = 1 3s = s = The lines intersect at into equation to solve for s $, " # 31 31% &. Chapters 1 to 8 Course Review Question 91 Page 514 " " Let P 1 (2, 7, 3), P 2 (3, 4, 2), m1 = [ 3, 10, 1 ], and m2 = [ 1, 6, 1] " P 1 P 2 n " = m = "# 1, 11, 5$ % " 1 m = #$ "16, " 2, 28% & 2. d = " # 1, 11, 5$ % & "# 16, 2, 28$ % d = 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
55 d = " # 3, 5, 6$ % & "# 1, 5, 7$ % ( 5) = 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
56 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
57 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 = t 1 Substitute x = + 2 t and z = t into equation " # 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
Calculus I Sample Exam #01
Calculus I Sample Exam #01 1. Sketch the graph of the function and define the domain and range. 1 a) f( x) 3 b) g( x) x 1 x c) hx ( ) x x 1 5x6 d) jx ( ) x x x 3 6 . Evaluate the following. a) 5 sin 6
More informationFinal practice, Math 31A - Lec 1, Fall 2013 Name and student ID: Question Points Score Total: 90
Final practice, Math 31A - Lec 1, Fall 13 Name and student ID: Question Points Score 1 1 1 3 1 4 1 5 1 6 1 7 1 8 1 9 1 Total: 9 1. a) 4 points) Find all points x at which the function fx) x 4x + 3 + x
More informationFormulas that must be memorized:
Formulas that must be memorized: Position, Velocity, Acceleration Speed is increasing when v(t) and a(t) have the same signs. Speed is decreasing when v(t) and a(t) have different signs. Section I: Limits
More informationAB CALCULUS SEMESTER A REVIEW Show all work on separate paper. (b) lim. lim. (f) x a. for each of the following functions: (b) y = 3x 4 x + 2
AB CALCULUS Page 1 of 6 NAME DATE 1. Evaluate each it: AB CALCULUS Show all work on separate paper. x 3 x 9 x 5x + 6 x 0 5x 3sin x x 7 x 3 x 3 5x (d) 5x 3 x +1 x x 4 (e) x x 9 3x 4 6x (f) h 0 sin( π 6
More informationCurriculum Correlation
Curriculum Correlation Ontario Grade 12(MCV4U) Curriculum Correlation Rate of Change Chapter/Lesson/Feature Overall Expectations demonstrate an understanding of rate of change by making connections between
More informationCalculus I Exam 1 Review Fall 2016
Problem 1: Decide whether the following statements are true or false: (a) If f, g are differentiable, then d d x (f g) = f g. (b) If a function is continuous, then it is differentiable. (c) If a function
More informationDATE: MATH ANALYSIS 2 CHAPTER 12: VECTORS & DETERMINANTS
NAME: PERIOD: DATE: MATH ANALYSIS 2 MR. MELLINA CHAPTER 12: VECTORS & DETERMINANTS Sections: v 12.1 Geometric Representation of Vectors v 12.2 Algebraic Representation of Vectors v 12.3 Vector and Parametric
More informationChapter 3 Prerequisite Skills. Chapter 3 Prerequisite Skills Question 1 Page 148. a) Let f (x) = x 3 + 2x 2 + 2x +1. b) Let f (z) = z 3 6z 4.
Chapter 3 Curve Sketching Chapter 3 Prerequisite Skills Chapter 3 Prerequisite Skills Question 1 Page 148 a) Let f (x) = x 3 + 2x 2 + 2x +1. f (1) = 6 f (Ğ1) = 0 (x +1) is a factor. x 3 + 2x 2 + 2x +1
More informationFree Response Questions Compiled by Kaye Autrey for face-to-face student instruction in the AP Calculus classroom
Free Response Questions 1969-010 Compiled by Kaye Autrey for face-to-face student instruction in the AP Calculus classroom 1 AP Calculus Free-Response Questions 1969 AB 1 Consider the following functions
More informationSANDERSON HIGH SCHOOL AP CALCULUS AB/BC SUMMER REVIEW PACKET
SANDERSON HIGH SCHOOL AP CALCULUS AB/BC SUMMER REVIEW PACKET 017-018 Name: 1. This packet is to be handed in on Monday August 8, 017.. All work must be shown on separate paper attached to the packet. 3.
More informationPart A: Short Answer Questions
Math 111 Practice Exam Your Grade: Fall 2015 Total Marks: 160 Instructor: Telyn Kusalik Time: 180 minutes Name: Part A: Short Answer Questions Answer each question in the blank provided. 1. If a city grows
More informationFall 2009 Math 113 Final Exam Solutions. f(x) = 1 + ex 1 e x?
. What are the domain and range of the function Fall 9 Math 3 Final Exam Solutions f(x) = + ex e x? Answer: The function is well-defined everywhere except when the denominator is zero, which happens when
More informationCalculus I 5. Applications of differentiation
2301107 Calculus I 5. Applications of differentiation Chapter 5:Applications of differentiation C05-2 Outline 5.1. Extreme values 5.2. Curvature and Inflection point 5.3. Curve sketching 5.4. Related rate
More informationMath 112 (Calculus I) Midterm Exam 3 KEY
Math 11 (Calculus I) Midterm Exam KEY Multiple Choice. Fill in the answer to each problem on your computer scored answer sheet. Make sure your name, section and instructor are on that sheet. 1. Which of
More informationMath Fall 08 Final Exam Review
Math 173.7 Fall 08 Final Exam Review 1. Graph the function f(x) = x 2 3x by applying a transformation to the graph of a standard function. 2.a. Express the function F(x) = 3 ln(x + 2) in the form F = f
More informationLimits and Continuity. 2 lim. x x x 3. lim x. lim. sinq. 5. Find the horizontal asymptote (s) of. Summer Packet AP Calculus BC Page 4
Limits and Continuity t+ 1. lim t - t + 4. lim x x x x + - 9-18 x-. lim x 0 4-x- x 4. sinq lim - q q 5. Find the horizontal asymptote (s) of 7x-18 f ( x) = x+ 8 Summer Packet AP Calculus BC Page 4 6. x
More informationMath 121 Test 3 - Review 1. Use differentials to approximate the following. Compare your answer to that of a calculator
Math Test - Review Use differentials to approximate the following. Compare your answer to that of a calculator.. 99.. 8. 6. Consider the graph of the equation f(x) = x x a. Find f (x) and f (x). b. Find
More informationMATH 2053 Calculus I Review for the Final Exam
MATH 05 Calculus I Review for the Final Exam (x+ x) 9 x 9 1. Find the limit: lim x 0. x. Find the limit: lim x + x x (x ).. Find lim x (x 5) = L, find such that f(x) L < 0.01 whenever 0 < x
More informationAP Calculus Free-Response Questions 1969-present AB
AP Calculus Free-Response Questions 1969-present AB 1969 1. Consider the following functions defined for all x: f 1 (x) = x, f (x) = xcos x, f 3 (x) = 3e x, f 4 (x) = x - x. Answer the following questions
More informationSpring 2015 Sample Final Exam
Math 1151 Spring 2015 Sample Final Exam Final Exam on 4/30/14 Name (Print): Time Limit on Final: 105 Minutes Go on carmen.osu.edu to see where your final exam will be. NOTE: This exam is much longer than
More informationApplied Calculus I Practice Final Exam Solution Notes
AMS 5 (Fall, 2009). Solve for x: 0 3 2x = 3 (.2) x Taking the natural log of both sides, we get Applied Calculus I Practice Final Exam Solution Notes Joe Mitchell ln 0 + 2xln 3 = ln 3 + xln.2 x(2ln 3 ln.2)
More informationAS PURE MATHS REVISION NOTES
AS PURE MATHS REVISION NOTES 1 SURDS A root such as 3 that cannot be written exactly as a fraction is IRRATIONAL An expression that involves irrational roots is in SURD FORM e.g. 2 3 3 + 2 and 3-2 are
More informationCulminating Review for Vectors
Culminating Review for Vectors 0011 0010 1010 1101 0001 0100 1011 An Introduction to Vectors Applications of Vectors Equations of Lines and Planes 4 12 Relationships between Points, Lines and Planes An
More informationMATH 100 and MATH 180 Learning Objectives Session 2010W Term 1 (Sep Dec 2010)
Course Prerequisites MATH 100 and MATH 180 Learning Objectives Session 2010W Term 1 (Sep Dec 2010) As a prerequisite to this course, students are required to have a reasonable mastery of precalculus mathematics
More informationMth Review Problems for Test 2 Stewart 8e Chapter 3. For Test #2 study these problems, the examples in your notes, and the homework.
For Test # study these problems, the examples in your notes, and the homework. Derivative Rules D [u n ] = nu n 1 du D [ln u] = du u D [log b u] = du u ln b D [e u ] = e u du D [a u ] = a u ln a du D [sin
More informationMath 611b Assignment #6 Name. MULTIPLE CHOICE. Choose the one alternative that best completes the statement or answers the question.
Math 611b Assignment #6 Name MULTIPLE CHOICE. Choose the one alternative that best completes the statement or answers the question. Find a formula for the function graphed. 1) 1) A) f(x) = 5 + x, x < -
More informationMTH Calculus with Analytic Geom I TEST 1
MTH 229-105 Calculus with Analytic Geom I TEST 1 Name Please write your solutions in a clear and precise manner. SHOW your work entirely. (1) Find the equation of a straight line perpendicular to the line
More information4.1 Analysis of functions I: Increase, decrease and concavity
4.1 Analysis of functions I: Increase, decrease and concavity Definition Let f be defined on an interval and let x 1 and x 2 denote points in that interval. a) f is said to be increasing on the interval
More informationBlue Pelican Calculus First Semester
Blue Pelican Calculus First Semester Student Version 1.01 Copyright 2011-2013 by Charles E. Cook; Refugio, Tx Edited by Jacob Cobb (All rights reserved) Calculus AP Syllabus (First Semester) Unit 1: Function
More informationTable of Contents. Module 1
Table of Contents Module Order of operations 6 Signed Numbers Factorization of Integers 7 Further Signed Numbers 3 Fractions 8 Power Laws 4 Fractions and Decimals 9 Introduction to Algebra 5 Percentages
More informationa) An even function is symmetric with respect to the y-axis. An odd function is symmetric with respect to the origin.
Course Review Course Review Question 1 Page 479 a) An even function is symmetric with respect to the y-axis. An odd function is symmetric with respect to the origin. b) Substitute x for x in f(x). If f(
More informationCalculus I Review Solutions
Calculus I Review Solutions. Compare and contrast the three Value Theorems of the course. When you would typically use each. The three value theorems are the Intermediate, Mean and Extreme value theorems.
More informationCalculus III: Practice Final
Calculus III: Practice Final Name: Circle one: Section 6 Section 7. Read the problems carefully. Show your work unless asked otherwise. Partial credit will be given for incomplete work. The exam contains
More informationHarbor Creek School District
Unit 1 Days 1-9 Evaluate one-sided two-sided limits, given the graph of a function. Limits, Evaluate limits using tables calculators. Continuity Evaluate limits using direct substitution. Differentiability
More informationa x a y = a x+y a x a = y ax y (a x ) r = a rx and log a (xy) = log a (x) + log a (y) log a ( x y ) = log a(x) log a (y) log a (x r ) = r log a (x).
You should prepare the following topics for our final exam. () Pre-calculus. (2) Inverses. (3) Algebra of Limits. (4) Derivative Formulas and Rules. (5) Graphing Techniques. (6) Optimization (Maxima and
More informationDr. Sophie Marques. MAM1020S Tutorial 8 August Divide. 1. 6x 2 + x 15 by 3x + 5. Solution: Do a long division show your work.
Dr. Sophie Marques MAM100S Tutorial 8 August 017 1. Divide 1. 6x + x 15 by 3x + 5. 6x + x 15 = (x 3)(3x + 5) + 0. 1a 4 17a 3 + 9a + 7a 6 by 3a 1a 4 17a 3 + 9a + 7a 6 = (4a 3 3a + a + 3)(3a ) + 0 3. 1a
More informationUnit 1 PreCalculus Review & Limits
1 Unit 1 PreCalculus Review & Limits Factoring: Remove common factors first Terms - Difference of Squares a b a b a b - Sum of Cubes ( )( ) a b a b a ab b 3 3 - Difference of Cubes a b a b a ab b 3 3 3
More information3. Find the slope of the tangent line to the curve given by 3x y e x+y = 1 + ln x at (1, 1).
1. Find the derivative of each of the following: (a) f(x) = 3 2x 1 (b) f(x) = log 4 (x 2 x) 2. Find the slope of the tangent line to f(x) = ln 2 ln x at x = e. 3. Find the slope of the tangent line to
More informationAPPM 1350 Exam 2 Fall 2016
APPM 1350 Exam 2 Fall 2016 1. (28 pts, 7 pts each) The following four problems are not related. Be sure to simplify your answers. (a) Let f(x) tan 2 (πx). Find f (1/) (5 pts) f (x) 2π tan(πx) sec 2 (πx)
More informationAP Calculus Summer Prep
AP Calculus Summer Prep Topics from Algebra and Pre-Calculus (Solutions are on the Answer Key on the Last Pages) The purpose of this packet is to give you a review of basic skills. You are asked to have
More informationMLC Practice Final Exam. Recitation Instructor: Page Points Score Total: 200.
Name: PID: Section: Recitation Instructor: DO NOT WRITE BELOW THIS LINE. GO ON TO THE NEXT PAGE. Page Points Score 3 20 4 30 5 20 6 20 7 20 8 20 9 25 10 25 11 20 Total: 200 Page 1 of 11 Name: Section:
More informationMath 142 Week-in-Review #11 (Final Exam Review: All previous sections as well as sections 6.6 and 6.7)
Math 142 Week-in-Review #11 (Final Exam Review: All previous sections as well as sections 6.6 and 6.7) Note: This review is intended to highlight the topics covered on the Final Exam (with emphasis on
More informationFinal Exam Review Packet
1 Exam 1 Material Sections A.1, A.2 and A.6 were review material. There will not be specific questions focused on this material but you should know how to: Simplify functions with exponents. Factor quadratics
More informationFinal Exam Review Packet
1 Exam 1 Material Sections A.1, A.2 and A.6 were review material. There will not be specific questions focused on this material but you should know how to: Simplify functions with exponents. Factor quadratics
More informationFinal Exam 2011 Winter Term 2 Solutions
. (a Find the radius of convergence of the series: ( k k+ x k. Solution: Using the Ratio Test, we get: L = lim a k+ a k = lim ( k+ k+ x k+ ( k k+ x k = lim x = x. Note that the series converges for L
More informationM408 C Fall 2011 Dr. Jeffrey Danciger Exam 2 November 3, Section time (circle one): 11:00am 1:00pm 2:00pm
M408 C Fall 2011 Dr. Jeffrey Danciger Exam 2 November 3, 2011 NAME EID Section time (circle one): 11:00am 1:00pm 2:00pm No books, notes, or calculators. Show all your work. Do NOT open this exam booklet
More information3. On the grid below, sketch and label graphs of the following functions: y = sin x, y = cos x, and y = sin(x π/2). π/2 π 3π/2 2π 5π/2
AP Physics C Calculus C.1 Name Trigonometric Functions 1. Consider the right triangle to the right. In terms of a, b, and c, write the expressions for the following: c a sin θ = cos θ = tan θ =. Using
More information1) The line has a slope of ) The line passes through (2, 11) and. 6) r(x) = x + 4. From memory match each equation with its graph.
Review Test 2 Math 1314 Name Write an equation of the line satisfying the given conditions. Write the answer in standard form. 1) The line has a slope of - 2 7 and contains the point (3, 1). Use the point-slope
More information(a) The best linear approximation of f at x = 2 is given by the formula. L(x) = f(2) + f (2)(x 2). f(2) = ln(2/2) = ln(1) = 0, f (2) = 1 2.
Math 180 Written Homework Assignment #8 Due Tuesday, November 11th at the beginning of your discussion class. Directions. You are welcome to work on the following problems with other MATH 180 students,
More information1 + x 2 d dx (sec 1 x) =
Page This exam has: 8 multiple choice questions worth 4 points each. hand graded questions worth 4 points each. Important: No graphing calculators! Any non-graphing, non-differentiating, non-integrating
More information" $ CALCULUS 2 WORKSHEET #21. t, y = t + 1. are A) x = 0, y = 0 B) x = 0 only C) x = 1, y = 0 D) x = 1 only E) x= 0, y = 1
CALCULUS 2 WORKSHEET #2. The asymptotes of the graph of the parametric equations x = t t, y = t + are A) x = 0, y = 0 B) x = 0 only C) x =, y = 0 D) x = only E) x= 0, y = 2. What are the coordinates of
More information2.1 Limits, Rates of Change and Slopes of Tangent Lines
2.1 Limits, Rates of Change and Slopes of Tangent Lines (1) Average rate of change of y f x over an interval x 0,x 1 : f x 1 f x 0 x 1 x 0 Instantaneous rate of change of f x at x x 0 : f x lim 1 f x 0
More informationUNIT 3: DERIVATIVES STUDY GUIDE
Calculus I UNIT 3: Derivatives REVIEW Name: Date: UNIT 3: DERIVATIVES STUDY GUIDE Section 1: Section 2: Limit Definition (Derivative as the Slope of the Tangent Line) Calculating Rates of Change (Average
More information5 t + t2 4. (ii) f(x) = ln(x 2 1). (iii) f(x) = e 2x 2e x + 3 4
Study Guide for Final Exam 1. You are supposed to be able to determine the domain of a function, looking at the conditions for its expression to be well-defined. Some examples of the conditions are: What
More informationHigher Mathematics Course Notes
Higher Mathematics Course Notes Equation of a Line (i) Collinearity: (ii) Gradient: If points are collinear then they lie on the same straight line. i.e. to show that A, B and C are collinear, show that
More informationHello Future Calculus Level One Student,
Hello Future Calculus Level One Student, This assignment must be completed and handed in on the first day of class. This assignment will serve as the main review for a test on this material. The test will
More informationMath 2413 General Review for Calculus Last Updated 02/23/2016
Math 243 General Review for Calculus Last Updated 02/23/206 Find the average velocity of the function over the given interval.. y = 6x 3-5x 2-8, [-8, ] Find the slope of the curve for the given value of
More informationReview Guideline for Final
Review Guideline for Final Here is the outline of the required skills for the final exam. Please read it carefully and find some corresponding homework problems in the corresponding sections to practice.
More information(1) Recap of Differential Calculus and Integral Calculus (2) Preview of Calculus in three dimensional space (3) Tools for Calculus 3
Math 127 Introduction and Review (1) Recap of Differential Calculus and Integral Calculus (2) Preview of Calculus in three dimensional space (3) Tools for Calculus 3 MATH 127 Introduction to Calculus III
More information( ) - 4(x -3) ( ) 3 (2x -3) - (2x +12) ( x -1) 2 x -1) 2 (3x -1) - 2(x -1) Section 1: Algebra Review. Welcome to AP Calculus!
Welcome to AP Calculus! Successful Calculus students must have a strong foundation in algebra and trigonometry. The following packet was designed to help you review your algebra skills in preparation for
More informationx f(x)
1. Name three different reasons that a function can fail to be differentiable at a point. Give an example for each reason, and explain why your examples are valid. 2. Given the following table of values,
More informationx f(x)
1. Name three different reasons that a function can fail to be differential at a point. Give an example for each reason, and explain why your examples are valid. 2. Given the following table of values,
More informationc) xy 3 = cos(7x +5y), y 0 = y3 + 7 sin(7x +5y) 3xy sin(7x +5y) d) xe y = sin(xy), y 0 = ey + y cos(xy) x(e y cos(xy)) e) y = x ln(3x + 5), y 0
Some Math 35 review problems With answers 2/6/2005 The following problems are based heavily on problems written by Professor Stephen Greenfield for his Math 35 class in spring 2005. His willingness to
More informationCALCULUS ASSESSMENT REVIEW
CALCULUS ASSESSMENT REVIEW DEPARTMENT OF MATHEMATICS CHRISTOPHER NEWPORT UNIVERSITY 1. Introduction and Topics The purpose of these notes is to give an idea of what to expect on the Calculus Readiness
More informationMath 8 Winter 2010 Midterm 2 Review Problems Solutions - 1. xcos 6xdx = 4. = x2 4
Math 8 Winter 21 Midterm 2 Review Problems Solutions - 1 1 Evaluate xcos 2 3x Solution: First rewrite cos 2 3x using the half-angle formula: ( ) 1 + cos 6x xcos 2 3x = x = 1 x + 1 xcos 6x. 2 2 2 Now use
More information1) Find the equations of lines (in point-slope form) passing through (-1,4) having the given characteristics:
AP Calculus AB Summer Worksheet Name 10 This worksheet is due at the beginning of class on the first day of school. It will be graded on accuracy. You must show all work to earn credit. You may work together
More informationDIFFERENTIATION RULES
3 DIFFERENTIATION RULES DIFFERENTIATION RULES Before starting this section, you might need to review the trigonometric functions. DIFFERENTIATION RULES In particular, it is important to remember that,
More informationlim4 4. By the definition of a limit, there is a positive real number such that if 0 x 2. The largest valid value of is
ACTM Regional Calculus Competition 017 Begin by removing the three tie breaker sheets at the end of the exam and writing your name on all three pages. Work the multiple-choice questions first, choosing
More information( ) as a fraction. If both numerator and denominator are
A. Limits and Horizontal Asymptotes What you are finding: You can be asked to find lim f x x a (H.A.) problem is asking you find lim f x x ( ) and lim f x x ( ). ( ) or lim f x x ± ( ). Typically, a horizontal
More informationMath 113 HW #10 Solutions
Math HW #0 Solutions 4.5 4. Use the guidelines of this section to sketch the curve Answer: Using the quotient rule, y = x x + 9. y = (x + 9)(x) x (x) (x + 9) = 8x (x + 9). Since the denominator is always
More informationand Calculus and Vectors
and Calculus and Vectors Autograph is spectacular dynamic software from the UK that allows teachers to visualise many of the mathematical topics that occur in the Ontario Grade 12 CALCULUS and VECTORS
More informationFinal Exam Review Exercise Set A, Math 1551, Fall 2017
Final Exam Review Exercise Set A, Math 1551, Fall 2017 This review set gives a list of topics that we explored throughout this course, as well as a few practice problems at the end of the document. A complete
More informationMath 180, Final Exam, Fall 2012 Problem 1 Solution
Math 80, Final Exam, Fall 0 Problem Solution. Find the derivatives of the following functions: (a) ln(ln(x)) (b) x 6 + sin(x) e x (c) tan(x ) + cot(x ) (a) We evaluate the derivative using the Chain Rule.
More informationAP Calculus Summer Homework MULTIPLE CHOICE. Choose the one alternative that best completes the statement or answers the question.
AP Calculus Summer Homework 2015-2016 Part 2 Name Score MULTIPLE CHOICE. Choose the one alternative that best completes the statement or answers the question. Find the distance d(p1, P2) between the points
More informationUse a graphing utility to approximate the real solutions, if any, of the equation rounded to two decimal places. 4) x3-6x + 3 = 0 (-5,5) 4)
Advanced College Prep Pre-Calculus Midyear Exam Review Name Date Per List the intercepts for the graph of the equation. 1) x2 + y - 81 = 0 1) Graph the equation by plotting points. 2) y = -x2 + 9 2) List
More informationCalculus AB Topics Limits Continuity, Asymptotes
Calculus AB Topics Limits Continuity, Asymptotes Consider f x 2x 1 x 3 1 x 3 x 3 Is there a vertical asymptote at x = 3? Do not give a Precalculus answer on a Calculus exam. Consider f x 2x 1 x 3 1 x 3
More informationMath 121: Calculus 1 - Fall 2013/2014 Review of Precalculus Concepts
Introduction Math 121: Calculus 1 - Fall 201/2014 Review of Precalculus Concepts Welcome to Math 121 - Calculus 1, Fall 201/2014! This problems in this packet are designed to help you review the topics
More information= π + sin π = π + 0 = π, so the object is moving at a speed of π feet per second after π seconds. (c) How far does it go in π seconds?
Mathematics 115 Professor Alan H. Stein April 18, 005 SOLUTIONS 1. Define what is meant by an antiderivative or indefinite integral of a function f(x). Solution: An antiderivative or indefinite integral
More information10550 PRACTICE FINAL EXAM SOLUTIONS. x 2 4. x 2 x 2 5x +6 = lim x +2. x 2 x 3 = 4 1 = 4.
55 PRACTICE FINAL EXAM SOLUTIONS. First notice that x 2 4 x 2x + 2 x 2 5x +6 x 2x. This function is undefined at x 2. Since, in the it as x 2, we only care about what happens near x 2 an for x less than
More informationAim: How do we prepare for AP Problems on limits, continuity and differentiability? f (x)
Name AP Calculus Date Supplemental Review 1 Aim: How do we prepare for AP Problems on limits, continuity and differentiability? Do Now: Use the graph of f(x) to evaluate each of the following: 1. lim x
More informationREQUIRED MATHEMATICAL SKILLS FOR ENTERING CADETS
REQUIRED MATHEMATICAL SKILLS FOR ENTERING CADETS The Department of Applied Mathematics administers a Math Placement test to assess fundamental skills in mathematics that are necessary to begin the study
More informationBC Exam 1 - Part I 28 questions No Calculator Allowed - Solutions C = 2. Which of the following must be true?
BC Exam 1 - Part I 8 questions No Calculator Allowed - Solutions 6x 5 8x 3 1. Find lim x 0 9x 3 6x 5 A. 3 B. 8 9 C. 4 3 D. 8 3 E. nonexistent ( ) f ( 4) f x. Let f be a function such that lim x 4 x 4 I.
More informationMath Exam 02 Review
Math 10350 Exam 02 Review 1. A differentiable function g(t) is such that g(2) = 2, g (2) = 1, g (2) = 1/2. (a) If p(t) = g(t)e t2 find p (2) and p (2). (Ans: p (2) = 7e 4 ; p (2) = 28.5e 4 ) (b) If f(t)
More information3.Applications of Differentiation
3.Applications of Differentiation 3.1. Maximum and Minimum values Absolute Maximum and Absolute Minimum Values Absolute Maximum Values( Global maximum values ): Largest y-value for the given function Absolute
More informationLearning Objectives These show clearly the purpose and extent of coverage for each topic.
Preface This book is prepared for students embarking on the study of Additional Mathematics. Topical Approach Examinable topics for Upper Secondary Mathematics are discussed in detail so students can focus
More informationReview for the Final Exam
Math 171 Review for the Final Exam 1 Find the limits (4 points each) (a) lim 4x 2 3; x x (b) lim ( x 2 x x 1 )x ; (c) lim( 1 1 ); x 1 ln x x 1 sin (x 2) (d) lim x 2 x 2 4 Solutions (a) The limit lim 4x
More informationCALCULUS Exercise Set 2 Integration
CALCULUS Exercise Set Integration 1 Basic Indefinite Integrals 1. R = C. R x n = xn+1 n+1 + C n 6= 1 3. R 1 =ln x + C x 4. R sin x= cos x + C 5. R cos x=sinx + C 6. cos x =tanx + C 7. sin x = cot x + C
More informationFinal Exam Study Guide
Final Exam Study Guide Final Exam Coverage: Sections 10.1-10.2, 10.4-10.5, 10.7, 11.2-11.4, 12.1-12.6, 13.1-13.2, 13.4-13.5, and 14.1 Sections/topics NOT on the exam: Sections 10.3 (Continuity, it definition
More informationSolution: It could be discontinuous, or have a vertical tangent like y = x 1/3, or have a corner like y = x.
1. Name three different reasons that a function can fail to be differentiable at a point. Give an example for each reason, and explain why your examples are valid. It could be discontinuous, or have a
More informationMath. 151, WebCalc Sections December Final Examination Solutions
Math. 5, WebCalc Sections 507 508 December 00 Final Examination Solutions Name: Section: Part I: Multiple Choice ( points each) There is no partial credit. You may not use a calculator.. Another word for
More informationUNIVERSITY of LIMERICK
UNIVERSITY of LIMERICK OLLSCOIL LUIMNIGH Faculty of Science and Engineering Department of Mathematics & Statistics END OF SEMESTER ASSESSMENT PAPER MODULE CODE: MA4701 SEMESTER: Autumn 2011/12 MODULE TITLE:
More informationIntroduction to Calculus
Introduction to Calculus Contents 1 Introduction to Calculus 3 11 Introduction 3 111 Origin of Calculus 3 112 The Two Branches of Calculus 4 12 Secant and Tangent Lines 5 13 Limits 10 14 The Derivative
More informationMath 1120, Section 1 Calculus Final Exam
May 7, 2014 Name Each of the first 17 problems are worth 10 points The other problems are marked The total number of points available is 285 Throughout the free response part of this test, to get credit
More informationNo calculators, cell phones or any other electronic devices can be used on this exam. Clear your desk of everything excepts pens, pencils and erasers.
Name: Section: Recitation Instructor: READ THE FOLLOWING INSTRUCTIONS. Do not open your exam until told to do so. No calculators, cell phones or any other electronic devices can be used on this exam. Clear
More informationJim Lambers MAT 169 Fall Semester Practice Final Exam
Jim Lambers MAT 169 Fall Semester 2010-11 Practice Final Exam 1. A ship is moving northwest at a speed of 50 mi/h. A passenger is walking due southeast on the deck at 4 mi/h. Find the speed of the passenger
More informationNote: Unless otherwise specified, the domain of a function f is assumed to be the set of all real numbers x for which f (x) is a real number.
997 AP Calculus BC: Section I, Part A 5 Minutes No Calculator Note: Unless otherwise specified, the domain of a function f is assumed to be the set of all real numbers for which f () is a real number..
More information7.1. Calculus of inverse functions. Text Section 7.1 Exercise:
Contents 7. Inverse functions 1 7.1. Calculus of inverse functions 2 7.2. Derivatives of exponential function 4 7.3. Logarithmic function 6 7.4. Derivatives of logarithmic functions 7 7.5. Exponential
More informationVectors, dot product, and cross product
MTH 201 Multivariable calculus and differential equations Practice problems Vectors, dot product, and cross product 1. Find the component form and length of vector P Q with the following initial point
More informationSecond Midterm Exam Name: Practice Problems Septmber 28, 2015
Math 110 4. Treibergs Second Midterm Exam Name: Practice Problems Septmber 8, 015 1. Use the limit definition of derivative to compute the derivative of f(x = 1 at x = a. 1 + x Inserting the function into
More informationMLC Practice Final Exam
Name: Section: Recitation/Instructor: INSTRUCTIONS Fill in your name, etc. on this first page. Without fully opening the exam, check that you have pages through. Show all your work on the standard response
More information