Friday, October 7, 2011

A curve is described by the equations {x = 2 + 2 cos t and y = 1 – 3 sin^2 t} with 0 ≤ t

We have to mention that the given equations are called
parametric equations:


x = f(t) and y = g(t), where t is the
parameter.


x = 2 + 2 cos t
(1)


y = 1 – 3 sin^2 t (2)


To
determine the equation of the line, we'll have to eliminate the parameter
t.


Since the equation for y contains the term 3 sin^2 t,
we'll try to obtain the same term, but with opposite sign, in the equation for
x.


The first step will be to isolate the term in t to the
right side and to square raise, both sides, the equation
(1):


(x-2)^2 = (2 cos
t)^2


We'll expand the square from the right
side:


x^2 - 4x + 4 = 4 (cos t)^2
(3)


 We'll isolate 3(sin t)^2 to the right side, in
equation (2):


1 - y = 3(sin t)^2
(4)


We'll multiply (3) by 3 and (4) by
4:


3x^2 - 12x + 12 = 12 (cos t)^2
(5)


4 - 4y = 12(sin t)^2
(6)


We'll add (5)+(6):


3x^2 -
12x + 12+4 - 4y=12 (cos t)^2+12(sin t)^2


We'll factorize by
12 to the right side:


3x^2 - 12x + 12+4 - 4y=12[(sin t)^2 +
(cos t)^2]


From the fundamental formula of trigonometry,
we'll have:


(sin t)^2 + (cos t)^2 =
1


3x^2 - 12x + 12+4 -
4y=12


We'll subtract 12 both
sides:


3x^2 - 12x + 12 + 4 - 4y - 12 =
0


We'll combine and eliminate like
terms:


3x^2 - 12x - 4y + 4 =
0


We'll add 4y both sides and we'll use symmetric
property:


4y = 3x^2 - 12x +
4


We'll divide by 4:


y
=  3x^2/4 - 3x + 1


The equation of the line
described by the parametric equations, x = f(t) and y = g(t),
is:


y =  3x^2/4 - 3x +
1

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