1.) Which of the following polar equations is equivalent to the parametric equations below?
x=t^2
y=2t
A.) r=4cot(theta)csc(theta)
B.) r=4tan(theta)sec(theta)
C.) r=tan(theta)sec(theta)/4
D.) r=16cot(theta)csc(theta)
2.) Which polar equation is equivalent to the parametric equations below? x=sin(theta)cos(theta)+cos(theta)
y=sin^2(theta)+sin(theta)
A.) r=cos(theta)+1
B.) r=sin(theta)+1
C.) r=sin^2(theta)+1
D.) r=1-cos(theta)
nup
To find the equivalent polar equation given a set of parametric equations, we need to eliminate the parameter (t or theta) and express x and y solely in terms of r and theta.
1.) For the given parametric equations:
x = t^2
y = 2t
To eliminate t, we can express t in terms of r and theta. From the first equation, we have t = √x. Substitute this into the second equation:
y = 2√x
Next, rewrite the equation in terms of r and theta:
r sin(theta) = 2 √(r^2 cos^2(theta))
Now, square both sides of the equation to eliminate the square root:
(r sin(theta))^2 = 4 r^2 cos^2(theta)
Expanding and simplifying:
r^2 sin^2(theta) = 4 r^2 cos^2(theta)
Finally, divide both sides by r^2 to get the equivalent polar equation:
sin^2(theta) = 4 cos^2(theta)
Comparing this with the options:
A.) r = 4 cot(theta) csc(theta)
B.) r = 4 tan(theta) sec(theta)
C.) r = tan(theta) sec(theta)/4
D.) r = 16 cot(theta) csc(theta)
None of the options match the equivalent polar equation sin^2(theta) = 4 cos^2(theta). Therefore, none of the given options are correct.
2.) For the given parametric equations:
x = sin(theta)cos(theta) + cos(theta)
y = sin^2(theta) + sin(theta)
To eliminate theta, we can express sin(theta) and cos(theta) in terms of r and theta using trigonometric identities.
From x = sin(theta)cos(theta) + cos(theta), we can rewrite it as:
x = cos(theta)(sin(theta) + 1)
Now, express sin(theta) in terms of r and theta using the identity: sin(theta) = r sin(theta):
x = r cos(theta) sin(theta) + r cos(theta)
From y = sin^2(theta) + sin(theta), we can rewrite it as:
y = sin(theta)(sin(theta) + 1)
Next, substitute sin(theta) = r sin(theta) into the equation:
y = r sin^2(theta) + r sin(theta)
Now, use the identity sin^2(theta) = 1 - cos^2(theta):
y = r(1 - cos^2(theta)) + r sin(theta)
Simplifying:
y = r - r cos^2(theta) + r sin(theta)
Finally, rearrange the equation to get the equivalent polar equation:
r cos^2(theta) + r sin(theta) = r - y
Comparing this with the options:
A.) r = cos(theta) + 1
B.) r = sin(theta) + 1
C.) r = sin^2(theta) + 1
D.) r = 1 - cos(theta)
The equivalent polar equation is r = 1 - cos(theta), which matches option D. Therefore, the correct answer is D.) r = 1 - cos(theta).
like t = y/2
then x = t^2 = y^2/4
so
y^2 = 4 x
but
y = r sin T
x = r cos T
so
r^2 sin^2 T = 4 r cos T
r sin^2 T = 4 cos T
r =4 cos T/sin^2T = 4 cot T csc T