chapters 1 to 8 course review question 1 page...

MHR • Calculus and Vectors 12 Solutions 1020

Chapters 1 to 8 Course Review Chapters 1 to 8 Course Review Question 1 Page 509

a) i)

[2(16)!12 + 4]! [2 ! 3+ 4]

4 !1=

21

3

= 7

ii)

[2(2.25)! 4.5+ 4]! [2 ! 3+ 4]

1.5!1=

1

0.5

= 2

iii)

[2(1.21)! 3.3+ 4]! [2 ! 3+ 4]

1.1!1=

0.12

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 + 3h2! 3a

2

h

=6ah + 3h

2

h

= 6a + 3h

6(2) + 3(0.01) = 12.03

ii) 3 3 3 2 2 3 3 2 2 3

2 2

( ) 3 3 3 3

3 3

a h a a a h ah h a a h ah h

h h h

a ah h

+ ! + + + ! + += =

= + +

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.


ii) This is an approximation of the value of the slope of the tangent to f (x) = x

3 at x = 2.


a)

!4.9[(2 + h)2! 22 ]+ 6h

h=!4.9(4h + h

2 ) + 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.

!4.9(2.12!1.92 ) + 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).

c)

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.


Chapters 1 to 8 Course Review Question 6 Page 509 a)

limx!2

(3x2" 4x +1) = 5

b)

limx!"8

5x + 40

x + 8= lim

x!"8

5(x + 8)

x + 8

= 5

c)

limx!6+

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)

limx!3

f (x) = 17

ii) The limit does not exist. The limits on the left and right sides are unequal.


a)

dy

dx= lim

h!0

f (x + h)" f (x)

h

= limh!0

4(x + h)2" 3( )" 4x

2" 3( )

h

= limh!0

8xh + h2

h

= limh!0

(8x + h)

= 8x


At x = 2,

dy

dx = 16 and y = 13.

The equation of the tangent is: 13 16( 2)

16 19

y x

y x

! = !

= !

b)

!f (x) = limh"0

f (x + h)# f (x)

h

= limh"0

(x + h)3# 2(x + h)2

( )# x3# 2x

2( )

h

= limh"0

3x2h + 3xh

2+ h

3# 4xh # 2h

2

h

= limh"0

3x2

+ 3xh + h2# 4x # 2h

= 3x2# 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) = limh"0

f (x + h)# f (x)

h

= limh"0

3

x + h#

3

x

$%&

'()

h

= limh"0

3x # 3x # 3h

x(x + h)

$%&

'()

h

= limh"0

#3h

xh(x + h)

= limh"0

#3

x(x + h)

= #3

x2

At x = 2, !g (x) =

!

3

4 and g(x) =

3

2.



y !3

2= !

3

4(x ! 2)

y = !3

4x + 3

d)

!h (x) = limh"0

f (x + h)# f (x)

h

= limh"0

(2 x + h # 2 x )

h

= limh"0

(2 x + h # 2 x )(2 x + h + 2 x )

h(2 x + h + 2 x )

= limh"0

=(4(x + h)# 4x)

h(2 x + h + 2 x )

= limh"0

4h

h(2 x + h + 2 x )

= limh"0

4

(2 x + h + 2 x )

=1

x

At x = 2, !h (x) =

1

2 and h(x) = 2 2 .


y ! 2 2 =1

2(x ! 2)

y =1

2x + 2

Chapters 1 to 8 Course Review Question 9 09 a)


b)


a)

dy

dx= !6x + 4

b)

!f (x) = "6x"2

+10x"3

c)

!f (x) = 2x"

1

2

=2

x

d)

dy

dx= (3x

2 ! 4x)1

2x!

1

2"

#$%

&'+ x !1( )(6x ! 4)

e)

dy

dx=

3x(2x)! (x2

+ 4)(3)

9x2

=6x

2! 3x

2!12

9x2

=(x ! 2)(x + 2)

3x2

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) ") = (x2!2)

= 4x

!f (x) " !g (x) = 2x "0

= 0

L.S. ≠ R.S.

Therefore, the statement is false.


b) Let f(x) = 2x3 + x

2 and g(x) = x.

f (x)

g(x)

!

"#$

%&'

=2x

3+ x

2

x

!

"#$

%&'

= 2x2

+ x( )'

= 4x +1

!f (x)

!g (x)=

6x2

+ 2x

1

= 6x2

+ 2x

L.S. ≠ R.S

Therefore, the statement is false.


!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

2x"

1

2

At x = 4, !g (x) =

1

4 and g(x) = 7.

Find b.

y = mx + b

b = 7 !1

b = 6

The equation of the tangent is y =

1

4x – 6.


v(t) = !s (t)

= 6t2"14t + 4

b)

v(5) = 150 ! 70 + 4

= 84


The velocity is 84 m/s.


The slope is 0 when !f (x) = 0 .

Set f(x) = 0.

3x2 – 4x – 4 = 0

x = 2 or x = !

2

3

Evaluate for these values of x.

f (2) = 8! 8! 8 + 4

= !4

f !2

3

"

#$%

&'= !

8

27!

8

9+

8

3+ 4

=40

27+ 4

= 513

27

The points are (2, –4) and

!2

3, 5

13

27

"

#$%

&'.


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.


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.



a)

dy

dx= 4(3x ! x

!1)(3+ x!2 )

b)

!g (x) =1

2(2x + 5)

"1

2 (2)

=1

2x + 5

c)

dy

dx= !

1

2(3x !1)

!3

2 (3)

= !3

2 3x !1( )3

d)

y =!1

x2

+ 3x3

= !(x2

+ 3x)!

1

3

dy

dx=

1

3(x

2+ 3x)

!4

3 (2x + 3)

=(2x + 3)

3 x2

+ 3x3

( )4



!f (x) = x2 (3(x

3" 3x)2 )(3x

2" 3) + 2x(x

3" 3x)3

= (x3" 3x)2 (3x

2 )(3x2" 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.


C(2000) = 15+ 0.1 2000

= 19.47

The cost is $19.47.

!C (w) = 0.11

2

"

#$%

&'w

(1

2

!C (2000) = 0.11

2

"

#$%

&'(2000)

(1

2

= 0.0011

The rate of change of the cost is $0.0011/L.

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.


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.


Let s = gt

2+ v

0t + s

0, where t is time in seconds and g, v0, and s0 and constants.

v(t) = 2gt + v

0

a(t) = 2g

At t = 0, v0 = 120 km/h or

100

3 m/s and s0 = 0.

a = –10 km/h/s or !

25

9 m/s

2 and g =

!

25

18.

Find the value of t when v(t) = 0.

0 = !25

9t +

100

3

t = 12

After 12 s, the car has stopped.

s = !25

18(144) +

100

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, limx!"

(2x3# 5x

2+ 9x # 8) = " .


b) Use the laws of limits.

limx!"

x +1

x #1= lim

x!"

1+1

x

1#1

x

=

limx!"

1+1

x

$%&

'()

limx!"

1+1

x

$%&

'()

=1

1

= 1

Divide the numerator and denominator by x.

c) Use the laws of limits.

limx!"#

x2 " 3x +1

x2

+ 4x + 8= lim

x!"#

1"3

x+

1

x2

1+4

x+

8

x2

=

limx!"#

1"3

x+

1

x2

$%&

'()

limx!"#

1+4

x+

8

x2

$%&

'()

=1

1

= 1

Divide the numerator and denominator by x2.


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])



f (x) = x3+ 2x

2! 4x +1

"f (x) = 3x2

+ 4x ! 4

""f (x) = 6x + 4

Find the critical numbers.

3x2

+ 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

!!f2

3

#

$%&

'(= 8

Therefore, (–2, 9) is a local maximum point and 2 59

,3 27

! "# $% &

is a local minimum.

Find the possible points of inflection.

6x + 4 = 0

x = !2

3

Have already tested the interval around this value.

Therefore, 2 115

,3 27

! "#$ %& '

is a point of inflection.

Therefore, f is increasing for2

2 and 3

x x< ! > and decreasing for2

23

x! < < .

Also, f is concave down for2

3x < ! and concave up for

2

3x > ! .

b)

f (x) = !3x4! 2x

3+15x

2!12x + 2

"f (x) = !12x3! 6x

2+ 30x !12

""f (x) = !36x2!12x + 30


!12x3! 6x

2+ 30x !12 = 0

!6(2x3+ x

2! 5x + 2) = 0

!6(x !1)(2x2

+ 3x ! 2) = 0

!6(x !1)(2x !1)(x + 2) = 0


x = 1 or x = –2 or x =

1

2


!!f ("2) = "90

!!f1

2

#

$%&

'(= 15

!!f (1) = "18

Therefore, (–2, 54) and (1, 0) are local maximum points and 1

, 0.68752

! "#$ %& '

is a local minimum.


!36x2!12x + 30 = 0

!6(6x2

+ 2x ! 5) = 0

x =!2 ± 124

12

x =!1± 31

6

x ! –1.09 or x ! 0.76

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 2 and 0.5 1x x< ! < < and decreasing for 2 0.5 and 1x x! < < > .

Also, f is concave down for x < !1.09 and x > 0.76 and concave up for !1.09 < x < 0.76 .

c)

f (x) =3

x2

+1

= 3(x2

+1)!1

!f (x) = "3(x2

+1)"2 (2x)

= "6x(x2

+1)"2

!!f (x) = "6(x2

+1)"2+ "6x(–2)(x

2+1)"3(2x)

= ("6x2" 6 + 24x

2 )(x2

+1)"3

= 6(3x2"1)(x

2+1)"3


!6x(x2

+1)!2= 0

x = 0


!!f (0) = "6


Therefore, (0, 3) is a local maximum.


6(3x2!1)(x

2+1)!3

= 0

x = ±1

3

Therefore, 1 9 1 9

, and ,4 43 3

! " ! "#$ % $ %

& ' & ' are points of inflection.

Therefore, f is increasing for 0x < and decreasing for 0x > .

Also, f is concave down for1 1

3 3x! < < and concave up for

1 1 and

3 3x x< ! > .

Chapters 1 to 8 Course Review Question 25 Page 512 a) Follow the six step plan.

f (x) = 3x4! 8x

3+ 6x

2

"f (x) = 12x3! 24x

2+12x

""f (x) = 36x2! 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 0y = .

3x4! 8x

3+ 6x

2= 0

x2 (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.

12x3! 24x

2+12x = 0

12x(x2! 2x +1) = 0

12x(x !1)(x !1) = 0

x = 0, x =1



!!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.

36x2! 48x +12 = 0

12(3x2! 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 1 and 1x x< < > and decreasing for 0x < .

From Step 4, f is concave down for 1

13

x< < and concave up for 1

and 13

x x< > .

Step 6. Sketch the graph.

b) Follow the six step plan.

( )

( )

( )

3 2

2

8 3

3 2 8

6 2

f x x x x

f x x x

f x x

= ! + + !

" = ! + +

"" = ! +

Step 1. Since this is a polynomial function, the domain is {x !!} .

Step 2. f (0) = !3

The y-intercept is –3.


!x3+ x

2+ 8x ! 3 = 0 .

This expression is not factorable. There may be up to three x-intercepts.



!3x2

+ 2x + 8 = 0

(!3x ! 4)(x ! 2) = 0

x = 2 or x = !

4

3


!!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.

Step 4. From Step 3, f is increasing for 4

23

x! < < and decreasing for 4

and 23

x x< ! > .

From Step 4, f is concave down for 1

3x > and concave up for

1

3x < .


(x ! [–5, 10], Xscl = 2, y! [–6, 6], Yscl = 20) c) Follow the six step plan.

f (x) =x

x2

+1

= x(x2

+1)!1


!f (x) = 1(x2

+1)"1+ x(–1)(x

2+1)"2 (2x)

= (x2

+1" 2x2 )(x

2+1)"2

= ("x2

+1)(x2

+1)"2

!!f (x) = "2x("x2

+1)"2+ ("x

2+1)(–2)(x

2+1)"3(2x)

= ("x3" 2x + 4x

3" 4x)(x

2+1)"3

= (3x3" 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



x

x2

+1= 0

x = 0



(!x2

+1)(x2

+1)!2= 0

!x2

+1= 0

x = ±1


!!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.

(3x3! 6x)(x

2+1)!3

= 0

3x3! 6x = 0

3x(x2! 2) = 0

x = 0 or x = 2 or x = –2

Test the intervals.

!!f ("2) =12

729

!!f (2) = "12

729


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 0.375x! < < and decreasing for x < –0.375 and

x > 0.375.

From Step 4, f is concave down for 2 and 0 2x x< < < and concave up for

2 0 and 2x x! < < > .


(x ! [–5, 5], y! [–1, 1], Yscl = 0.25) Chapters 1 to 8 Course Review Question 26 Page 512

a)

P(v) = 100v !3

16v

3

"P (v) = 100 !9

16v

2

To find the maximum, let !P (v) = 0 .

100 !9

16v

2= 0

v2

=1600

9

v = ±40

3

Since linear velocity cannot be negative, the maximum occurs when 13.3v = m/s.

b)

P40

3

!

"#$

%&= 100

40

3

!

"#$

%&'

3

16

40

3

!

"#$

%&

3

! 888.9

The maximum power is about 889 A.



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.

d(t) = (9t)2+ (15!12t)2

= (225t2! 360t + 225)

1

2

!d (t) =1

2

"

#$%

&'(225t

2 ( 360t + 225)(

1

2 (450t ( 360)

= (225t (180)(225t2 ( 360t + 225)

(1

2

For a minimum, the derivative must equal zero.

(225t !180)(225t2! 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+ (15!12(0.8))2

= 9

The closest distance of approach of the two ships is 9 km.


C(x) =0.000 25x

2+ 8x +10

x

= 0.000 25x + 8 +10x!1

C '(x) = 0.000 25!10x

!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

d

9t

15 – 12t


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 25!10x!2

= 0

0.000 25x2!10 = 0

x2

= 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.


a)

dy

dx= 3cos2

x(! sin x)

= !3(cos2x)(sin x)

b)

dy

dx= 3x

2 cos (x3)

c)

!f (x) = "5sin (5x " 3)

d)

!f (x) = sin2 x " sin x

2

#

$%&

'(1

2

#

$%&

'(#

$%&

'(+ 2(sin x)(cos x) cos

x

2

#

$%&

'(#

$%&

'(

= 2(sin x)(cos x) cosx

2

#

$%&

'(#

$%&

'("

1

2

#

$%&

'(sin2 x sin

x

2

#

$%&

'(#

$%&

'(

e)

!f (x) = 2(cos 4x2 )(" sin (4x

2 )(8x))

= "16x(cos 4x2 )(sin 4x

2 )


f)

!g (x) =(cos x " sin x)(" sin x)" cos x(" sin x " cos x)

(cos x " sin x)2

=– sin xcos x + sin2

x + cos x sin x + cos2x)

(cos x " sin x)2

=1

(cos x " sin x)2


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.


y = sec x

= (cos x)!1

dy

dx= !(cos x)!2 (! sin x)

=sin x

cos2x

= sec x tan x


y = csc x

= (sin x)!1

dy

dx= !(sin x)!2 (cos x)

= !cos x

sin2x

= – csc x cot x

y = tan x

=sin x

cos x

dy

dx=

cos x(cos x)! sin x(! sin x)

cos2x

=cos2

x + sin2x

cos2x

=1

cos2x

= sec2x


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!+!

6, for k !!

x = k!+!

12 and 2x = (2k "1)!"

!

6 for k !!

x = k!"7!

12

d2y

dx2

= 2cos2x is positive at x =

!

12,

13!

12,

25!

12, ..., k!+

!

12 and negative at

x =

5!

12,

17!

12,

29!

12,

41!

12, ..., k!!

7!

12.

There are local minima at x =

!

12,

13!

12,

25!

12, ..., k!+

!

12.


There are local maxima at x =

5!

12,

17!

12,

29!

12,

41!

12, ...k!!

7!

12.

The points of inflection are given by:

d2y

dx2

= 2cos2x

= 0

Points of inflection are 2x =

(2k +1)!

2 or

x =

(2k +1)!

4 for k !! .


!h (t) =2!

3cos

2!

30t

"

#$%

&'

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.


!d (t) = 6cos 6t + 24sin 6t

!d (1) = 6cos 6 + 24sin 6

= "0.945

The rate of change is –0.945 cm/s.


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.


y = e

x increases faster as x increases. Both graphs have the same horizontal asymptote and

pass through (0, 1).

b)


The graphs are !f (x) = e

x and !g (x) = ln 2(2x ) . !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 2

e = 2

c) (ln 2)e = 1

Chapters 1 to 8 Course Review Question 37 Page 511 a) ln ( )x

e = x

b) ln x

e = x

c)

2ln 2xe x=


a)

dy

dx= !2e

x

b)

!g (x) = 5(ln 10)(10x )

c)

h(x) = !e

x sin (ex )

d)

!f (x) = "xe" x

+ e" x

= e" x (1" x)



!f (x) =

1

2e

x+1

At x = 0, f (0) =

e

2 and !f (0) =

e

2.

The equation of the line perpendicular to f (x) =

1

2e

x+1 has a slope !

2

e and passes through

0, e

2

!

"#$

%&:

y !e

2= !

2

ex

y = !2

ex +

e

2

Chapters 1 to 8 Course Review Question 40 Page 512 The equation for exponential decay is 0

tN N e

!"= where 0N = 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.



!f (x) = x2e

x+ 2xe

x

Let !f (x) = 0.

x2e

x+ 2xe

x= 0

ex (x

2+ 2x) = 0

x = 0 or x = –2 since ex > 0.


!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.


a)

dy

dx= !20.96(0.0329)

e0.0329x ! e

!0.0329x

2

"

#$%

&'

= !0.344 792(e0.0329x ! e!0.0329x )

b)

dy

dx= !0.344 792(e0.0329(2)

! e!0.0329(2) )

= !0.0454


c) The width can be found by letting y = 0.

!20.96e

0.0329x+ e

!0.0329x

2!10.06

"

#$%

&'= 0

e0.0329x

+ e!0.0329x

2= 10.06

e0.0329x

+ 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(e0.0329x! e

!0.0329x ) = 0

e0.0329x

= e!0.0329x

x = 0

Find the value of y when x = 0.

y = !20.96e

0.0329(0)+ e

!0.0329(0)

2!10.06

"

#$%

&'

= 20.96(9.06)

= 189.9

The height of the arch is 189.9 m.



a)

dy

dx= e

! x cos x ! e! xsin x

Let

dy

dx = 0.

cos x = sin x

tan x = 1

x =!

4,

5!

4,

9!

4, ...

d2y

dx2

= !e! x cos x ! e

! x sin x + e! xsin x ! e

! xcos x

= !2e! xcos x

d2y

dx2

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!"4

1

2

#

$%&

'(

= 1.55

When x =

9!

4:

y = e!

9"4

1

2

#

$%&

'(

!= 0.0006.



a)

b)


Since BD||AE and BD = AE, ABDE is a parallelogram and opposite sides are equal and parallel.


a) u!

!

b) v!

!

c) 1

5u

!

d)

AD! "!!

= AE! "!!

+ ED! "!!

= u

"

+ v

"

e)

CD! "!!

= CB! "!!

+ BD! "!!

= !1

5u

"

+ v

"

f)

CE! "!!

= CD! "!!

+ DE! "!!

= !1

5u

"

+ v

"

+ !u

"

( )

= !6

5u

"

+ v

"


a)

3u

!

+ 5u

!

! 7u

!

! 6u

!

= (3+ 5! 7 ! 6)u!

= !5u

!

b)

!5 c

!

+ d

"!

( )! 8 c

!

! 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.


2 212 15

19.2

R = +

!"

#

tan ! =15

12

! = tan"1 15

12

#

$%&

'(

! ! 51.3o

The resultant displacement is about 19.2 km in a direction N51.3ºE.

b) Draw a diagram. Use the Pythagorean theorem and trigonometry.

2 260 40

72.1

R = +

!"

#

tan ! =40

60

! = tan"1 40

60

#

$%&

'(

! ! 33.7o

The resultant force is about 72.1 N in a direction 56.3° up from the horizontal.


Chapters 1 to 8 Course Review Question 49 Page 512 a) Draw a diagram. Use the Pythagorean theorem and trigonometry.

2 2450 15

450.2

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 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 !200

=sin 160o

340

sin ! =200sin 160o

340

! = sin"1 200sin 160o

340

#

$%&

'(

! ! 11.6o


To find the magnitude of the second force, use the sine law again.

F

!"

sin 8.4o=

340

sin 160o

F

!"

=340sin 8.4o

sin 160o

F

!"

# 145.2

The second force has a magnitude of 145.2 N at an angle of 11.6o 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 1 2 and T T

!" !!"as shown.

To find 1T

!", use the sine law.

T1

!"

sin 20o=

2450

sin 140o

T1

!"

=2450sin 20o

sin 140o

T1

!"

# 1303.6

By symmetry the two tensions are identical.

Each of the four tensions is approximately 1303.6 N at 70º to the horizontal.


Chapters 1 to 8 Course Review Question 52 Page 512 Draw a diagram. Use the Pythagorean theorem.

2 2v 58 40

42

F = !

=

!"

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 40o! 91.9

120sin 40o! 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 48o

# 66.9

F

!"

perp. = 100sin 48o

# 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.


Chapters 1 to 8 Course Review Question 55 Page 513 a) 5 6 4i j k+ !

! ! !

b) 8 7j k! +

! !


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 70o , 30sin 70o!"

#$

" 10.3, 28.2!" #$

b) Assume standard x- and y-axes.

v

!

= 40cos 80o , 40sin 80o!"

#$

" 6.9, 39.4!" #$



The ship’s vector is !20cos 56o , ! 20sin 56o"#

$% .

The current’s vector is !11cos 3o , +11sin 3o"#

$% .

The resultant velocity is !20cos 56o !11cos 3o , ! 20sin 56o

+11sin 3o"#

$% ! !22.2, !16.0"# $% .

R

!"

= (–22.2)2+ (–16)2

# 27.4

! = tan"1 "16

"22.2

#

$%&

'(

! 35.8o

For the bearing, o o o270 35.8 234.2! = .

The resultant velocity is about 27 km on a bearing of 234.2°


a)

Let u!

= 1, 1, 1!" #$ , v

!

= 3, 2, 1!" #$ , and w"!

= 0, %1, 4!" #$ .

LS = u

!

+ v

!

( ) + w

"!

= 1, 1, 1!" #$ + 3, 2, 1!" #$( ) + 0, %1, 4!" #$

= 4, 3, 2!" #$ + 0, %1, 4!" #$

= 4, 2, 6!" #$

RS = u

!

+ v

!

+ w

"!

( )

= 1, 1, 1!" #$ + 3, 2, 1!" #$ + 0, %1, 4!" #$( )

= 1, 1, 1!" #$ + 3, 1, 5!" #$

= 4, 2, 6!" #$

L.S. = R.S.


b) Let a!

= 1, 1, 1!" #$ , b

!

= 3, 2, 1!" #$ , and k = %2 .

L.S. = k a

!

!b

!

( )= (–2)(1(3) +1(2) +1(1))

= (–2)(6)

= "12

R.S. = a

!

! kb

!

( )

= 1, 1, 1"# $% ! &6, & 4, & 2"# $%

= 1(–6) +1(–4) +1(–2)

= &12

L.S. = R.S.


u

!

!v

!

= u

!

v

!

cos "

= (12)(21)cos 20o

" 236.8

b)

s

!

! t

!

= s

!

t

!

cos "

= (115)(150)cos 42o

" 12 819.2


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#$ %&( )

= 3, " 4#$ %& (6(–9) +1(6))

= 3, " 4#$ %& (–48)

= "48 3, " 4#$ %&

= "144, 192#$ %&

d)

u

!

! v

!

( ) " u

!

+ v

!

( ) = 3, ! 4#$ %& + 6, 1#$ %&( ) " 3, ! 4#$ %& ! 6, 1#$ %&( )

= 9, ! 3#$ %& " !3, ! 5#$ %&

= 9(–3) + (–3)(–5)

= !12



and u v

! !are perpendicular if and only if 0u v! =

! !.

!3, 7"# $% & 6, k"# $% = 0

!18 + 7k = 0

k =18

7


AB! "!!

= !4, !1"# $% ; BC! "!!

= 7, ! 2"# $% ;AC! "!!

= 3, ! 3"# $%

AB! "!!

!BC! "!!

= "4, "1#$ %& ! 7, " 2#$ %&

= "26

' 0

AB! "!!

!AC! "!!

= "4, "1#$ %& ! 3, " 3#$ %&

= "9

' 0

BC! "!!

!AC! "!!

= 7, " 2#$ %& ! 3, " 3#$ %&

= 27

' 0

None of the vectors are perpendicular. The triangle is not a right triangle.


cos ! =u

!

"v!

u

!

v

!

cos ! =

4, 10, # 2$% &' " 1, 7, #1$% &'

42+102

+ (–2)2 12+ 72

+ (–1)2

cos ! =76

120 51

cos ! " 0.9715

! = cos#1(0.9175)

! = 13.7o

The angle between the vectors is about 14º.



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


projv

!u

!

= u

!

cos !

= 25cos 38o

" 19.7

The projection has magnitude 19.7 and has the same direction as v .


W = F

!"

! s

"

= F

!"

s

"

cos "

100 = F

!"

(3)cos 10o

F

!"

=100

3cos10o

F

!"

= 33.8

Roni applies a force with a magnitude of about 33.8 N.



a)

b)

c)



AB! "!!

= OB! "!!

!OA! "!!

OB! "!!

= AB! "!!

+ OA! "!!

= 6, 3, ! 2"# $% + 1, 4, 5"# $%

= 7, 7, 3"# $%

The terminal point is B (7, 7, 3).


cos ! =a

!

"b!

a

!

b

!

cos ! =

7, 2, 4#$ %& " '6, 3, 0#$ %&

72+ 22

+ 42 (–6)2+ 32

+ 02

cos ! ='36

69 45

! = cos'1 '36

69 45

(

)*+

,-

! " 130.2o

cos ! =a

!

"c!

a

!

c

!

cos ! =

7, 2, 4#$ %& " 4, 8, 6#$ %&

72+ 22

+ 42 42+ 82

+ 62

cos ! =68

69 116

! = cos'1 68

69 116

(

)*+

,-

! " 40.5o

cos ! =b

!

"c!

b

!

c

!

cos ! =

#6, 3, 0$% &' " 4, 8, 6$% &'

(–6)2+ 32

+ 02 42+ 82

+ 62


cos ! =0

45 116

! = cos"1 0( )

! = 90o

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°.


L.S. = u

!

! v

!

+ w

"!

( )

= u1, u

2, u

3"# $% ! v

1, v

2, v

3"# $% + w

1, w

2, w

3"# $%( )

= u1,u

2,u

3"# $% ! v

1+ w

1, v

2+ w

2, v

3+ w

3"# $%

= u1

v1+ w

1( ) + u2

v2

+ w2( ) + u

3v

3+ w

3( )

= u1v

1+ u

1w

1+ u

2v

2+ u

2w

2+ u

3v

3+ u

3w

3

R.S. = u

!

!v!

+ u

!

!w"!

= u1, u

2, u

3"# $% ! v

1, v

2, v

3"# $% + u

1, u

2, u

3"# $% ! w

1, w

2, w

3"# $%

= u1v

1+ u

2v

2+ u

3v

3+ u

1w

1+ u

2w

2+ u

3w

3

= u1v

1+ u

1w

1+ u

2v

2+ u

2w

2+ u

3v

3+ u

3w

3

L.S. = R.S.

Therefore, ( )u v w u v u w! + = ! + !

! ! "! ! ! ! "!.


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 14o

! 174.7 km/h.

A top-view sketch shows the velocities.

The vector for the plane is [ ]174.7, 0 .


The wind vector is

15cos 45o , 15sin 45o!"

#$ ! 10.61, 10.61!" #$ .

The resultant vector is [ ]185.31, 10.61 .

R

!"

= 185.31( )2

+ 10.61( )2

# 185.6

! = tan"1 10.61

185.31

#

$%&

'(

! 3.3o

The ground velocity is about 185.6 km/h on a bearing of 86.7º.


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

= 0

You do not have to check b a!

! ! since this vector is the opposite of a b!

! ! and has the same direction.


The area of a triangle is given by A =

1

2PQ! "!!

!QR! "!!

PQ! "!!

= OQ! "!!

!OP! "!!

= !2, 5, 7"# $% ! 3, 4, 8"# $%

= !5, 1, !1"# $%

QR! "!!

= OR! "!!

!OQ! "!!

= !5, !1, 6"# $% ! !2, 5, 7"# $%

= !3, ! 6, !1"# $%


A =1

2!5, 1, !1"# $% & !3, ! 6, !1"# $%

=1

21(–1)! (–6)(–1), (–1)(–3)! (–1)(–5), (–5)(–6)! (–3) 1( )

=1

2!7( ) , ! 2, 33"

#$%

=1

2(–7)2

+ (–2)2+ 332

! 16.9

The area of the triangle is approximately 16.9 units2.


Let [ ] [ ]1, 0, 1 , 0, 1, 2 , and 3u v k= = ! ! =

! !.

L.S. = ku

!

( )! v

!

= 3 1, 0, 1"# $%( )! 0, &1, & 2"# $%

= 3, 0, 3"# $% ! 0, &1, & 2"# $%

= 0(–2)& (–1)(3), 3(0)& (–2)(3), 3(–1)& 0(0)"# $%

= 3, 6, & 3"# $%

R.S. = u

!

! 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.


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 = w

!"

!u"

" 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 units3.


!

!

= r

!

F

"!

sin "

26 = (0.35)95sin "

sin " =26

0.35( )95

" # 51.4o

The force is applied to the wrench at about a 51.4º angle.


AB! "!!

= 7, 1, 17!" #$

A vector equation is

x, y, z!" #$ = 5, 0, 10!" #$ + t 7, 1, 17!" #$ , t %! .


Possible parametric equations are:

x = 5+ 7t

y = t

z = 10 +17t, t !!


b)

Chapters 1 to 8 Course Review Question 82 Page 514 a) The equation is of the form 4 8 0x y C+ + = . Substitute the coordinates of P0 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 6 7 0x y C! ! + = . Substitute the coordinates of P0 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 &! .


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 %! .


Choose arbitrary values for two of the coordinates and solve for the third.

If 0 and 0, then 2.

If 2 and 1, then 0.

x y z

x y z

= = = !

= = ! =

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

!

= m

"!

1 ! m

"!

2

= 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 9 10 2 0x y z D! ! + = .

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 #


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 #


!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!125

7

"

#$%

&'+ 4 !

6

7

"

#$%

&'

= !4


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 #


!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 !123

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.



AB! "!!

= !9 ! 5, 10 ! 2, 3! 8"# $%

= !14, 8, ! 5"# $%

BC! "!!

= !2 ! (–9), (–6)!10, 5! 3"# $%

= 7, !16, 2"# $%

n

!

= AB" !""

! BC" !""

= "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 64 7 168 0x y z D+ ! + = .

Use the point A(5, 2, 8) to determine C.

64(5) + 7(2)!168(8) + D = 0

D = 1010

The scalar equation is 64 7 168 1010 0x y z+ ! + = .

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.

[ ]7, 8, 5n = !

!

The vector equation of the line is

x, y, z!" #$ = 6, 1, % 2!" #$ + t 7, % 8, 5!" #$ , t &! .


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 "n!

2

n

!

1 n

!

2

cos ! =

8, 10, 3#$ %& " 2, ' 4, 6#$ %&8, 10, 3#$ %& 2, ' 4, 6#$ %&

cos ! =8(2) +10(–4) + 3(6)

82+102

+ 32 22+ (–4)2

+ (6)2

cos ! ='6

98.4276

! = cos'1 '6

98.4276

(

)*+

,-

! " 93.5o

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

3s + 4t = 1 !

!3+ 4s = 4 + 5t

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 4!!3"

t = –17

31


Substitute t = –

17

31 into equation ! to solve for s.

3s + 4 –17

31

!

"#$

%&= 1

3s –68

31= 1

3s =99

31

s =33

31

The lines intersect at

33

31,–

17

31

!

"#$

%&.

Chapters 1 to 8 Course Review Question 91 Page 514 Let P1(2, –7, 3), P2(3, 4,–2), [ ] [ ]1 23, 10, 1 , and 1, 6, 1m m= ! =

!" !".

P

1P

2

! "!!!

= 1, 11, ! 5"# $%

n

!

= m

"!

1 ! m

"!

2

= "16, " 2, 28#$ %&

d =

1, 11, ! 5"# $% & !16, ! 2, 28"# $%

!16, ! 2, 28"# $%

d =178

1044

d ! 5.51


The distance between a point A and a plane is given by

d =

n

!

!AQ" !""

n

! where Q is any point on the plane

with normal n

!.

Choose Q(3, 0, 0) on the plane.

AQ! "!!

= 3, 0, 0!" #$ % 2, 5, % 7!" #$

= 1, % 5, 7!" #$


d =

3, ! 5, 6"# $% & 1, ! 5, 7"# $%

32+ (–5)2

+ 62

=70

70

! 8.37

The distance is approximately 8.37 units.


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.


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.

2 3

2 2 4 17

2 0

x y z

x y z

x y z

+ + =

+ + =

+ + =

b) The planes must have identical equations (except for a scalar multiple).

2 3

2 2 4 6

2 3

x y z

x y z

x y z

+ + =

+ + =

! ! ! = !

c) The normals must be coplanar (e.g., 1 2 35 3n n n= +

! ! !) and the constant terms must satisfy the same

relationship as the normals (in this case, 1 2 35 3D D D= + ).

4 6 2 0

3 5 2 4 0

2 0

x y z

x y z

x y z

! + + =

! + + =

! + ! =


d) The normals must be non-coplanar. (i.e., 1 2 3 0n n n! " #

! ! !)

Choose n

!

1 = 1, 3, ! 2"# $% , n

!

2 = 1, 0, 0"# $% , n

!

3 = 0, 1, 0"# $% .

n

!

2 ! n

!

3 = 0, 0, 1"# $%

n

!

1 !n!

2 " n

!

3 = 1, 3, # 2$% &' ! 0, 0, 1$% &'

= #2

( 0

x + 3y ! 2 = 8

x = 14 and y = !12


a) First examine the normals. None are scalar multiples of each other.

Check if the normals are coplanar.

[ ] [ ] [ ]

[ ] [ ]

1 2 3 2, 4, 5 3, 2, 1 4, 3, 3

2, 4, 5 3, 5, 3

11

n n n! " = # ! # " #

= # ! #

= #

! ! !

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


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.

[ ] [ ] [ ]

[ ] [ ]

1 2 3 5, 4, 2 3, 1, 3 7, 7, 7

5, 4, 2 28, 42, 14

0

n n n! " = ! # "

= ! #

=

! ! !

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

Substitute 1

2 and 7

x t z t= + = into equation .

31

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 (! .

chapters 1 to 8 course review question 1 page...

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