# Thread: Curve of intersection with 2 surfaces

1. ## Curve of intersection with 2 surfaces

Show that the curve with vector equation $\displaystyle r(t)=<2cos^{2} t, sin(2t), 2sint>$
is the curve of intersection of the surfaces $(x-1)^2+y^2 = 1$ and $x^2 + y^2 + z^2 = 4$.

Attempt:

(1)
$\displaystyle (x-1)^2+y^2=1$
$\displaystyle y^2 = 1-(x-1)^2$

(2)

$\displaystyle x^2+y^2+z^2=4$

plug (1) into (2) for y

$\displaystyle x^2+1 -(x-1)^2+z^2=4$

Expanding and canceling

$\displaystyle x^2+1-x^2+2x-1+z^2=4$

(3)
$\displaystyle 2x+z^2=4$

vector equation $\displaystyle r(t)=<2cos^{2} t, sin(2t), 2sint>$

(4)
$\displaystyle x= 2cos^{2} t$
$\displaystyle y= sin(2t)$
$\displaystyle z= 2sint$

plug (4) into (3)

$\displaystyle 2(2cos^2(t)) +(2sint)^2=4$
$\displaystyle 4cos^2(t) +4sin^{2}t=4$

divide out the 4

$\displaystyle \frac{4cos^2(t) +4sin^{2}(t)=4}{4}$

$\displaystyle cos^2(t) +sin^{2}(t)=1$

identity
$\displaystyle cos^2(t) +sin^{2}(t) = 1$
therefore 1= 1

Did i even do this correct?
If not where did i go wrong?
Thank you

Hi,