2008-01-09 - More Parametrics

Today's nonsense
- Homework due tomorrow. Due 01-10 (tomorrow) pg 716 #4-32even, 33, 39-46

Sketch and write the corresponding rectangular equation. In these examples T can be anything, not just greater than or equal to zero.
9) x = sqrt(T)
y = T-2
T = x^{^(2)} so therefore y = x^{^(2)} - 2 and you are going to have to specify that x > = 0 if T can be any number, or possibly any number, so you have to ask if x can be any possible number.

15) x = e^{^(T)}
y = e^{^(3T)} +1
ln x = T so y = e^{^(3lnx) + 1 --> y = x^{^(3)} + 1 and e^T is always greater than zero so you have to say here that x > 0 for the domain restrictions

21) x = 4sin(2 theta)

y = 2cos(2 theta)

sin^2 2theta + cos^2 2theta = 1 so you can do

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

ellipse, and no domain restrictions. Look at your definition of x, x = 4 sin(2 theta) so sine goes from 1 to -1 so x would go from 4 to -4, and that's all that the ellipse did anyway ...

27) x = 4sec(theta)

y = 3tan(theta)

(tan(theta))^2 + 1 = (sec(theta))^2 from sin^2 + cos^2 = 1

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

hyperbola - because one side is negative (when you solve for the '1'). To sketch the hyperbola, you draw the box. The center is zero, zero in this case. Because you don't have anything like x plus two or x minus three. So in the x direction, you go over four, and in the x direction you go over three each way, and then you make an imaginary box where the end points are the sides, then you draw your diagonals and then you do left-and-right because the x^2 and so these are asymptotes that come and hit off of the side ... this is probably algebra-2 or maybe precal.

29) x = T^3

y = 3 ln T

x^(1/3) = T and so there for y = 3 ln x^(1/3) ... and then y = lnx and x must be greater than zero -- x > 0.

31) x = e^(-T)

y = e^(3T)

ln x = -T, so y = e^(ln(x^(-3))) --> y = x^(-3) ... and x > 0}