Vectors, Space Curves, and Motion

(a) Complete the square in each variable. Group the terms by variable and move the constant across:

$$(x^2-2x)+(y^2+6y)+(z^2-4z)=-5.$$

Add $1$, $9$ and $4$ to both sides to complete the three squares:

$$(x-1{)}^2+(y+3{)}^2+(z-2{)}^2=-5+1+9+4=9.$$

This is the equation of the sphere with centre $C(1,-3,2)$ and radius $r=\sqrt{9}=3$. The right-hand side came out positive, so it is a genuine sphere and not a point or the empty set.

(b) Compare a distance with the radius. By the distance formula, $|OC|=\sqrt{1^2+(-3{)}^2+2^2}=\sqrt{14}$. Since $14>9$, $\sqrt{14}>3=r$, so the origin is farther from the centre than the radius: it lies outside $S$.

Second justification: the original left-hand side equals $(x-1{)}^2+(y+3{)}^2+(z-2{)}^2-9=|PC{|}^2-9$, which is negative inside the sphere, zero on it and positive outside. At the origin it equals $0+0+0+5=5>0$, so outside. (Equivalently $|OC{|}^2=14=9+5$.)

(c) Slice with a horizontal plane. Substituting $z=k$ into the completed-square form gives

$$(x-1{)}^2+(y+3{)}^2=9-(k-2{)}^2.$$

This has solutions exactly when $9-(k-2{)}^2\ge 0$, that is $|k-2|\le 3$, so $-1\le k\le 5$. At the two extremes $k=-1$ and $k=5$ the plane is tangent and the intersection is a single point; at $k=2$ it passes through the centre and cuts a great circle of radius $3$.

For $k=4$ the equation is $(x-1{)}^2+(y+3{)}^2=9-4=5$: a circle lying in the plane $z=4$, with centre $(1,-3,4)$ (directly above $C$) and radius $\sqrt{5}$.

(d) Read the inequalities geometrically. The first inequality is $|PC{|}^2\le 9$, the solid ball of radius $3$ centred at $C(1,-3,2)$. The plane $y=-3$ passes through the centre, and $y\ge -3$ keeps the half of the ball on one side of it. The region is a solid hemisphere (half-ball) of radius $3$: its flat face is the disc of radius $3$ centred at $C$ in the plane $y=-3$, and its curved face bulges toward positive $y$.

(a) Normalise. $|\mathbf{a}|=\sqrt{4+1+4}=\sqrt{9}=3$. The unit vector in the direction of $\mathbf{a}$ is

$$\mathbf{u}=\frac{\mathbf{a}}{|\mathbf{a}|}=⟨\frac{2}{3},-\frac{1}{3},\frac{2}{3}⟩.$$

A vector of length $6$ in the opposite direction is $-6\mathbf{u}=⟨-4,2,-4⟩$. Check: $\sqrt{16+4+16}=\sqrt{36}=6$, and it is a negative multiple of $\mathbf{a}$, so it points the opposite way.

(b) Solve for the unknown vector exactly as for a number. From $\mathbf{a}+2\mathbf{c}=\mathbf{b}$, $\mathbf{c}=\frac{1}{2}(\mathbf{b}-\mathbf{a})=\frac{1}{2}⟨1-2,2+1,-2-2⟩=\frac{1}{2}⟨-1,3,-4⟩=⟨-\frac{1}{2},\frac{3}{2},-2⟩$.

Check: $\mathbf{a}+2\mathbf{c}=⟨2-1,-1+3,2-4⟩=⟨1,2,-2⟩=\mathbf{b}$.

(c) Magnitude is not additive. $\mathbf{a}+\mathbf{b}=⟨3,1,0⟩$, so $|\mathbf{a}+\mathbf{b}|=\sqrt{9+1+0}=\sqrt{10}$. Meanwhile $|\mathbf{b}|=\sqrt{1+4+4}=3$, so $|\mathbf{a}|+|\mathbf{b}|=6$. Since $10<36$, $\sqrt{10}<6$: the sum of the lengths is larger.

This is the triangle inequality $|\mathbf{a}+\mathbf{b}|\le |\mathbf{a}|+|\mathbf{b}|$, with equality only when the vectors point the same way, i.e. one is a positive scalar multiple of the other. Here $\mathbf{a}$ and $\mathbf{b}$ are not parallel at all (the ratio of first components is $2$, of second components $-\frac{1}{2}$), so strict inequality was guaranteed before any arithmetic.

(d) Head equals tail plus displacement. $2\mathbf{a}-\mathbf{b}=⟨4,-2,4⟩-⟨1,2,-2⟩=⟨3,-4,6⟩$. Since $\overrightarrow{PQ}=Q-P$, $Q=P+⟨3,-4,6⟩=(3+3,0-4,-1+6)=(6,-4,5)$.

(a) Expand both squares with the dot product. Using $|\mathbf{v}{|}^2=\mathbf{v}\cdot \mathbf{v}$, distributivity and commutativity,

$$|\mathbf{a}+\mathbf{b}{|}^2=(\mathbf{a}+\mathbf{b})\cdot (\mathbf{a}+\mathbf{b})=\mathbf{a}\cdot \mathbf{a}+2\mathbf{a}\cdot \mathbf{b}+\mathbf{b}\cdot \mathbf{b}=|\mathbf{a}{|}^2+2\mathbf{a}\cdot \mathbf{b}+|\mathbf{b}{|}^2,$$

and in the same way $|\mathbf{a}-\mathbf{b}{|}^2=|\mathbf{a}{|}^2-2\mathbf{a}\cdot \mathbf{b}+|\mathbf{b}{|}^2$. Adding, the cross terms cancel and

$$|\mathbf{a}+\mathbf{b}{|}^2+|\mathbf{a}-\mathbf{b}{|}^2=2|\mathbf{a}{|}^2+2|\mathbf{b}{|}^2.■$$

Geometrically, $\mathbf{a}+\mathbf{b}$ and $\mathbf{a}-\mathbf{b}$ are the two diagonals of the parallelogram with sides $\mathbf{a}$ and $\mathbf{b}$, so the identity says: the sum of the squares of the two diagonals equals the sum of the squares of the four sides.

(b) Subtract the same two expansions instead of adding. Directly,

$$(\mathbf{a}+\mathbf{b})\cdot (\mathbf{a}-\mathbf{b})=\mathbf{a}\cdot \mathbf{a}-\mathbf{a}\cdot \mathbf{b}+\mathbf{b}\cdot \mathbf{a}-\mathbf{b}\cdot \mathbf{b}=|\mathbf{a}{|}^2-|\mathbf{b}{|}^2,$$

because $\mathbf{a}\cdot \mathbf{b}=\mathbf{b}\cdot \mathbf{a}$. $■$

For a genuine parallelogram $\mathbf{a}$ and $\mathbf{b}$ are non-parallel, so both diagonals are nonzero vectors. Nonzero vectors are perpendicular if and only if their dot product is zero, and by the identity that happens if and only if $|\mathbf{a}{|}^2=|\mathbf{b}{|}^2$, i.e. $|\mathbf{a}|=|\mathbf{b}|$. Opposite sides of a parallelogram are equal anyway, so $|\mathbf{a}|=|\mathbf{b}|$ says all four sides are equal: the diagonals are perpendicular exactly when the parallelogram is a rhombus.

(c) Feed the data into (a) and into the expansion. From the parallelogram identity,

$$|\mathbf{a}-\mathbf{b}{|}^2=2\cdot 9+2\cdot 25-49=68-49=19,|\mathbf{a}-\mathbf{b}|=\sqrt{19}.$$

For the angle, the expansion of $|\mathbf{a}+\mathbf{b}{|}^2$ gives $49=9+2\mathbf{a}\cdot \mathbf{b}+25$, so $\mathbf{a}\cdot \mathbf{b}=\frac{15}{2}$. By the geometric form of the dot product,

$$\cos \theta =\frac{\mathbf{a}\cdot \mathbf{b}}{|\mathbf{a}||\mathbf{b}|}=\frac{15/2}{15}=\frac{1}{2},\theta =\frac{\pi }{3}.$$

Cross-check with the other expansion: $|\mathbf{a}-\mathbf{b}{|}^2=9+25-2\cdot \frac{15}{2}=19$, agreeing with the first calculation.

(a) Expand along the first row.

$$\mathbf{a}\times \mathbf{b}=\left|\begin{array}{ccc}\mathbf{i}& \mathbf{j}& \mathbf{k}\\ 2& 1& -1\\ 1& -1& 3\end{array}\right|=(1\cdot 3-(-1)(-1))\mathbf{i}-(2\cdot 3-(-1)\cdot 1)\mathbf{j}+(2\cdot (-1)-1\cdot 1)\mathbf{k}=⟨2,-7,-3⟩.$$

The middle component carries the minus sign: $-(6+1)=-7$. Check perpendicularity: $⟨2,-7,-3⟩\cdot \mathbf{a}=4-7+3=0$ and $⟨2,-7,-3⟩\cdot \mathbf{b}=2+7-9=0$.

$|\mathbf{a}\times \mathbf{b}|=\sqrt{4+49+9}=\sqrt{62}$, so the two unit vectors perpendicular to both are $\pm \frac{1}{\sqrt{62}}⟨2,-7,-3⟩$. There are exactly two because the perpendicular direction to a plane is unique up to sign.

(b) Area is the length of the cross product. The parallelogram has area $|\mathbf{a}\times \mathbf{b}|=\sqrt{62}$. The triangle with those two sides from the common vertex is half of it: $\frac{\sqrt{62}}{2}$.

(c) First way: use the product already computed. Since $\mathbf{a}\cdot (\mathbf{b}\times \mathbf{c})=(\mathbf{a}\times \mathbf{b})\cdot \mathbf{c}$,

$$(\mathbf{a}\times \mathbf{b})\cdot \mathbf{c}=⟨2,-7,-3⟩\cdot ⟨0,3,-7⟩=0-21+21=0.$$

Second way: the determinant. Expanding along the first row,

$$\left|\begin{array}{ccc}2& 1& -1\\ 1& -1& 3\\ 0& 3& -7\end{array}\right|=2\left|\begin{array}{cc}-1& 3\\ 3& -7\end{array}\right|-1\left|\begin{array}{cc}1& 3\\ 0& -7\end{array}\right|+(-1)\left|\begin{array}{cc}1& -1\\ 0& 3\end{array}\right|=2(7-9)-(-7-0)-(3-0)=-4+7-3=0.$$

(Or compute $\mathbf{b}\times \mathbf{c}=⟨7-9,-(-7-0),3-0⟩=⟨-2,7,3⟩$ and dot with $\mathbf{a}$: $-4+7-3=0$.)

The parallelepiped has volume $|0|=0$: it is flat, so the three vectors are coplanar. Being coplanar, and with $\mathbf{a}$, $\mathbf{b}$ not parallel, $\mathbf{c}$ must be a combination of the other two. Solve $\mathbf{c}=s\mathbf{a}+t\mathbf{b}$ componentwise:

$$2s+t=0,s-t=3,-s+3t=-7.$$

Adding the first two gives $3s=3$, so $s=1$ and $t=-2$. The third equation is the check: $-1-6=-7$. Hence $\mathbf{c}=\mathbf{a}-2\mathbf{b}$; verify: $⟨2,1,-1⟩-⟨2,-2,6⟩=⟨0,3,-7⟩$.

(a) Compare directions, then try to solve for a common point. The direction vectors are ${\mathbf{v}}_1=⟨1,1,2⟩$ and ${\mathbf{v}}_2=⟨2,-1,1⟩$. Neither is a scalar multiple of the other (the first components would need the factor $2$, the second components $-1$), so the lines are not parallel.

If they meet, some $t$ and $s$ satisfy all three coordinate equations:

$$1+t=2s,2+t=1-s,2t=3+s.$$

Subtracting the first equation from the second eliminates $t$: $1=1-3s$, so $s=0$, and then $t=2s-1=-1$. Test these values in the third equation: the left side is $2t=-2$, the right side is $3+s=3$. They disagree, so no common point exists. Not parallel and not intersecting: the lines are skew.

(b) Distance via a common normal. The lines lie in a pair of parallel planes whose common normal is perpendicular to both directions:

$$\mathbf{n}={\mathbf{v}}_1\times {\mathbf{v}}_2=⟨1\cdot 1-2\cdot (-1),-(1\cdot 1-2\cdot 2),1\cdot (-1)-1\cdot 2⟩=⟨3,3,-3⟩,$$

and any nonzero multiple will do, so take $\mathbf{n}=⟨1,1,-1⟩$. (Check: $\mathbf{n}\cdot {\mathbf{v}}_1=1+1-2=0$, $\mathbf{n}\cdot {\mathbf{v}}_2=2-1-1=0$.) Pick $P_1(1,2,0)$ on $L_1$ ($t=0$) and $P_2(0,1,3)$ on $L_2$ ($s=0$), so $\overrightarrow{P_1{P}_2}=⟨-1,-1,3⟩$. The distance is the absolute scalar projection of this vector onto $\mathbf{n}$:

$$D=\frac{|\mathbf{n}\cdot \overrightarrow{P_1{P}_2}|}{|\mathbf{n}|}=\frac{|-1-1-3|}{\sqrt{3}}=\frac{5}{\sqrt{3}}=\frac{5\sqrt{3}}{3}.$$

(c) A plane through $L_1$ with normal $\mathbf{n}$. A plane containing $L_1$ and parallel to $L_2$ must have a normal perpendicular to both ${\mathbf{v}}_1$ and ${\mathbf{v}}_2$, which is exactly $\mathbf{n}$. Through $P_1(1,2,0)$:

$$1(x-1)+1(y-2)-1(z-0)=0⟺x+y-z=3.$$

It contains $L_1$: substituting gives $(1+t)+(2+t)-2t=3$ for every $t$. It is parallel to $L_2$: substituting gives $2s+(1-s)-(3+s)=-2$ for every $s$, a constant, so $L_2$ never crosses the plane and in fact lies in the parallel plane $x+y-z=-2$.

Second route to (b): the distance from any point of $L_2$, say $(0,1,3)$, to the plane $x+y-z-3=0$ is

$$\frac{|0+1-3-3|}{\sqrt{1+1+1}}=\frac{5}{\sqrt{3}},$$

the same value. (Equivalently the two parallel planes $x+y-z=3$ and $x+y-z=-2$ are $|3-(-2)|/\sqrt{3}=5/\sqrt{3}$ apart.)

(d) Substitute the parametric equations into the plane.

$$2(1+t)-(2+t)+2t=7⟹2+2t-2-t+2t=7⟹3t=7,t=\frac{7}{3}.$$

The point is $\left(1+\frac{7}{3},2+\frac{7}{3},\frac{14}{3}\right)=\left(\frac{10}{3},\frac{13}{3},\frac{14}{3}\right)$. Check in the plane: $\frac{20}{3}-\frac{13}{3}+\frac{14}{3}=\frac{21}{3}=7$. (A single solution was guaranteed because ${\mathbf{v}}_1\cdot ⟨2,-1,1⟩=2-1+2=3\ne 0$: the line is not parallel to the plane.)

(a) Complete the square in $x$ and in $y$. Group and factor the leading coefficient out of the $x$-terms:

$$4(x^2-2x)+(y^2+2y)-z^2+1=0⟹4[(x-1{)}^2-1]+[(y+1{)}^2-1]-z^2+1=0.$$

Collecting constants, $4(x-1{)}^2+(y+1{)}^2-z^2=4$, and dividing by $4$,

$$(x-1{)}^2+\frac{(y+1{)}^2}{4}-\frac{z^2}{4}=1.$$

Three squared terms, exactly one negative, and $1$ on the right: a hyperboloid of one sheet. Its centre is $(1,-1,0)$ and its axis is the line $x=1$, $y=-1$, parallel to the $z$-axis — the variable carrying the minus sign.

(b) Slice. Substitute the constant into the standard form.

$z=0$: $(x-1{)}^2+\frac{(y+1{)}^2}{4}=1$, an ellipse centred at $(1,-1,0)$ with semi-axes $1$ in the $x$-direction and $2$ in the $y$-direction. This is the waist, the smallest horizontal cross-section.

$z=\pm 2$: $(x-1{)}^2+\frac{(y+1{)}^2}{4}=1+1=2$, ellipses centred at $(1,-1,\pm 2)$ with semi-axes $\sqrt{2}$ and $2\sqrt{2}$ — the same shape as the waist, scaled by $\sqrt{2}$.

$x=1$: $\frac{(y+1{)}^2}{4}-\frac{z^2}{4}=1$, i.e. $(y+1{)}^2-z^2=4$: a hyperbola in the plane $x=1$ opening in the $\pm y$ directions, vertices at $(1,1,0)$ and $(1,-3,0)$, asymptotes $z=\pm (y+1)$.

$y=-1$: $(x-1{)}^2-\frac{z^2}{4}=1$: a hyperbola in the plane $y=-1$ opening in the $\pm x$ directions, vertices at $(2,-1,0)$ and $(0,-1,0)$, asymptotes $z=\pm 2(x-1)$.

(c) The drawing. Draw the axis as a vertical line through $(1,-1,0)$ and mark the centre. Draw the waist ellipse at height $0$ (width $2$ in $x$, $4$ in $y$), and the larger ellipses at heights $2$ and $-2$ (width $2\sqrt{2}$ in $x$, $4\sqrt{2}$ in $y$). Join them with the two hyperbolic profiles: in the front-to-back plane $y=-1$ the outline passes through $(0,-1,0)$ and $(2,-1,0)$ and flares outward above and below; in the side plane $x=1$ it passes through $(1,-3,0)$ and $(1,1,0)$ and flares likewise. The surface opens along the $z$-axis in both directions, getting wider without bound as $|z|$ grows (the ellipses at height $z$ have semi-axes $\sqrt{1+z^2/4}$ and $2\sqrt{1+z^2/4}$). It is connected because every horizontal plane meets it in a nonempty ellipse: $1+z^2/4>0$ for all $z$, so there is no gap.

(d) Vary the constant. With constant $k$ the completed-square computation gives

$$4(x-1{)}^2+(y+1{)}^2-z^2=5-k.$$

If $k=5$ the right-hand side is $0$: $z^2=4(x-1{)}^2+(y+1{)}^2$ is an elliptic cone with vertex $(1,-1,0)$, and the horizontal traces pinch to a point there. If $k>5$ the right-hand side is negative; dividing by it flips the signs to $-\frac{4(x-1{)}^2}{k-5}-\frac{(y+1{)}^2}{k-5}+\frac{z^2}{k-5}=1$, two negative squared terms: a hyperboloid of two sheets, with no points where $z^2<k-5$. If $k<5$ the right-hand side is positive and the surface is a hyperboloid of one sheet, as in (a). The cone at $k=5$ is the boundary case between the two hyperboloids, and it is their common asymptotic surface.

(a) Intersect the component domains; take the limit componentwise. $\ln (t+1)$ needs $t>-1$; $\sqrt{4-t^2}$ needs $-2\le t\le 2$; $\frac{1}{t-1}$ needs $t\ne 1$. The domain is the intersection, $(-1,1)\cup (1,2]$.

For the limit, each component is its own single-variable problem. $\frac{t^2-1}{t-1}=t+1$ for $t\ne 1$, so the first component tends to $2$; $\ln t\to \ln 1=0$ by continuity; $\sqrt{t+3}\to \sqrt{4}=2$ by continuity. The limit is $⟨2,0,2⟩$.

(b) Project onto the $xy$-plane, then lift. A point on both surfaces satisfies $x^2+y^2=2y$, that is $x^2+(y-1{)}^2=1$ after completing the square. So the projection of $C$ onto the $xy$-plane is the circle of radius $1$ centred at $(0,1)$. Parametrise it by $x=\cos t$, $y=1+\sin t$, and let the plane supply the height $z=2y$:

$$\mathbf{r}(t)=⟨\cos t,1+\sin t,2+2\sin t⟩,0\le t\le 2\pi .$$

Verification on the paraboloid: $x^2+y^2={\cos }^{2}t+1+2\sin t+{\sin }^{2}t=2+2\sin t=z$. On the plane: $2y=2+2\sin t=z$. Both hold for every $t$, and as $t$ runs over $[0,2\pi )$ the projection goes once round its circle, so $C$ is traversed once. (Since $C$ lies in the plane $z=2y$ and projects to a circle, it is an ellipse.)

The height is $z=2+2\sin t$, largest when $\sin t=1$ and smallest when $\sin t=-1$. Highest point: $t=\frac{\pi }{2}$, $\mathbf{r}=(0,2,4)$. Lowest point: $t=\frac{3\pi }{2}$, $\mathbf{r}=(0,0,0)$. Check: $(0,2,4)$ has $0+4=4=2\cdot 2$, and the origin lies on both surfaces trivially.

(c) Solve $\cos t=1+\sin t$ and watch for extraneous roots. On $C$, $x=y$ reads $\cos t=1+\sin t$. Squaring both sides,

$${\cos }^{2}t=1+2\sin t+{\sin }^{2}t⟹1-{\sin }^{2}t=1+2\sin t+{\sin }^{2}t⟹0=2\sin t(1+\sin t).$$

So $\sin t=0$ or $\sin t=-1$. Squaring may have added solutions, so test each candidate in the original equation. $\sin t=0$ gives $t=0$ or $t=\pi$: at $t=0$, $\cos 0=1=1+0$ ✓; at $t=\pi$, $\cos \pi =-1\ne 1$ ✗, extraneous (it solves $\cos t=-(1+\sin t)$ instead). $\sin t=-1$ gives $t=\frac{3\pi }{2}$: $\cos \frac{3\pi }{2}=0=1-1$ ✓.

The curve meets the plane $x=y$ at exactly two points: $\mathbf{r}(0)=(1,1,2)$ and $\mathbf{r}\left(\frac{3\pi }{2}\right)=(0,0,0)$. Both have $x=y$, and both satisfy $z=x^2+y^2$ and $z=2y$: $2=1+1=2\cdot 1$ and $0=0=0$. (A closed curve crossing a plane generically does so an even number of times, which is a comforting check.)

(a) Product rule for the dot, then for the cross. By the dot product rule,

$$\frac{d}{dt}[\mathbf{r}\cdot ({\mathbf{r}}^{\prime }\times {\mathbf{r}}^{\prime \prime })]={\mathbf{r}}^{\prime }\cdot ({\mathbf{r}}^{\prime }\times {\mathbf{r}}^{\prime \prime })+\mathbf{r}\cdot \frac{d}{dt}({\mathbf{r}}^{\prime }\times {\mathbf{r}}^{\prime \prime }).$$

The first term is zero because ${\mathbf{r}}^{\prime }\times {\mathbf{r}}^{\prime \prime }$ is perpendicular to both of its factors, in particular to ${\mathbf{r}}^{\prime }$, and the dot product of perpendicular vectors is $0$. For the second, the cross product rule (factors kept in order) gives

$$\frac{d}{dt}({\mathbf{r}}^{\prime }\times {\mathbf{r}}^{\prime \prime })={\mathbf{r}}^{\prime \prime }\times {\mathbf{r}}^{\prime \prime }+{\mathbf{r}}^{\prime }\times {\mathbf{r}}^{\prime \prime \prime }.$$

Here ${\mathbf{r}}^{\prime \prime }\times {\mathbf{r}}^{\prime \prime }=0$, because the cross product of a vector with itself has length $|{\mathbf{r}}^{\prime \prime }{|}^2\sin 0=0$ (a vector is parallel to itself). What survives is

$$\frac{d}{dt}[\mathbf{r}\cdot ({\mathbf{r}}^{\prime }\times {\mathbf{r}}^{\prime \prime })]=\mathbf{r}\cdot ({\mathbf{r}}^{\prime }\times {\mathbf{r}}^{\prime \prime \prime }).■$$

(b) Differentiate the squared length. Let $f(t)=|\mathbf{r}(t){|}^2=\mathbf{r}(t)\cdot \mathbf{r}(t)$. By the dot product rule,

$$f^{\prime }(t)={\mathbf{r}}^{\prime }\cdot \mathbf{r}+\mathbf{r}\cdot {\mathbf{r}}^{\prime }=2\mathbf{r}(t)\cdot {\mathbf{r}}^{\prime }(t)=0$$

for every $t$, by hypothesis. A real function with zero derivative on an interval is constant, so $|\mathbf{r}(t){|}^2=c$ for some constant $c\ge 0$, and therefore $|\mathbf{r}(t)|=\sqrt{c}$ is constant. $■$ (The hypothesis says the velocity is always tangent to the sphere through the current point; the conclusion is that the point never leaves that sphere.)

(c) Chain rule on the square root. Where $\mathbf{r}\ne 0$ the function $|\mathbf{r}|=(\mathbf{r}\cdot \mathbf{r}{)}^{1/2}$ is a composition of the differentiable function $u\mapsto \sqrt{u}$ (for $u>0$) with $\mathbf{r}\cdot \mathbf{r}$, so by the chain rule and the dot product rule,

$$\frac{d}{dt}|\mathbf{r}|=\frac{1}{2\sqrt{\mathbf{r}\cdot \mathbf{r}}}\cdot 2\mathbf{r}\cdot {\mathbf{r}}^{\prime }=\frac{\mathbf{r}\cdot {\mathbf{r}}^{\prime }}{|\mathbf{r}|}.■$$

For $\mathbf{r}(t)=⟨t,t^2,2⟩$: ${\mathbf{r}}^{\prime }(t)=⟨1,2t,0⟩$. At $t=1$, $\mathbf{r}(1)=⟨1,1,2⟩$, ${\mathbf{r}}^{\prime }(1)=⟨1,2,0⟩$, $\mathbf{r}\cdot {\mathbf{r}}^{\prime }=1+2+0=3$ and $|\mathbf{r}(1)|=\sqrt{6}$. So the distance from the origin is changing at the rate

$$\frac{d}{dt}|\mathbf{r}|{|}_{t=1}=\frac{3}{\sqrt{6}}=\frac{\sqrt{6}}{2},$$

while the speed is $|{\mathbf{r}}^{\prime }(1)|=\sqrt{1+4+0}=\sqrt{5}$.

They differ because the formula just proved is the scalar projection of ${\mathbf{r}}^{\prime }$ onto $\mathbf{r}$: $\frac{\mathbf{r}\cdot {\mathbf{r}}^{\prime }}{|\mathbf{r}|}={\mathrm{comp}}_{\mathbf{r}}{\mathbf{r}}^{\prime }=|{\mathbf{r}}^{\prime }|\cos \varphi$, where $\varphi$ is the angle between the position and the velocity. Only the radial part of the velocity changes the distance from the origin; the part perpendicular to $\mathbf{r}$ moves the point sideways around the origin without changing how far away it is. Since $|\cos \varphi |\le 1$, the rate of change of distance can never exceed the speed, and indeed $\frac{\sqrt{6}}{2}<\sqrt{5}$ because $\frac{6}{4}<5$.

(a) Integrate componentwise. Each component is an ordinary integral:

$${\int }_0^1\frac{dt}{1+t^2}=[\arctan t{]}_0^1=\frac{\pi }{4},{\int }_0^1\frac{2tdt}{1+t^2}=[\ln (1+t^2){]}_0^1=\ln 2,$$

and, integrating by parts with $u=t$, $dv=e^{t}dt$,

$${\int }_0^{1}t{e}^{t}dt=[t{e}^t-e^t{]}_0^1=(e-e)-(0-1)=1.$$

The integral is $⟨\frac{\pi }{4},\ln 2,1⟩$.

(b) Match all three coordinates with independent parameters. A common point needs $t$ and $s$ with

$$t=3-s,1-t=s-2,3+t^2=s^2.$$

Substituting the first into the second gives $1-(3-s)=s-2$, which is $s-2=s-2$: an identity, so the first two equations impose only one condition, $t=3-s$. (Both curves project onto the same line $x+y=1$ in the $xy$-plane, which is why.) The third equation decides everything: $3+(3-s{)}^2=s^2$ gives $3+9-6s+s^2=s^2$, so $6s=12$, $s=2$ and $t=1$. There is exactly one solution, hence exactly one meeting point:

$${\mathbf{r}}_1(1)=(1,0,4)={\mathbf{r}}_2(2).$$

Tangent vectors: ${\mathbf{r}}_1^{\prime }(t)=⟨1,-1,2t⟩$ so ${\mathbf{r}}_1^{\prime }(1)=⟨1,-1,2⟩$; ${\mathbf{r}}_2^{\prime }(s)=⟨-1,1,2s⟩$ so ${\mathbf{r}}_2^{\prime }(2)=⟨-1,1,4⟩$. By the geometric form of the dot product,

$$\cos \theta =\frac{⟨1,-1,2⟩\cdot ⟨-1,1,4⟩}{|⟨1,-1,2⟩||⟨-1,1,4⟩|}=\frac{-1-1+8}{\sqrt{6}\cdot \sqrt{18}}=\frac{6}{\sqrt{108}}=\frac{6}{6\sqrt{3}}=\frac{1}{\sqrt{3}},$$

so $\theta =\arccos \frac{1}{\sqrt{3}}=\arccos \frac{\sqrt{3}}{3}$, an acute angle (the dot product is positive).

(c) Point plus direction. The tangent line passes through $(1,0,4)$ with direction ${\mathbf{r}}_1^{\prime }(1)=⟨1,-1,2⟩$:

$$x=1+u,y=-u,z=4+2u.$$

It meets the $xy$-plane where $z=0$: $4+2u=0$, $u=-2$, giving the point $(-1,2,0)$.

(d) Find a linear relation among the coordinates. On ${\mathbf{r}}_1$, $x+y=t+(1-t)=1$ for every $t$; on ${\mathbf{r}}_2$, $x+y=(3-s)+(s-2)=1$ for every $s$. So both curves lie in the plane $x+y=1$, whose normal is $\mathbf{n}=⟨1,1,0⟩$. A vector is parallel to the plane exactly when it is perpendicular to $\mathbf{n}$: $⟨1,-1,2⟩\cdot \mathbf{n}=1-1+0=0$ and $⟨-1,1,4⟩\cdot \mathbf{n}=-1+1+0=0$, as they must be, since the tangent to a curve lying in a plane lies in that plane. (Within the plane $x+y=1$ each curve is a parabola in the height $z$; two parabolas of the same width $z\sim t^2$, shifted, cross exactly once, which is the geometric reason behind (b).)

(a) Check the equation of the cone. $x^2+y^2=e^{2t}{\cos }^{2}t+e^{2t}{\sin }^{2}t=e^{2t}=z^2$, so every point of the curve satisfies $z^2=x^2+y^2$. Since $z=e^t>0$ the curve lies on the upper nappe $z=\sqrt{x^2+y^2}$, spiralling outward and upward as $t$ increases (the distance from the $z$-axis is $\sqrt{x^2+y^2}=e^t$, which grows, while the angle in the $xy$-plane is $t$).

(b) Speed, length, arc length function. Differentiate each component with the product rule:

$${\mathbf{r}}^{\prime }(t)=⟨e^t(\cos t-\sin t),e^t(\sin t+\cos t),e^t⟩.$$

Squaring and adding, the mixed terms cancel:

$$|{\mathbf{r}}^{\prime }(t){|}^2=e^{2t}[(\cos t-\sin t{)}^2+(\sin t+\cos t{)}^2+1]=e^{2t}[(1-2\sin t\cos t)+(1+2\sin t\cos t)+1]=3e^{2t},$$

so $|{\mathbf{r}}^{\prime }(t)|=\sqrt{3}{e}^t$. The speed is never zero and ${\mathbf{r}}^{\prime }$ is continuous, so the curve is smooth and the arc length formula applies:

$$L={\int }_0^{\ln 2}\sqrt{3}{e}^{t}dt=\sqrt{3}[e^t{]}_0^{\ln 2}=\sqrt{3}(2-1)=\sqrt{3}.$$

The arc length function from $t=0$ is $s(t)={\int }_0^t\sqrt{3}{e}^{u}du=\sqrt{3}(e^t-1)$. Solving, $e^t=1+\frac{s}{\sqrt{3}}$, so $t=\ln \left(1+\frac{s}{\sqrt{3}}\right)$ for $s\ge 0$. (Check: $s=\sqrt{3}$ gives $t=\ln 2$, matching $L$.)

(c) Curvature by the cross product formula. Differentiate ${\mathbf{r}}^{\prime }$ once more:

$${\mathbf{r}}^{\prime \prime }(t)=⟨e^t(\cos t-\sin t)+e^t(-\sin t-\cos t),e^t(\sin t+\cos t)+e^t(\cos t-\sin t),e^t⟩=⟨-2e^t\sin t,2e^t\cos t,e^t⟩.$$

Factor $e^t$ out of each vector and write $c=\cos t$, $s=\sin t$ for brevity: ${\mathbf{r}}^{\prime }\times {\mathbf{r}}^{\prime \prime }=e^{2t}⟨c-s,s+c,1⟩\times ⟨-2s,2c,1⟩$. Expanding the determinant,

$$⟨c-s,s+c,1⟩\times ⟨-2s,2c,1⟩=⟨(s+c)-2c,-[(c-s)+2s],2c(c-s)+2s(s+c)⟩=⟨s-c,-s-c,2⟩,$$

since $2c^2-2cs+2s^2+2cs=2$. Its squared length is $(s-c{)}^2+(s+c{)}^2+4=2+4=6$, so $|{\mathbf{r}}^{\prime }\times {\mathbf{r}}^{\prime \prime }|=\sqrt{6}{e}^{2t}$. (Check that $⟨s-c,-s-c,2⟩$ is perpendicular to $⟨c-s,s+c,1⟩$: $-(s-c{)}^2-(s+c{)}^2+2=-2+2=0$.) Hence

$$\kappa (t)=\frac{\sqrt{6}{e}^{2t}}{(\sqrt{3}{e}^t{)}^3}=\frac{\sqrt{6}}{3\sqrt{3}}{e}^{-t}=\frac{\sqrt{2}}{3}{e}^{-t},\kappa (0)=\frac{\sqrt{2}}{3}.$$

As $t\to \infty$, $\kappa \to 0$: the spiral's turns get ever larger in radius (the distance from the axis is $e^t$), and a bend of larger radius is gentler. The curvature falls off like the reciprocal of that radius, exactly as for circles.

(d) Dot the tangent with $\mathbf{k}$. By the geometric form of the dot product the angle $\varphi$ between ${\mathbf{r}}^{\prime }(t)$ and $\mathbf{k}=⟨0,0,1⟩$ satisfies

$$\cos \varphi =\frac{{\mathbf{r}}^{\prime }(t)\cdot \mathbf{k}}{|{\mathbf{r}}^{\prime }(t)||\mathbf{k}|}=\frac{e^t}{\sqrt{3}{e}^t}=\frac{1}{\sqrt{3}},$$

independent of $t$. So the tangent makes the same angle $\arccos \frac{1}{\sqrt{3}}$ with the axis of the cone at every point: the curve climbs the cone at a constant pitch, the way a helix climbs a cylinder. (The same computation with $\mathbf{r}$ in place of $\mathbf{k}$ gives $\mathbf{r}\cdot {\mathbf{r}}^{\prime }=2e^{2t}$ and $|\mathbf{r}|=\sqrt{2}{e}^t$, so the tangent also makes a constant angle with the position vector, with cosine $\sqrt{2/3}$.)

(a) Steps 1 and 2 are correct; Steps 3 and 5 are wrong; Step 4 inherits the error of Step 3.

Step 3 confuses turning per unit time with turning per unit arc length. Curvature is defined as $\kappa =\left|\frac{d\mathbf{T}}{ds}\right|$, and converting to the parameter $t$ costs a factor of the speed: $\kappa =\frac{|{\mathbf{T}}^{\prime }(t)|}{|{\mathbf{r}}^{\prime }(t)|}$. The formula $\kappa =|{\mathbf{T}}^{\prime }(t)|$ is valid only when $t$ is arc length, i.e. when the speed is $1$; here the speed is $2\sqrt{13}$. The computation of ${\mathbf{T}}^{\prime }$ itself is right (differentiating $\mathbf{v}/(2\sqrt{13})$ is legitimate precisely because the speed is constant), but the last equality of the step is not.

Step 5 draws the wrong conclusion from a failed check. The identity $|\mathbf{a}{|}^2=a_T^2+a_N^2$ has no extra hypothesis: it holds for every smooth motion, because $\mathbf{a}=a_T\mathbf{T}+a_N\mathbf{N}$ with $\mathbf{T}$ and $\mathbf{N}$ perpendicular unit vectors, and Pythagoras does the rest. A check that fails is evidence of an error upstream, never evidence against the theorem. The student computed $|\mathbf{a}|=12$ correctly and should have gone back to look for the mistake.

(b) Divide by the speed. With $|{\mathbf{T}}^{\prime }|=\frac{6}{\sqrt{13}}$ and $|{\mathbf{r}}^{\prime }|=2\sqrt{13}$,

$$\kappa =\frac{|{\mathbf{T}}^{\prime }(t)|}{|{\mathbf{r}}^{\prime }(t)|}=\frac{6/\sqrt{13}}{2\sqrt{13}}=\frac{6}{26}=\frac{3}{13},$$

a constant, as it should be for a helix. Then

$$a_N=\kappa v^2=\frac{3}{13}\cdot 52=12.$$

The check now passes: $a_T^2+a_N^2=0+144=144=|\mathbf{a}{|}^2$. With $a_T=0$ the whole acceleration is normal, and indeed $\mathbf{a}=⟨-12\cos 2t,-12\sin 2t,0⟩$ points horizontally toward the axis, which is the direction of $\mathbf{N}$.

(c) The cross product route. With ${\mathbf{r}}^{\prime }=⟨-6\sin 2t,6\cos 2t,4⟩$ and ${\mathbf{r}}^{\prime \prime }=⟨-12\cos 2t,-12\sin 2t,0⟩$,

$${\mathbf{r}}^{\prime }\times {\mathbf{r}}^{\prime \prime }=⟨6\cos 2t\cdot 0-4(-12\sin 2t),-[(-6\sin 2t)\cdot 0-4(-12\cos 2t)],(-6\sin 2t)(-12\sin 2t)-(6\cos 2t)(-12\cos 2t)⟩$$