Q.14

Question

Show that the parametrization x=2t+1,y=4t2-4 for t[-1,) has the same graph as the one we plotted point by point in the reading.

Step-by-Step Solution

Verified
Answer

The required parameterization of the curves x=2t+1,y=4t2-4 traces exactly the same curve as that of the equations x=sint+1,y=sin2t-4.

1Step: 1 Given information

Consider the parametric curves x=2f+1,y=4s2-4,  t[-1,).

2Step; 2 Calculation

Consider the parametric curves x=2t+1,y=4t2-4,t[-1,).

The objective is to parametrize the curves.

To parameterize the curves eliminate the parameter t from the parametric equations.

Take x=2 t+1

Add -1 on both sides of the equation.

x-1=2t+1-1x-1=2t+1-1x-1=2t

Now divide the equation by 2 .


x-12=2t2x-12=2t2x-12=t


Now substitute t=x-12 in y=4t2-4 then the equation becomes,


y=4x-122-4y=4(x-1)24-4y=4(x-1)24-4

On further simplification,

y=(x-1)2-4y=x2+1-2x-4 since (x-1)2=x2-2x+1y=x2-2x-3


3Step: 3 Further simplification

Since the parametrization of the curves gives the same equation that of x=sint+1,y=sin2t-4.

Now check the parametrization of the curves x=sint+1,y=sin2t-4.

To parameterize the curves eliminate the parameter t from the parametric equations.

Takex=sint+1

Add -1 on both sides of the equation.

x-1=sint+1-1x-1=sint+1t-1x-1=sint

Now substitutesint=x-1 in the equation y=sin2t-4.

Then,

y=(x-1)2-4y=x2-2x+1-4y=x2-2x-3

Thus, the parameterization of the curves x=2t+1,y=4t2-4 traces exactly the same curve as that of the equations x=sint+1,y=sin2t-4.

Hence the required explanation.