Provide a parametrization with the given properties:
(a) The curve is circled at point (a, b). It is traced once counterclockwise, starting at the point (a+r, b) with $t \in [0, 2 \pi]$
(b) The line segment, connecting points in the direction from $(1,4,0)$ to $(6,7,-2)$
(c) The curve of intersection of the cone $z = \sqrt{x^2 + y^2}$ and the ellipsoid $3x^2 + y^2 + z^2 = 2y$ in counterclockwise direction
attempt:
(a) $x = a + r cos(t), y = a+rsin(t)$ for $t \in [0, 2\pi]$
(b) $x = (1-t(1)) + 6(t) = 1 + 5t$, $y = (1-t)(4) + 7t = 4+3t, z = (1-t)(6) - 2(t) = 6-8t$
therefore $(1+5t, 4+3t, 6-8t)$ for $0 \leq t \leq 1$
(c)
$3x^2 + y^2 + z^2 = 2y \to 4x^2+2y^2 = 2y \to 2x^2 + y^2 - y = 0$
$\to 2x^2 + y^2 - y + 1/4 - 1/4 = 0 \to 2x^2 + (y-1/2)^2 = 1/4 \to 8x^2 + 4(y-1/2)^2 = 1$
therefore $x = cos(t)\sqrt{8}$, $y = 1/2 + sin(t)/2$
hence, $(\sqrt{8}cos(t), 1/2 + 1/2 sin(t))$ for $0 \leq t \leq 2 \pi$
Is this correct?
(a) $y = b + \sin t,$ I think that was just a typo.
(b) $z(0)$ should be $0$; you looked at $x(1)$ instead of $z(0).$
(c) Since $z=\sqrt{x^2+y^2} > 0,$ we can say that the intersection of the two is given by $$ \sqrt{x^2 + y^2} = \sqrt{2y - y^2 - 3x^2} \implies 4x^2 +2(y^2-y) = 0 $$ We can rewrite this to form a perfect square as $$ 4x^2+2 \left( y - \frac{1}{2} \right)^2 = \frac{1}{2} \implies \frac{x^2}{\left(8^{-1/2}\right)^2}+\frac{\left( y - \frac{1}{2} \right)^2}{\left(2^{-1}\right)^2} = 1, $$ which is the equation of an ellipse centered at $(0,1/2)$. Thus we have the parameterization for $t \in (0, 2\pi)$ as $$ \vec{r}(t) = \begin{pmatrix} \displaystyle \frac{\cos t}{\sqrt{8}} \\ \displaystyle \frac{1}{2} + \frac{\sin t}{2} \end{pmatrix}. $$