math

1

SOLVED PROBLEMS FOR BASIC

MATHEMATICS By DR Hamdy Abu Basha Egyptian Aviation Academy January 2013

2

Algebra 1- Solve the equation : (a) 15 x 2 – 12 = - 8 x −b ± b −4ac 2



x=

∴x =

2a

−8 −28 = 30

15 χ 2 +8 χ – 12 = 0 −(8) ± (8) −4(15)(−12)

a = 15 b = 8

2

=

or

2(15) −8 +28 x= 30

=

c = -12

−8 ± 64 +720 30

=

−8 ±28 30

(b) 15 x2 – 14 = 29x 15 x2 – 29 x – 14 = 0 continue as in (a) ____________________________________________________________ (c) 2 x ( 4x + 15 ) = 27 8x2 + 30 x – 27 = 0 continue as in (a) ____________________________________________________________ (d) x (3x + 10 ) = 77 3x2 + 10 x – 77 = 0 continue as in (a) ____________________________________________________________ (e) x2 + 4x + 2 = 0 a = 1 , b = 4 , c = 2 continue as in (a) ______________________________________________________________ (f) x2 – 6x - 3 = 0 see problem (e) 2 χ χ (g) 2 -3 -4 =0 see problem (e) 2 (h) 3x + 5x + 1 = 0 see problem (e) 2- Solve the inequality and express the solution in terms of intervals whenever possible . (a)

2 χ + 5 < 3χ - 7

- χ < - 12

(b) 3 ≤

χ > 12

2 χ −3 5

(f)

χ +1 2 χ- 3

> 2

1

χ −2

-2x ≥-7

(i) (j)

≥

15 ≤ 2 x – 3 < 35 Add 3 for all terms

divide by 2

9 ≤ x < 19

x+1 >2 (2–x)

∴χ ∈[ 9 , 18 [

x+1 >4–2x

x+2x > 4-1

x > 1 ∴χ ∈ ] 1 , ∞ [

3x> 3

(g)

∴χ ∈ ] 12 , ∞ [

< 7

18 ≤ 2x < 38

2 χ - 3 χ < - 7-5

3

x+1 ≥ 3( x – 2 ) x + 1≥ 3 x - 6 2x ≤7 x ≤ 7/2 ∴χ ∈ ] − ∞ , 7 / 2 ]

χ +1

1 3 ≥ χ −2 χ +1 2 2 ≤ 2 χ +3 χ −5

See Problem (g) See Problem (g)

x - 3x ≥ - 6 - 1

3 (k) -2 <

4 χ +1 ≤0 3

-6–1<4x ≤-1

(l) 3-

χ −2 ≤4 3χ +5

Multiply by 3

divide by 4

- 6 < 4 x +1 ≤ 0 - 7/ 4 < x ≤ -1/4

Add – 1 for each term

∴χ ∈] - 7/ 4 , - 1/4 ]

See Problem (f)

Sketch the graph of the equation

System of Linear Equations

4 Solve the equations 1-

 y = 3x + 1   4x + 2 y = 7

[ Ans. ( 0.5 , 2.5) ]

Solution

Arrange the equations in the form 3x - y = -1 4x + 2y = 7 The solution requires the values of three determinants: The denominator, D, formed by writing the coefficients of x and y in order

The numerator of x, Nx, formed from D by replacing the coefficients of x by the constant terms

The numerator of y, Ny, formed from D by replacing the coefficients of y by the constant terms

Solve the equations 2-

 2x + y = 4   3x + 4 y = 1

[ Ans. ( 3 , - 2 ) ]

Solution: See Problem ( 1 )

3-

 5x + 2 y = 2   3x − 5 y = 26

[ Ans. ( 2 , - 4 ) ]

4 - Solve the equations

 x + 3 y + 2 z = − 13   2 x − 6 y + 3z = 32  3x − 4 y − z = 12 

[ Ans. ( -2 , - 5 , 2 ) ]

Solution The solution requires the values of four determinants: The denominator,

5

The numerator of x,

The numerator of y,

The numerator of z,

Then

5 - Solve the equations

 2x − 3y + 2z = 6   x + 8 y + 3z = − 31  3x − 2 y + z = − 5  Solution: We evaluate

[ Ans. ( -5 , - 4 , 2 ) ]

6

Then

6 - Solve the equations

 2x + y = 2   z − 4y = 0  4x + z = 6  Solution:

[ Ans. ( 0.5 , 1 , 4 ) ]

____________________________________________________________ Solve the equations

 x + 2 y + 2z = 4  7 -  3x − y + 4 z = 25  3x + 2 y − z = − 4 

 2x − 3 y + 5z = 4  8 -  3x − 2 y + 2 z = 3  4x + y − 4z = − 6 

[ Ans. ( 2 , - 3 , 4 ) ]

Solution: See Problems ( 4.5.6 )

[ Ans. ( -

1 2 4 , , 3 3 3

    9-     

) ]

1 2 1 + + =2 x y z 3 4 2 − − =1 x y z 2 5 2 + − =3 x y z

[ Ans. ( 1 , 3 , 3 ) ]

7

Analytical Geometry 1- Given A ( -2, 3 ) and B ( 4, -2 ) find (a) d ( A, B ) (b) The midpoint of the segment A B

Solution: (a) [6]

d ( A, B ) = ( x 2 − x1 ) 2 + ( y 2 − y1 ) 2 2

+[ −5 ]

=

[ ( 4) −( −2 ) ]

2

+[ ( −2) −(3) ]

2

=

2

=

36 +25

=

61

( b ) The midpoint of the segment A B is

(

x1 + x2 y1 + y 2 , 2 2

)=(

− 2 + 4 3 + ( −2) , 2 2

)

2 - Find an equation of the circle that has center ( -2, 3 ) and contain the point ( 4, 5 ) .

Solution: The radius of the required circle is the distance between the two points ( -2, 3 ) and ( 4, 5 ) which given by r = d ( A, B ) = ( x 2 − x1 ) 2 + ( y 2 − y1 ) 2 = [ ( 4) −( −2 ) ] 2 +[ ( 5) −( 3) ] 2 = = 36 +4 = 40 The equation of the circle that has center ( h , k ) and radius r is given by ( x − h ) 2 +( y − k ) 2 = r 2 then The equation of the required circle is ( x − ( −2) ) 2 +( y − 3 ) 2 = ( 40 ) 2 or (x+2)2 + (y-3)2 = 40

[6]

2

+[ 2 ]2

3 - Find the linear equation for the line through ( 1. 7 ) and ( -3, 2 ) .

Solution: The linear equation for the line passing through two points A ( x 1 . y 1 ) and B ( x 2 . y 2 ) is given by

y2 − y1 y − y1 = x −x1 x2− x1

then the equation of the required line is

y −7 x −1

=

2 −7 − 3 −1

_______________________________________________________ 4 - Find the linear equations for lines through ( 3, 4 ) that are parallel to the Line 3x - 4 y = 9 .

Solution: The linear equation for the line passing through the point A( x1. y1 ) and a slope m is given by y- y1 = m ( x – x1 ) and the required line has the same slope of the given line whose slope is m

=

−A −3 −3 3 = = = B ( −4) ( −4) 4

then the equation of the required line y- 4 =

3 4

( x – 3)

5 - If A ( 4, -3 ) and B ( 6, 2 ) find (a) d ( A, B ) (b) The midpoint of A B

Solution: See Problems ( 1 ) _____________________________________________________________

8 8 - Find an equation of the circle that has center ( -4, 6 ) and passing through the point ( 1, 2 ) .

Solution:

See Problems ( 2 )

11- Find an equation of the line whose χ - intercept is 4 and y-intercept is -3 .

Solution: The required line passing through the points (4.0) and (0.-3) then continue as problems ( 3 )

12 - Find an equation the line passing through the points ( 5, 2 ) and ( -1, 4 ) .

Solution:

See Problems ( 2 )

_________________________________________________________________________________________________________________________

13 - Find an equation of the line passing through ( 2. -4 ) and parallel to the line 5 χ -2 y = 4 .

Solution: See Problems ( 4 ) 14 - Find an equation of the line passing through ( 7, -3 ) and perpendicular to the line 2 χ - 5 y = 8 .

Solution: The linear equation for the line passing through the point A( x1. y1 ) and a slope m is given by

y- y1 = m ( x – x1 ) and the required line has a slope m1 = line whose slope is m2 = required line y- (-3)=

−5 2

−A −2 2 = = B ( −5) 5

then m1 =

−1 m2 where m2 is the slope of the given −1 −5 = then the equation of the m2 2

( x – 7)

15 - Find an equation for the perpendicular bisector of the segment AB where A ( 3. -1 ) and B ( -2, 6 )

Solution: The equation of the required line passing through the midpoint of AB which given by (x3 x1 + x2 y1 + y 2 3 + ( −2) ( −1) + (6) 1 5 , , )=( ) = ( 2 , 2 ) and has a slope 2 2 2 2 y − y 6 − ( − 1 ) 7 −1 2 1 m1 = m where m2 = x − x = − 2 − 3 = − 5 the slope of the given segment , then 2 2 1

, y3 ) = (

the slope of the given line is m1 = (y - y3) = m1 ( x – x3 ) Then

−1 m2

=

5 7

y-

5 2

=

and the equation of the required line is 5 7

(x-

Functions

1 2

)

9 If f ( χ ) = χ 2 - 1 and g ( χ ) = 3 χ + 5 , find ( f o g ) ( χ ) and ( g o f ) ( χ ) ( 1 ) and . the domain of each

Solution: f o g ) ( χ ) = f ( g ) = g2 – 1 = ( 3 x + 5 ) 2 – 1 g o f ) ( χ ) = g ( f ) = 3 f + 5 = 3 ( x2 – 1 ) + 5 If f ( χ ) = χ 2 -16 and g ( χ ) = . the domain of each

its domain is R ) its domain is R )

find ( f o g ) ( χ ) and

χ

( g o f ) ( χ ) and ( 2 )

Solution: ] ∞ , f o g ) ( χ ) = f ( g ) = g2 – 16 = ( ] ∞ .g o f ) (χ )= g (f ) = f = If f ( χ ) = χ 3 +1 and g ( χ ) = . the domain of each

3

χ )2 x

χ −1

2

– 16 = x – 16 its domain is [ 0 ) - 16 its domain is ] - ∞ , - 4 ] U [ 4 )

find ( f o g ) ( χ ) and ( g o f ) ( χ ) and ( 3 )

Solution: f o g ) ( χ ) = f ( g ) = g3 + 1 = ( 3 χ −1 )3 + 1 = (x – 1) + 1 = x its domain is R ) g o f ) ( χ ) = g ( f ) = 3 f −1 = 3 ( x 3 + 1 ) −1 = x its domain is R ) If f ( χ ) =

χ 3 χ +2

2

and g ( χ ) = χ . and the domain of each

find ( f o g ) ( χ ) and (g o f ) ( χ ) ( 10 )

Solution: 2 ) 2 1 x χ {f o g ) ( ) = f ( g ) = = = = 6 + 2 x 3 + x its domain is R – {-3 , 0 ) 2 3 ( ) +2 x 2 2 2( 3 x + 2 ) 6x + 4 2 gof)(χ )=g(f)= f = ( x ) = = x its domain is R – { - 3 , ) x 3x + 2 (

g 3 g +2

{0

If f ( χ ) =

χ χ −2

3

and g ( χ ) = χ find ( f o g ) ( χ ) . and the domain of each

and ( g o f ) ( χ ) ( 11 )

Solution: {fog)(χ )=f(g)=

g g −2

(

=

3 ) x

=

3 3 −2 x

its domain is R – {

3 ( ) −2 x 3 3( x − 2) 3 3x − 6 {gof)(χ )=g(f)= x = ( x ) = = x x x −2

3 2

,0 )

its domain is R – { 0 , 2 )

10

Logarithms and Exponents 1- Solve the equation then

(

1 2

)5 −x

-5+x=1

(2 )

= 2

−1

5 −X

= 2

( 2 ) − 5 +X =2

x=6

_____________________________________________________________________________________________________________________

2 - Solve the equation

(2)

−100 x

= (0.5)

χ−4

(2)

−100 x

χ−4

(2)

−100 x

=

− x +4

(2)

then

- 100 x = - x + 4 2

3 - Solve the equation

log X

4 - Solve the equation

log χ 2

x2 + 3 x + 2 = 0

4 =

x=1

5

5

.5

−1

+5

5 2x (

2x

5

2 x

− 126 ( 5 )

χ2

-3 χ - 2

=

x = - 2 and x = - 1 4

x

.4 +4

x

. 4 −1 =17

. 5 =130

− 126 ( 5 )

x

x = - 4/ 99

Solve the equation – 6 5 2 x ( 5 −1 + 5 ) = 130

x=1 2

25

= ( 2 −1 )

X = 100

log ( -3 χ - 2 )

4 x +1 +4 x −1 =17

26 ) = 130 5

x=2

−100 x

4x

5 2 x −1 +5 2 x +1 =130 2x

(2)

- 99 x = 4

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

5 – Solve the equation

χ−4

10 4 = X 2

= 4

=

1

= (2 )

x

x −1

52

+5 = 0

5 2 x ( 5 −1 + 5 ) = 130

=5 2x

Solve the equation – 7

. ( 5 ) −1 +5 = 0

Multiply by 5 then 5 . 5 2 x − 126 ( 5 ) x +25 = 0 consider 5 x = y then our equation will be 5 y 2 – 126 y + 25 = 0 ( 5 y -1 ) ( y – 25 ) = 0 y=

y=1 Or

y = 25

1 = 5 −1 5

5

x

y = 52

8 - Solve the equation

x=-1 5

=5 −1

x=2

5 x =5 2

5 (5)

χ−2

5x

= 1

−2

=5

−1

x–2=-1

x=2–1=

1

9 - Solve the equation [9]

−

3 2

log χ+1 3

+ x + 1 = [ 9 ]−2

2

9=- 3

x = -1

( x +1 ) [ ( x +1 )

−

2 3

−

2 3

=9 ]

−

3 2

=[ 9 ]

−

3 2

11 10 - Solve the equation x 2 - x - 12 = 0 x–4=0 OR x+3=0

log 2

3

2 (χ - χ ) = 2

3 ) 2 = x 2 −x

(2

(x–4)(x+3)=0 x=4 x=-3

11 - Solve the equation

log 8 0.25 = 9 χ 2 – 5 χ

8 (9 x

x 2 - 15 x = - 2 27 x 2 - 15 x + 2 = 0 27 Here a = 27 . b = - 15 , c = 2

2

−5 x )

= 0.25

2 −2 2 3 ( 9 x

2

−5 x )

=

−b ± b 2 −4ac −( −15) ± ( −15) 2 −4( 27)( 2) 15 ± 225 − 21 6 15 ±3 = = = 2a 2( 27) 54 54 15 +3 15 −3 18 1 12 2 ∴x = = or x= ∴x = = or x= = 54 54 54 3 54 9 

x=

12 - Solve the equation x2– 3 x – 1 = 0 1 4

x=-

(x2

log 2

3 x) = -2 4

(4x+1)(x–1)=0 OR

x–1=0

13 - Solve the equation

log

x = 2-2 =

4x+1=0

1 4

4x=-14

x=1

χ = - 2 3

0.064

3 4

x2 –

( 0.064 )

−

2 3

( 0.04 )

=x

−2

=x

x=

16

14- Solve the equation log 3

(3 )

1 x 2

=3

2

15- Solve the equation

(3 )

27

3 x 2

log 3

16- Solve the equation log

9 = χ

(

3 2

=32

3

10

=

x

=9

x=2

χ = -2 3

10

27 )

(3

1 χ 6

( 27 )

x= 3)

( 10 )

−2

1 x 6

1 x 2

=9

4 3

x=

=x 1

= 10 3

1 6

1 27

x=

1 3

=2

17- Solve the equation

log χ

1 = -1 6

x

−1

=

1 6

1 1 = x 6

x=6

x

12

Trigonometric Identities 1 - Verify the identity L.H.S. = ( 1 − sin 2 t

2

2

2

t sec 2 t = cos 2 t

1 = 1 = R.H.S cos 2 t

sec β − cos β = tan β sin β

1 1 −cos 2 β sin 2 β sin β −cos β = = = sin β = tan β sin β cos β cos β cos β cos β

-Verify the identity

sec 2 θ 1 + cot 2 θ

1 cos 2 θ csc 2 θ

sec 2 θ = L.H.S. = 1 + cot 2 θ

4-Verify the identity L.H.S. =

2

) ( 1 + tan t ) = cos

2-Verify the identity sec β − cos β =

( 1 − sin t ) ( 1 + tan t ) = 1

cot χ + tan χ =

2

=

tan θ

3

=

1 cos 2 θ 1 sin 2 θ

=

sin 2 θ cos 2 θ

=

2

tan θ

= .L.H.S

= R.H.S

cot χ + tan χ = sec χ csc χ cos x sin x cos 2 x +sin 2 x 1 + = = = sec χ csc χ sin x cos x sin x cos x sin x cos x

=

R.H.S

6-Verify the identity

csc 2 θ = cot θ − cot 2 θ

cos θ cos 2θ sin 2 θ cos θ −cos 2θ sin θ − = sin θ sin 2 θ sin θ sin 2 θ sin ( 2θ −θ ) sin θ 1 = = = =csc 2 θ = sin θ sin 2 θ sin θ sin 2 θ sin 2 θ

cot θ − cot 2 θ =

L.H.S

7- Verify the identity

= .R.H.S

sin 2 θ sec θ + cos 2 θ csc θ = csc θ

sin 2 θ cos 2 θ cos 2 θ cos θ +sin 2 θ sin θ + = cos θ sin θ sin θ cos θ cos ( 2θ −θ) cos θ 1 = = = = csc θ sin θ cos θ sin θ cos θ sin θ

sin 2 θ sec θ + cos 2 θ csc θ =

.R.H.S =

10- Verify the identity

csc 2 A +cot 2 A =cot A

1 cos 2 A 1 +cos 2 A 1 +cos 2 A −sin 2 A csc 2 A +cot 2 A = + = = sin 2 A sin 2 A sin 2 A sin 2 A

.R.H.S

= .L.H.S

= .L.H.S

cos 2 A + cos 2 A 2 cos 2 A cos A = = 2 sin A cos A 2 sin A cos A = sin A =cot A

11- Verify the identity L.H.S = cos .=R.H.S

4

A − sin

4

A =( cos

13- Verify the identity tan χ + cot χ =

cos 2

4

A −sin

A − sin

2

4

A =cos 2 A

A ) (cos

2

A + sin

2

A ) = cos 2 A −sin

2 csc 2 χ=tan χ+cot χ

sin χ cos χ sin 2 χ + cos 2 χ 1 + = = cos χ sin χ sin χ cos χ sin χ cos χ

= .R.H.S

2

A =cos 2 A

13 2

2

L.H.S = = 2 sin χ cos χ = sin 2 χ =2 csc 2 χ

14- Verify the identity

cot A −tan A =2 cot 2 A

cos A sin A cos 2 A −sin 2 A cos 2 A cot A −tan A = − = = sin A cos A sin A cos A sin A cos A 2 cos 2 A 2 cos 2 A R.H.S = =2 sin A cos A = sin 2 A =2 cot 2 A

16- Verify the identity

1 −tan cos 2 A = 1 +tan

2 2

= .L.H.S

A A

sin 2 A 1− 2 1 − tan 2 A cos 2 A − sin 2 A cos 2 A cos A = = = = cos 2 A = L.H.S .R.H.S. = 2 2 2 2 sin A cos A + sin A 1 + tan A 1 1+ cos 2 A

________________________________________________________________

Trigonometric Equations 1- Solve the equation 2 sin u =1 −sin u 2 sin 2 u + sin u -1 = 0 sin u – 1 ) ( sin u + 1 ) = 0 2 sin u = 1 sin u = 1/2 u = 30 0 2 ) OR sin u + 1 = 0 sin u = - 1 u = 270 0 OR u = - 90 0 2

3- Solve the equation 2 tan t −sec t =0 2 tan t – ( tan 2 t + 1 ) = 0 tan 2 t – 2 tan t + 1 = 0 ( tan t – 1 ) 2 = 0 tan t = 1 t = 45 0 2

4- Solve the equation sin 2 t +sin t =0 sin t ( 2 cos t + 1 ) = 0 sin t = 0 OR 2 cos t + 1 = 0

2 sin t cos t + sin t = 0 t=0 cos t = - 1/2 t = 120 0

5- Solve the equation cos u +cos 2 u =0 cos u + 2 cos 2 u – 1 = 0 cos 2 u + cos u – 1 = 0 ( 2 cos u – 1 ) ( cos u + 1 ) = 0 2 cos u – 1 = 0 cos u = 1/2 u = 60 0 2 OR cos u + 1 = 0 cos u = – 1 u = 180 0 8- Solve the equation cos 2 θ−cos θ+1 =0 2 cos 2 θ – 1 – cos ө = 0 cos 2 θ – cos θ – 1 = 0 ( 2 cos θ +1 ) ( cos θ – 1 ) = 0 2 θ = 120 0 2 cos θ + 1 = 0 cos θ = – 1/2

14 OR

cos

θ

–1=0

cos

θ

θ

= 1

=00

12- Solve the equation cos θ −cos 2 θ =0 cos2 θ – ( 2 cos 2 θ – 1 ) = 0 cos 2 θ + 1 = 0 cos 2 θ – 1 = 0 θ= 0 0 ) cos θ – 1 ) ( cos θ + 1 ) = 0 cos θ – 1 = 0 cos θ = 1 θ = 180 0 OR cos θ + 1 = 0 cos θ = – 1 2

14- Solve the equation 3 tan 2 x + 1 = 2

2 tan

x +sec

2

2

2 tan 2 x + tan 2 x + 1 = 2 tan 2 x = - 1/3 tan x = ±

x =2

3 tan 2 x = - 1

15- Solve the equation sin 2 C − 3 cos C =0 2 sin C cos C - 3 cos C = 0 cos C ( 2 sin C - 3 ) = 0 cos C = 0 C = 90 0 OR 2 sin C - 3 = 0 2 sin C = 3 sin C = 3 / 2 C = 60 0 16- Solve the equation 2 sin y =1 +cos y 2 ( 1 – cos 2 y ) = 1 + cos y 2 – 2 cos 2 y = 1 + cos y 2 cos 2 y + cos y – 1 = 0 cos y – 1 ) ( cos y + 1 ) = 0 2 ) cos y – 1 = 0 cos y = 1/2 y = 60 0 2 OR cos y + 1 = 0 cos y = – 1 y = 180 0 2

17- Solve the equation

1 2 sin β +1 = sin β

2 sin β+1 =csc β

2 sin 2 β + sin β = 1

2 sin 2 β + sin β −1 = 0

sin β – 1 ) ( sin β + 1 ) = 0 2 sin β = 1 sin β = 1/2 β = 30 0 2 ) OR sin β + 1 = 0 sin β = - 1 β = 270 0 OR β = - 90 18- Solve the equation tan 2 x + sec x =1 sec 2 x – 1 + sec x = 1 sec 2 x + sec x – 2 = 0 ( sec x – 1 ) ( sec x + 2 ) = 0 1 =1 sec x – 1 = 0 sec x = 1 cos x = 1 x = 00 cos x sec x + 2 = 0

20- Solve the equation sin α− 2

α =0

1 =2 cos x

sec x = 2

sin αsec α = 2

sin αcos α=0

sin α =0

sin α =

α = 45 0

21- Solve the equation

1= 2

χ +1 −4 = 0 ( tan χ − 1 ) ( tan χ + 1 ) = 0 3 tan

OR

2

tan χ +1 =0

sin α= 2

1 2 2 tan

sin α

sin α

sin α( 1 − 2

2

sin αcos α

1 = 2 sin α cos a

sin α )= 0

sin α

χ +sec 2 χ =4 3 tan 2 χ − 3 = 0 tan χ −1 =0

tan x = - 1

x = 60 0

cos x = 1/2

1− 2

sin α= 0

OR

χ +tan 2 χ +1 =4 tan 2 χ − 1 = 0

2 tan

2

tan x = 1 x = - 45 0

x = 45 0

1 3

15

Complex Numbers − 1

= Prove that ( 6 – 2 j ) + ( 2 + 3 j ) = 8 + j where j - 1

L.H.S. = ( 6 – 2 j ) + ( 2 + 3 j ) = ( 6 + 2 ) + ( - 2 j + 3 j ) = ( 6 + 2 ) + ( - 2 + 3) j = 8 + j = R.H.S − 1

= Prove that ( 6 – 2 j ) – ( 2 + 3 j ) = 4 – 5 j

where j - 2

L.H.S. = ( 6 – 2 j ) – ( 2 + 3 j ) = ( 6 + 2 ) – ( - 2 j + 3 j ) = ( 6 – 2 ) + ( - 2 – 3) j = 4 – 5 j = R.H.S

Prove that ( 2 + 3 j ) ( 1 + 2 j ) = - 4 + 7 j

where j =

L.H.S. - 3 (= (2+3j)(1+2j)=(2)(1)+(2)(2j)+(3j)(1)+(3j)(2j j +3 j + 6 j 2 = 2 + 7 j + 6 ( −1 ) = 2 + 7 j + 6(-1)= 2 + 7 j – 6 4 + 2 = j = R.H.S 7 + 4 - = − 1

2

= Prove that ( 2 - 3 j ) ( 5 + 2 j ) = 16 – 11 j where j - 4 See Problem 3 − 1 = Prove that ( 3 - 2 j ) ( - 4 + j ) = - 10 + 11 j where j - 5 See Problem 3 − 1 = Prove that ( 2 + 3 j ) ( 3 + 2 j ) = 13 j where j - 6 See Problem 3 Prove that ( 2 – j ) 2 = 3 – 4 j - 7 Multiply ( 2 – j ) ( 2 – j ) and continue as in Problem 3 − 1 = Prove that ( 4 + 2 j ) 2 = 12 + 16 j where j - 8 See Problem 7 − 1 = Prove that ( 1 + j ) 2 ( 2 + 3 j ) = - 6 + 4 j where j - 9 2 ( L.H.S. = ( 1 + j ) ( 2 + 3 j ) =( 1 + j ) ( 1 + j ) ( 2 + 3 j ( j + j 2 ) ( 2 + 3 j ) = ( 1 + 2 j + ( −1 ) ) ( 2 + 3 j 2 + 1 ) = ( −1 ) j – 1 ) ( 2 + 3 j ) = 2 j ( 2 + 3 j ) = 4 j + 6 j 2 = 4 j + 6 2 + 1 ) = j = R.H.S 4 + 6 - = − 1

2

2

2 +3 j 1+j

1 j where 2 2 +3 j 1 −j 2 −2 j +3 j −3 j 2 2 + j −3 ( −1 ) 2 × = = 1+ j 1 −j 1−j2 1 − ( −1 ) 2 5 + j = 2 + j +3 = 2 + j −3 ( −1 ) 1 5 j 2 + 2 = = 2 1 +1 1 − ( −1 ) − 1

− 1

= Prove that

= Prove that

3 −2 j 3 −4 j

=

=

5 2

+

17 25

+

6 25

j

j - 10 =

2 +3 j 1+j

where j - 11

= .L.H.S

16 =

3 −2 j 3 −4 j

= .L.H.S

3 −2 j 3 +4 j 9 +12 j − 6 j −8 j × = 3 −4 j 3 +4 j 9 −16 j 2

− 1

=

=

17 + 6 j 25

= Prove that

3 −2 j 2 +3 j

6 25

j= R.H.S

3 −2 j 2 +3 j

17 25

+

=

=

9 + 6 j −8 ( −1 ) 2 9 −16 ( −1 ) 2 9 + 6 j −8 ( −1 ) 9 + 6 j +8 = = 9 +16 9 −16 ( −1 ) =

=-j

where j - 12

= .L.H.S

3 −2 j 2 −3 j 6 −9 j − 4 j +6 j × = 2 +3 j 2 −3 j 4 −9 j 2

j -=

2

−13 j 13

=

6 −13 j − 6 4 +9

6 −13 j +6 ( −1 ) 2 4 −9 ( −1 ) 2 6 −13 j +6 ( −1) = 4 −9 ( −1) 2

=

13 - Show that 3 ( cos 25 0 + j sin 25 0 ) 8 ( cos 200 0 + j sin 200 0 ) = - 12 where j = −1

2

- 12

2

j

L.H.S. = 3 ( cos 25 0 + j sin 25 0 ) 8 ( cos 200 0 + j sin 200 0 ) = ( 3 ) ( 8 ) [ cos ( 200 0 + 25 0 ) + j sin ( 200 0 + 25 0 ) ] = 24 ( cos 225 + j sin 225 ) = 24 [ -

1 2

-

1 2

2 2

j ] = 24 [ -

2 2

-

j ] = - 12

2

- 12

2

j = R.H.S

14 - Show that 4 ( cos 50 0 + j sin 50 0 ) 2 ( cos 100 0 + j sin 100 0 ) = - 4 where j = −1

3

+4j

L.H.S. = 4 ( cos 50 0 + j sin 50 0 ) 2 ( cos 100 0 + j sin 100 0 ) = ( 4 ) ( 2 ) [ cos ( 50 0 + 100 0 ) + j sin ( 50 0 + 100 0 ) ] = 8 ( cos 150 + j sin 150 ) =8[-

3 2

15 - Show that L.H.S. =

1 j ] = - 4 3 + 4 j = R.H.S 2 4 ( cos190 0 + j sin 190 0 ) =-1+ 2 ( cos 70 0 + j sin 70 0 )

+

4 ( cos190 0 + j sin 190 0 ) 2 ( cos 70 0 + j sin 70 0 )

=

4 2

L.H.S. =

1 2

+

3 2

12 ( cos 200 0 + j sin 200 0 ) = 3 ( cos 350 0 + j sin 350 0 )

-2

12 ( cos 200 0 + j sin 200 0 ) 3 ( cos 350 0 + j sin 350 0 )

=

where j =

3

− 1

[ cos ( 190 0 – 70 0 ) + j sin ( 190 0 – 70 0 )]

= 2 ( cos 120 0 + j sin 120 0 ) = 2 ( -

16 - Show that

j

12 3

j)=-1+ j

= R.H.S

where j =

-2 j

3

3

− 1

[ cos ( 200 0 – 350 0 ) + j sin ( 200 0 – 350 0 )]

= 4 [ cos ( - 120 0 ) + j sin ( - 120 0 ) ] = 4 ( –

3 2

–

1 2

j)=-2

3

- 2 j = R.H.S

17 – Convert the coplex number Z = 3 + 3 j into polar form Here x = 3 and y = 3 Ө = tan – 1 (

y x

r=

) = tan – 1 (

3 3

x

2

+ y

2

=

( 3)

2

+ (3)

2

=3

) = tan – 1 ( 1 ) = 45 0

the coplex number in polar form is Z = 3 2 ( cos 45 0 + j sin 45 0 )

Z = r ( cos Ө + j sin Ө )

2

17 18 – Convert the coplex number Z = 1 + Here x = 1 and y = Ө = tan – 1 (

y x

r=

3

) = tan – 1 (

x

3 1

3

+ y

2

19 – Convert the coplex number Z = 2

Ө = tan – 1 (

and y = y x

) = tan – 1 (

2

r=

2

=

(1)

) = tan – 1 (

the coplex number in polar form is Z = 2 ( cos 60 0 + j sin 60 0 )

Here x =

2

j into polar form

−

2

3

2

+(

3)

2

=2

) = 60 0

Z = r ( cos Ө + j sin Ө )

x

2

2 + y

j into polar form 2

=

(

2)

2

+(−

2)

) = tan – 1 ( - 1 ) = - 45 0

2

the coplex number in polar form is Z = 2 [ cos ( - 45 0) + j sin (- 45 0 )]

Z = r ( cos Ө + j sin Ө )

20 – Convert the coplex number Z = - 2 j into polar form Here x = 0 and y = - 2 Ө = tan – 1 (

y x

) = tan – 1 (

r=

−2 0

x

2

+ y

2

=

(0)

2

+ (− 2)

2

=2

) = tan – 1 ( ∞ ) = - 90 0

the coplex number in polar form is Z = 2 [ cos ( - 90 0) + j sin (- 90 0 )]

Limits

Z = r ( cos Ө + j sin Ө )

2

=2

18 Lim

1 - Evaluate 2 - Evaluate χ 5 - Evaluate 7 - Evaluate

( 2 χ− 4 x

χ → − 2

3 x 2 − x −10 x 2 − x −2

Lim

→ 2

1 χ−3

Lim +

χ → 3

Lim

χ → 2

9 - Evaluate Lim

x  → − 3

= 10 - Evaluate

2

3

x − 2 χ −2

= χ Lim

Lim

3

Lim

x−→ −2

2

6 −7 χ (3 + 2 χ ) 4

− 4 − 14

(

1 ( χ −3 χ +9 )

Lim x −→ −3

χ → ∞

=

=

+(−2)

x− 2 x − 2 )( x + 2 )

→ 2

x +3 x 2 +27

2

=

→ 2

1 3 −3

2 ( −2 ) − 4 ( −2 )

( 3 χ+5 ) ( χ−2 ) Lim ( 3 χ +5 ) = χ → 2 ( χ+1 ) ( χ−2 ) ( χ +1 ) 1 = ∞ Limits Does Not Exist 0

= χLim

=

3

=

+χ )

χ +3 3

χ + 27

=

3

=

( −3 )

3

2

1

= χ Lim

→ 2

Lim x −→ −3

(

x + 2)

=

=

11 3

1 2

2

χ +3 ( χ +3 ) ( χ 2 −3 χ +9 )

1 −3 ( −3 ) +9 )

1 3

=

=0

Because the power of numerator less than that of denominator 11 - Evaluate

Lim

χ → ∞

χ − 100 2

χ + 100

=

( χ −100 ) 2

Lim

χ → ∞

χ +100 2

=

Lim

x −→∞

( χ −100 ) 2 χ 2 +100

=

1 1

=1

Because the power of numerator is equal to that of denominator 13 - Evaluate 14 - Evaluate

Lim

χ → 0+

Lim

χ → ∞

x−

1 x

=

8 +χ 2 χ (χ +1 )

0−

1 0

=

Lim

=-

x −→∞

1 0

= - ∞ Limit Does Not Exist

8 +χ 2 χ 2 +χ

1 1

=

=1

Because the power of numerator is equal to that of denominator

15 - Evaluate =

16 1

Lim

χ → − ∞

4 x −3 x 2 +1

=

Lim

χ → − ∞

( 4 χ −3 ) 2

χ +1 2

=

Lim

x −→− ∞

16 χ

−24 χ +9 χ 2 +1 2

=-4

Because the power of numerator is equal to that of denominator , and the minus sign for limit to - ∞

19

Continuity 3 χ −5 2 χ2 −χ −3

1- Find all numbers at which the function f ( x ) =

is continuous

To find the domain of the function , Put the denominator 2 x 2 – x – 3 = 0 (2x–3)(x+1)=0 2x–3=0 x = 3/2 OR x+1=0 x=-1 { The function f ( x ) is continuous at R – { - 1 , 3/2 2 - Find all numbers at which the function f ( x ) =

+ x 2 is continuous

2 χ−3

To find the domain of the function solve the inequality 2x – 3 > 0 ] ∞ , x > 3/2 the domain of the function is ] 3/2 ] ∞ , The function f ( x ) is continuous at ] 3/2 . Find all numbers at which the function f ( x ) =

3

χ χ −4

is continuous - 3

the domain of the function f ( x ) is R – { 4 } , therefore , The function { f ( x ) is continuous at R – { 4 . Find all numbers at which the function f ( x ) =

5 3

χ −χ

2

is continuous - 6

To find the domain of the function , Put the denominator x 3 – x 2 = 0 x2( x – 1 ) = 0 x=0 OR x – 1 = 0 x=1 the domain of the function f ( x ) is R – { 0 , 1 } , therefore , The function { f ( x ) is continuous at R – { 0 , 1 12 - Find all numbers at which the function f ( x ) =

x

2

x −1 + x −2

is continuous

To find the domain of the function , Put the denominator x 2 – x – 2 = 0 (x–2)(x+1)=0 x–2=0 x=2 OR

x+1= 0

x=-1

the domain of the function f ( x ) is R – { -1 , 2 } , therefore , the function { f ( x ) is continuous at R – { -1 , 2 . Find all numbers at which the function f ( x ) =

x

2

x −4 − x − 12

is continuous - 13

To find the domain of the function , Put the denominator x 2 – x – 12 = 0 (x–4)(x+3)=0 x–4=0 x=4 OR

x+3= 0

x=-3

the domain of the function f ( x ) is R – { -3 , 4 } , therefore , The function { f ( x ) is continuous at R – { -3 , 4

Derivatives Find the first derivatives for the function

20 2

x2 f (x) = 3 x −5 3

(1)

f ' (x) =

2 3 −1 x −3 3 ( 3x −5 ) 2

( 3x −5 )

3

x2 3

x −3 x

y =

(2)

−1 3 x ( 2 x −3 ) −( x −3 x ) x 2 ' 2 y = 3 2 ( x ) ( 1 −cos θ ) ( cos θ ) − sin θ [ 0 −( −sin θ ) ] ' f (θ ) = ( 1 −cos θ ) 2 3

2

x3

sin θ f ( θ )= 1 −cos θ

(3)

(1+x f ' ( x)=

tan x 1 +x 2

(4)

f ( x)=

(5)

1 +sec x y= tan x +sin x

2

2

) sec 2 x − tan x ( 0 + 2 x ) (1+x 2 ) 2

( tan x +sin x ) ( 0 + sec x tan x ) −( 1 +sec x ) ( sec 2 x +cos x ) y'= ( tan x +sin x ) 2 h( θ ) =

(6)

1 +sec θ 1 −sec θ

( 1 −sec θ ) ( 0 +sec θ tan θ ) −( 1 +sec θ ) ( 0 −sec θ tan θ ) ' h (θ ) = ( 1 −sec θ ) 2 1 1 y= y'= [ cos x ( −csc 2 x ) + cot x ( −sin x ) ] 2 (7) cos χcot χ ( cos x cot x )

(9)

g ( r ) = 1 +cos 2r

(12)

K( Φ ) =

0 −sin 2 r ( 2 ) g '( r ) = 2 1 +cos 2 r

sin Φ cos Φ −sin Φ

( cos Φ − sin Φ ) cos Φ − sin Φ ( −sin Φ − cos Φ ) K ' ( Φ) = ( cos Φ − sin Φ ) 2 csc u +1 G (u ) = cot u +1

(13)

( cot u +1 ) ( − csc u cot u + 0 ) −( csc u +1 ) ( −csc 2 u + 0 ) ' G (u) = ( cot u +1 ) 2 1 1 + z sec z

(16)

g (z) =csc

(20)

h (x) =(sec x +tan x) 5

(21)

sin 2θ K (θ) = 1 +cos 2θ

g ' ( z ) =− csc

1 1 1 1 cot [ − 2 ] + [ − ] sec z tan z z z z sec 2 z

h ' ( x ) =5 (sec x +tan x )

4

(sec x tan x +sec 2 x )

( 1 + cos 2 θ ) cos 2 θ ( 2 ) −sin 2θ [ 0 −sin 2θ( 2 ) ] K ' (θ ) = ( 1 + cos 2 θ ) 2 sec 2 x f (x) = 1 +tan 2 x

(22) f

'

( x) =

(27) r ' (t ) =

( 1 + tan 2 x ) sec 2 x tan 2 x ( 2 ) −sec 2 x [ 0 +sec 2 2 x ( 2 ) ] ( 1 + tan 2 x ) 2

r( t ) =

sin 2 t − cos 2 t

1 [ cos 2t ( 2 ) −( − sin 2t ) ( 2 ) ] 2 sin 2 t − cos 2 t

21

f ( x ) = sin

(28)

f ' ( χ ) = cos χ [

1

sin x

1 [ cos χ ] 2 sin χ

]+

2 χ

+

x

(29) h (Φ) =(tan 2 Φ−sec 2 Φ) 2 2 h ' (Φ ) = 3 (tan 2 Φ − sec 2 Φ ) [ sec 2 Φ ( 2 ) − sec 2 Φ tan 2 Φ ( 2 ) ] 3

cos 4 w h ( w) = 1 −sin 4 w

(30)

( 1 −sin 4 w ) ( −sin 4 x ) ( 4 ) − cos 4 w [ 0 −cos 4 w ( 4 ) ] h ' ( w) = ( 1 −sin 4 w ) 2

(32)

N (x) =(sin 5 x −cos 5 x) 5

N ' ( x ) =5 (sin 5 x −cos 5 x ) 4 [cos 5 x ( 5 ) − ( sin 5 x ) ( 5 ) ]

(40)

K ( z ) =csc ( z 2 +4 )

(41)

P (v) =sin 4 v csc 4 v

K ' ( z ) =−csc ( z 2 +4 ) cot ( z 2 +4 ) [ 2 z +0 ]

P ( v ) =sin 4 v csc 4 v =sin 4 v

1 =1 sin 4 v

P ' (v ) = 0

g ( χ) = ( χ +

(44)

1

χ

1 4 1 g ' ( χ ) = 5( χ + ) (1 − 2 )

)5

χ

χ

___________________________________________________ Implicit Differentiation 1 - If

sin 2 3 y =χ+y −1

find

y'

Solution : Differentiate both sides sin 3y cos 3y ( 3 y ' ) = 1 + sin 3y cos 3y ( 3 y ' ) - y ' = 1 y'

1 6 sin 3y cos 3y - 1

2 - If

=

y'

-0

Solve for 2 y ' ( 6 sin 3y cos 3y – 1 ) = 1 2

y'

find

x = sin ( x y )

y'

Solution : Differentiate both sides cos ( x y ) [ x y ' + y ( 1 ) ] ( x y ' cos ( x y ) + y cos ( x y ) y'

1 - y cos ( x y ) x cos ( x y )

3 - If y'

=

Solve for = 1 x y ' cos ( x y ) = 1 - y cos ( x y = 1

y'

y = csc ( x y ) find

csc ( x y ) cot ( x y ) [ x

y'

y'

+y(1)]

Solution : Differentiate both sides - =

22 y ' = - x y ' csc ( x y ) cot ( x y ) - y csc ( x y ) cot ( x y ( Solve for y ' ( x y ' csc ( x y ) cot ( x y ) = - y csc ( x y ) cot ( x y + y ' y'[

1 + x csc ( x y ) cot ( x y ) ] = - y csc ( x y ) cot ( x y )

- y csc ( x y ) cot ( x y ) [ 1 + x csc ( x y ) cot ( x y ) ]

=

y'

y 2 + 1 = x 2 sec y

4 - if

find

Solution : Differentiate both sides 2 y Solve for 2 x sec y 2 - x 2 sec y tan y

5 - If (

y')

2

y'

=

y'

y'

+ 0 = x 2 sec y tan y ( - x 2 sec y tan y ( y ' ) = 2 x sec y y'

x 2 sec y tan y ) = 2 x sec y - 2 )

y'

x y = tan y

find

-y ( x - sec

2

=

y)

8 - If

x

y'

- sec 2 y (

y'

+ 2 x sec y

y'

y'

Solution : Differentiate both sides x

Solve for

y')

y')

y'+

=-y

y ( 1 ) = sec 2 y y ' ( x - sec 2 y ) = - y

y'

sin y + y = x

find

y'

Solution : Differentiate both sides cos y ( 1 = y' cos y + 1 ) = 1 ) y ' cos y + 1

y'

12 - If x y 2 = sin ( x + 2 y ) find

y')

+

y'

=1

Solve for

y'

[ y ' Solution : Differentiate both sides x ( 2 y y ' ) + y 2 =sin ( x + 2 y )[ 1 + 2 ( Solve for y ' 2xy y ' + y 2 = sin ( x + 2 y ) + 2 y ' sin ( x + 2 y 2xy y ' - 2 y ' sin ( x + 2 y ) = sin ( x + 2 y ) - y 2 sin ( x + 2 y ) - y 2 2xy - 2 sin ( x + 2 y )

13 - If

=

2xy - 2 sin ( x + 2 y )] = sin ( x + 2 y ) - y 2 ]

y'

y = cot x y

find

y'

y'

[ ( Solution : Differentiate both sides y ' = - csc 2 x y [ x y ' + y ( 1 y ' = - x y ' csc 2 x y - y csc 2 x y y ' + x y ' csc 2 x y = - y csc 2 x y Solve for y ' - y csc 2 x y 1 + x csc 2 x y

=

y'

x csc 2 x y ) = - y csc 2 x y + 1 )

14 - If x 2 + 4 x y – y 2 = 8

find

y'

y'

Solution : Differentiate both sides 2 x + 4[x y ' + y ( 1 ) ] – 2 y Solve for

2x+4x

y'

-4y-2x 4x -2y

=

y'

y'

+4 y – 2 y

y'

=0

y'(

=0 4 x– 2 y ) = - 4 y – 2 x y'

23 15 - if 5 x 3 - 2 x 2 y 2 + 4 y 3 = 7

find

y'

Solution : Differentiate both sides 15 x 2 -2 [ x 2 ( 2y y ' ) + y 2 ( 2 x ) ] +12 y 2 Solve for 2

2

2

15 x - 4 x y

y' 2

4 y x - 15 x - 4 x 2 y + 12 y

2

=

y'

y'-

2

4 y x +12 y

2

y'=

y '=

0

0

x 2 y +12 y 2) = 4 y 2 x – 15 x 2 4 - )

y'

Geometric Applications 1 - Find the x coordinates of all points on the graph of the function y = x 3 + 2 x 2 – 4 x + 5 at which the tangent line is parallel to the line 2y + 8 x = 5 Solution : The slope of the tangent line m1 = The slope of the given line m2 = - A/B = - 8/2 = - 4 . But the slope of the tangent line = y ' = m1 x 2 + 4 x – 4 = – 4 Solve the equation 3 x 2 + 4 x – 4 = – 4 3 = y ' 4 -= y' 2 3x +4x= 0 x(3x+4)=0 x = 0 OR 3 x + 4 = 0 x = - 4/3

3 - If y = 3 + 2 sin x , find the χ coordinates of all points on the graph at which the tangent line is parallel to the line y = 2 χ−5 Solution : The slope of the tangent line m1 = The slope of the given line 2 = y' m2 = 2 But the slope of the tangent line = y ' = m1 2cos x = 2 Solve the equation 2cos x = 2 cos x = 2 / 2 x= +0 = y' 45 0

4 - If y = 1 + 2 cos x find the χ coordinates of all points on the graph at 1 which the tangent line is perpendicular to the line y = 3 χ +4 Solution : The slope of the tangent line m1 = - 1/ m2 where m2 is the slope of the given line , here m2 = 1/ 3 m1 = - 1/ m2 = - 3 . But the slope of the y'= y ' = 0 – 2 sin x 3 - = tangent line = y ' = m1 3 Solve the equation – 2 sin x = - 3 sin x = 3 / 2 x = 60 0 5 - Find the equations of the tangent line and normal line of the equation y =3 χ+sin 3 χ at the point (0.0) Solution : The slope of the tangent line m = y ' at ( 0 , 0 ) • m = y ' = 3 + 3 cos 3x = 3 + 3 cos 3(0) = 6 .

The equations of the tangent line is y – y1 = m ( x – x1 ) OR y – 0 = 6 ( x – 0 ) The equations of the normal line is y – y1 = -1/m ( x – x1 ) OR y – 0 =-1/6 ( x – 0 )

6 - Find an equation of a line through P (-2.3) that is tangent to the

24 ellipse

5 x 2 + 4 y 2 = 56

Solution : The slope of the tangent line m = y ' at ( - 2 , 3 ) . Differentiate both sides of the given equation 10 x + 8 y y ' = 0 10 ( -2 ) + 8 ( 3 ) y ' = 0 , find y ' y '= 20/24 = 5/6 = m The equations of the tangent line is y – y1 = m ( x – x1 ) OR y – 3 =5/6 ( x – ( -2) )

8 - Find the equation of the line through (-2, -2) that is tangent to the hyperbola x 2 - 4 y 2 = - 12 Solution : See Problem 6 9 - Find the equation of the tangent line and normal line to the hyperbola 3 y 2 - 2 x 2 = 40 at the point (2 , - 4) Solution : The slope of the tangent line m = y ' at (2 , - 4 ) . Differentiate both sides of the given equation 6 x - 4 y y' = 0 6 (2 ) - 4 ( -4 ) y ' = 0 , find y ' y '= 12/16 = 3/4 = m The equations of the tangent line is y – y1 = m ( x – x1 ) OR y – (-4) =3/4 ( x – 2 ) The equations of the normal line is y – y1 = -1/m ( x – x1 ) OR y (-4) =-4/3 ( x – 2 )

10 - Find the equation of the tangent line and normal line to the ellipse 9 x 2 + 4 y 2 = 72 at the point (2, 3) Solution : See Problem 9 11 - Find the equation of the tangent line and normal line to the hyperbola 2 x 2 - 5 y 2 = 3 at the point P ( -2 , 1 ) Solution : See Problem 9

Mechanical Applications 1 - A rescue helicopter drops a crate of supplies from a height of 160 ft.

25 After t seconds, the crate is 160 -16 t2 feet above the ground. With what velocity does the crate strike the ground?

Solution : The distance above ground as a function of time is s( t ) = 160 -16 t2 , the velocity as a function of time is v( t ) = s ' ( t ) = 0 – 32 t . At ground s = 0 160 -16 t2 = 0 Solve for t

t=

160 16

=

10

Sec

The velocity does the crate strike the ground is v ( 10 ) = -32 10 ft / sec 2 - A projectile is fired directly upward from the ground with an initial velocity of 112 ft/sec. Its distance above the ground after t seconds is 112 t- 16 t2 feet. Find the velocity at the instant the projectile strikes the ground.

Solution : The distance above ground as a function of time is s( t ) = 112 t -16 t2 , the velocity as a function of time is v( t ) = s ' ( t ) = 112 – 32 t . At ground s = 0 112 t -16 t2= 0 Solve for t t = 112/16 = 7 Sec . The velocity does the projectile strike the ground is v (7)=112 – 32(7) = - 112 ft / sec 3 - An athlete run the hundred-meter dash in such away that the distance 1 is given by s (t) = 5 t2 + 8 t . Find the athlete velocity at the finish line.

Solution : The distance of the athlete as a function of time is 1 s (t) = 5 t2 + 8 t , the velocity as a function of time is v( t ) = s ' ( t ) =

2 5

t + 8 . At at the finish line s = 100 ft

1 5

t2 + 8 t = 100 t2 + 40 t – 500 = 0 ( t + 50 ) ( t – 10 ) = 0 t = 10 Sec .

The velocity of the athlete at the finish line is 2 v ( 10 )= 5 ( 10 ) + 8 = 12 m / sec 6 - A projectile is fired straight upward with a velocity of 400 ft / sec , Its distance above the ground after t seconds is s ( t ) = -16 t2 + 400 t .

Find the maximum altitude achieved by the projectile and velocity at which it hits the ground.

26 Solution : The maximum altitude achieved by the projectile is at v = 0 v = s ' ( t ) = - 32 t + 400 = 0 Solve for t t = 400/32 = 12.5 Sec 2 The maximum altitude is s ( 12.5 ) = - 16 ( 12.5 ) + 400 ( 12.5 ) = 2656.25 ft The projectile hits the ground when s = 0 -16 t2 + 400 t = 0 t = 400/16 = 25 sec velocity at which the projectile hits the ground. is v ( 25 ) = - 32 ( 25 ) + 400 = - 400 ft / sec

Related Rates 1 - A ladder 20 ft long leans against the wall of a vertical building. If the bottom of the ladder slides away from the building horizontally at a

27 rate of 2 ft/s, how fast is the ladder sliding down the building when the top of the ladder is 12 ft above ground?

Solution : From the geometry of the figure x 2 + y 2 = ( 20 ) 2 Differentiate both sides with respect to time

2x

dx dt

+ 2y

Here x = 12 ft , y = 16 ft .

dx dt

the above equation and find ( 12 ) ( 2 )+( 16 )

dy dt

=0

dy dt

=0,

= 2 ft / sec . Apply in dy dt dy dt

24

3

= 16 = 2 ft / sec

3 - A water tank has the shape of an inverted right circular cone of altitude 12 ft and base of radius 6 ft. If water is being pumped into the tank at a rate of 1.337 ft3 /min, approximate the rate at which the water level is rising when the water is 3 ft deep .

Solution : The volume of the cone is given by 1 Vcone = 3 π r 2 h From the geometry of the figure r 6

h 12 1 Vcone = 3 π

=

1 h substitute in the above 2 1 1 1 = 3 π [ 2 h ] 2 h = 12 π h 3

r= r2h

eqn

Differentiate both sides with respect to time dv dt

=

/min. Find (3)2

dh dt

1 12 dh dt

π 3 h2

dh dt

. Here

dv dt

= 1.337 ft3

when h=3ft by substituting in the last eqn . Then

, calculate

1.337 =

1 12

π3

dh dt

5 - The top a silo has the shape of a hemispheres of diameter 20 ft. If it is coated uniformly with a layer of ice and if the thickness is decreasing 1 at a rare of 4 in/hr how fast is the volume of the ice changing when the ice is 2 in. thick ?

28 Solution : The volume of the hemisphere is given by

V=

2 3

π r3 .

Here the radius is 10 ft ×12 in = 120 inch for the smallest hemisphere and 120 + x for the larger .

The volume of the hemisphere of ice is given by

V=

2 3

π ( 120 + x ) 3 +

2 3

π ( 120 ) 3

Differentiate both sides with respect to time dV dt

2 3

=

π 3 ( 120 + x ) 2

dx dt

+ 0 . When x = 2 inches and

by substituting in the last equation . Then

dV dt

=

2 3

π 3 ( 120 + 2 ) 2 (

1 dx = dt 4 1 ) , 4

,

6 - As sand leaks out of a hole in a container, it forms a conical pile whose altitude is always the same as its radius. If the height of the pile is increasing at a rate of 6 in/min, find the rate at which the sand is leaking out when the altitude is 10 in.

Solution : The volume of the cone is given by 1 Vcone = 3 π r 2 h From the geometry of the V=

figure r = h

V=

1 3

dv dt

=

1 3

π ( h2) h

π h 3 Differentiate both sides with respect to

time 1 3

π 3 h2

dh dt

calculate

Optimization Find the maximum volume of – 1 a right circular cylinder that can be inscribed in a cone of altitude 12 cm and base radius 4 cm if the axes of the cylinder and cone

dv dt

when h = 10 in and

dh dt

= 6 in/min

29 . coincide Solution : The volume of the cylinder . of radius r and height h is V = π r 2 h From the geometry of triangle

h = 12 - 3 r

12 − h 12

=

r 4

V = π r 2 ( 12 – 3 r ) = 12 π r 2- 3 π r 3 Differentiate both sides with respect to r dV dr

= 24 π r – 9 π r 2 = 0 . Solve for r

r=

8 3

, find V

2 – A window has a shape of a rectangle semicircle. If the perimeter of the window is 15 feet, find the dimensions that will allow the maximum amount of light to enter.

Solution :To find a relation between x and y , the perimeter of window is the perimeter of semicircle plus The perimeter of the rectangle without the side of length 2x because it is common with the semicircle and the rectangle perimeter P = π x + 2 x + 2 y = 15 ft ( y = 15 - π x + 2 x (1 The area of window is the area of semicircle plus the area of the rectangle A =

1 2

π x 2+ 2 x y

( substitute about y from eqn ( 1 1 2

1

π x 2 + 2 x ( 15 - π x + 2 x ) = 2 π x 2 + 2 x ( ( 15 - π x + 2 x π x 2 + 30 x - π x 2 + 4 x 2 Differentiate both sides with respect to x A=

1 2

( π x +30 - 2 π x + 8 x = 0 solve for x then find y from equation ( 1 =

= dA dx

3 – Find the dimensions of the rectangle of maximum area that can be .inscribed in a semicircle of radius α , if two vertices lie on the diameter

30

Solution :The area of the required rectangle ( is A = 2 x y y = a −x a −x A=2x Differentiate with respect to x 2

2

2x

2

1 2

(1

2

a

2

−x

2

( −2 x ) +2

a 2 −x 2

=0 =

Solve for x , multiply both sides by (x=

a 3

dA dx

- 2 x2 + 2 ( a2 – x 2 ) = 0

a 2 −x 2

, then find y from equation ( 1

8 - A window has the shape of a rectangle surmounted by an equilateral triangle if the perimeter of the window is 12 ft. Find the dimensions of the rectangle that will produce the largest area for the window .

Solution :To find a relation between x and y , the perimeter of window is the perimeter of triangle plus the perimeter of the rectangle without the side of length x because it is common with the triangle and the rectangle perimeter P = 2 x + 2 y + x =3 x + 2 y = 12 ft (y=

12 − 3 x 2

=6-

3 2

x

(1

The area of window is the area of triangle plus the area of the rectangle A = A =

3 4

x 2 + x (6 -

3 2

x)=

Differentiate with respect to x

=0

solve for x then ( find y from equation ( 1

1 2

x 3 4

3 2

x + x y therefore

x2+ 6 x dA dx

=

3 2

3 2

x2

x + 6 - 3x

31 5- Find the dimensions of the rectangle of maximum area that can be inscribed in an equilateral triangle of side α , if two vertices of the . rectangle lie on one of the sides of the triangle

Solution : Consider the dimensions of the required rectangle are 2 x and y ,then , the area of the required rectangle is A = 2 x y . To find a relation between x and y from the geometry of the figure

x 1 a 2

=

3 a −y 2 3 a 2

y=

3 a 2

-

3

x

((1 3 a 2

A=2x( (a - 4

3

-

x =0

x )=

3

3

ax-2

3

x 2 Differentiate with respect to x

solve for x then find y from equation ( 1

3

=

dA dx

6- Of all possible right circular cones that can be inscribed in a sphere of radius α . Find the volume of the one that has maximum volume .

Solution : The volume of the cone is given by 1 Vcone = 3 π r 2 h From the geometry of the figure r2=a2 - (h – a)2 = a2 – h2 – a2 + 2ha =2ha-h2 1 • V = 3 π [2ha-h2] h V=

2 3

π ah2 -

respect to h

1 3

π h3 Differentiate with dV dh

= = 3a - 4 3x = 0

4 3

π ah - π h 2 = 0 solve for h

dA dx

32

Integration Evaluate the integral (1)

2 ∫ ( x + 4) dx

(2)

( x + 3) 2 dx = ∫ x

x

1 1 − +1 2 1 1 − +1 2

+ 6x +

∫( x

=

9x

1 − +1 2

sin4x

2

=

+8 x +16 ) dx

x +6

1 − +1 2

(3) ∫ cos2x dx = ∫ (4) ∫ (sin x + cos x )

∫

2

x 3 8x + 3 2

x +9 dx = ∫ ( x x

1−

1 2

2

+16 x +C

+ 6 +9 x

−

1 2

) dx

=

+C

2 sin 2 x cos 2 x dx cos 2 x

= ∫ 2 sin 2 xdx =

−2 cos 2 x 2

+C

dx = ∫ (sin 2 x + 2 sin x cos x + cos 2 x ) dx = ∫(1 +sin 2 x) dx =

x+

cos 2 x 2

+C

(10) ∫ ( 2 + 5 cos t )

3

=

sin t dt

1 ( 2 +5 cos t ) 3 ( −5 sin t ) dt −5 ∫

1 ( 2 + 5 cos t ) 4 −5 4

=

1

(11) ∫cos 3x (13) ∫ sec

2

3

sin 3 x

dx

3x tan 3x dx

(14) ∫(3 cos 2π t −5 sin dx

(15) ∫ tan 4 x sin 4 x

=

1 3 cos 3x ( sin 3x ) 3∫

1 3

dx

=

1 ( sin 3x ) 3 1 3 +1 3

+1

+C

1 ( tan 3 x ) 2 +C 3 2 3 sin 2π t 5 ( − cos 4 π t ) − +C 4 π t ) dt = 2π 4π − csc 4 x +C = ∫csc 4 x cot 4 x dx = 4 1

= 3 ∫ 3 sec

2

cos 3 x cos 3 x dx = ∫ dx 3 3x sin 3 x sin 2 3 x

(18) ∫ sin

=

3 x tan 3 x dx

= ∫csc

2

3 x cot 3 x dx

=

1 −3 csc 2 3 x cot 3 x dx −3 ∫

=

1 ( cot 3x ) 2 +C −3 2

(22)

2 ∫ ( 4 x +1) (4 x

(23)

3 ∫( x +1)

dx

+ 2 x − 7 ) 2 dx

=

( x + 1) 4 4

(24) ∫cos 3x sin 3x dx = (26) ∫

3

5 x −1

dx

1 1 = 2 ∫ 2( 4 x + 1) (4 x 2 + 2 x − 7) 2 dx = 2

1 5 ( 5 x −1 ) 5∫ cos ϑ

1 3

(28) ∫csc ϑ cot ϑsec ϑ dϑ= ∫ sin ϑ sin ϑ cos ϑ tan 2 ϑ + C

2

+ 2 x − 7) 3 +C 3

+ C

1 3 cos 3 x sin 3 x dx 3∫

=

(4 x

= dx

dϑ

1 ( sin 3 x ) 2 + C 3 2

= =∫

1 ( 5 x − 1) 1 +1 5 3 1 dϑ sin 2 ϑ

1 +1 3

+C

= ∫ sec

2

ϑ dϑ

=

+C

33 (29) ∫

csc w cos w sin w

dw

= ∫csc w cot w

dw

= - csc w + C

The Definite Integral x dx x2+9

4

(1) ∫ 0

4

=

1

2 − ∫ x ( x + 9 ) 2 dx

=

0

4

1 1 2 −2 2 x ( x + 9 ) dx = 2 ∫0 ………… = 1/2 [ 9 + 2 ( 0 ) ] -

1/2

[9 +2(4 )]=

----------------------------------------------------------------------------------------------Π/ 6

∫

( x −sin 5 x ) dx

(2)

−Π/ 6

-----------------------------------------------------------------------------------------------Π/3 ∫ ( 4 sin 2 θ − 6 cos θ ) dθ (4) Π/4

= ---------------------------------------------------------------------------------------------[ −2 cos 2( Π 3 ) −6 sin( Π 3 ) ] − [ −2 cos 2( Π 4 ) −6 sin( Π 4 ) ] 3

(6) ∫ 1

3

2x3 − 4x2 + 5 dx x2

2x2 5 x −1 −4 x + 2 −1

= ∫ 1

(

2x3 4x2 5 − 2 + 2 ) dx 2 x x x

3

= ∫ ( 2 x − 4 + 5 x −2 )

dx

=

1

1

= = [ ( 3 )2– 4 ( 3 ) – 5 / 3 ] – [ ( 1 ) 2– 4 ( 1 ) – 5 / 1 ] -----------------------------------------------------------------------------------------------π/2

(9) ∫ cos x 0

2 ( 3 −5 sin x ) 3

= 2

(13) ∫ 1

π /2

3 − 5 sin x dx = ∫

dx

=

(3 − 5 sin x ) 1 +1 2

0

3 2

π 2

   0 

=

π2

x +1 x 2 +2 x 2

1 +1 2

  0

2 π (3 − 5 sin ) 3 2

2[ ( 2 )

− 5 cos x (3 − 5 sin x )

1 2

dx

3 2

-

2 (3 − 5 sin 0 ) 3

3 2

2

=

+2( 2) ]

1 2

1 2 ( x +1 ) ( x 2 + 2 x ) ∫ 21

− 2[ ( 2 )

2

+2( 2) ]

1 2

1 − 2

−

dx

=

1 +1 2

( x2 +2x ) 1 − +1 2

2

    1

=

2 ( x 2 +2 x )

1 2

2

  1

34 2

2 (12) ∫ x 0

3

x 3 + 1 dx =

2

1 3

1 3 x 2 ( x 3 +1 ) 3 ∫0

dx

=

1 +1 3

1 ( x 3 +1 ) 1 3 +1 3

2

  1 ( x 3 +1 )  = 4 3  3 0

4 3

2

    0

continue as last problems

Differential Equations 1 - Solve the differential equation f '' ( x ) = 5 + 2 sin x subjected to the initial conditions f ( 0 ) = 3 and f ' (0) = 4 .

If f '' ( x ) = 5 + 2 sin x Integrate f ' ( x ) = 5 x – 2 cos x + C 1 To find the constant C 1 , substitute by the given initial conditions • f ' ( 0 ) = 5 ( 0 ) – 2 cos ( 0 ) + C 1 = 4 C1 = 6 • f ' ( x ) = 5 x – 2 cos x + 6 Integrate 5 2

f(x)=

x 2 – 2 sin x +6x + C 2

To find the constant C 2 , substitute by the given initial conditions f(0)=

5 2

f(x)=

( 0 ) 2 – 2 sin ( 0 ) +6( 0 ) + C 2 = 3 5 2

C2 = 3

x 2 – 2 sin x +6x + 3

2 - Solve the equation f '' ( x ) = 6x - 4 where f (2) = 5, f ' (2) = 4 .

If f '' ( x ) = 6x - 4 Integrate f ' ( x ) = 3 x 2– 4 x + C1 To find the constant C 1 , substitute by the given initial conditions •f ' ( 2 ) = 3 ( 2 )2– 4 ( 2 ) + C1 = 4 C1= 0 2 • f ' ( x ) = 3 x – 4 x + 0 Integrate f ( x ) = x 3 – 2 x2+ C2 To find the constant C 2 , substitute by the given initial conditions f ( 2 ) = ( 2 ) 3 – 2 ( 2 )2 + C2 = 5 C2 = 5 3 2 f(x)= x –2x +5 3 - Solve the equation f ' ( x ) = 9 x 2 + 8

where f (-1) =1

If f ' ( x ) = 9 x 2 + 8 Integrate f ( x ) = 3 x 3 + 8 x+ C To find the constant C , substitute by the given initial conditions • f ( -1 ) = 3 ( -1 ) 3 + 8 ( -1 ) + C = 1 C = 12 3 • f ( x ) = 3 x + 8 x+ 12 5 - Solve the equation f ' ( x ) = If

f'(x)=

3

3χ + 2

3

= ( 3χ + 2 )

3χ + 2

1 3

where Integrate

f (2) = 9 . f ( x ) = 3 ( 3χ + 2 ) 4 3

4 3

+C

35 To find the constant C , substitute by the given initial conditions

• f (2)=

4

3 [ 3( 2 ) + 2 ) 3 + C = 9 4

6 - Solve the differential equation

d2y dx 2

4

C=-3

d2y dx 2

=2 cos x

Integrate

3 ( 3χ + 2 ) 3 - 3 4

=2 cos x subjected to the

conditions y=2+6 π and y ' =3 when If

f(x)=

dy dx

x=π . = y ' = 2 sin x+ C 1

To find the constant C 1 , substitute by the given initial conditions • y ' ( π ) = 2 sin π + C 1 = 3 C1= 3 • y ' ( x ) = 2 sin x+ 3 Integrate y = - 2 cos x + 3 x + C 2 To find the constant C 2 , substitute by the given initial conditions y ( π ) = - 2 cos π + 3 π + C 2= 2+6 π C2 = 3 π y = - 2 cos x + 3 x + 3 π 7 - Solve the differential equation

d2y dx 2

the conditions y = 7 and y ' = 2 if If

d2y dx 2

= 3sin x - 4 cos x subjected to x = 0.

=3sin x - 4 cos x Integrate

dy dx

= y ' = - 3 cos x - 4sin x + C 1

To find the constant C 1 , substitute by the given initial conditions • y ' ( 0 ) = - 3 cos 0 - 4sin 0 + C 1 = 2 C1= 5 •y ' ( x ) = - 3 cos x - 4sin x + 5 Integrate y = - 3 sin x + 4 cos x + 5 x + C 2 To find the constant C 2 , substitute by the given initial conditions y ( 0 ) = - 3 sin 0 + 4 cos 0 + 5 ( 0 ) + C 2= 7 C2 = 3 y = - 3 sin x + 4 cos x + 5 x + 3 8 - Solve the equation f '' ( x ) = 4 sin 2x+ 16 cos 4x and f ( 0 ) = 6 , f ' ( 0 ) = 1 .

If f '' ( x ) = 4 sin 2x + 16 cos 4x Integrate f ' ( x ) = - 2 cos 2x + 4 sin 4x + C 1 To find the constant C 1 , substitute by the given initial conditions • f ' ( 0 ) = - 2 cos 2 ( 0 ) + 4 sin 4( 0 ) + C 1 = 1 C1 = 3 • f ' ( x ) = - 2 cos 2x + 4 sin 4x + 3 Integrate f ( x ) = - sin 2x - cos 4x +3 x + C 2 To find the constant C 2 , substitute by the given initial conditions f ( 0 ) = - sin 2 ( 0 ) - cos 4 ( 0 ) + 3 ( 0 ) + C 2 = 3 C2 = 4 f ( x ) = - sin 2x - cos 4x +3 x + 4 9 - Solve the equation f '' ( x ) = 4 sin 2x + 16 cos 4x and f ( 0 ) = 6 , f ' ( 0 ) = 4 See problem 8

36

With Best Wishes : Dr Hamdy Abubasha