1.6 Curves and their Tangent Vectors

Find the value of $t$ at which the three points occur on the curve.

$(1,25)$, $(-1/\sqrt2,0)$, $(0,25)$.

The curve “crosses itself" when $(\sin t,t^2)$ gives the same coordinate for different values of $t$. When these crossings occur will depend on which crossing you're referring to, so your answers should all depend on $t$.

The curve crosses itself at all points $(0,(\pi n)^2)$ where $n$ is an integer. It passes such a point twice, $2\pi n$ time units apart.

Draw sketches. Don't forget the range that the parameter runs over.

(a) $\vr(y)=\sqrt{a^2-y^2}\,\hi+ y\,\hj$, $0\le y\le a$

(b) $\big(x(\phi),y(\phi)\big) =\big(a\sin \phi ,-a\cos \phi \big)$, $\frac{\pi}{2}\le\phi\le\pi$

(c) $\big(x(s),y(s)\big) =\big(a\cos(\tfrac{\pi}{2}-\frac{s}{a}), a\sin(\tfrac{\pi}{2}-\tfrac{s}{a})\big)$, $0\le s\le\tfrac{\pi}{2}a$

For part (b), find the position of $P$ relative to the centre of the circle. Then combine your answer with part (a).

(a) $(a+a\theta,a)$ (b)$(a+a\theta+a\sin\theta,a+a\cos\theta)$

We aren't concerned with $x$, so we can eliminate it by solving one equation for $x$ as a function of $y$ and $z$ and plugging the result into the other equation.

$z=-\frac12\sqrt{1-\frac{y^2}{2}}-\frac{y}{4}$

To determine whether the particle is rising or falling, we only need to consider its $z$-coordinate.

The particle is moving upwards from $t=1$ to $t=2$, and from $t=3$ onwards. The particle is moving downwards from $t=0$ to $t=1$, and from $t=2$ to $t=3$.

The particle is moving faster when $t=1$ than when $t=3$.

This is the setup from Lemma 1.6.12 in the CLP-3 text. The two quantities you're labelling are related, but different.

The red vector is $\vr(t+h)-\vr(t)$. The arclength of the segment indicated by the blue line is the (scalar) $s(t+h)-s(t)$.

Remark: as $h$ approaches 0, the curve (if it's differentiable at $t$) starts to resemble a straight line, with the length of the vector $\vr(t+h)-\vr(t)$ approaching the scalar $s(t+h)-s(t)$. This step is crucial to understanding Lemma 1.6.12 in the CLP-3 text.

See the note just before Example 1.6.14 in the CLP-3 text.

Velocity is a vector-valued quantity, so it has both a magnitude and a direction. Speed is a scalar–the magnitude of the velocity. It does not include a direction.

To simplify your answer, remember: the cross product of $\va$ and $\vb$ is a vector orthogonal to both $\va$ and $\vb$; the cross product of a vector with itself is zero; and two orthogonal vectors have dot product 0.

(c)

Just compute $|\vv(t)|$. Note that $\big(e^{at}+e^{-at}\big)^2 =e^{2at} + 2 + e^{-2at}$.

(d)

To figure out what the curves look like, first detemine what curve $\big(x(t),y(t)\big)$ traces out. For part (b) this will be easier if trig identities are first used to express $x(t)$ and $y(t)$ in terms of $\sin(2t)$ and $\cos(2t)$.

(a)

The path is a helix with radius $a$ and with each turn having height $2\pi c$.

(b)

The $(x,y)$ coordinates go around a circle of radius $\frac{a}{2}$ and centre $\big(0,\frac{a}{2}\big)$ counterclockwise. At the same time the $z$ coordinate oscillates over the interval between $1$ and $-1$ half as fast. In addition, the curve lies on the intersection of the cylinder $x^2+\big[y-\tfrac{a}{2}\big]^2=\tfrac{a^2}{4}$ and the sphere $x^2+y^2+z^2=a^2$.

Review Lemma 1.6.12 in the CLP-3 text. The arc length should be positive.

(a) $\hat\vT(1) = \frac{(2,0,1)}{\sqrt{5}}$ (b) $\frac{1}{3}\big[5^{3/2}-8\big]$

From Lemma 1.6.12 in the CLP-3 text, we know the arclength from $t=0$ to $t=1$ will be

The notation looks a little confusing at first, but we can break it down piece by piece: $\diff{\vr}{t}(t)$ is a vector, whose components are functions of $t$. If we take its magnitude, we'll get one big function of $t$. That function is what we integrate. Before integrating it, however, we should simplify as much as possible.

2

$\vr(t)$ is the position of the particle, so its acceleration is $\vr''(t)$.

1

Review §1.5 and Lemma 1.6.12 in the CLP-3 text.

(a) $\nicefrac{20}{3}$ (b) $x(t) = -2\pi -2t,\ y(t) = -2\pi t,\ z(t) = \nicefrac{\pi^3}{3} + \pi^2 t$

Review Lemma 1.6.12 in the CLP-3 text.

(a) $\vr'(t) = \big(-3 \sin t, 3 \cos t, 4\big)$ (b) 5

If you got the answer $0$ in part (b), you dropped some absolute value signs.

(a) $\frac{1}{27}\big(10\sqrt{10}-1\big)$ (b) $\frac{2}{27}\big(10\sqrt{10}-1\big)$

$s(t)=\frac{t^3}{3} +\frac{t}{2}$

The integral you get can be evaluated with a simple substitution. You may want to factor the integrand first.

$\frac{8}{27}\Big[\Big(2 + \frac{9}{4}b^m\Big)^{3/2} -\Big(2 + \frac{9}{4}a^m\Big)^{3/2}\Big]$

Given the position of a particle, you can find its velocity.

$|t|$

(a) $\vr(2)=2\hj+4\hk$ (b) any nonzero multiple of $\vr'(2)=2\pi\,\hi+\hj+4\,\hk$

(c) $\Ga$ and $E$ do not intersect at right angles.

$\diff{}{t}\big[|\vr(t)|^2+|\vr'(t)|^2\big]=0$

If $\vr(u)$ is the parametrization of $\cC$ by $u$, then the position of the particle at time $t$ is $\vR(t) = \vr\big(u(t)\big)$.

(a) $\vr(u)=u^3\,\hi+3u^2\,\hj+6u\,\hk$ (b) $7$ (c) $2$ (d) $1$

By Newton's law, $\vF=m\va$.

(a) $\vr(t) = \big(\frac{\pi^2 t}{2}-\frac{t^3}{2}\big)\,\hi + (t- \sin t)\,\hj + \left(\frac{1}{2}e^{2t}-t\right)\,\hk$ (b) $t=\pi$

(c) $-\pi^2\,\hi +2\,\hj + \big(e^{2\pi}-1\big)\,\hk$

Denote by $\vr(x)$ the parametrization of $C$ by $x$. If the $x$–coordinate of the particle at time $t$ is $x(t)$, then the position of the particle at time $t$ is $\vR(t)=\vr\big(x(t)\big)$. Also, though the particle is moving at a constant speed, it doesn't necessarily have a constant value of $\diff{\vx}{t}$.

(a) $21$ (b) $6$ (c) $2\hi+4\,\hj+4\,\hk$ (d) $-\frac{8}{3}\big(2\hi+\,\hj-2\,\hk\big)$

The question is already set up as an $xy$–plane, with the camera at the origin, so the vector in the direction the camera is pointing is $(x(t),y(t))$. Let $\theta$ be the angle the camera makes with the positive $x$-axis (due east). The tangent function gives a clean-looking relation between $\theta(t)$, $x(t)$, and $y(t)$.

$\frac{x(t)y'(t)-y(t)x'(t)}{x^2+y^2}$

$\vr(t)=\vr_0-\frac{e^{-\alpha t}-1}{\alpha}\vv_0 +g\frac{1-\alpha t-e^{-\alpha t}}{\alpha^2}\hk$

(a) $\vr(t)=\llt -\cos t, -\sin t, t\rgt$ (b) $\vv(t)\cdot\va(t)=0$

(c) $\vr(u)=\llt 0,1,-\frac{\pi}{2}\rgt+u\llt -1,0,1\rgt$ (d) True

(a) $x(t)^2+y(t)^2=z(t)^2$ for all $t$

(b) $\text{velocity}= \big[\cos\big(\tfrac{\pi t}{2}\big) -\tfrac{\pi t}{2}\sin\big(\tfrac{\pi t}{2}\big)\big]\hi +\big[\sin\big(\tfrac{\pi t}{2}\big) +\tfrac{\pi t}{2}\cos\big(\tfrac{\pi t}{2}\big)\big]\hj +\hk$ $\text{speed}=\sqrt{2+\frac{\pi^2 t^2}{4}}$

(c) $\llt x,y,z\rgt = \llt 0,1,1\rgt +(t-1)\llt -\frac{\pi}{2},1,1\rgt$

(d) $\frac{2}{\pi}$ seconds

(a) $90^\circ$ (b) $2\sqrt{3}$