MATH Assignments (a) Fill out the given table using the parametric equations (I encourage you to use a calculator on some of these). (b) Sketch the curve, using arrows to indicate the direction of the curve as t increases. (c) Find a rectangular-coordinate equation for the curve by eliminating the parameter. Question 1 x=3t+1; y=3-2t

Answer 

x=3t+1;      y=3-2t

At  t = -2; 

x=3(-2)+1;      y=3-2(-2)

x=-5 ;      y=7

At  t = -1; 

x=3(-1)+1;      y=3-2(-1)

x=-2 ;      y=5

t x y

-2 -5 7

-1 -2 5

0 1 3

1 4 1

2 7 -1

Sketch the curve

Rectangular-coordinate equation

x=3t+1

   y=3-2t

3t=x-1

t=(x-1)/3

Put the value of t in y equation 

      y=3-2t

      y=3-2((x-1)/3)

y=3-(2x-2)/6

y=(18-2x-2)/6

6y=18-2x-2

6y+2x-16=0

2x+6y-16=0

Question 2 

x=4-6t;      y=-3t

t x Y

0 4 0

1 -2 -3

2 -8 -6

3 -14 -9

4 -20 12

Sketch the curve

Rectangular-coordinate equation

x=4-6t

   y=-3t

-6t=x-4

-t=(x-4)/6

t=(-x+4)/6

Put the value of t in y equation 

      y=-3t

      y=-3((-x+4)/6)

y=(3x-12)/18

18y=3x-12

3x-18y-12=0

Question 3 

x=t+1 ;   y=t/(t+1)

t x y

-10 -9 1.11

-2 -1 2

-1.01 -0.01 101

-0.99 0.01 -99

0 1 0

1 2 0.5

10 11 0.90

Sketch the curve

Rectangular-coordinate equation

x=t+1 

   y=t/(t+1)

t=x-1

Put the value of t in y equation 

        y=t/(t+1)

   y=(x-1)/(x-1+1)

   y=(x-1)/x

yx=x-1

x-xy-1=0

Question 4

x=t^4+1 ,           y=t^2

Solution 

t x y

-2 17 4

-1.5 6.06 2.25

-1 2 1

0 1 0

1 2 1

1.5 6.06 2.25

2 17 4

Sketch the curve

Rectangular-coordinate equation

x=t^4+1  

   y=t^2

t^4=x-1

〖(t^2)〗^2=x-1

Taking square root on both side 

t^2=√(x-1)

Again Taking square root on both side 

t=x-1

Put the value of t in y equation 

        y=t^2

y=(x-1)^2

y=x^2-2x+1

y-x^2+2x-1=0

Question 5

x=sec^2 t ,           y=cost

-π/2<t<π/2

cos^(-1) (y)= t

x=1/(Cos^2 (cos^(-1) (y)))

x=1/y^2 

Solution 

t x y

-5π/12 16 0.25

-π/3 4 0.5

-π/4 2.04 0.70

-π/6 1.35 0.86

0 1 1

π/6 1.35 0.86

π/4 2.04 0.707

π/3 4 0.5

5π/12 16 0.25

Sketch the curve

Question 6

x=1+3Sint ,                  y=3Cost-4

Solution 

t x y

0 1 -1

π/6 2.5 -1.40

π/4 3.12 -1.87

π/3 3.59 -2.5

π/2 4 -4

3π/4 3.12 -6.12

π 1 -7

5π/4 -1.12 -6.12

3π/2 -2 -4

7π/4 -1.12 -1.87

Sketch Curve 

Rectangular-coordinate equation

x=1+3Sint ,   

x-1=3Sint  

Sint =(x-1)/3

  y=3Cost-4

y+4=3Cost

Cost=(y+4)/3

sin^2 t+Cos^2 t=1

((x-1)/3)^2+((y+4)/3)^2=1

(x^2-2x+1)/9+(y^2+8y+16)/9=1

(x^2-2x+1+y^2+8y+16)/9=1

x^2-2x+1+y^2+8y+16=9

x^2+y^2-2x+8y+17=9

Chapter 8

Question 66 

Find the indicated power by using De-Moivers Theorem 

(1-i)^10 

Solution 

z=(1-i)^10

r= √((1)^2+(-1)^2 )

r= √2

θ= arctan(-1/1)

θ=-π/4

DeMoviers Theorem

z^n= [r(Cosθ+iSinθ)]^n=r^n (cosnθ+iSin(nθ)

z^n= [√2 (Cos(-π/4)+iSin(-π/4))]^10=〖√2〗^10 (cos⁡(10) (π/4)+iSin(-10)(π/4)

(1-i)^10= 32i

Question 67

Find the indicated power by using De-Moivers Theorem 

(1+i)^7 

Solution 

z=(1+i)^7

r= √((1)^2+(1)^2 )

r= √2

θ= arctan(1/1)

θ=π/4

DeMoviers Theorem

z^n= [r(Cosθ+iSinθ)]^n=r^n (cosnθ+iSin(nθ)

z^n= [√2 (Cos(-π/4)+iSin(-π/4))]^7=〖√2〗^7 (cos⁡(7) (π/4)+iSin(7)(π/4)

z^n=〖√2〗^7 (√2/2+i(-√2/2)

(1+i)^7==11.31(√2/2+i(-√2/2)

(1+i)^7==7.99-7.99i

Question 72  

       (-1/2-√3/2 i)^15  

Solution 

z=(-1/2-√3/2 i)^15

r= √((-1/2)^2+(-√3/2)^2 )

r= 1

θ= arctan((-√3/2)/(-1/2))

θ=π/3

DeMoviers Theorem

z^n= [r(Cosθ+iSinθ)]^n=r^n (cosnθ+iSin(nθ)

z^n= [1(Cos(π/3)+iSin(π/3))]^15=1^15 (cos⁡(15) (π/3)+iSin(15)(π/3)

=1(-1+0)

(1+i)^15= -1

Question 38 

Path of projectile 

v_o=2048ft/s;     θ=30°

Solution 

After how many seconds will bullet hit ground?  

Formula of  Component V_x  ,V_y of velocity and components x, y of the displacement is given by; 

V_y=V_o Sin(θ)-16t^2

Height of the projectile is given by the component y, and it reaches its maximum value when the component Vy is equal to zero.

V_o Sin(θ)-16t^2=0 

t^2=(V_o Sin(θ))/16

t^2=(2048ft/s)Sin(30°)/16

1ft= 0.3048

t^2=19.5s

t=4.41sec

What is the maximum height attained by bullet?

y(4.41sec)=(2048ft/s)Sin(30°)-16〖(4.41sec)〗^2

y(4.41sec)=312.11-16〖(4.41sec)〗^2

y(4.41sec)=312.11-274.23

y(4.41sec)=37.87

y(4.41sec)=37.87

maximum height= y=8.58m

How far from the gun will bullet hit the ground? 

S = V_o Cost

S = (2048ft/s)Cos(4.41sec)

S = 624.23Cos(4.41sec)

S = 622.38