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Vectors and Geometry in Two and Three Dimensions

1.6 Curves and their Tangent Vectors

30 problems · hints, answers and solutions shown beside each one

1Stage 1Conceptual

Do you understand the idea? Usually little or no calculation.

Questions 1 through 5 provide practice with curve parametrization. Being comfortable with the algebra and interpretation of these descriptions are essential ingredients in working effectively with parametrizations.

Q1Stage 1

Consider the following time-parametrized curve:

r(t)=(cos⁡(π4t) , (t−5)2)\vr(t)=\left( \cos\left(\frac{\pi}{4}t\right)~,~(t-5)^2\right)

List the three points (−1/2,0)(-1/\sqrt{2},0), (1,25)(1,25), and (0,25)(0,25) in chronological order.

Hint

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

Answer

(1,25)(1,25), (−1/2,0)(-1/\sqrt2,0), (0,25)(0,25).

Full solution

We can find the time at which the curve hits a given point by considering the two equations that arise from the two coordinates. For the yy-coordinate to be 0, we must have (t−5)2=0(t-5)^2=0, i.e. t=5t=5. So, the point (−1/2,0)(-1/\sqrt{2},0) happens when t=5t=5.

Similarly, for the yy-coordinate to be 2525, we need (t−5)2=25(t-5)^2=25, so (t−5)=±5(t-5)=\pm 5. When t=0t=0, the curve hits (1,25)(1, 25); when t=10t=10, the curve hits (0,25)(0,25).

So, in order, the curve passes through the points (1,25)(1,25), (−1/2,0)(-1/\sqrt2,0), and (0,25)(0,25).

Q2Stage 1

At what points in the xyxy-plane does the curve (sin⁡t,t2)(\sin t, t^2) cross itself? What is the difference in tt between the first time the curve crosses through a point, and the last?

Hint

The curve “crosses itself" when (sin⁡t,t2)(\sin t,t^2) gives the same coordinate for different values of tt. When these crossings occur will depend on which crossing you're referring to, so your answers should all depend on tt.

Answer

The curve crosses itself at all points (0,(πn)2)(0,(\pi n)^2) where nn is an integer. It passes such a point twice, 2πn2\pi n time units apart.

Full solution

The curve “crosses itself" when the same coordinates occur for different values of tt, say t1t_1 and t2t_2. So, we want to know when sin⁡t1=sin⁡t2\sin t_1=\sin t_2 and also t12=t22t_1^2=t_2^2. Since t1t_1 and t2t_2 should be different, the second equation tells us t2=−t1t_2=-t_1. Then the first equation tells us sin⁡t1=sin⁡t2=sin⁡(−t1)=−sin⁡t1\sin t_1=\sin t_2=\sin(-t_1)=-\sin t_1. That is, sin⁡t1=−sin⁡t1\sin t_1 = -\sin t_1, so sin⁡t1=0\sin t_1=0. That happens whenever t1=πnt_1=\pi n for an integer nn.

So, the points at which the curve crosses itself are those points (0,(πn)2)(0,(\pi n)^2) where nn is an integer. It passes such a point at times t=πnt=\pi n and t=−πnt=-\pi n. So, the curve hits this point 2πn2\pi n time units apart.

Q3Stage 1

Find the specified parametrization of the first quadrant part of the circle x2+y2=a2x^2+y^2=a^2.

  1. In terms of the yy coordinate.

  2. In terms of the angle between the tangent line and the positive xx-axis.

  3. In terms of the arc length from (0,a)(0,a).

Hint

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

Answer

(a) r(y)=a2−y2 ı^+y ȷ^\vr(y)=\sqrt{a^2-y^2}\,\hi+ y\,\hj, 0≤y≤a0\le y\le a

(b) (x(ϕ),y(ϕ))=(asin⁡ϕ,−acos⁡ϕ)\big(x(\phi),y(\phi)\big) =\big(a\sin \phi ,-a\cos \phi \big), π2≤ϕ≤π\frac{\pi}{2}\le\phi\le\pi

(c) (x(s),y(s))=(acos⁡(π2−sa),asin⁡(π2−sa))\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≤s≤π2a0\le s\le\tfrac{\pi}{2}a

Full solution

(a) Since, on the specified part of the circle, x=a2−y2x=\sqrt{a^2-y^2} and yy runs from 00 to aa, the parametrization is r(y)=a2−y2 ı^+y ȷ^\vr(y)=\sqrt{a^2-y^2}\,\hi+ y\,\hj, 0≤y≤a0\le y\le a.

(b) Let θ\theta be the angle between

  • the radius vector from the origin to the point (acos⁡θ,asin⁡θ)(a\cos\theta,a\sin\theta) on the circle and

  • the positive xx-axis.

The tangent line to the circle at (acos⁡θ,asin⁡θ)(a\cos\theta,a\sin\theta) is perpendicular to the radius vector and so makes angle ϕ=π2+θ\phi=\frac{\pi}{2}+\theta with the positive xx axis. (See the figure on the left below.) As θ=ϕ−π2\theta =\phi-\frac{\pi}{2}, the desired parametrization is

(x(ϕ),y(ϕ))=(acos⁡(ϕ−π2),asin⁡(ϕ−π2))=(asin⁡ϕ,−acos⁡ϕ), π2≤ϕ≤π\begin{equation*} \big(x(\phi),y(\phi)\big) =\big(a\cos(\phi-\tfrac{\pi}{2}),a\sin(\phi-\tfrac{\pi}{2})\big) =\big(a\sin \phi ,-a\cos \phi \big),\ \tfrac{\pi}{2}\le\phi\le\pi \end{equation*}

Figure from prob_s1.6, line 88

Figure from prob_s1.6, line 88

Figure from prob_s1.6, line 88

Figure from prob_s1.6, line 88

(c) Let θ\theta be the angle between

  • the radius vector from the origin to the point (acos⁡θ,asin⁡θ)(a\cos\theta,a\sin\theta) on the circle and

  • the positive xx-axis.

The arc from (0,a)(0,a) to (acos⁡θ,asin⁡θ)(a\cos\theta,a\sin\theta) subtends an angle π2−θ\frac{\pi}{2}-\theta and so has length s=a(π2−θ)s=a\big(\frac{\pi}{2}-\theta\big). (See the figure on the right above.) Thus θ=π2−sa\theta=\frac{\pi}{2}-\frac{s}{a} and the desired parametrization is

(x(s),y(s))=(acos⁡(π2−sa) , asin⁡(π2−sa)), 0≤s≤π2a\begin{equation*} \big(x(s),y(s)\big) =\left(a\cos\left(\frac{\pi}{2}-\frac{s}{a}\right)\,,\, a\sin\left(\frac{\pi}{2}-\frac{s}{a}\right)\right) ,\ 0\le s\le\frac{\pi}{2}a \end{equation*}
Q4Stage 1

Figure from prob_s1.6, line 2

Figure from prob_s1.6, line 2

A circle of radius aa rolls along the xx-axis in the positive direction, starting with its centre at (a,a)(a,a). In that position, we mark the topmost point on the circle PP. As the circle moves, PP moves with it. Let θ\theta be the angle the circle has rolled–see the diagram below.

  1. Give the position of the centre of the circle as a function of θ\theta.

  2. Give the position of PP a function of θ\theta.

Figure from prob_s1.6, line 2

Figure from prob_s1.6, line 2

Hint

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

Answer

(a) (a+aθ,a)(a+a\theta,a) (b)(a+aθ+asin⁡θ,a+acos⁡θ)(a+a\theta+a\sin\theta,a+a\cos\theta)

Full solution

Pretend that the circle is a spool of thread. As the circle rolls it dispenses the thread along the ground. When the circle rolls θ\theta radians it dispenses the arc length θa\theta a of thread and the circle advances a distance θa\theta a. So centre of the circle has moved θa\theta a units to the right from its starting point, x=ax=a. The centre of the circle always has yy-coordinate aa. So, after rolling θ\theta radians, the centre of the circle is at position c(θ)=(a+aθ,a)\vc(\theta)=(a+a\theta,a).

Now, let's consider the position of PP on the circle, after the circle has rolled θ\theta radians.

Figure from prob_s1.6, line 2

Figure from prob_s1.6, line 2

From the diagram, we see that PP is acos⁡θa\cos \theta units above the centre of the circle, and asin⁡θa\sin \theta units to the right of it. So, the position of PP is (a+aθ+asin⁡θ,a+acos⁡θ)(a+a\theta+a\sin\theta,a+a\cos\theta).

Remark: this type of curve is known as a cycloid.

Q5Stage 1

The curve CC is defined to be the intersection of the hyperboloid

x2−14y2+3z2=1x^2-\frac{1}{4}y^2+3z^2=1

and the plane

x+y+z=0.x+y+z=0.

When yy is very close to 0, and zz is negative, find an expression giving zz in terms of yy.

Hint

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

Answer

z=−121−y22−y4z=-\frac12\sqrt{1-\frac{y^2}{2}}-\frac{y}{4}

Full solution

We aren't concerned with xx, so we can eliminate it by solving for it in one equation, and plugging that into the other. Since CC lies on the plane, x=−y−zx=-y-z, so:

1=x2−14y2+3z2=(−y−z)2−14y2+3z2=34y2+4z2+2yz\begin{align*}1&=x^2-\frac{1}{4}y^2+3z^2=(-y-z)^2-\frac14y^2+3z^2\\ &=\frac{3}{4}y^2+4z^2+2yz\end{align*}

Completing the square,

1=12y2+(2z+y2)21−y22=(2z+y2)2\begin{align*}1&=\frac{1}{2}y^2+\left(2z+\frac{y}{2}\right)^2\\ 1-\frac{y^2}{2}&=\left(2z+\frac{y}{2}\right)^2\end{align*}

Since yy is small, the left hand is close to 11 and the right hand side is close to (2z)2(2z)^2. So (2z2)≈1(2z^2)\approx 1. Since zz is negative, z≈−12z\approx -\frac{1}{2} and 2z+y2<02z+\frac{y}{2}<0. Also, 1−y221-\frac{y^2}{2} is positive, so it has a real square root.

−1−y22=2z+y2−121−y22−y4=z\begin{align*}-\sqrt{1-\frac{y^2}{2}}&=2z+\frac{y}{2}\\ -\frac12\sqrt{1-\frac{y^2}{2}}-\frac{y}{4}&=z\end{align*}
Q6Stage 1

A particle traces out a curve in space, so that its position at time tt is

r(t)=e−t ı^+1t ȷ^+(t−1)2(t−3)2 k^\vr(t)=e^{-t}\,\hi+\frac{1}{t}\,\hj+(t-1)^2(t-3)^2\,\hk

for t>0t > 0.

Let the positive zz axis point vertically upwards, as usual. When is the particle moving upwards, and when is it moving downwards? Is it moving faster at time t=1t=1 or at time t=3t=3?

Hint

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

Answer

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

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

Full solution

To determine whether the particle is rising or falling, we only need to consider its zz-coordinate: z(t)=(t−1)2(t−3)2z(t)=(t-1)^2(t-3)^2. Its derivative with respect to time is z′(t)=4(t−1)(t−2)(t−3)z'(t)=4(t-1)(t-2)(t-3). This is positive when 1<t<21<t<2 and when 3<t3<t, so the particle is increasing on (1,2)∪(3,∞)(1,2) \cup (3,\infty) and decreasing on (0,1)∪(2,3)(0,1) \cup (2,3).

If r(t)\vr(t) is the position of the particle at time tt, then its speed is ∣r′(t)∣|\vr'(t)|. We differentiate:

r′(t)=−e−t ı^−1t2 ȷ^+4(t−1)(t−2)(t−3)k^\vr'(t)=-e^{-t}\,\hi-\frac{1}{t^2}\,\hj+4(t-1)(t-2)(t-3)\hk

So, r(1)=−1e ı^−1 ȷ^\vr(1)=-\frac{1}{e}\,\hi-1\,\hj and r(3)=−1e3 ı^−19 ȷ^\vr(3)=-\frac{1}{e^3}\,\hi-\frac{1}{9}\,\hj. The absolute value of every component of r(1)\vr(1) is greater than or equal to that of the corresponding component of r(3)\vr(3), so ∣r(1)∣>∣r(3)∣|\vr(1)|>|\vr(3)|. That is, the particle is moving more swiftly at t=1t=1 than at t=3t=3.

Note: We could also compute the sizes of both vectors directly: ∣r′(1)∣=(1e)2+(−1)2|\vr'(1)|=\sqrt{\left(\frac{1}{e}\right)^2+(-1)^2}, and ∣r′(3)∣=(1e3)2+(−19)2|\vr'(3)|=\sqrt{\left(\frac{1}{e^3}\right)^2+\left(-\frac{1}{9}\right)^2}.

Q7Stage 1

Below is the graph of the parametrized function r(t)\vr(t). Let s(t)s(t) be the arclength along the curve from r(0)\vr(0) to r(t)\vr(t).

Figure from prob_s1.6, line 2

Figure from prob_s1.6, line 2

Indicate on the graph s(t+h)−s(t)s(t+h)-s(t) and r(t+h)−r(t)\vr(t+h)-\vr(t). Are the quantities scalars or vectors?

Hint

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

Answer

Figure from prob_s1.6, line 2

Figure from prob_s1.6, line 2

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

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

Full solution

Figure from prob_s1.6, line 2

Figure from prob_s1.6, line 2

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

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

Q8Stage 1

What is the relationship between velocity and speed in a vector-valued function of time?

Hint

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

Answer

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.

Full solution

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.

Q9Stage 1Past exam · M317 2005D

Let r(t)\vr(t) be a vector-valued function. Let r′\vr', r′′\vr'' , and r′′′\vr''' denote drdt\diff{\vr}{t}, d2rdt2\difftwo{\vr}{t} and d3rdt3\frac{\mathrm{d^3}\vr}{\mathrm{d}{t}^3}, respectively. Express

ddt[(r×r′)⋅r′′]\begin{equation*} \diff{}{t}\big[ (\vr \times \vr')\cdot\vr'' \big] \end{equation*}

in terms of r\vr, r′\vr' , r′′\vr'' , and r′′′\vr'''. Select the correct answer.

  1. (r′×r′′)⋅r′′′(\vr'\times\vr'' )\cdot\vr'''

  2. (r′×r′′)⋅r+(r×r′)⋅r′′′(\vr'\times\vr'' )\cdot\vr + (\vr\times\vr' )\cdot\vr'''

  3. (r×r′)⋅r′′′(\vr\times\vr' )\cdot\vr'''

  4. 00

  5. None of the above.

Hint

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

Answer

(c)

Full solution

By the product rule

ddt[(r×r′)⋅r′′]=(r′×r′)⋅r′′+(r×r′′)⋅r′′+(r×r′)⋅r′′′\begin{align*} \diff{}{t}\big[ (\vr \times \vr')\cdot\vr'' \big] &= (\vr' \times \vr')\cdot\vr'' +(\vr \times \vr'')\cdot\vr'' +(\vr \times \vr')\cdot\vr''' \end{align*}

The first term vanishes because r′×r′=0\vr'\times\vr'=\vZero. The second term vanishes because r×r′′\vr \times \vr'' is perpendicular to r′′\vr''. So

ddt[(r×r′)⋅r′′]=(r×r′)⋅r′′′\begin{equation*} \diff{}{t}\big[ (\vr \times \vr')\cdot\vr'' \big] = (\vr \times \vr')\cdot\vr''' \end{equation*}

which is (c).

2Stage 2Procedural

Practising the skill itself, until applying it is automatic.

Q10Stage 2Past exam · M317 2005D

Find the speed of a particle with the given position function

r(t)=52 t ı^+e5t ȷ^−e−5t k^\begin{equation*} \vr(t) = 5 \sqrt{2}\,t\,\hi + e^{5t}\,\hj - e^{-5t}\,\hk \end{equation*}

Select the correct answer:

  1. ∣v(t)∣=(e5t+e−5t)|\vv(t)| = \big(e^{5t} + e^{-5t}\big)

  2. ∣v(t)∣=10+5et+5e−t|\vv(t)| = \sqrt{10 + 5e^{t} + 5e^{-t}}

  3. ∣v(t)∣=10+e10t+e−10t|\vv(t)| = \sqrt{10 + e^{10t} + e^{-10t}}

  4. ∣v(t)∣=5(e5t+e−5t)|\vv(t)| = 5\big(e^{5t} + e^{-5t}\big)

  5. ∣v(t)∣=5(et+e−t)|\vv(t)| = 5\big(e^t + e^{-t}\big)

Hint

Just compute ∣v(t)∣|\vv(t)|. Note that (eat+e−at)2=e2at+2+e−2at\big(e^{at}+e^{-at}\big)^2 =e^{2at} + 2 + e^{-2at}.

Answer

(d)

Full solution

We have

v(t)=r′(t)=52 ı^+5e5t ȷ^+5e−5t k^\begin{equation*} \vv(t) = \vr'(t) = 5 \sqrt{2}\,\hi + 5e^{5t}\,\hj +5 e^{-5t}\,\hk \end{equation*}

and hence

∣v(t)∣=∣r′(t)∣=5∣2 ı^+e5t ȷ^+e−5t k^∣=52+e10t+e−10t\begin{equation*} |\vv(t)| = |\vr'(t)| = 5 \big|\sqrt{2}\,\hi + e^{5t}\,\hj + e^{-5t}\,\hk\big| = 5\sqrt{2+ e^{10t}+ e^{-10t}} \end{equation*}

Since 2+e10t+e−10t=(e5t+e−5t)22+ e^{10t}+ e^{-10t} = \big(e^{5t}+e^{-5t}\big)^2, that's (d).

Q11Stage 2

Find the velocity, speed and acceleration at time tt of the particle whose position is r(t)\vr(t). Describe the path of the particle. Let a>0a>0.

  1. r(t)=acos⁡t ı^+asin⁡t ȷ^+ct k^\vr(t)= a \cos t\,\hi + a\sin t\,\hj + ct\,\hk

  2. r(t)=acos⁡tsin⁡t ı^+asin⁡2t ȷ^+acos⁡t k^\vr(t)= a \cos t\sin t\,\hi + a\sin^2 t\,\hj + a\cos t\,\hk

Hint

To figure out what the curves look like, first detemine what curve (x(t),y(t))\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)x(t) and y(t)y(t) in terms of sin⁡(2t)\sin(2t) and cos⁡(2t)\cos(2t).

Answer

(a)

v(t)=−asin⁡t ı^+acos⁡t ȷ^+c k^dsdt(t)=a2+c2a(t)=−acos⁡t ı^−asin⁡t ȷ^\begin{align*} \vv(t)&= -a \sin t\,\hi+a\cos t\,\hj+c\,\hk\\ \diff{s}{t}(t)&= \sqrt{a^2+c^2}\\ \va(t)&= -a \cos t\,\hi-a\sin t\,\hj \end{align*}

The path is a helix with radius aa and with each turn having height 2πc2\pi c.

(b)

v(t)=acos⁡2t ı^+asin⁡2t ȷ^−asin⁡t k^dsdt(t)=a1+sin⁡2ta(t)=−2asin⁡2t ı^+2acos⁡2t ȷ^−acos⁡t k^\begin{align*} \vv(t)&= a \cos 2t\,\hi +a\sin 2t\,\hj-a\sin t\,\hk\\ \diff{s}{t}(t)&= a\sqrt{1+\sin^2t}\\ \va(t)&= -2a \sin 2t\,\hi +2a\cos 2t\,\hj-a\cos t\,\hk \end{align*}

The (x,y)(x,y) coordinates go around a circle of radius a2\frac{a}{2} and centre (0,a2)\big(0,\frac{a}{2}\big) counterclockwise. At the same time the zz coordinate oscillates over the interval between 11 and −1-1 half as fast. In addition, the curve lies on the intersection of the cylinder x2+[y−a2]2=a24x^2+\big[y-\tfrac{a}{2}\big]^2=\tfrac{a^2}{4} and the sphere x2+y2+z2=a2x^2+y^2+z^2=a^2.

Full solution

(a) By definition,

r(t)=acos⁡t ı^+asin⁡t ȷ^+ct k^v(t)=r′(t)=−asin⁡t ı^+acos⁡t ȷ^+c k^dsdt(t)=∣v(t)∣=a2+c2a(t)=r′′(t)=−acos⁡t ı^−asin⁡t ȷ^\begin{align*} \vr(t)&= a \cos t\,\hi+a\sin t\,\hj+ct\,\hk\\ \vv(t)=\vr'(t)&= -a \sin t\,\hi+a\cos t\,\hj+c\,\hk\\ \diff{s}{t}(t)=|\vv(t)|&= \sqrt{a^2+c^2}\\ \va(t)=\vr''(t)&= -a \cos t\,\hi-a\sin t\,\hj \end{align*}

The (x,y)=a(cos⁡t,sin⁡t)(x,y)= a(\cos t,\sin t) coordinates go around a circle of radius aa and centre (0,0)(0,0) counterclockwise. One circle is completed for each increase of tt by 2π2\pi. At the same time, the zz coordinate increases at a constant rate. Each time the (x,y)(x,y) coordinates complete one circle, the zz coordinate increases by 2πc2 \pi c. The path is a helix with radius aa and with each turn having height 2πc2\pi c.

(b) By definition,

r(t)=acos⁡tsin⁡t ı^+asin⁡2t ȷ^+acos⁡t k^=a2sin⁡2t ı^+a1−cos⁡2t2 ȷ^+acos⁡t k^v(t)=r′(t)=acos⁡2t ı^+asin⁡2t ȷ^−asin⁡t k^dsdt(t)=∣v(t)∣=a1+sin⁡2ta(t)=r′′(t)=−2asin⁡2t ı^+2acos⁡2t ȷ^−acos⁡t k^\begin{align*} \vr(t)&= a \cos t\sin t\,\hi +a\sin^2 t\,\hj+a\cos t\,\hk\\ &= \frac{a}{2} \sin 2t\,\hi+a\frac{1-\cos 2t}{2}\,\hj+a\cos t\,\hk\\ \vv(t)=\vr'(t)&= a \cos 2t\,\hi +a\sin 2t\,\hj-a\sin t\,\hk\\ \diff{s}{t}(t)=|\vv(t)|&= a\sqrt{1+\sin^2t}\\ \va(t)=\vr''(t)&= -2a \sin 2t\,\hi +2a\cos 2t\,\hj-a\cos t\,\hk \end{align*}

Write

x(t)=a2sin⁡2ty(t)=a1−cos⁡2t2z(t)=acos⁡t\begin{align*} x(t)=\tfrac{a}{2} \sin 2t \qquad y(t)=a\tfrac{1-\cos 2t}{2} \qquad z(t)=a\cos t \end{align*}

Then

x(t)2+[y(t)−a2]2=a24sin⁡22t+a24cos⁡22t=a24\begin{align*} x(t)^2+\big[y(t)-\tfrac{a}{2}\big]^2&=\tfrac{a^2}{4} \sin^2 2t + \tfrac{a^2}{4} \cos^2 2t =\tfrac{a^2}{4} \end{align*}

Thus the (x,y)(x,y) coordinates go around a circle of radius a2\frac{a}{2} and centre (0,a2)\big(0,\frac{a}{2}\big) counterclockwise. At the same time the zz coordinate oscillates over the interval between 11 and −1-1 half as fast. In fact, one can say more about the curve.

x(t)2+y(t)2=a24sin⁡22t+a24(1−cos⁡2t)2=a24sin⁡22t+a24−a22cos⁡2t+a24cos⁡22t=a22−a22(1−2sin⁡2t)=a2sin⁡2tx(t)2+y(t)2+z(t)2=a2sin⁡2t+a2cos⁡2t=a2\begin{align*} x(t)^2+y(t)^2&=\tfrac{a^2}{4} \sin^2 2t + \tfrac{a^2}{4}\big(1-\cos 2t\big)^2 \\ &=\tfrac{a^2}{4} \sin^2 2t + \tfrac{a^2}{4} - \tfrac{a^2}{2}\cos 2t + \tfrac{a^2}{4} \cos^2 2t \\ &= \tfrac{a^2}{2} - \tfrac{a^2}{2} (1- 2\sin^2 t) \\ &= a^2\sin^2 t \\ x(t)^2+y(t)^2+z(t)^2 &= a^2\sin^2 t + a^2\cos^2 t =a^2 \end{align*}

So our curve lies on the intersection of the cylinder
x2+[y−a2]2=a24x^2+\big[y-\tfrac{a}{2}\big]^2=\tfrac{a^2}{4} and the sphere x2+y2+z2=a2x^2+y^2+z^2=a^2. (Don't worry if you didn't think of trying to evaluate x(t)2+y(t)2+z(t)2x(t)^2+y(t)^2+z(t)^2.)

Q12Stage 2Past exam · M317 2013D

  1. Let

    r(t)=(t2,3,13t3)\begin{equation*} \vr(t) = \left(t^2 , 3, \tfrac{1}{3} t^3 \right) \end{equation*}

    Find the unit tangent vector to this parametrized curve at t=1t = 1, pointing in the direction of increasing tt.

  2. Find the arc length of the curve from (a) between the points (0,3,0)(0, 3, 0) and (1,3,−13)(1, 3, -\frac{1}{3}).

Hint

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

Answer

(a) T^(1)=(2,0,1)5\hat\vT(1) = \frac{(2,0,1)}{\sqrt{5}} (b) 13[53/2−8]\frac{1}{3}\big[5^{3/2}-8\big]

Full solution

(a) Since r′(t)=(2t,0,t2)\vr'(t) = (2t,0,t^2), the specified unit tangent at t=1t=1 is

T^(1)=(2,0,1)5\begin{equation*} \hat\vT(1) = \frac{(2,0,1)}{\sqrt{5}} \end{equation*}

(b) We are to find the arc length between r(0)\vr(0) and r(−1)\vr(-1). As dsdt=4t2+t4\diff{s}{t}=\sqrt{4t^2+t^4}, the

arc length=∫−104t2+t4 dt\begin{align*} \text{arc length} &= \int_{-1}^0 \sqrt{4t^2+t^4}\ \dee{t} \end{align*}

The integrand is even, so

arc length=∫014t2+t4 dt=∫01t4+t2 dt=[13(4+t2)3/2]01=13[53/2−8]\begin{align*} \text{arc length} &= \int_0^1 \sqrt{4t^2+t^4}\ \dee{t} =\int_0^1 t\sqrt{4+t^2}\ \dee{t} =\Big[\tfrac{1}{3}{(4+t^2)}^{3/2}\Big]_0^1 =\tfrac{1}{3}\big[5^{3/2}-8\big] \end{align*}
Q13Stage 2

Using Lemma 1.6.12 in the CLP-3 text, find the arclength of r(t)=(t,32t2,t3)\vr(t)=\left(t,\sqrt{\frac{3}{2}}t^2,t^3\right) from t=0t=0 to t=1t=1.

Hint

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

∫01∣drdt(t)∣dt\int_{0}^1\left| \diff{\vr}{t}(t)\right|\dee t

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

Answer

2

Full solution

By Lemma 1.6.12 in the CLP-3 text,
the arclength of r(t)\vr(t) from t=0t=0 to t=1t=1 is ∫01∣drdt(t)∣dt\int_{0}^1\left| \diff{\vr}{t}(t)\right|\dee t. We'll calculate this in a few pieces to make the steps clearer.

r(t)=(t,32t2,t3)drdt(t)=(1,6t,3t2)∣drdt(t)∣=12+(6t)2+(3t2)2=1+6t2+9t4=(3t2+1)2=3t2+1∫01∣drdt(t)∣dt=∫01(3t2+1)dt=2\begin{align*} \vr(t)&=\left(t,\sqrt{\frac{3}{2}}t^2,t^3\right)\\ \diff{\vr}{t}(t)&=\left(1,\sqrt{6}t,3t^2\right)\\ \left|\diff{\vr}{t}(t)\right|&=\sqrt{1^2+(\sqrt{6}t)^2+(3t^2)^2}=\sqrt{1+6t^2+9t^4}=\sqrt{(3t^2+1)^2}=3t^2+1\\ \int_{0}^1\left| \diff{\vr}{t}(t)\right|\dee t&=\int_0^1\left(3t^2+1 \right)\dee t=2 \end{align*}
Q14Stage 2

A particle's position at time tt is given by r(t)=(t+sin⁡t,cos⁡t)\vr(t)=(t+\sin t, \cos t) (The particle traces out a cycloid–see Question 4). What is the magnitude of the acceleration of the particle at time tt?

Hint

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

Answer

1

Full solution

Since r(t)\vr(t) is the position of the particle, its acceleration is r′′(t)\vr''(t).

r(t)=(t+sin⁡t,cos⁡t)r′(t)=(1+cos⁡t,−sin⁡t)r′′(t)=(−sin⁡t,−cos⁡t)∣r′′(t)∣=sin⁡2t+cos⁡2t=1\begin{align*} \vr(t)&=(t+\sin t, \cos t)\\ \vr'(t)&=(1+\cos t,-\sin t)\\ \vr''(t)&=(-\sin t,-\cos t)\\ |\vr''(t)|&=\sqrt{\sin^2t+\cos^2t}=1 \end{align*}

The magnitude of acceleration is constant, but its direction is changing, since r′′(t)\vr''(t) is a vector with changing direction.

Q15Stage 2Past exam · M317 2011D

A curve in R3\bbbr^3 is given by the vector equation r(t)=(2tcos⁡t,2tsin⁡t,t33)\vr(t) = \left(2t \cos t, 2t \sin t,\frac{t^3}{3}\right)

  1. Find the length of the curve between t=0t = 0 and t=2t = 2.

  2. Find the parametric equations of the tangent line to the curve at t=πt = \pi.

Hint

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

Answer

(a) 20 ⁣/3\nicefrac{20}{3} (b) x(t)=−2π−2t, y(t)=−2πt, z(t)=π3 ⁣/3+π2tx(t) = -2\pi -2t,\ y(t) = -2\pi t,\ z(t) = \nicefrac{\pi^3}{3} + \pi^2 t

Full solution

(a) The speed is

dsdt(t)=∣r′(t)∣=∣(2cos⁡t−2tsin⁡t , 2sin⁡t+2tcos⁡t , t2)∣=(2cos⁡t−2tsin⁡t)2+(2sin⁡t+2tcos⁡t)2+t4=4+4t2+t4=2+t2\begin{align*} \diff{s}{t}(t) =\big|\vr'(t)\big| & = \left|\left(2\cos t - 2t\sin t\,,\, 2\sin t + 2t \cos t\,,\, t^2\right)\right| \\ &=\sqrt{\big(2\cos t - 2t\sin t\big)^2 +\big(2\sin t + 2t \cos t\big)^2 +t^4} \\ &= \sqrt{4+ 4t^2 +t^4} \\ &= 2+t^2 \end{align*}

so the length of the curve is

length =∫02dsdt dt=∫02(2+t2) dt=[2t+t33]02=203\begin{align*} \text{length } &=\int_0^2 \diff{s}{t}\,\dee{t} =\int_0^2 (2+t^2)\,\dee{t} = \left[2t +\frac{t^3}{3}\right]_0^2 =\frac{20}{3} \end{align*}

(b) A tangent vector to the curve at r(π)=(−2π , 0 , π3 ⁣/3)\vr(\pi)=\big(-2\pi\,,\, 0\,,\, \nicefrac{\pi^3}{3}\big) is

r′(π)=(2cos⁡π−2πsin⁡π , 2sin⁡π+2πcos⁡π , π2)=(−2 , −2π , π2)\begin{equation*} \vr'(\pi) = \left(2\cos\pi - 2\pi\sin\pi\,,\, 2\sin\pi + 2\pi \cos\pi\,,\, \pi^2\right) = (-2\,,\,-2\pi\,,\,\pi^2) \end{equation*}

So parametric equations for the tangent line at r(π)\vr(\pi) are

x(t)=−2π−2ty(t)=−2πtz(t)=π3 ⁣/3+π2t\begin{align*} x(t) &= -2\pi -2t \\ y(t) &= -2\pi t \\ z(t) &= \nicefrac{\pi^3}{3} + \pi^2 t \end{align*}
Q16Stage 2Past exam · M317 2010D

Let r(t)=(3cos⁡t,3sin⁡t,4t)\vr(t) = \big(3 \cos t, 3 \sin t, 4t\big) be the position vector of a particle as a function of time t≥0t \ge 0.

  1. Find the velocity of the particle as a function of time tt.

  2. Find the arclength of its path between t=1t = 1 and t=2t = 2.

Hint

Review Lemma 1.6.12 in the CLP-3 text.

Answer

(a) r′(t)=(−3sin⁡t,3cos⁡t,4)\vr'(t) = \big(-3 \sin t, 3 \cos t, 4\big) (b) 5

Full solution

(a) As r(t)=(3cos⁡t,3sin⁡t,4t)\vr(t) = \big(3 \cos t, 3 \sin t, 4t\big), the velocity of the particle is

r′(t)=(−3sin⁡t,3cos⁡t,4)\begin{equation*} \vr'(t) = \big(-3 \sin t, 3 \cos t, 4\big) \end{equation*}

(b) As dsdt\diff{s}{t}, the rate of change of arc length per unit time, is

dsdt(t)=∣r′(t)∣=∣(−3sin⁡t,3cos⁡t,4)∣=5\begin{equation*} \diff{s}{t}(t) = |\vr'(t)| = \big|\big(-3 \sin t, 3 \cos t, 4\big)\big| =5 \end{equation*}

the arclength of its path between t=1t = 1 and t=2t = 2 is

∫12dt dsdt(t)=∫12dt 5=5\begin{equation*} \int_1^2\dee{t}\ \diff{s}{t}(t) =\int_1^2\dee{t}\ 5 =5 \end{equation*}
Q17Stage 2Past exam · M317 2007A

Consider the curve

r(t)=13cos⁡3t ı^+13sin⁡3t ȷ^+sin⁡3t k^\begin{equation*} \vr(t) = \frac{1}{3}\cos^3 t\,\hi +\frac{1}{3} \sin^3 t\,\hj + \sin^3 t\,\hk \end{equation*}

  1. Compute the arc length of the curve from t=0t = 0 to t=π2t = \frac{\pi}{2}.

  2. Compute the arc length of the curve from t=0t = 0 to t=πt = \pi.

Hint

If you got the answer 00 in part (b), you dropped some absolute value signs.

Answer

(a) 127(1010−1)\frac{1}{27}\big(10\sqrt{10}-1\big) (b) 227(1010−1)\frac{2}{27}\big(10\sqrt{10}-1\big)

Full solution

(a) As

r′(t)=−sin⁡tcos⁡2t ı^+sin⁡2tcos⁡t ı^+3sin⁡2tcos⁡t k^=sin⁡tcos⁡t(−cos⁡t ı^+sin⁡t ȷ^+3sin⁡t k^)dsdt(t)=∣sin⁡tcos⁡t∣cos⁡2t+sin⁡2t+9sin⁡2t=∣sin⁡tcos⁡t∣1+9sin⁡2t\begin{align*} \vr'(t) & = -\sin t\cos^2 t\,\hi + \sin^2 t\cos t\,\hi + 3\sin^2 t\cos t\,\hk = \sin t\cos t\big(-\cos t\,\hi +\sin t\,\hj +3\sin t\,\hk\big) \\ \diff{s}{t}(t) & = |\sin t\cos t|\sqrt{\cos^2 t + \sin ^2 t + 9\sin^2 t} = |\sin t\cos t|\sqrt{1+ 9\sin^2 t} \end{align*}

the arclength from t=0t = 0 to t=π2t = \frac{\pi}{2} is

∫0π/2dsdt(t) dt=∫0π/2sin⁡tcos⁡t1+9sin⁡2t dt=118∫110u duwith u=1+9sin⁡2t, du=18sin⁡tcos⁡t dt=118[23u3/2]110=127(1010−1)\begin{align*} \int_0^{\pi/2} \diff{s}{t}(t)\,\dee{t} &=\int_0^{\pi/2} \sin t\cos t \sqrt{1+ 9\sin^2 t}\,\dee{t} \\ &=\frac{1}{18}\int_1^{10} \sqrt{u}\ \dee{u} \qquad \text{with } u = 1+ 9\sin^2 t,\ \dee{u} = 18\sin t\cos t\,\dee{t}\\ &=\frac{1}{18}\Big[\frac{2}{3}u^{3/2}\Big]_1^{10}\\ &=\frac{1}{27}\big(10\sqrt{10}-1\big) \end{align*}

(b) The arclength from t=0t = 0 to t=πt = \pi is

∫0πdsdt(t) dt=∫0π∣sin⁡tcos⁡t∣1+9sin⁡2t dtDon’t forget the absolute value signs!=2∫0π/2∣sin⁡tcos⁡t∣1+9sin⁡2t dt=2∫0π/2sin⁡tcos⁡t1+9sin⁡2t dt\begin{align*} \int_0^{\pi} \diff{s}{t}(t)\,\dee{t} &=\int_0^{\pi} |\sin t\cos t| \sqrt{1+ 9\sin^2 t}\,\dee{t} \qquad\text{Don't forget the absolute value signs!}\\ &=2\int_0^{\pi/2} |\sin t\cos t| \sqrt{1+ 9\sin^2 t}\,\dee{t} = 2\int_0^{\pi/2} \sin t\cos t \sqrt{1+ 9\sin^2 t}\,\dee{t} \end{align*}

since the integrand is invariant under t→π−tt\rightarrow\pi-t. So the arc length from t=0t = 0 to t=πt = \pi is just twice the arc length from part (a), namely 227(1010−1)\frac{2}{27}\big(10\sqrt{10}-1\big).

Q18Stage 2Past exam · M317 2017D

Let r(t)=(13t3,12t2,12t)\vr(t)=\big(\frac{1}{3}t^3,\frac{1}{2}t^2,\frac{1}{2}t\big), t≥0t\ge 0. Compute s(ts(t), the arclength of the curve at time tt.

Answer

s(t)=t33+t2s(t)=\frac{t^3}{3} +\frac{t}{2}

Full solution

Since

r(t)=t33 ı^+t22 ȷ^+t2 k^r′(t)=t2 ı^+t ȷ^+12 k^dsdt(t)=∣r′(t)∣=t4+t2+14=(t2+12)2=t2+12\begin{align*} \vr(t)&= \frac{t^3}{3}\,\hi + \frac{t^2}{2}\,\hj + \frac{t}{2}\,\hk\\ \vr'(t)&= t^2\,\hi + t\,\hj + \frac{1}{2}\,\hk \\ \diff{s}{t}(t)=|\vr'(t)|&=\sqrt{t^4+t^2+\frac{1}{4}} =\sqrt{\Big(t^2+\frac{1}{2}\Big)^2}=t^2+\frac{1}{2} \end{align*}

the length of the curve is

s(t)=∫0tdsdt(u) du=∫0t(u2+12) du=t33+t2\begin{align*} s(t)=\int_0^t \diff{s}{t}(u)\,\dee{u} =\int_0^t \Big(u^2+\frac{1}{2}\Big)\,\dee{u} =\frac{t^3}{3} +\frac{t}{2} \end{align*}
Q19Stage 2Past exam · M317 2011A

Find the arc length of the curve r(t)=(tm , tm , t3m/2)\vr(t) = \big(t^m\,,\, t^m\,,\, t^{3m/2}\big) for 0≤a≤t≤b0 \le a \le t \le b, and where m>0m > 0. Express your result in terms of mm, aa, and bb.

Hint

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

Answer

827[(2+94bm)3/2−(2+94am)3/2]\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]

Full solution

Since

r(t)=tm ı^+tm ȷ^+t3m/2 k^r′(t)=mtm−1 ı^+mtm−1 ȷ^+3m2t3m/2−1 k^dsdt=∣r′(t)∣=2m2t2m−2+9m24t3m−2=mtm−12+94tm\begin{align*} \vr(t) & = t^m\,\hi + t^m\,\hj + t^{3m/2}\,\hk \\ \vr'(t) &= mt^{m-1}\,\hi + mt^{m-1}\,\hj +\frac{3m}{2}t^{3m/2-1}\,\hk \\ \diff{s}{t} = |\vr'(t)| & = \sqrt{ 2m^2 t^{2m-2} +\frac{9m^2}{4} t^{3m-2} } = mt^{m-1}\sqrt{2 + \frac{9}{4}t^m } \end{align*}

the arc length is

∫abdsdt(t) dt=∫abmtm−12+94tm dt=49∫2+94am2+94bmu duwith u=2+94tm, du=9m4tm−1=49[23u3/2]2+94am2+94bm=827[(2+94bm)3/2−(2+94am)3/2]\begin{align*} \int_a^b \diff{s}{t}(t)\,\dee{t} &=\int_a^b mt^{m-1}\sqrt{2 + \frac{9}{4}t^m }\ \dee{t} \\ &=\frac{4}{9}\int_{2 + \frac{9}{4}a^m}^{2 + \frac{9}{4}b^m}\sqrt{u}\,\dee{u} \qquad\text{with } u = 2 + \frac{9}{4}t^m,\ \dee{u} = \frac{9m}{4}t^{m-1}\\ &=\frac{4}{9}\Big[\frac{2}{3}u^{3/2}\Big]_{2 + \frac{9}{4}a^m} ^{2 + \frac{9}{4}b^m} \\ &=\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] \end{align*}
Q20Stage 2

If a particle has constant mass mm, position r\vr, and is moving with velocity v\vv, then its angular momentum is L=m(r×v)\vL=m(\vr\times\vv).

For a particle with mass m=1m=1 and position function r=(sin⁡t,cos⁡t,t)\vr=(\sin t, \cos t, t), find ∣dLdt∣\left|\diff{\vL}{t} \right|.

Hint

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

Answer

∣t∣|t|

Full solution

Given the position of the particle, we can find its velocity:

v(t)=r′(t)=(cos⁡t,−sin⁡t,1)\vv(t)=\vr'(t)=(\cos t, -\sin t , 1)

Applying the given formula,

L(t)=r×v=(sin⁡t,cos⁡t,t)×(cos⁡t,−sin⁡t,1).\vL(t)=\vr \times \vv=(\sin t , \cos t , t) \times (\cos t, -\sin t , 1).

  • We can first compute the cross product, then differentiate:

    L(t)=(cos⁡t+tsin⁡t)ı^+(tcos⁡t−sin⁡t)ȷ^−k^L′(t)=tcos⁡t ı^−tsin⁡t ȷ^∣L′(t)∣=t2(sin⁡2t+cos⁡2t)=t2=∣t∣\begin{align*} \vL(t)& = (\cos t + t\sin t)\hi + (t\cos t - \sin t)\hj-\hk\\ \vL'(t)&=t\cos t\,\hi -t\sin t\, \hj\\ |\vL'(t)|&=\sqrt{t^2(\sin^2 t + \cos^2 t)}=\sqrt{t^2}=|t| \end{align*}

  • Using the product rule:

    L′(t)=r′(t)×v(t)+r(t)×v′(t)=r′(t)×r′(t)⏟0+r(t)×v′(t)=(sin⁡t,cos⁡t,t)×(−sin⁡t,−cos⁡t,0)=tcos⁡t ı^−tsin⁡t ȷ^∣L′(t)∣=t2cos⁡2t+t2sin⁡t2=∣t∣\begin{align*} \vL'(t)&=\vr'(t)\times \vv(t) + \vr(t) \times \vv'(t)\\ &=\underbrace{\vr'(t)\times \vr'(t)}_{0} + \vr(t) \times \vv'(t)\\ &= (\sin t ,\cos t,t)\times(-\sin t , -\cos t, 0)\\ &=t\cos t\,\hi-t\sin t\,\hj\\ |\vL'(t)|&=\sqrt{t^2\cos^2 t + t^2\sin t^2 }=|t| \end{align*}
Q21Stage 2Past exam · M200 2002A

Consider the space curve Γ\Ga whose vector equation is

r(t)=tsin⁡(πt) ı^+tcos⁡(πt) ȷ^+t2k^ 0≤t<∞\begin{equation*} \vr(t)=t\sin(\pi t)\,\hi+t\cos(\pi t)\,\hj +t^2\hk\,\qquad 0\le t<\infty \end{equation*}

This curve starts from the origin and eventually reaches the ellipsoid EE whose equation is 2x2+2y2+z2=242x^2+2y^2+z^2=24.

  1. Determine the coordinates of the point PP where Γ\Ga intersects EE.

  2. Find the tangent vector of Γ\Ga at the point PP.

  3. Does Γ\Ga intersect EE at right angles? Why or why not?

Answer

(a) r(2)=2ȷ^+4k^\vr(2)=2\hj+4\hk (b) any nonzero multiple of r′(2)=2π ı^+ȷ^+4 k^\vr'(2)=2\pi\,\hi+\hj+4\,\hk

(c) Γ\Ga and EE do not intersect at right angles.

Full solution

(a) The curve intersects EE when

2(tsin⁡(πt))2+2(tcos⁡(πt))2+(t2)2=24  ⟺  2t2+t4=24  ⟺  (t2−4)(t2+6)=0\begin{equation*} 2\big(t\sin(\pi t)\big)^2+2\big(t\cos(\pi t)\big)^2 +\big(t^2\big)^2=24 \iff 2t^2+t^4=24 \iff (t^2-4)(t^2+6)=0 \end{equation*}

Since we need t>0t>0, the desired time is t=2t=2 and the corresponding point is r(2)=2 ȷ^+4 k^\vr(2)=2\,\hj+4\,\hk.

(b) Since

r′(t)=[sin⁡(πt)+πtcos⁡(πt)]ı^+[cos⁡(πt)−πtsin⁡(πt)]ȷ^+2tk^\begin{equation*} \vr'(t)=\big[\sin(\pi t)+\pi t\cos(\pi t)\big]\hi +\big[\cos(\pi t)-\pi t\sin(\pi t)\big]\hj + 2t\hk \end{equation*}

a tangent vector to Γ\Ga at PP is any nonzero multiple of

r′(2)=2π ı^+ȷ^+4 k^\begin{equation*} \vr'(2)=2\pi\,\hi+\hj+4\,\hk \end{equation*}

(c) A normal vector to EE at PP is

∇(2x2+2y2+z2)∣(0,2,4)=<4x,4y,2z>∣(0,2,4)=<0,8,8>\begin{equation*} \vnabla(2x^2+2y^2+z^2)\big|_{(0,2,4)}= \llt 4x,4y,2z\rgt\big|_{(0,2,4)} =\llt 0,8,8\rgt \end{equation*}

Since r′(2)\vr'(2) and <0,8,8>\llt 0,8,8\rgt are not parallel, Γ\Ga and EE do not intersect at right angles.

Q22Stage 2Past exam · M200 2001A

Suppose a particle in 3-dimensional space travels with position vector r(t)\vr(t), which satisfies r′′(t)=−r(t)\vr''(t)=-\vr(t). Show that the “energy” ∣r(t)∣2+∣r′(t)∣2|\vr(t)|^2+|\vr'(t)|^2 is constant (that is, independent of tt).

Answer

ddt[∣r(t)∣2+∣r′(t)∣2]=0\diff{}{t}\big[|\vr(t)|^2+|\vr'(t)|^2\big]=0

Full solution
ddt[∣r(t)∣2+∣r′(t)∣2]=ddt[r(t)⋅r(t)+r′(t)⋅r′(t)]=2r(t)⋅r′(t)+2r′(t)⋅r′′(t)=2r′(t)⋅[r(t)+r′′(t)]=0 since r′′(t)=−r(t)\begin{align*} \diff{}{t}\big[|\vr(t)|^2+|\vr'(t)|^2\big] &=\diff{}{t}\big[\vr(t)\cdot\vr(t)+\vr'(t)\cdot\vr'(t)\big]\\ &=2\vr(t)\cdot\vr'(t)+2\vr'(t)\cdot\vr''(t)\\ &=2\vr'(t)\cdot\big[\vr(t)+\vr''(t)\big]\\ &=0\qquad \text{ since }\vr''(t)=-\vr(t) \end{align*}

Since ddt[∣r(t)∣2+∣r′(t)∣2]=0\diff{}{t}\big[|\vr(t)|^2+|\vr'(t)|^2\big]=0 for all tt, ∣r(t)∣2+∣r′(t)∣2|\vr(t)|^2+|\vr'(t)|^2 is independent of tt.

3Stage 3Application

Further than practice: several ideas at once, or an unfamiliar situation.

Q23Stage 3Past exam · M317 2000A

A particle moves along the curve C\cC of intersection of the surfaces z2=12yz^2=12y and 18x=yz18x=yz in the upward direction. When the particle is at (1,3,6)(1,3,6) its velocity v\vv and acceleration a\va are given by

v=6 ı^+12 ȷ^+12 k^a=27 ı^+30 ȷ^+6 k^\vv =6\,\hi+12\,\hj+12\,\hk\qquad \va = 27\,\hi+30\,\hj+6\,\hk

  1. Write a vector parametric equation for C\cC using u=z6u=\frac{z}{6} as a parameter.

  2. Find the length of C\cC from (0,0,0)(0,0,0) to (1,3,6)(1,3,6).

  3. If u=u(t)u=u(t) is the parameter value for the particle's position at time tt, find dudt\diff{u}{t} when the particle is at (1,3,6)(1,3,6).

  4. Find d2udt2\difftwo{u}{t} when the particle is at (1,3,6)(1,3,6).

Hint

If r(u)\vr(u) is the parametrization of C\cC by uu, then the position of the particle at time tt is R(t)=r(u(t))\vR(t) = \vr\big(u(t)\big).

Answer

(a) r(u)=u3 ı^+3u2 ȷ^+6u k^\vr(u)=u^3\,\hi+3u^2\,\hj+6u\,\hk (b) 77 (c) 22 (d) 11

Full solution

(a) Since z=6uz=6u, y=z212=3u2y=\frac{z^2}{12}=3u^2 and x=yz18=u3x=\frac{yz}{18}=u^3,

r(u)=u3 ı^+3u2 ȷ^+6u k^\begin{align*} \vr(u)&=u^3\,\hi+3u^2\,\hj+6u\,\hk\cr \end{align*}

(b)

r′(u)=3u2 ı^+6u ȷ^+6 k^r′′(u)=6u ı^+6 ȷ^dsdu(u)=∣r′(u)∣=9u4+36u2+36=3(u2+2)\begin{align*} \vr'(u)&=3u^2\,\hi+6u\,\hj+6\,\hk\cr \vr''(u)&=6u\,\hi+6\,\hj\cr \diff{s}{u}(u)=|\vr'(u)| &=\sqrt{9u^4+36u^2+36}=3\big(u^2+2\big)\cr \end{align*}
∫C ds=∫01dsdu du=∫013(u2+2) du=[u3+6u]01=7\int_\cC \ ds =\int_0^1 \diff{s}{u}\ \dee{u} =\int_0^1 3\big(u^2+2\big)\ \dee{u} =\big[u^3+6u\big]_0^1 =7

(c) Denote by R(t)\vR(t) the position of the particle at time tt. Then

R(t)=r(u(t))  ⟹  R′(t)=r′(u(t))dudt\vR(t)=\vr\big(u(t)\big)\implies \vR'(t)=\vr'\big(u(t)\big)\diff{u}{t}

In particular, if the particle is at (1,3,6)(1,3,6) at time t1t_1, then u(t1)=1u(t_1)=1 and

6 ı^+12 ȷ^+12 k^=R′(t1)=r′(1)dudt(t1)=(3 ı^+6 ȷ^+6 k^)dudt(t1)6\,\hi+12\,\hj+12\,\hk=\vR'(t_1)=\vr'(1)\diff{u}{t}(t_1) =\big(3\,\hi+6\,\hj+6\,\hk\big)\diff{u}{t}(t_1)

which implies that dudt(t1)=2\diff{u}{t}(t_1)=2.

(d) By the product and chain rules,

R′(t)=r′(u(t))dudt  ⟹  R′′(t)=r′′(u(t))(dudt)2+r′(u(t))d2udt2\vR'(t)=\vr'\big(u(t)\big)\diff{u}{t}\implies \vR''(t)=\vr''\big(u(t)\big)\Big(\diff{u}{t}\Big)^2 +\vr'\big(u(t)\big)\difftwo{u}{t}

In particular,

27 ı^+30 ȷ^+6 k^=R′′(t1)=r′′(1)(dudt(t1))2+r′(1)d2udt2(t1)=(6 ı^+6 ȷ^)22+(3 ı^+6 ȷ^+6 k^)d2udt2(t1)\begin{align*} 27\,\hi+30\,\hj+6\,\hk &=\vR''(t_1)=\vr''(1)\Big(\diff{u}{t}(t_1)\Big)^2 +\vr'\big(1\big)\difftwo{u}{t}(t_1) \\ &=\big(6\,\hi+6\,\hj\big)2^2+\big(3\,\hi+6\,\hj+6\,\hk\big)\difftwo{u}{t}(t_1) \end{align*}

Simplifying

3 ı^+6 ȷ^+6 k^=(3 ı^+6 ȷ^+6 k^)d2udt2(t1)  ⟹  d2udt2(t1)=13\,\hi+6\,\hj+6\,\hk =\big(3\,\hi+6\,\hj+6\,\hk\big)\difftwo{u}{t}(t_1) \implies \difftwo{u}{t}(t_1)=1
Q24Stage 3Past exam · M317 2008A

A particle of mass m=1m = 1 has position r0=12 k^\vr_0 = \frac{1}{2}\,\hk and velocity v0=π22 ı^\vv_0 =\frac{\pi^2}{2}\,\hi at time 00. It moves under a force

F(t)=−3t ı^+sin⁡t ȷ^+2e2t k^.\begin{equation*} \vF(t) = -3t\,\hi + \sin t\,\hj + 2e^{2t}\,\hk. \end{equation*}

  1. Determine the position r(t)\vr(t) of the particle depending on tt.

  2. At what time after time t=0t = 0 does the particle cross the plane x=0x = 0 for the first time?

  3. What is the velocity of the particle when it crosses the plane x=0x = 0 in part (b)?

Hint

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

Answer

(a) r(t)=(π2t2−t32) ı^+(t−sin⁡t) ȷ^+(12e2t−t) k^\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=πt=\pi

(c) −π2 ı^+2 ȷ^+(e2π−1) k^-\pi^2\,\hi +2\,\hj + \big(e^{2\pi}-1\big)\,\hk

Full solution

(a) According to Newton,

mr′′(t)=F(t)so thatr′′(t)=−3t ı^+sin⁡t ȷ^+2e2t k^\begin{equation*} m\vr''(t) = \vF(t)\qquad\text{so that}\qquad \vr''(t) = -3t\,\hi + \sin t\,\hj + 2e^{2t}\,\hk \end{equation*}

Integrating once gives

r′(t)=−3t22 ı^−cos⁡t ȷ^+e2t k^+c\begin{align*} \vr'(t) = -3\frac{t^2}{2}\,\hi - \cos t\,\hj + e^{2t}\,\hk +\vc \end{align*}

for some constant vector c\vc. We are told that r′(0)=v0=π22 ı^\vr'(0)=\vv_0 =\frac{\pi^2}{2}\,\hi. This forces c=π22 ı^+ȷ^−k^\vc=\frac{\pi^2}{2}\,\hi+\hj-\hk so that

r′(t)=(π22−3t22) ı^+(1−cos⁡t) ȷ^+(e2t−1) k^\begin{align*} \vr'(t) = \left(\frac{\pi^2}{2}-\frac{3t^2}{2}\right)\,\hi +(1- \cos t)\,\hj + \big(e^{2t}-1\big)\,\hk \end{align*}

Integrating a second time gives

r(t)=(π2t2−t32) ı^+(t−sin⁡t) ȷ^+(12e2t−t) k^+c\begin{align*} \vr(t) = \left(\frac{\pi^2 t}{2}-\frac{t^3}{2}\right)\,\hi +(t- \sin t)\,\hj + \left(\frac{1}{2}e^{2t}-t\right)\,\hk + \vc \end{align*}

for some (other) constant vector c\vc. We are told that r(0)=r0=12 k^\vr(0)=\vr_0 =\frac{1}{2}\,\hk. This forces c=0\vc=\vZero so that

r(t)=(π2t2−t32) ı^+(t−sin⁡t) ȷ^+(12e2t−t) k^\begin{align*} \vr(t) = \left(\frac{\pi^2 t}{2}-\frac{t^3}{2}\right)\,\hi +(t- \sin t)\,\hj + \left(\frac{1}{2}e^{2t}-t\right)\,\hk \end{align*}

(b) The particle is in the plane x=0x=0 when

0=(π2t2−t32)=t2(π2−t2)  ⟺  t=0,±π\begin{align*} 0=\left(\frac{\pi^2 t}{2}-\frac{t^3}{2}\right) =\frac{t}{2}(\pi^2-t^2) \iff t=0, \pm\pi \end{align*}

So the desired time is t=πt=\pi.

(c) At time t=πt=\pi, the velocity is

r′(π)=(π22−3π22) ı^+(1−cos⁡π) ȷ^+(e2π−1) k^=−π2 ı^+2 ȷ^+(e2π−1) k^\begin{align*} \vr'(\pi) &= \left(\frac{\pi^2}{2}-\frac{3\pi^2}{2}\right)\,\hi +(1- \cos\pi)\,\hj + \big(e^{2\pi}-1\big)\,\hk \\ &= -\pi^2\,\hi +2\,\hj + \big(e^{2\pi}-1\big)\,\hk \end{align*}
Q25Stage 3Past exam · M317 1999A

Let CC be the curve of intersection of the surfaces y=x2y=x^2 and z=23x3z=\frac{2}{3}x^3. A particle moves along CC with constant speed such that dxdt>0\diff{x}{t}>0. The particle is at (0,0,0)(0,0,0) at time t=0t=0 and is at (3,9,18)(3,9,18) at time t=72t=\frac{7}{2}.

  1. Find the length of the part of CC between (0,0,0)(0,0,0) and (3,9,18)(3,9,18).

  2. Find the constant speed of the particle.

  3. Find the velocity of the particle when it is at (1,1,23)\big(1,1,\frac{2}{3}\big).

  4. Find the acceleration of the particle when it is at (1,1,23)\big(1,1,\frac{2}{3}\big).

Hint

Denote by r(x)\vr(x) the parametrization of CC by xx. If the xx–coordinate of the particle at time tt is x(t)x(t), then the position of the particle at time tt is R(t)=r(x(t))\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 dxdt\diff{\vx}{t}.

Answer

(a) 2121 (b) 66 (c) 2ı^+4 ȷ^+4 k^2\hi+4\,\hj+4\,\hk (d) −83(2ı^+ ȷ^−2 k^)-\frac{8}{3}\big(2\hi+\,\hj-2\,\hk\big)

Full solution

(a) Parametrize CC by xx. Since y=x2y=x^2 and z=23x3z=\frac{2}{3}x^3,

r(x)=x ı^+x2 ȷ^+23x3 k^r′(x)=ı^+2x ȷ^+2x2 k^r′′(x)=2 ȷ^+4x k^dsdx=∣r′(x)∣=1+4x2+4x4=1+2x2\begin{align*} \vr(x)&=x\,\hi+x^2\,\hj+\frac{2}{3}x^3\,\hk\cr \vr'(x)&=\hi+2x\,\hj+2x^2\,\hk\cr \vr''(x)&=2\,\hj+4x\,\hk\cr \diff{s}{x} &=|\vr'(x)| =\sqrt{1+4x^2+4x^4}=1+2x^2\cr \end{align*}

and

∫C ds=∫03dsdx dx=∫03(1+2x2) dx=[x+23x3]03=21\int_C \ \dee{s} =\int_0^3 \diff{s}{x}\ \dee{x} =\int_0^3 \big(1+2x^2\big)\ \dee{x} ={\Big[x+\frac{2}{3}x^3\Big]}_0^3 =21

(b) The particle travelled a distance of 21 units in 72\frac{7}{2} time units. This corresponds to a speed of 217/2=6\frac{21}{7/2}=6.

(c) Denote by R(t)\vR(t) the position of the particle at time tt. Then

R(t)=r(x(t))  ⟹  R′(t)=r′(x(t))dxdt\vR(t)=\vr\big(x(t)\big)\implies \vR'(t)=\vr'\big(x(t)\big)\diff{x}{t}

By parts (a) and (b) and the chain rule

6=dsdt=dsdxdxdt=(1+2x2)dxdt  ⟹  dxdt=61+2x26=\diff{s}{t}=\diff{s}{x}\diff{x}{t}=(1+2x^2)\diff{x}{t} \implies \diff{x}{t}=\frac{6}{1+2x^2}

In particular, the particle is at (1,1,23)\big(1,1,\frac{2}{3}\big) at x=1x=1. At this time dxdt=61+2×1=2\diff{x}{t}=\frac{6}{1+2\times 1}=2 and

R′=r′(1)dxdt=(ı^+2 ȷ^+2 k^)2=2ı^+4 ȷ^+4 k^\vR'=\vr'\big(1\big)\diff{x}{t} =\big(\hi+2\,\hj+2\,\hk\big)2 =2\hi+4\,\hj+4\,\hk

(d) By the product and chain rules,

R′(t)=r′(x(t))dxdt  ⟹  R′′(t)=r′′(x(t))(dxdt)2+r′(x(t))d2xdt2\vR'(t)=\vr'\big(x(t)\big)\diff{x}{t}\implies \vR''(t)=\vr''\big(x(t)\big){\Big(\diff{x}{t}\Big)}^2 +\vr'\big(x(t)\big)\difftwo{x}{t}

Applying ddt\diff{}{t} to 6=(1+2x(t)2)dxdt(t)6=\big(1+2x(t)^2\big)\diff{x}{t}(t) gives

0=4x(dxdt)2+(1+2x2)d2xdt20=4x{\Big(\diff{x}{t}\Big)}^2+(1+2x^2)\difftwo{x}{t}

In particular, when x=1x=1 and dxdt=2\diff{x}{t}=2, 0=4×1(2)2+(3)d2xdt20=4\times 1\big(2\big)^2+(3)\difftwo{x}{t} gives d2xdt2=−163\difftwo{x}{t}=-\frac{16}{3} and

R′′=(2 ȷ^+4 k^)(2)2−(ı^+2 ȷ^+2 k^)163=−83(2ı^+ ȷ^−2 k^)\vR''=\big(2\,\hj+4\,\hk\big)\big(2\big)^2 -\big(\hi+2\,\hj+2\,\hk\big)\frac{16}{3} = -\frac{8}{3}\big(2\hi+\,\hj-2\,\hk\big)
Q26Stage 3

A camera mounted to a pole can swivel around in a full circle. It is tracking an object whose position at time tt seconds is x(t)x(t) metres east of the pole, and y(t)y(t) metres north of the pole.

In order to always be pointing directly at the object, how fast should the camera be programmed to rotate at time tt? (Give your answer in terms of x(t)x(t) and y(t)y(t) and their derivatives, in the units rad/sec.)

Hint

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

Answer

x(t)y′(t)−y(t)x′(t)x2+y2\frac{x(t)y'(t)-y(t)x'(t)}{x^2+y^2}

Full solution

The question is already set up as an xyxy–plane, with the camera at the origin, so the vector in the direction the camera is pointing is (x(t),y(t))(x(t),y(t)). Let θ\theta be the angle the camera makes with the positive xx-axis (due east). The camera, the object, and the due-east direction (positive xx-axis) make a right triangle.

Figure from prob_s1.6, line 2

Figure from prob_s1.6, line 2

tan⁡θ=yx\begin{align*}\tan\theta&=\frac{y}{x}\end{align*}

Differentiating implicitly with respect to tt:

sec⁡2θ dθdt=xy′−yx′x2dθdt=cos⁡2θ(xy′−yx′x2)=(xx2+y2)2(xy′−yx′x2)=xy′−yx′x2+y2\begin{align*}\sec^2\theta\,\diff{\theta}{t}&=\frac{xy'-yx'}{x^2}\\ \diff{\theta}{t}&=\cos^2\theta\left(\frac{xy'-yx'}{x^2}\right)=\left(\frac{x}{\sqrt{x^2+y^2}}\right)^2\left(\frac{xy'-yx'}{x^2}\right)=\frac{xy'-yx'}{{x^2+y^2}}\end{align*}
Q27Stage 3

A projectile falling under the influence of gravity and slowed by air resistance proportional to its speed has position satisfying

d2rdt2=−gk^−αdrdt\frac{d^2\vr}{dt^2}=-g\hk-\alpha\frac{d\vr}{dt}

where α\alpha is a positive constant. If r=r0\vr=\vr_0 and drdt=v0\frac{d\vr}{dt}=\vv_0 at time t=0t=0, find r(t)\vr(t). (Hint: Define u(t)=eαtdrdt(t)\vu(t)=e^{\alpha t}\frac{d\vr}{dt}(t) and substitute drdt(t)=e−αtu(t)\frac{d\vr}{dt}(t)=e^{-\alpha t}\vu(t) into the given differential equation to find a differential equation for u\vu.)

Answer

r(t)=r0−e−αt−1αv0+g1−αt−e−αtα2k^\vr(t)=\vr_0-\frac{e^{-\alpha t}-1}{\alpha}\vv_0 +g\frac{1-\alpha t-e^{-\alpha t}}{\alpha^2}\hk

Full solution

Define u(t)=eαtdrdt(t)\vu(t)=e^{\alpha t}\frac{d\vr}{dt}(t). Then

dudt(t)=αeαtdrdt(t)+eαtd2rdt2(t)=αeαtdrdt(t)−geαtk^−αeαtdrdt(t)=−geαtk^\begin{align*} \frac{d\vu}{dt}(t)&=\alpha e^{\alpha t}\frac{d\vr}{dt}(t) +e^{\alpha t}\frac{d^2\vr}{dt^2}(t)\cr &=\alpha e^{\alpha t}\frac{d\vr}{dt}(t) -ge^{\alpha t}\hk -\alpha e^{\alpha t}\frac{d\vr}{dt}(t)\cr &=-ge^{\alpha t}\hk\cr \end{align*}

Integrating both sides of this equation from t=0t=0 to t=Tt=T gives

u(T)−u(0)=−geαT−1αk^  ⟹  u(T)=u(0)−geαT−1αk^=drdt(0)−geαT−1αk^=v0−geαT−1αk^\begin{align*} &\vu(T)-\vu(0)=-g\frac{e^{\alpha T}-1}{\alpha}\hk \\ &\implies \vu(T)=\vu(0)-g\frac{e^{\alpha T}-1}{\alpha}\hk =\frac{d\vr}{dt}(0)-g\frac{e^{\alpha T}-1}{\alpha}\hk =\vv_0-g\frac{e^{\alpha T}-1}{\alpha}\hk \end{align*}

Substituting in u(T)=eαtdrdt(T)\vu(T)=e^{\alpha t}\frac{d\vr}{dt}(T) and multiplying through by e−αTe^{-\alpha T}

drdt(T)=e−αTv0−g1−e−αTαk^\frac{d\vr}{dt}(T) =e^{-\alpha T}\vv_0-g\frac{1-e^{-\alpha T}}{\alpha}\hk

Integrating both sides of this equation from T=0T=0 to T=tT=t gives

r(t)−r(0)=e−αt−1−αv0−gtαk^+ge−αt−1−α2k^  ⟹  r(t)=r0−e−αt−1αv0+g1−αt−e−αtα2k^\begin{align*} &\vr(t)-\vr(0) =\frac{e^{-\alpha t}-1}{-\alpha}\vv_0-g\frac{ t}{\alpha}\hk +g\frac{e^{-\alpha t}-1}{-\alpha^2}\hk \\ &\implies \vr(t)=\vr_0-\frac{e^{-\alpha t}-1}{\alpha}\vv_0 +g\frac{1-\alpha t-e^{-\alpha t}}{\alpha^2}\hk \end{align*}
Q28Stage 3Past exam · M200 2003A

At time t=0t=0 a particle has position and velocity vectors r(0)=<−1,0,0>\vr(0)=\llt -1,0,0\rgt and v(0)=<0,−1,1>\vv(0)=\llt 0,-1,1\rgt. At time tt, the particle has acceleration vector

a(t)=<cos⁡t,sin⁡t,0>\begin{equation*} \va(t)=\llt \cos t,\sin t,0\rgt \end{equation*}

  1. Find the position of the particle after tt seconds.

  2. Show that the velocity and acceleration of the particle are always perpendicular for every tt.

  3. Find the equation of the tangent line to the particle's path at t=−π/2t=-\pi/2.

  4. True or False: None of the lines tangent to the path of the particle pass through (0,0,0)(0,0,0). Justify your answer.

Answer

(a) r(t)=<−cos⁡t,−sin⁡t,t>\vr(t)=\llt -\cos t, -\sin t, t\rgt (b) v(t)⋅a(t)=0\vv(t)\cdot\va(t)=0

(c) r(u)=<0,1,−π2>+u<−1,0,1>\vr(u)=\llt 0,1,-\frac{\pi}{2}\rgt+u\llt -1,0,1\rgt (d) True

Full solution

(a) By definition,

v′(t)=a(t)=<cos⁡t,sin⁡t,0>  ⟹  v(t)=<sin⁡t+c1,−cos⁡t+c2,c3>\begin{equation*} \vv'(t)=\va(t)=\llt \cos t,\sin t,0\rgt \implies \vv(t)=\llt \sin t+c_1, -\cos t+c_2, c_3\rgt \end{equation*}

for some constants c1c_1, c2c_2, c3c_3. To satisfy v(0)=<0,−1,1>\vv(0)=\llt 0,-1,1\rgt, we need c1=0c_1=0, c2=0c_2=0 and c3=1c_3=1. So v(t)=<sin⁡t,−cos⁡t,1>\vv(t)=\llt \sin t, -\cos t, 1\rgt. Similarly,

r′(t)=v(t)=<sin⁡t,−cos⁡t,1>  ⟹  r(t)=<−cos⁡t+d1,−sin⁡t+d2,t+d3>\begin{equation*} \vr'(t)=\vv(t)=\llt \sin t,-\cos t,1\rgt \implies \vr(t)=\llt -\cos t+d_1, -\sin t+d_2, t+d_3\rgt \end{equation*}

for some constants d1d_1, d2d_2, d3d_3. To satisfy r(0)=<−1,0,0>\vr(0)=\llt -1,0,0\rgt, we need d1=0d_1=0, d2=0d_2=0 and d3=0d_3=0. So r(t)=<−cos⁡t,−sin⁡t,t>\vr(t)=\llt -\cos t, -\sin t, t\rgt.

(b) To test for orthogonality, we compute the dot product

v(t)⋅a(t)=<sin⁡t,−cos⁡t,1>⋅<cos⁡t,sin⁡t,0>=sin⁡tcos⁡t−cos⁡tsin⁡t+1×0=0\begin{equation*} \vv(t)\cdot\va(t)=\llt \sin t, -\cos t, 1\rgt\cdot\llt \cos t,\sin t,0\rgt =\sin t\cos t-\cos t \sin t+1\times 0=0 \end{equation*}

so v(t)⊥a(t)\vv(t)\perp\va(t) for all tt.

(c) At t=−π2t=-\frac{\pi}{2} the particle is at r(−π2)=<0,1,−π2>\vr\big(-\frac{\pi}{2}\big)=\llt 0,1,-\frac{\pi}{2}\rgt and has velocity v(−π2)=<−1,0,1>\vv\big(-\frac{\pi}{2}\big)=\llt -1,0,1\rgt. So the tangent line must pass through <0,1,−π2>\llt 0,1,-\frac{\pi}{2}\rgt and have direction vector <−1,0,1>\llt -1,0,1\rgt. Here is a vector parametric equation for the tangent line.

r(u)=<0,1,−π2>+u<−1,0,1>\begin{equation*} \vr(u)=\llt 0,1,-\frac{\pi}{2}\rgt+u\llt -1,0,1\rgt \end{equation*}

(d) True. Look at the path followed by the particle from the top so that we only see xx and yy coordinates. The path we see (call this the projected path) is x(t)=−cos⁡tx(t)=-\cos t, y(t)=−sin⁡ty(t)=-\sin t, which is a circle of radius one centred on the origin. Any tangent line to any circle always remains outside the circle. So no tangent line to the projected path can pass through the (0,0)(0,0). So no tangent line to the path followed by the particle can pass through the zz–axis and, in particular, through (0,0,0)(0,0,0).

Q29Stage 3Past exam · M200 2002D

The position of a particle at time tt (measured in seconds s) is given by

r(t)=tcos⁡(πt2)ı^+tsin⁡(πt2)ȷ^+t k^\begin{equation*} \vr(t)=t\cos\left(\frac{\pi t}{2}\right)\hi +t\sin\left(\frac{\pi t}{2}\right)\hj +t\,\hk \end{equation*}

  1. Show that the path of the particle lies on the cone z2=x2+y2z^2=x^2+y^2.

  2. Find the velocity vector and the speed at time tt.

  3. Suppose that at time t=1t=1s the particle flies off the path on a line LL in the direction tangent to the path. Find the equation of the line LL.

  4. How long does it take for the particle to hit the plane x=−1x=-1 after it started moving along the straight line LL?

Answer

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

(b) velocity=[cos⁡(πt2)−πt2sin⁡(πt2)]ı^+[sin⁡(πt2)+πt2cos⁡(πt2)]ȷ^+k^\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 speed=2+π2t24\text{speed}=\sqrt{2+\frac{\pi^2 t^2}{4}}

(c) <x,y,z>=<0,1,1>+(t−1)<−π2,1,1>\llt x,y,z\rgt = \llt 0,1,1\rgt +(t-1)\llt -\frac{\pi}{2},1,1\rgt

(d) 2π\frac{2}{\pi} seconds

Full solution

(a) Since

x(t)2+y(t)2=t2cos⁡2(πt2)+t2sin⁡2(πt2)=t2andz(t)2=t2\begin{equation*} x(t)^2+y(t)^2=t^2\cos^2\big(\tfrac{\pi t}{2}\big) +t^2\sin^2\big(\tfrac{\pi t}{2}\big)=t^2\qquad\text{and}\qquad z(t)^2=t^2 \end{equation*}

are the same, the path of the particle lies on the cone z2=x2+y2z^2=x^2+y^2.

(b) By definition,

velocity=r′(t)=[cos⁡(πt2)−πt2sin⁡(πt2)]ı^+[sin⁡(πt2)+πt2cos⁡(πt2)]ȷ^+k^speed=∣r′(t)∣=[cos⁡(πt2)−πt2sin⁡(πt2)]2+[sin⁡(πt2)+πt2cos⁡(πt2)]2+12=[cos⁡2(πt2)−2πt2cos⁡(πt2)sin⁡(πt2)+(πt2)2sin⁡2(πt2)+sin⁡2(πt2)+2πt2cos⁡(πt2)sin⁡(πt2)+(πt2)2cos⁡2(πt2)+1]1/2=2+π2t24\begin{alignat*}{3} \text{velocity}&=\vr'(t)&&= \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}&=|\vr'(t)|&&= \sqrt{\big[\cos\big(\tfrac{\pi t}{2}\big) -\tfrac{\pi t}{2}\sin\big(\tfrac{\pi t}{2}\big)\big]^2 +\big[\sin\big(\tfrac{\pi t}{2}\big) +\tfrac{\pi t}{2}\cos\big(\tfrac{\pi t}{2}\big)\big]^2 +1^2} \\ & &&= \Big[\cos^2\big(\tfrac{\pi t}{2}\big) -2\tfrac{\pi t}{2}\cos\big(\tfrac{\pi t}{2}\big) \sin\big(\tfrac{\pi t}{2}\big) +\big(\tfrac{\pi t}{2}\big)^2\sin^2\big(\tfrac{\pi t}{2}\big) \\ & &&\hskip.2in +\sin^2\big(\tfrac{\pi t}{2}\big) +2\tfrac{\pi t}{2}\cos\big(\tfrac{\pi t}{2}\big) \sin\big(\tfrac{\pi t}{2}\big) +\big(\tfrac{\pi t}{2}\big)^2\cos^2\big(\tfrac{\pi t}{2}\big) +1\Big]^{1/2}\\ & &&=\sqrt{2+\frac{\pi^2 t^2}{4}} \end{alignat*}

(c) At t=1t=1, the particle is at r(1)=(0,1,1)\vr(1)=(0,1,1) and has velocity r′(1)=<−π2,1,1>\vr'(1)=\llt -\frac{\pi}{2},1,1\rgt. So for t≥1t\ge 1, the particle is at

<x,y,z>=<0,1,1>+(t−1)<−π2,1,1>\begin{equation*} \llt x,y,z\rgt = \llt 0,1,1\rgt+(t-1)\llt -\frac{\pi}{2},1,1\rgt \end{equation*}

This is also a vector parametric equation for the line.

(d) Assume that the particle's speed remains constant as it flies along LL. Then the xx-coordinate of the particle at time tt (for t≥1t\ge 1) is −π2(t−1)-\frac{\pi}{2}(t-1). This takes the value −1-1 when t−1=2πt-1=\frac{2}{\pi}. So the particle hits x=−1x=-1, 2π\frac{2}{\pi} seconds after it flew off the cone.

Q30Stage 3Past exam · M200 2001D

  1. The curve r1(t)=<1+t,t2,t3>\vr_1(t)=\llt 1+t, t^2, t^3\rgt and r2(t)=<cos⁡t,sin⁡t,t>\vr_2(t)=\llt \cos t, \sin t, t\rgt intersect at the point P(1,0,0)P(1,0,0). Find the angle of intersection between the curves at the point PP.

  2. Find the distance between the line of intersection of the planes x+y−z=4x+y-z=4 and 2x−z=42x-z=4 and the line r(t)=<t,−1+2t,1+3t>\vr(t)=\llt t, -1+2t, 1+3t\rgt.

Answer

(a) 90∘90^\circ (b) 232\sqrt{3}

Full solution

(a) The tangent vectors to the two curves are

r1′(t)=<1,2t,3t2>r2′(t)=<−sin⁡t,cos⁡t,1>\begin{equation*} \vr'_1(t)=\llt 1, 2t, 3t^2\rgt\qquad \vr'_2(t)=\llt-\sin t, \cos t, 1\rgt \end{equation*}

Both curves pass through PP at t=0t=0 and then the tangent vectors are

r1′(0)=<1,0,0>r2′(0)=<0,1,1>\begin{equation*} \vr'_1(0)=\llt 1, 0, 0\rgt\qquad \vr'_2(0)=\llt 0, 1, 1\rgt \end{equation*}

So the angle of intersection, θ\theta, is determined by

r1′(0)⋅r2′(0)=∣r1′(0)∣ ∣r2′(0)∣ cos⁡θ  ⟹  <1,0,0>⋅<0,1,1>=1⋅2⋅cos⁡θ  ⟹  cos⁡θ=0  ⟹  θ=90∘\begin{align*} \vr'_1(0)\cdot\vr'_2(0)=|\vr'_1(0)|\,|\vr'_2(0)|\,\cos\theta &\implies \llt 1, 0, 0\rgt\cdot\llt 0, 1, 1\rgt=1\cdot\sqrt{2}\cdot\cos\theta \\ &\implies\cos\theta=0\implies \theta=90^\circ \end{align*}

(b) Our strategy will be to

  • find a vector v\vv whose tail is on one line and whose head is on the other line and then

  • find a vector n\vn that is perpendicular to both lines.

  • Then, if we denote by θ\theta the angle between v\vv and n\vn, the distance between the two lines is ∣v∣cos⁡θ=∣v⋅n∣∣n∣|\vv|\cos\theta = \frac{|\vv\cdot\vn|}{|\vn|}

Here we go

  • So the first step is to find a v\vv.

    • One point on the line r(t)=<t,−1+2t,1+3t>\vr(t)=\llt t, -1+2t, 1+3t\rgt is r(0)=<0,−1,1>\vr(0)=\llt 0,-1,1\rgt.

    • (x,y,z)(x,y,z) is on the other line if and only if x+y−z=4x+y-z=4 and 2x−z=42x-z=4. In particular, if z=0z=0 then x+y=4x+y=4 and 2x=42x=4 so that x=2x=2 and y=2y=2.

    • So the vector v=<2−0 , 2−(−1) , 0−1>=<2,3,−1>\vv=\llt 2-0\,,\,2-(-1)\,,\,0-1\rgt=\llt 2,3,-1\rgt has its head on one line and its tail on the other line.

  • Next we find a vector n\vn that is perpendicular to both lines.

    • First we find a direction vector for the line x+y−z=4x+y-z=4, 2x−z=42x-z=4. We already know that x=y=2x=y=2, z=0z=0 is on that line. We can find a second point on that line by choosing, for example, z=2z=2 and then solving x+y=6x+y=6, 2x=62x=6 to get x=3x=3, y=3y=3. So one direction vector for the line x+y−z=4x+y-z=4, 2x−z=42x-z=4 is d1=<3−2 , 3−2 , 2−0>=<1,1,2>\vd_1=\llt3-2\,,\,3-2\,,\,2-0 \rgt = \llt 1,1,2\rgt.

    • A second way to get a direction vector for the line x+y−z=4x+y-z=4, 2x−z=42x-z=4 is to observe that <1,1,−1>\llt 1,1,-1\rgt is normal to x+y−z=4x+y-z=4 and so is perpendicular to the line and <2,0,−1>\llt 2,0,-1\rgt is normal to 2x−z=42x-z=4 and so is also perpendicular to the line. So <1,1,−1>×<2,0,−1>\llt 1,1,-1\rgt\times\llt 2,0,-1\rgt is a direction vector for the line.

    • A direction vector for the line r(t)=<t,−1+2t,1+3t>\vr(t)=\llt t, -1+2t, 1+3t\rgt is d2=r′(t)=<1,2,3>\vd_2=\vr'(t)=\llt 1,2,3\rgt.

    • So

      n=d2×d1=det⁡[ı^ȷ^k^123112]=ı^+ȷ^−k^\begin{equation*} \vn=\vd_2\times\vd_1 =\det\left[\begin{matrix} \hi & \hj & \hk \\ 1 & 2 & 3 \\ 1 & 1 & 2 \end{matrix}\right] = \hi + \hj -\hk \end{equation*}

      is perpendicular to both lines.

The distance between the two lines is then

∣v∣cos⁡θ=∣v⋅n∣∣n∣=<2,3,−1>⋅<1,1,−1>∣<1,1,−1>∣=63=23\begin{equation*} |\vv|\cos\theta = \frac{|\vv\cdot\vn|}{|\vn|} =\frac{\llt 2,3,-1\rgt \cdot\llt 1,1,-1\rgt}{|\llt 1,1,-1\rgt|} =\frac{6}{\sqrt{3}} =2\sqrt{3} \end{equation*}

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From the UBC Math 200 open textbook project, © Joel Feldman, Andrew Rechnitzer, Elyse Yeager and others. Licensed CC BY-NC-SA 4.0; this HTML adaptation is released under the same licence. Full attribution.