Space Curves, Tangent Vectors & Arc Length

The cylinder is a circle in $x, y$: $x = 2\cos t$, $y = 2\sin t$.

$z = 3 - x = 3 - 2\cos t$.

$\mathbf r(t) = \langle 2\cos t,\ 2\sin t,\ 3 - 2\cos t\rangle$, $0 \le t \le 2\pi$. Check: $x^2 + y^2 = 4$.

$u = \dfrac{y}{2}$, so $y = 2u$.

$x = \dfrac{y^2}{4} = u^2$ and $z = y^3 = 8u^3$.

$\mathbf r(u) = \langle u^2,\ 2u,\ 8u^3\rangle$, all real $u$. Check: $(2u)^2 = 4u^2 = 4x$.

$z = 1$ gives $x^2 + y^2 = 9$: $x = 3\cos t$, $y = 3\sin t$.

$z = 1$.

$\mathbf r(t) = \langle 3\cos t,\ 3\sin t,\ 1\rangle$, $0 \le t \le 2\pi$. Check: $9 + 1 = 10$.

$\mathbf v = \langle 2t, 3t^2, 1\rangle$.

$|\mathbf v| = \sqrt{4t^2 + 9t^4 + 1}$.

$\mathbf a = \langle 2, 6t, 0\rangle$.

At $t = 1$: $\mathbf v = \langle 2, 3, 1\rangle$, speed $\sqrt{14}$, $\mathbf a = \langle 2, 6, 0\rangle$.

$\mathbf v = \langle -\sin t,\ \cos t,\ 3t^2 - 3\rangle$.

$|\mathbf v| = \sqrt{\sin^2 t + \cos^2 t + (3t^2 - 3)^2} = \sqrt{1 + 9(t^2 - 1)^2}$.

$\mathbf a = \langle -\cos t,\ -\sin t,\ 6t\rangle$.

$z'(t) = 3(t^2 - 1)$: negative for $0 \le t < 1$, positive for $t > 1$. Speeds: $\sqrt{10}$ at $t = 0$, $\sqrt{82}$ at $t = 2$.

Product rule in each component: $\mathbf v = \langle \cos t - t\sin t,\ \sin t + t\cos t\rangle$.

The cross terms cancel: $|\mathbf v|^2 = 1 + t^2$, so $|\mathbf v| = \sqrt{1 + t^2}$.

$\mathbf a = \langle -2\sin t - t\cos t,\ 2\cos t - t\sin t\rangle$.

At $t = \pi$: $\mathbf v = \langle -1, -\pi\rangle$, speed $\sqrt{1 + \pi^2}$, $\mathbf a = \langle \pi, -2\rangle$.

$z$: $2t = -4$, so $t_0 = -2$. Check: $x = 4$, $y = -8$.

$\mathbf r'(t) = \langle 2t, 3t^2, 2\rangle$.

$\mathbf r'(-2) = \langle -4, 12, 2\rangle$.

$|\mathbf r'(-2)| = \sqrt{164} = 2\sqrt{41}$, so $\hat{\mathbf T} = \dfrac{1}{\sqrt{41}}\langle -2, 6, 1\rangle$. Line: $\langle 4, -8, -4\rangle + u\langle -4, 12, 2\rangle$.

$t_0 = \tfrac{\pi}{4}$ is given; the point is $\mathbf r\left(\tfrac{\pi}{4}\right) = \left(\tfrac{\sqrt 2}{2},\ 1,\ \tfrac{\pi}{4}\right)$.

$\mathbf r'(t) = \langle -\sin t,\ 2\cos 2t,\ 1\rangle$.

$\mathbf r'\left(\tfrac{\pi}{4}\right) = \left\langle -\tfrac{\sqrt 2}{2},\ 0,\ 1\right\rangle$.

$x = \tfrac{\sqrt 2}{2} - \tfrac{\sqrt 2}{2}u$, $y = 1$, $z = \tfrac{\pi}{4} + u$.

$\mathbf r_1(0) = \mathbf 0$ at $t_0 = 0$; $\mathbf r_2(0) = \mathbf 0$ at $s_0 = 0$.

$\mathbf r_1'(t) = \langle 1, 2t, 3t^2\rangle$, $\mathbf r_2'(s) = \langle \cos s,\ 2\cos 2s,\ 1\rangle$.

$\mathbf r_1'(0) = \langle 1, 0, 0\rangle$, $\mathbf r_2'(0) = \langle 1, 2, 1\rangle$.

$\cos\theta = \dfrac{1}{1 \cdot \sqrt 6}$, so $\theta = \arccos\dfrac{1}{\sqrt 6} \approx 65.9^\circ$.

$\mathbf r'(t) = \left\langle 2,\ 2t,\ \tfrac{1}{t}\right\rangle$.

$|\mathbf r'|^2 = 4t^2 + 4 + \tfrac{1}{t^2} = \left(2t + \tfrac{1}{t}\right)^2$, so $|\mathbf r'| = 2t + \tfrac{1}{t}$ (positive for $t > 0$).

$1 \le t \le e$.

$\displaystyle\int_1^e \left(2t + \tfrac{1}{t}\right)\,dt = \Big[t^2 + \log t\Big]_1^e = (e^2 + 1) - 1 = e^2$.

$\mathbf r'(t) = e^t\langle \cos t - \sin t,\ \sin t + \cos t,\ 1\rangle$.

$(\cos t - \sin t)^2 + (\sin t + \cos t)^2 = 2$, so $|\mathbf r'| = e^t\sqrt{2 + 1} = \sqrt 3\,e^t$.

$(1, 0, 1)$ is $t = 0$; $z = e^t = 2$ at $t = \log 2$.

$\displaystyle\int_0^{\log 2}\sqrt 3\,e^t\,dt = \sqrt 3\,(2 - 1) = \sqrt 3$.

$\mathbf r'(t) = \langle 1 - \cos t,\ \sin t\rangle$.

$|\mathbf r'|^2 = 2 - 2\cos t = 4\sin^2\tfrac{t}{2}$, so $|\mathbf r'| = 2\sin\tfrac{t}{2}$ (not negative for $0 \le t \le 2\pi$).

From $0$ up to $t$; one arch ends at $t = 2\pi$.

$s(t) = \displaystyle\int_0^t 2\sin\tfrac{\tau}{2}\,d\tau = 4 - 4\cos\tfrac{t}{2}$, and $s(2\pi) = 8$.

$\mathbf v(t) = \langle 0, 0, -10t\rangle + \mathbf C$.

$\mathbf v(0) = \mathbf C = \langle 3, 4, 20\rangle$, so $\mathbf v(t) = \langle 3, 4, 20 - 10t\rangle$.

$\mathbf r(t) = \langle 3t, 4t, 20t - 5t^2\rangle + \mathbf D$.

$\mathbf r(0) = \mathbf D = \langle 0, 0, 2\rangle$.

$\mathbf v(t) = \langle 3t^2,\ -e^{-t},\ 2t\rangle + \mathbf C$.

$\mathbf v(0) = \langle 0, -1, 0\rangle + \mathbf C = \langle 1, 0, 0\rangle$, so $\mathbf C = \langle 1, 1, 0\rangle$ and $\mathbf v = \langle 3t^2 + 1,\ 1 - e^{-t},\ 2t\rangle$.

$\mathbf r(t) = \langle t^3 + t,\ t + e^{-t},\ t^2\rangle + \mathbf D$.

$\mathbf r(0) = \langle 0, 1, 0\rangle + \mathbf D = \langle 0, 1, 0\rangle$, so $\mathbf D = \mathbf 0$.

$\mathbf a = \tfrac{1}{2}\mathbf F = \langle -\sin t, 0, 2\rangle$, so $\mathbf v(t) = \langle \cos t, 0, 2t\rangle + \mathbf C$.

$\mathbf v(0) = \langle 1, 0, 0\rangle + \mathbf C = \langle 1, 3, 0\rangle$, so $\mathbf C = \langle 0, 3, 0\rangle$ and $\mathbf v = \langle \cos t, 3, 2t\rangle$.

$\mathbf r(t) = \langle \sin t, 3t, t^2\rangle + \mathbf D$.

$\mathbf r(0) = \mathbf D = \mathbf 0$.