Practice Exam 2

Vector Functions & Motion in Space

Questions10

Time90 min

Points100

SketchRequired for Q1, 7

1Domain & identifying curves8 pts

(a) Find the domain of \(\mathbf{r}(t) = \left\langle \sqrt{t},\ \ln(4 - t),\ \dfrac{1}{t-1}\right\rangle\). (b) Describe and sketch the curve \(\mathbf{r}(t) = \langle 2\cos t,\ 2\sin t,\ 3\rangle\), and the curve \(\mathbf{r}(t) = \langle t,\ t^2,\ t^2\rangle\).

Solution

(a) \(\sqrt t\) needs \(t\ge0\); \(\ln(4-t)\) needs \(t < 4\); \(1/(t-1)\) needs \(t\ne1\). Domain: \([0, 1)\cup(1, 4)\).

(b) \(x^2 + y^2 = 4\) with \(z = 3\) constant: a circle of radius 2 in the plane \(z = 3\), centred on the \(z\)-axis, traversed counterclockwise seen from above.

For the second, \(y = x^2\) and \(z = x^2\), so \(y = z\): the curve lies on the parabolic cylinder \(y = x^2\) and on the plane \(y = z\). It is a parabola in the tilted plane \(y = z\), opening away from the origin.

2Derivative & tangent line10 pts

For \(\mathbf{r}(t) = \langle t^2,\ e^{2t},\ \cos\pi t\rangle\), find \(\mathbf{r}'(t)\), the unit tangent \(\mathbf{T}(0)\), and parametric equations of the tangent line at \(t = 0\).

Solution

\[ \mathbf{r}'(t) = \langle 2t,\ 2e^{2t},\ -\pi\sin\pi t\rangle,\qquad \mathbf{r}(0) = \langle 0, 1, 1\rangle,\qquad \mathbf{r}'(0) = \langle 0, 2, 0\rangle \] \[ \mathbf{T}(0) = \frac{\langle 0,2,0\rangle}{2} = \langle 0, 1, 0\rangle,\qquad \text{Tangent line: } x = 0,\quad y = 1 + 2s,\quad z = 1 \]

At \(t = 0\) the curve is momentarily moving purely in the \(y\)-direction. Note \(\tfrac{d}{dt}\cos\pi t = -\pi\sin\pi t\) — the chain-rule \(\pi\) is the usual casualty.

3Product rule10 pts

Let \(\mathbf{u}(t) = \langle t, t^2, t^3\rangle\) and \(\mathbf{v}(t) = \langle 1, t, e^t\rangle\). Compute \(\dfrac{d}{dt}\big[\mathbf{u}\cdot\mathbf{v}\big]\) at \(t = 1\) using the product rule, then confirm by first forming the dot product.

Solution

\[ \mathbf{u}' = \langle 1, 2t, 3t^2\rangle,\qquad \mathbf{v}' = \langle 0, 1, e^t\rangle \] \[ \frac{d}{dt}[\mathbf{u}\cdot\mathbf{v}] = \mathbf{u}'\cdot\mathbf{v} + \mathbf{u}\cdot\mathbf{v}' = (1 + 2t^2 + 3t^2e^t) + (0 + t^2 + t^3e^t) \] \[ \text{At } t = 1:\quad (1 + 2 + 3e) + (1 + e) = 4 + 4e\approx 14.87 \]

Direct: \(\mathbf{u}\cdot\mathbf{v} = t + t^3 + t^3e^t\), derivative \(1 + 3t^2 + 3t^2e^t + t^3e^t\); at \(t = 1\): \(1 + 3 + 3e + e = 4 + 4e\). ✓

4Projectile from acceleration10 pts

A projectile is launched from the origin at 20 m/s, \(30^\circ\) above horizontal, in the \(xz\)-plane, with \(\mathbf{a} = \langle 0, 0, -9.8\rangle\) m/s². Find \(\mathbf{r}(t)\), the time of flight, the range and the maximum height.

Solution

\[ \mathbf{v}(0) = \langle 20\cos30^\circ,\ 0,\ 20\sin30^\circ\rangle = \langle 17.32,\ 0,\ 10\rangle \] \[ \mathbf{v}(t) = \langle 17.32,\ 0,\ 10 - 9.8t\rangle,\qquad \mathbf{r}(t) = \langle 17.32t,\ 0,\ 10t - 4.9t^2\rangle \]

Flight time: \(z = 0\Rightarrow t(10 - 4.9t) = 0\Rightarrow t = 10/4.9\approx 2.04\) s. Range: \(17.32\times2.04\approx 35.3\) m (or \(v_0^2\sin60^\circ/g = 400(0.866)/9.8 = 35.3\) ✓). Max height: at \(t = 10/9.8 = 1.02\) s, \(z = 10(1.02) - 4.9(1.02)^2\approx 5.10\) m (or \(v_z^2/2g = 100/19.6 = 5.10\) ✓).

5Arc length10 pts

Find the length of \(\mathbf{r}(t) = \left\langle 2t,\ t^2,\ \tfrac13 t^3\right\rangle\) for \(0\le t\le3\).

Solution

\[ \mathbf{r}'(t) = \langle 2,\ 2t,\ t^2\rangle,\qquad |\mathbf{r}'| = \sqrt{4 + 4t^2 + t^4} = \sqrt{(t^2 + 2)^2} = t^2 + 2 \] \[ L = \int_0^3 (t^2 + 2)\,dt = \left[\frac{t^3}{3} + 2t\right]_0^3 = 9 + 6 = 15 \]

The perfect square under the root is the giveaway that this was designed to work — the same trick as Calculus II arc length. Sanity: the chord from \((0,0,0)\) to \((6, 9, 9)\) has length \(\sqrt{36 + 81 + 81}\approx 14.1 < 15\). ✓

6Curvature12 pts

(a) Find the curvature of \(y = x^2\) at \(x = 0\) and at \(x = 1\). (b) Find the curvature of the helix \(\mathbf{r}(t) = \langle 3\cos t,\ 3\sin t,\ 4t\rangle\) and the radius of its osculating circle.

Solution

(a) \(f' = 2x\), \(f'' = 2\):

\[ \kappa(x) = \frac{|f''|}{(1 + f'^2)^{3/2}} = \frac{2}{(1 + 4x^2)^{3/2}},\qquad \kappa(0) = 2,\qquad \kappa(1) = \frac{2}{5^{3/2}}\approx 0.179 \]

Tightest at the vertex, flattening out as the parabola steepens.

(b)

\[ \mathbf{r}' = \langle -3\sin t, 3\cos t, 4\rangle,\quad |\mathbf{r}'| = 5,\qquad \mathbf{r}'' = \langle -3\cos t, -3\sin t, 0\rangle \] \[ \mathbf{r}'\times\mathbf{r}'' = \begin{vmatrix}\mathbf{i}&\mathbf{j}&\mathbf{k}\\ -3\sin t&3\cos t&4\\ -3\cos t&-3\sin t&0\end{vmatrix} = \langle 12\sin t,\ -12\cos t,\ 9\rangle,\qquad |\mathbf{r}'\times\mathbf{r}''| = \sqrt{144 + 81} = 15 \] \[ \kappa = \frac{15}{5^3} = \frac{3}{25} = 0.12,\qquad \rho = \frac{1}{\kappa} = \frac{25}{3}\approx 8.33 \]

Constant — every point of a helix is like every other. Note the osculating radius \(8.33\) is larger than the cylinder radius 3: the helix bends less sharply than its shadow circle because it is also climbing. The general formula is \(\kappa = a/(a^2 + b^2)\).

7The TNB frame10 pts

For \(\mathbf{r}(t) = \langle \cos t,\ \sin t,\ t\rangle\), find \(\mathbf{T}\), \(\mathbf{N}\) and \(\mathbf{B}\) at \(t = \pi/2\). Sketch them on the curve.

Solution

\[ \mathbf{r}' = \langle -\sin t, \cos t, 1\rangle,\quad |\mathbf{r}'| = \sqrt2,\qquad \mathbf{T}(t) = \tfrac{1}{\sqrt2}\langle -\sin t, \cos t, 1\rangle \] \[ \mathbf{T}'(t) = \tfrac{1}{\sqrt2}\langle -\cos t, -\sin t, 0\rangle,\qquad \mathbf{N}(t) = \frac{\mathbf{T}'}{|\mathbf{T}'|} = \langle -\cos t, -\sin t, 0\rangle \]

At \(t = \pi/2\): \(\mathbf{T} = \tfrac{1}{\sqrt2}\langle -1, 0, 1\rangle\), \(\mathbf{N} = \langle 0, -1, 0\rangle\).

\[ \mathbf{B} = \mathbf{T}\times\mathbf{N} = \tfrac{1}{\sqrt2}\begin{vmatrix}\mathbf{i}&\mathbf{j}&\mathbf{k}\\ -1&0&1\\ 0&-1&0\end{vmatrix} = \tfrac{1}{\sqrt2}\langle 0 + 1,\ -(0 - 0),\ 1 - 0\rangle = \tfrac{1}{\sqrt2}\langle 1, 0, 1\rangle \]

Checks: \(\mathbf{T}\cdot\mathbf{N} = 0\) ✓; all three are unit ✓. \(\mathbf{N}\) points from the point \((0, 1, \pi/2)\) straight back toward the \(z\)-axis — the inside of the bend — exactly as the lab shows for the helix.

8Components of acceleration10 pts

For \(\mathbf{r}(t) = \langle t^2,\ 2t,\ \ln t\rangle\), find \(a_T\) and \(a_N\) at \(t = 1\), and verify \(a_T^2 + a_N^2 = |\mathbf{a}|^2\).

Solution

\[ \mathbf{r}' = \langle 2t, 2, 1/t\rangle\to\langle 2, 2, 1\rangle,\quad |\mathbf{r}'| = 3;\qquad \mathbf{r}'' = \langle 2, 0, -1/t^2\rangle\to\langle 2, 0, -1\rangle \] \[ a_T = \frac{\mathbf{r}'\cdot\mathbf{r}''}{|\mathbf{r}'|} = \frac{4 + 0 - 1}{3} = 1 \] \[ \mathbf{r}'\times\mathbf{r}'' = \begin{vmatrix}\mathbf{i}&\mathbf{j}&\mathbf{k}\\ 2&2&1\\ 2&0&-1\end{vmatrix} = \langle -2 - 0,\ -(-2 - 2),\ 0 - 4\rangle = \langle -2, 4, -4\rangle,\quad |\cdot| = 6,\qquad a_N = \frac{6}{3} = 2 \]

Check: \(|\mathbf{a}|^2 = 4 + 0 + 1 = 5 = 1^2 + 2^2\). ✓ Also \(\kappa = 6/27 = 2/9\) and \(a_N = \kappa|\mathbf{v}|^2 = (2/9)(9) = 2\). ✓

9Uniform circular motion10 pts

A particle moves on a circle of radius 0.5 m at constant speed 3 m/s. Find its angular speed, period, and acceleration (magnitude and direction). Then write \(\mathbf{r}(t)\) and confirm the acceleration from \(\mathbf{r}''\).

Solution

\[ \omega = \frac{v}{R} = 6\ \text{rad/s},\qquad T = \frac{2\pi R}{v} = \frac{2\pi}{6}\approx 1.05\ \text{s},\qquad a_T = 0,\quad a_N = \frac{v^2}{R} = \frac{9}{0.5} = 18\ \text{m/s}^2 \]

The acceleration is entirely normal — \(18\) m/s² pointed at the centre — even though the speed never changes.

\[ \mathbf{r}(t) = \langle 0.5\cos6t,\ 0.5\sin6t\rangle,\qquad \mathbf{r}'' = \langle -18\cos6t,\ -18\sin6t\rangle = -36\,\mathbf{r}(t) \]

\(|\mathbf{r}''| = 36\times0.5 = 18\) ✓, and \(\mathbf{r}'' = -\omega^2\mathbf{r}\) points opposite to the position vector, i.e. toward the centre. ✓

10Electron in a magnetic field · EE10 pts

An electron (\(m = 9.11\times10^{-31}\) kg, \(|q| = 1.60\times10^{-19}\) C) enters a uniform field \(\mathbf{B} = 0.05\,\mathbf{k}\) T with velocity \(\langle 4\times10^6,\ 0,\ 2\times10^6\rangle\) m/s. Find the cyclotron frequency, the radius of the orbit, the period, and the pitch of the helix. Which way does it circulate, viewed from \(+z\)?

Solution

\[ \omega = \frac{|q|B}{m} = \frac{(1.6\times10^{-19})(0.05)}{9.11\times10^{-31}}\approx 8.78\times10^{9}\ \text{rad/s},\qquad f = \frac{\omega}{2\pi}\approx 1.40\ \text{GHz} \] \[ R = \frac{v_\perp}{\omega} = \frac{4\times10^6}{8.78\times10^9}\approx 4.6\times10^{-4}\ \text{m} = 0.46\ \text{mm},\qquad T = \frac{2\pi}{\omega}\approx 7.2\times10^{-10}\ \text{s} \] \[ \text{pitch} = v_z T = (2\times10^6)(7.2\times10^{-10})\approx 1.4\ \text{mm} \]

Direction: at the entry point \(\mathbf{v}\times\mathbf{B} = \langle 4\times10^6, 0, \cdot\rangle\times\langle 0,0,0.05\rangle\) has \(y\)-component \(-(4\times10^6)(0.05)\), i.e. \(\mathbf{v}\times\mathbf{B}\) points along \(-\mathbf{j}\). For a positive charge the force would be along \(-\mathbf{j}\) and the orbit clockwise from above; the electron's negative charge reverses it: counterclockwise viewed from \(+z\).

Note that \(\omega\) needed only \(q, m, B\) — a faster electron would trace a bigger circle in exactly the same time.

Scoring

Score	Where you are
90–100	Solid. On to partial derivatives.
75–89	Formulas are right, algebra is leaking — usually in \(\mathbf{r}'\times\mathbf{r}''\). Check \(a_T^2 + a_N^2 = \|\mathbf{a}\|^2\) every time; it catches most of it.
60–74	Confusion between \(\kappa\) forms or between \(\mathbf{N}\) and \(\mathbf{r}''\). Re-read Section 4 with the lab open.
< 60	Go back to Section 2 — componentwise differentiation has to be automatic before anything else in this unit works.

Where the points usually go

Forgetting to normalise \(\mathbf{T}'\) to get \(\mathbf{N}\) (Q7); using \(|\mathbf{r}''|\) as \(a_N\) instead of \(|\mathbf{r}'\times\mathbf{r}''|/|\mathbf{r}'|\) (Q8); claiming zero acceleration at constant speed (Q9); and reading \(\omega = qB/m\) as if it depended on the entry speed (Q10).