# Thread: Vector Function that represents the Curve

1. ## Vector Function that represents the Curve

I just need to check if my work has been done accordingly.

Q: Find the vector function that represents the curve of the intersection of the 2 surfaces.
The cylinder $x^2 + y^2 = 4$ and $z= xy$

Solution:

$x^2 + y^2 = 4 \rightarrow x+y = 2$

$x = cos(t) y=sin(t) 0 \lesseq t \lesseq \frac{pi}{2}$

2. Originally Posted by lllll
I just need to check if my work has been done accordingly.

Q: Find the vector function that represents the curve of the intersection of the 2 surfaces.
The cylinder $x^2 + y^2 = 4$ and $z= xy$

Solution:

$x^2 + y^2 = 4 \rightarrow x+y = 2$

$x = \cos (t) y= \sin (t) 0 \leq t \leq \frac{pi}{2}$
I think you want

$x = {\color{red}2} \cos (t)$

$y= {\color{red}2} \sin (t)$

${\color{red}z = 4 \sin (t) \, \cos (t) = 2 \sin (2t)}$

for $0 \leq t \leq {\color{red}2 \pi}$

3. $x^2 + y^2 = 4 \rightarrow x+y = 2$

$x = {\color{red}2}cos(t) \ \ y={\color{red}2}sin(t)$ $0 \leq t \leq {\color{red}2\pi}$

$z = xy = {\color{red}4}cos(t)sin(t)$

$\therefore x = 2cos(t) \ \ y=2sin(t) \ \ z = 4cos(t)sin(t)$ $0 \leq t \leq {\color{red}2\pi}$

$r(t) = 2cos(t)$ i + $2sin(t)$ j + $4cos(t)sin(t)$ k

$r(t) = 2cos(t)$ i + $2sin(t)$ j + $2sin(2t)$ k

So here it is completed with the changes is this correct?. And sorry for the double post.

4. Originally Posted by lllll
$x^2 + y^2 = 4 \rightarrow x+y = 2$

$x = {\color{red}2}cos(t) \ \ y={\color{red}2}sin(t)$ $0 \leq t \leq \frac{\pi}{2}$

$z = xy = {\color{red}4}cos(t)sin(t)$

$\therefore x = 2cos(t) \ \ y=2sin(t) \ \ z = 4cos(t)sin(t)$ $0 \leq t \leq \frac{\pi}{2}$

$r(t) = 2cos(t)$ i + $2sin(t)$ j + $4cos(t)sin(t)$ k

$r(t) = 2cos(t)$ i + $2sin(t)$ j + $2sin(2t)$ k

So here it is completed with the changes is this correct?. And sorry for the double post.
Yes. By the way, I made an edit on the value of the parameter t .....

If you consider the level curves of the surface z = xy you'll quickly see that it cuts the cylinder in all quadrants .....

5. Originally Posted by mr fantastic
Yes. By the way, I made an edit on the value of the parameter t .....

If you consider the level curves of the surface z = xy you'll quickly see that it cuts the cylinder in all quadrants .....
*Ahem* ....But not on the axes .... The values ${\color{red}t = 0, ~ \frac{\pi}{2}, ~ \frac{3\pi}{2}, ~ 2 \pi} \,$ need to be excluded.

6. Originally Posted by mr fantastic
... you'll quickly see that it cuts the cylinder in all quadrants .....
It looks like that:

7. Originally Posted by earboth
It looks like that:
Holy cow!