Convert the curve to an equation in rectangular coordinates:
x=sin theta + 1
y=2cos theta - 1
sinθ = (x-1)
cosθ = (y+1)/2
sin^2 θ = (x-1)^2
cos^2 θ = (y+1)^2/4
(x-1)^2 + (y+1)^2/4 = 1
Looks like an ellipse with center (1,-1)
Steve, thank you so very much!!! I am *trying* to wrap up my studying/homework and am having trouble with one more... It's asking the exact same question for the curve:
x=ln(t), y=ln sqrt(t)
A BIG THANK YOU AHEAD OF TIME!!!!! =)
x=ln(t)
y = ln(sqrt(t)) = 1/2 ln(t)
so, y = 1/2 x
Find the arc length of the curve described by the parametric equation over the given interval:
x=t^(2) + 1
y=2t - 3
...0 < t < 1
To convert the curve given by parametric equations to an equation in rectangular coordinates, we need to eliminate the parameter theta and express x and y only in terms of x and y.
We have the parametric equations:
x = sin(theta) + 1 ---(1)
y = 2cos(theta) - 1 ---(2)
To eliminate theta, we can use a trigonometric identity: sin^2(theta) + cos^2(theta) = 1
We can square equation (1) and square equation (2), then substitute these squared equations into the above trigonometric identity.
Squared Equation (1):
x^2 = (sin(theta) + 1)^2
= sin^2(theta) + 2sin(theta) + 1
Squared Equation (2):
y^2 = (2cos(theta) - 1)^2
= 4cos^2(theta) - 4cos(theta) + 1
Now, substitute the squared equations into the trigonometric identity:
x^2 - 1 = sin^2(theta) + 2sin(theta) ---(3)
y^2 + 1 = 4cos^2(theta) - 4cos(theta) ---(4)
We can rearrange equation (3) to get sin(theta) in terms of x:
sin^2(theta) + 2sin(theta) = x^2 - 1
(sin(theta))^2 + 2sin(theta) - (x^2 - 1) = 0
Now, let's solve this quadratic equation for sin(theta) using the quadratic formula:
sin(theta) = (-b ± √(b^2 - 4ac)) / (2a)
where a = 1, b = 2, and c = -(x^2 - 1)
sin(theta) = (-2 ± √(4 - 4(1)(-(x^2 - 1)))) / (2(1))
sin(theta) = (-2 ± √(4 + 4x^2 - 4)) / 2
sin(theta) = (-2 ± √(4x^2)) / 2
sin(theta) = (-2 ± 2x) / 2
sin(theta) = -1 ± x
Similarly, from equation (4), we can solve for cos(theta):
cos(theta) = (1 ± √(1 - y^2)) / 2
Now, we have expressions for sin(theta) and cos(theta) in terms of x and y.
Finally, we can convert these expressions back into rectangular coordinates:
sin(theta) + 1 = x
2cos(theta) - 1 = y
Substituting the expressions for sin(theta) and cos(theta):
(-1 ± x) + 1 = x
2(1 ± √(1 - y^2)) - 1 = y
Simplifying these equations, we can get the explicit equation of the curve in rectangular coordinates.