1. parameric equations

x=2-3cos(t)
y=1+2sin(t)

a)it asks for all points for vertical and horizontal tangency, which im not sure we covered in class or i dont understand it

b) eliminate the parameter

c) setup integral for area of surface by generated by revolving around line y=1.

these three questions, i am complety lost , any help will be greatly appreciated

2. Originally Posted by twofortwo
x=2-3cos(t)
y=1+2sin(t)

a)it asks for all points for vertical and horizontal tangency, which im not sure we covered in class or i dont understand it

vertical ... dy/dx = (dy/dt)/(dx/dt) is undefined.

horizontal ... dy/dx = (dy/dt)/(dx/dt) = 0.

b) eliminate the parameter

cos(t) = (2-x)/3

sin(t) = (y-1)/2

now use the fact that sin^2(t) + cos^2(t) = 1

cartesian equation will be an ellipse

c) setup integral for area of surface by generated by revolving around line y=1.

...

3. Originally Posted by skeeter
...
what number are you pluggin into the derivitive to get an undefined tangency and zero tangency...

thanx

4. Originally Posted by twofortwo
what number are you pluggin into the derivitive to get an undefined tangency and zero tangency...

thanx
you have to solve for t

vertical ... dx/dt = 0

horizontal ... dy/dt = 0

5. Originally Posted by twofortwo
x=2-3cos(t)
y=1+2sin(t)

c) setup integral for area of surface by generated by revolving around line y=1.
rotating the ellipse ...

$x = 3\cos{t}$

$y = 2\sin{t}$

about the x-axis (y = 0) will yield the same result.

might be easier to work.

6. if i solve dx/dt=3sint t for t, i get 0 and pi
and if i solve dy/dt for t, i get pi/2 and 3pi/2

my answers arent matching up with yours and im so sorry to keep peestering yu about the same problem but im honestly not getting it.

7. Originally Posted by twofortwo

points on the curve are (x,y) evaluated at those values of t ...

if i solve dx/dt=3sint t for t, i get 0 and pi

x(0) = ?
x(pi) = ?

and if i solve dy/dt for t, i get pi/2 and 3pi/2

y(pi/2) = ?
y(3pi/2) = ?
...