ohsx xm521 multivariable differential calculus: homework ... · homework solutions x14.3 (17)...

OHSx XM521 Multivariable Differential Calculus:Homework Solutions §14.1

(18) Describe the graph of the equation.

r = −2i + tj + (t2 − 1)k.

Solution: Let y(t) = t, so that z(t) = t2 − 1 = y2 − 1. In the yz-plane, this is just aparabola. Since the x-component of r is the constant −2, the parabola is shifted from theyz-plane by 2 in the negative x-direction.

(40) Show that the graph of

r = ti +1 + t

tj +

1− t2

tk, t > 0

lies in the plane x− y + z + 1 = 0.

Solution: r is represented by the parametric equations x(t) = t, y(t) = (1 + t)/t, z(t) =(1− t2)/t. Letting x, y, and z denote the components of r, we have:

x− y + z + 1 = t− 1 + t

t+

1− t2

t+ 1

= t+−1− t+ 1− t2

t+ 1

= t− 1− t+ 1

= 0.

Therefore the components of r satisfy the equation of the given plane, for t > 0. So thegraph of r must lie in the plane x− y + z + 1 = 0.

1

(42) Show that the graph of

r = 3 cos ti + 3 sin tj + 3 sin tk

is an ellipse, and find the lengths of the major and minor axes. [Hint: Show that the graphlies on both a circular cylinder and a plane and use the result in Exercise 60 of Sections12.4.]

Solution: The result of Exercise 60 in Section 12.4 states that if a plane is not parallelto the axis of a right circular cylinder, then the intersection of the plane and cylinderis an ellipse. Observe that if x(t), y(t), and z(t) denote the components of r(t), then(x(t))2 + (y(t))2 = 9, and y(t) = z(t). Therefore, r lies on the cylinder x2 + y2 = 9 and onthe plane y = z. Consequently, by Exercise 60, Section 12.4, the graph of r is an ellipse.

To find the lengths of the major and minor axes, consider r = 3 cos ti+ 3 sin tj+ 3 sin tk =3 cos ti + 3

√2 sin t( j+k√

2). Let u = j+k√

2so that r = 3 cos ti + 3

√2 sin tu. From the form

of this equation, it is clear this graph is an ellipse in the xu-plane where the u-axis isdefined by the direction of u. The minor axis is 2b = 2(3) = 6 and the major axis is2a = 2(3

√2) = 6

√2.

(43) For the helix r = a cos ti + a sin tj + ctk, find c (c > 0) so that the helix will make onecomplete turn in a distance of 3 units measured along the z-axis.

Solution: The above helix will make one complete turn when t = 2π since this is theperiod for the functions cos t and sin t. Therefore, when t = 2π we want z = 3, orequivalently c · 2π = 3. So c = 3/2π.

(45) Show that the curve r = t cos ti + t sin tj + tk, t ≥ 0, lies on the cone z =√x2 + y2.

Describe the curve.

Solution: Letting x, y, and z denote the components of r, we have:√x2 + y2 =

√(t cos t)2 + (t sin t)2

= t√

cos2 t+ sin2 t

= t

2

= z.

Since the components satisfy the cone equation, r lies on that cone. Qualitatively, thecurve is helical in nature with radius always increasing due to the multiplicative t factor inthe first two components of r(t). The curve starts at the origin and spirals upward in thepositive z-direction around the z-axis but with increasing distance away from this axis asit moves farther in the z-direction. The curve will wrap once around the z-axis every timet increases by 2π.

(47) In each part, match the vector equation with one of the accompanying graphs, and explainyour reasoning.

(a) r = ti− tj +√

2− t2k.

Solution: Let x(t) = t, y(t) = −t, and z(t) =√

2− t2. The graph must have x = −y,and at t = 0 it has the value (0, 0, 2). The only graph that has these characteristics is III.

(b) r = sinπti− tj + tk.

Solution: Let x(t) = sinπt, y(t) = −t, and z(t) = t. The graph must have y = −z, andat t = 0 it has the value (0, 0, 0). The only graph that has these characteristics is IV.

(c) r = sin ti + cos tj + sin 2tk.

Solution: Let x(t) = sin t, y(t) = cos t, and z(t) = sin 2t. Observe that (x(t))2+(y(t))2 =1, so r lies on the cylinder x2 + y2 = 1. Graph II is the only one with this property.

(d) r = 12 ti + cos 3tj + sin 3tk.

Solution: The form of the components of r are of a helix with rotation in planes parallelto the yz-plane and distance from the yz-plane (given by |x|) increasing as |t| increases.This corresponds to graph I.

3


(18) Find limt→1

⟨ 3

t2,

ln t

t2 − 1, sin 2t

⟩.

Solution: To find the limit of a vector function, take the limit of each individual compo-nent:

limt→1

⟨ 3

t2,

ln t

t2 − 1, sin 2t

⟩=⟨

limt→1

3

t2, limt→1

ln t

t2 − 1, limt→1

sin 2t⟩

=⟨

3,1

2, sin 2

⟩.

The y-component was evaluated using L’Hopital’s rule:

limt→1

ln t

t2 − 1= lim

t→1

1/t

2t=

1

2.

(20) Determine whether r(t) is continuous at t = 0. Explain your reasoning.

(a) r(t) = eti + j + csc tk.

Solution: r(t) is not continuous at t = 0 because csc t is undefined for t = 0.

(b) r(t) = 5i−√

3t+ 1j + e2tk.

Solution: r(t) is continuous at t = 0 because

limt→0

r(t) = limt→0

5i− limt→0

√3t+ 1j + lim

t→0e2tk = 5i− j + k = r(0).

We’ve used the facts that√

3t+ 1 and e2t are continuous at t = 0.

(46) Solve the vector initial-value problem for y(t) by integrating and using the initial conditionto find the constants of integration.

y′′(t) = 12t2i− 2tj, y(0) = 2i− 4j, y′(0) = 0.

Solution: Integrate y′′(t) to obtain y′(t):

y′(t) =

∫y′′(t)dt =

∫(12t2i− 2tj)dt = 4t3i− t2j + C1.

To satisfy the initial condition y′(0) = 0, we must have

0 = y′(0) = 4 · 03i− 02j + C1.

Therefore C1 = 0, and y′(t) = 4t3i− t2j. Now integrate y′(t) to obtain y(t):

y(t) =

∫y′(t)dt =

∫(4t3i− t2j)dt = t4i− t3

3j + C2.

To satisfy the initial condition y(0) = 2i − 4j, we must have C2 = 2i − 4j, implying that

y(t) = (t4 + 2)i− ( t3

3 + 4)j.

1

(52) Find where the tangent line to the curve

r(t) = e−2ti + cos tj + 3 sin tk

at the point (1, 1, 0) intersects the yz-plane.

Solution: The point (1, 1, 0) corresponds to t = 0 for r(t). To find the tangent line tor(t) at this point, first calculate r′(t) = −2e−2ti− sin tj+3 cos tk, so that r′(0) = −2i+3k.The tangent line can therefore be represented by the vector equation L(t) = L0 +vt whereL0 = 〈1, 1, 0〉 is a vector from the origin to a point on the line and v = r′(0) = 〈−2, 0, 3〉gives the direction of the tangent line. This line can be represented parametrically asx = 1 − 2t, y = 1, z = 3t. At the intersection of this line and the yz-plane, x = 0, whichforces t = 1/2. At this t value z = 3/2. Therefore, the intersection point is (0, 1, 3/2).

(54) Show that the graphs of r1(t) and r2(t) intersect at the point P . Find, to the nearestdegree, the acute angle between the tangent lines to the graphs of r1(t) and r2(t) at thepoint P (2, 1, 3).

r1(t) = 2e−ti + cos tj + (t2 + 3)k,r2(t) = (1− t)i + t2j + (t3 + 4)k.

Solution: r1 passes through P when t = 0, and r2 passes through P when t = −1. Thevectors in the directions of the respective tangent lines can be calculated by differentiatingr1(t) and r2(t) and evaluating at t = 0 and t = −1 respectively:

r′1(t) = −2e−ti− sin tj + 2tk, r′1(0) = −2i.

r′2(t) = −i + 2tj + 3t2k, r′2(−1) = −i− 2j + 3k.

Using a•b = ||a||||b|| cos θ, the angle between the above vectors is given by cos θ =[(−2)(−1)]/[(2)(

√14)] = 1/

√14, so that θ = arccos(1/

√14) ≈ 74.5o. Since this angle

is acute, it must be the acute angle between the corresponding tangent lines.

(57) Use Formula (9) to derive the differentiation formula

d

dt[r(t)× r′(t)] = r(t)× r′′(t).

Solution: Apply Formula (9) to ddt [r(t)× r′(t)] to get:

d

dt[r(t)× r′(t)] = r′(t)× r′(t) + r(t)× r′′(t).

Since a× a = 0 for all vectors a, the first term is 0, and we obtain the desired result.

(58) Let u = u(t), v = v(t), w = w(t) be differentiable vector-valued functions. Use Formulas(8) and (9) to show that

d

dt[u•(v ×w)] =

du

dt•[v ×w] + u•[dv

dt×w] + u•[v × dw

dt].

2

Solution: Let x = v ×w so that ddt [u•(v ×w)] = d

dt [u•x]. Apply Formula (8) to get:

d

dt[u•x] =

du

dt•x + u•dx

dt.

Now substitute v ×w for x and use Formula (9) to compute d[v×w]dt :

du

dt•[v ×w] + u•[dv

dt×w + v × dw

dt].

Finally, apply the distributive property of the dot product on the second term of thisexpression to get the desired formula.

(59) Let u1, u2, u3, v1, v2, v3, w1, w2, and w3 be differentiable functions of t. Use Exercise 58to show that

d

dt

∣∣∣∣∣∣u1 u2 u3v1 v2 v3w1 w2 w3

∣∣∣∣∣∣ =

∣∣∣∣∣∣u′1 u′2 u′3v1 v2 v3w1 w2 w3

∣∣∣∣∣∣+

∣∣∣∣∣∣u1 u2 u3v′1 v′2 v′3w1 w2 w3

∣∣∣∣∣∣+

∣∣∣∣∣∣u1 u2 u3v1 v2 v3w′1 w′2 w′3

∣∣∣∣∣∣Solution: For any vectors a = 〈a1, a2, a3〉, b = 〈b1, b2, b3〉, and c = 〈c1, c2, c3〉,

a • b× c =

∣∣∣∣∣∣a1 a2 a3b1 b2 b3c1 c2 c3

∣∣∣∣∣∣ .So the left-hand side of the desired equation can be written as d

dt (u • v × w), while the

right-hand side is dudt • v × w + u • dv

dt × w + u • v × dwdt . Therefore, the desired result

follows immediately from Exercise 58.

3


(17) Calculate dr/dτ by the chain rule, and then check your result by expressing r in terms ofτ and differentiating.

r = eti + 4e−tj; t = τ2.

Solution: dr/dt = eti− 4e−tj = eτ2

i− 4e−τ2

j and dt/dτ = 2τ . Using the chain rule:

dr

dτ=dr

dt

dt

dτ= (eτ

2

i− 4e−τ2

j) · 2τ = 2τeτ2

i− 8τe−t2

j.

To check this, writing r as a function of τ gives r = eti + 4e−tj = eτ2

i + 4e−τ2

j. Sodifferentiating and applying the chain rule to each component gives

dr

dτ= eτ

2

(2τ)i− 4e−τ2

(2τ)j = 2τeτ2

i− 8τe−τ2

j,

which agrees with the earlier computation of dr/dτ .

(22) (a) Find the arc length parametrization of the line

x = −5 + 3t, y = 2t, z = 5 + t

that has the same direction as the given line and has reference point (−5, 0, 5).

Solution: Since dx/dt = 3, dy/dt = 2, dz/dt = 1, and the point (−5, 0, 5) correspondsto t = 0, the arc length parameter for the line is:

s =

∫ t

0

√(dx/dt)2 + (dy/dt)2 + (dz/dt)2 dt =

∫ t

0

√32 + 22 + 12 dt =

√14t.

So t = s/√

14, and the arc length parametrization of the line is x = −5 + (3s/√

14), y =2s/√

14, z = 5 + s/√

14.

(b) Use the parametric equations obtained in part (a) to find the point on the line that is10 units from the reference point in the direction of increasing parameter.

Solution: Substituting s = 10 into the result of (a) gives

(−5 +30√14,

20√14, 5 +

10√14

).

(28) Find an arc length parametrization of the curve that has the same orientation as the givencurve and has t = 0 as the reference point.

r(t) = sin eti + cos etj +√

3etk; t ≥ 0.

1

Solution: dr(t)/dt = et cos eti− et sin etj +√

3etk, so that

||dr(t)/dt|| =√e2t cos2 et + e2t sin2 et + 3e2t =

√e2t(1 + 3) = 2et.

The arc length parameter s is given by:

s =

∫ t

0

||dr/du||du =

∫ t

0

2eudu = 2(et − 1),

implying that t = ln((s/2) + 1) = ln((s+ 2)/2). Using this to find an expression for et asa function of s:

et = eln((s+2)/2) =s+ 2

2.

So the desired arc length parametrization is:

r(s) = sin(s+ 2

2

)i + cos

(s+ 2

2

)j +

√3(s+ 2)

2k.

(32) Show that in cylindrical coordinates a curve given by the parametric equations r = r(t),θ = θ(t), and z = z(t) for a ≤ t ≤ b has arc length

L =

∫ b

a

√(dr/dt)2 + r2(dθ/dt)2 + (dz/dt)2 dt

[Hint: Use the relationships x = r cos θ and y = r sin θ.]

Solution: Let x(t), y(t), z(t), a ≤ t ≤ b, be a parametrization of the curve in rectangularcoordinates. We know that for all t, x(t) = r(t) cos θ(t) and y(t) = r(t) sin θ(t). Differenti-ating with respect to t, and using dot notation (popular with physicists and engineers) forthe derivatives with respect to t (e.g. dx/dt = x) gives

x = r cos θ − rθ sin θ, y = r sin θ + rθ cos θ.

This implies:

x2 = r2 cos2 θ − 2rrθ cos θ sin θ + r2θ2 sin2 θ,

y2 = r2 sin2 θ + 2rrθ cos θ sin θ + r2θ2 cos2 θ.

Adding these two expressions, grouping like terms, and using sin2 θ + cos2 θ = 1 givesx2 + y2 = r2 + r2θ2. Using this result in the formula for arc length L in 3-space, we obtain

L =

∫ b

a

√x2 + y2 + z2 dt =

∫ b

a

√r2 + r2θ2 + z2 dt,

which is the desired result in dot notation.

(38) What change of parameter g(τ) would you make if you wanted to trace the graph of r(t)(0 ≤ t ≤ 1) in the opposite direction with τ varying from 0 to 1.

Solution: The substitution t = g(τ) = 1 − τ will suffice. Simply note that as τ variesfrom 0 to 1, t varies from 1 to 0. In fact g can be any (smooth) strictly decreasing functionwith domain 0 ≤ τ ≤ 1 and range 0 ≤ t ≤ 1.

2

(40) Let x = cos t, y = sin t, z = t3/2. Find

(a) ||r′(t)||.

Solution: r′(t) = x′(t)i + y′(t)j + z′(t)k = − sin ti + cos tj + 32 t

1/2k. So:

||r′(t)|| =√

sin2 t+ cos2 t+9

4t =

1

2

√4 + 9t.

(b) ds/dt.

Solution: If s is an arc length parameter, then

s =

∫ t

t0

||r′(u)|| du.

Differentiating with respect to t gives ds/dt = ||r′(t)||. So from part (a), ds/dt = 12

√4 + 9t.

(c)∫ 2

0||r′(t)||dt.

Solution: Substituting the result from (a) gives∫ 2

0

||r′(t)||dt =

∫ 2

0

1

2

√4 + 9t dt =

1

27(4 + 9t)3/2

∣∣∣∣20

=1

27((22)3/2 − 8).

(42) Prove: If r(t) is a smoothly parameterized function, then the angles between r′(t) and thevectors i, j, and k are continuous functions of t.

Solution: Let r(t) = x(t)i + y(t)j + z(t)k. Since r(t) is a smoothly parameterizedfunction, r′ is continuous and r′ 6= 0. The continuity of r′ implies that x′, y′, and z′ mustbe continuous. The angle α between i and r′(t) can be found by using the dot product:

i•r′(t) = ||i|| ||r′(t)|| cos(α)

or

α = cos−1i•r′(t)

||i|| ||r′(t)||= cos−1

x′(t)

||r′(t)||.

(Note: In general, the equation cos(θ) = c does not necessarily imply that θ = cos−1(c)because θ may not lie between 0 and π. However, the angle between two vectors is alwaysbetween 0 and π, so in this case cos(θ) = c does imply θ = cos−1(c).) Since x′ and ||r′||are continuous and ||r′|| 6= 0, x′/||r′|| is a continuous function of t. Finally, since cos−1 iscontinuous, α is a continuous function of t. Similar arguments will hold for β and γ, theangles that r′(t) makes with j and k respectively.

3


(8) Find T(t) and N(t) at the given point.

r(t) = ti +1

2t2j +

1

3t3k; t = 0.

Solution: r′(t) = i + tj + t2k, and

T(t) =r′(t)

||r′(t)||=

1√1 + t2 + t4

(i + tj + t2k).

So T(0) = i.

To find N(t), first calculate T′(t):

T′(t) = −1

2(2t+ 4t3)(1 + t2 + t4)−

32 (i + tj + t2k) +

1√1 + t2 + t4

(j + 2tk).

So T′(0) = j (which makes sense since it must be orthogonal to T(0)), and N(0) =T′(0)/||T′(0)|| = j.

(17) Use the formula B(t) = T(t) ×N(t) to find B(t), and then check your answer by usingFormula (11) to find B(t) directly from r(t).

r(t) = (sin t− t cos t)i + (cos t+ t sin t)j + k.

Solution: r′(t) = t sin ti + t cos tj, ||r′(t)|| = t, and

T(t) =r′(t)

||r′(t)||= sin ti + cos tj.

So T′(t) = cos ti− sin tj, ||T′(t)|| = 1, and therefore N(t) = T′(t)/||T′(t)|| = cos ti− sin tj.This implies that

B(t) = T(t)×N(t) = (sin ti + cos tj)× (cos ti− sin tj) = − sin2 tk− cos2 tk = −k.

Formula (11) states that B(t) = (r′(t) × r′′(t))/||r′(t) × r′′(t)||, so first calculate r′′(t) =(sin t+ t cos t)i + (cos t− t sin t)j. Then

r′(t)× r′′(t) = (t sin ti + t cos tj)× ((sin t+ t cos t)i + (cos t− t sin t)j)

= −t2(sin2 t+ cos2 t)k

= −t2k,

and ||r′(t)× r′′(t)|| = t2. So B(t) = (r′(t)× r′′(t))/||r′(t)× r′′(t)|| = −k, which is the sameas the earlier result.

1

(20) Find T(t), N(t), and B(t) for the given value of t. Then find equations for the osculating,normal, and rectifying planes at the point that corresponds to that value of t.

r(t) = eti + et cos tj + et sin tk; t = 0.

Solution: The following calculations are easily checked:

r′(t) = eti + (et cos t− et sin t)j + (et sin t+ et cos t)k;

||r′(t)|| = et√

1 + (cos t− sin t)2 + (sin t+ cos t)2) =√

3et;

T(t) =r′(t)

||r′(t)||=

1√3

(i + (cos t− sin t)j + (sin t+ cos t)k);

T(0) =1√3

(i + j + k);

T′(t) =1√3

(−(sin t+ cos t)j + (cos t− sin t)k);

||T′(t)|| =1√3

√(sin t+ cos t)2 + (cos t− sin t)2 =

√2

3;

N(t) =T′(t)

||T′(t)||=

1√2

(−(sin t+ cos t)j + (cos t− sin t)k);

N(0) =1√2

(−j + k);

B(t) = T(t)×N(t) =1√6

(2i− (cos t− sin t)j− (sin t+ cos t)k);

B(0) =1√6

(2i− j− k).

With these calculations and r(0) = r0 = i + j, the various planes can be calculated usingthe point-normal form plane equation n•(r− r0) = 0.

The osculating plane has B(0) as a normal vector, so an equation for that plane is

1√6

(2i− j− k)•((x− 1)i + (y − 1)j + zk) = 0,

which simplifies to 2x− y − z − 1 = 0.

The normal plane has T(0) as a normal vector, so an equation for that plane is

1√3

(i + j + k)•((x− 1)i + (y − 1)j + zk) = 0,

which simplifies to x+ y + z − 2 = 0.

The rectifying plane has N(0) as a normal vector, so an equation for that plane is

1√2

(−j + k)•((x− 1)i + (y − 1)j + zk) = 0,

which simplifies to −y + z + 1 = 0.

(21) (a) Use the formula N(t) = B(t)×T(t) and Formulas (1) and (11) to show that N(t) canbe expressed in terms of r(t) as

N(t) =r′(t)× r′′(t)

||r′(t)× r′′(t)||× r′(t)

||r′(t)||.

2

Solution: To derive this result, just substitute T(t) = r′(t)/||r′(t)|| (which is Formula (1))and B(t) = r′(t)× r′′(t)/||r′(t)× r′′(t)|| (which is Formula (11)) into N(t) = B(t)×T(t).

(b) Use properties of cross products to show that the formula in part (a) can be expressedas

N(t) =r′(t)× r′′(t)× r′(t)

||r′(t)× r′′(t)× r′(t)||.

Solution: From part (a),

N(t) =r′(t)× r′′(t)

||r′(t)× r′′(t)||× r′(t)

||r′(t)||

=(r′(t)× r′′(t))× r′(t)

||r′(t)× r′′(t)|| ||r′(t)||.

The numerator of this expression is of the form (a× b)× a. But note that

(a× b)× a = −a× (a× b) = −a× (−(b× a)) = a× (b× a).

Therefore, the numerator in the expression for N(t) can be written as r′(t)× r′′(t)× r′(t)without any confusion regarding the order in which the cross products are performed. Inthe denominator, r′(t) must be orthogonal to r′(t)× r′′(t), so the angle between these twovectors is π/2, implying that

||r′(t)× r′′(t)× r′(t)|| = ||r′(t)× r′′(t)|| ||r′(t)|| sin(π/2) = ||r′(t)× r′′(t)|| ||r′(t)||.

Using this relation in the denominator of our expresssion for N(t) gives the desired result.

(c) Use the result in part (b) and Exercise 39 of Section 13.4 to show that N(t) can beexpressed directly in terms of r(t) as

N(t) =u(t)

||u(t)||

whereu(t) = ||r′(t)||2r′′(t)− (r′(t)•r′′(t))r′(t).

Solution: Starting with the result of (b), let u(t) = r′(t)× r′′(t)× r′(t), so that N(t) =u(t)||u(t)|| , and apply the identity (a× b)× c = (c•a)b− (c•b)a to obtain:

u(t) = (r′(t)•r′(t))r′′(t)− (r′(t)•r′′(t))r′(t)

= ||r′(t)||2r′′(t)− (r′(t)•r′′(t))r′(t),

as desired.

3


(17) (a) Use Formula (3) to show that in 2-space the curvature of a smooth parametric curvex(t), y(t) is

κ(t) =|x′y′′ − y′x′′|(x′2 + y′2)3/2

,

where primes denote differentiation with respect to t.

Solution: Formula (3) states:

κ(t) =||r′(t)× r′′(t)||||r′(t)||3

.

Let r = xi + yj, r′ = x′i + y′j and r′′ = x′′i + y′′j, (with the functional dependence on tomitted for notational convenience) . This implies r′ × r′′ = (x′i + y′j) × (x′′i + y′′j) =(x′y′′ − y′x′′)k, so that ||r′ × r′′|| = |x′y′′ − y′x′′|, which gives the desired form for the

numerator. Since ||r′|| =√x′2 + y′2, substituting this in the denominator yields the desired

result.

(b) Use the result in part (a) to show that in 2-space the curvature of the plane curve givenby y = f(x) is

κ(x) =|d2y/dx2|

[1 + (dy/dx)2]3/2.

[Hint: Express y = f(x) parametrically with x = t as the parameter.]

Solution: Expressing y = f(x) parametrically with x = t as the parameter gives y = f(t),y′ = dy/dt = (dy/dx)dx/dt = dy/dx (since dx/dt = 1), y′′ = d2y/dt2 = (d2y/dx2)dx/dt =d2y/dx2, x′ = 1, and x′′ = 0. Substituting these expressions into the result from (a) yieldsthe desired formula.

(34) The graphs of f(x) and the associated curvature function κ(x) are shown. Determine whichis which, and explain your reasoning.

Solution (a): The curve II flattens out as x goes to either ∞ or −∞, so curvaturemust go to 0 as x goes to either ∞ or −∞. Also curve II appears to have its maximumcurvature at x = 0, corresponding to where curve I reaches a maximum. Hence, curve IIis the function and I is the curvature function.

Solution (b): A circle has a constant curvature, so II is the function and I is the curvaturefunction.

(47) Find the radius of curvature of the parabola y2 = 4px at (0, 0).

Solution: Use the result of Exercise 17(a) with the given parabola expressed parametri-cally as y = t, x = t2/4p. It’s easy to check that x′ = t/2p, x′′ = 1/2p, y′ = 1, and y′′ = 0,which implies that

κ(t) =| − 1/2p|

((t2/4p2) + 1)3/2.

The point (0, 0) corresponds to t = 0, so the curvature at this point is κ(0) = |1/2p|, andthe radius of curvature is ρ = 1/κ = |2p| = 2|p|.

1

(48) At what point(s) does y = ex have maximum curvature?

Solution: It should be clear from the shape of the graph of y = ex that there is at leastone point of maximum curvature. (In other words the curvature does not get arbitrarilylarge.) Using d

dx (ex) = ex and the result of Exercise 17(b) gives

κ(x) =|d2y/dx2|

(1 + (dy/dx)2)3/2=

ex

(1 + e2x)3/2.

This function is differentiable everywhere, so any maximum must occur where the derivativeis zero:

dκ(x)

dx=ex(1 + e2x)3/2 − 3

2 (1 + e2x)1/2(2e2x)ex

(1 + e2x)3= 0.

Using simple algebra, the numerator of the above quotient can be written

ex(1 + e2x)1/2(1− 2e2x).

Therefore, dκ/dx = 0 when 1− 2e2x = 0, or x = − ln 22 and y = 1√

2. Moreover, dκ/dx > 0

(κ is increasing) for x < − ln 22 , while dκ/dx < 0 (κ is decreasing) for x > − ln 2

2 . Thus,

(x, y) = (− ln 22 , 1√

2) must be the point of maximal curvature.

(59) In Exercises 59, we will be concerned with the problem of creating a single smooth curve bypiecing together two separate smooth curves. If two smooth curves C1 and C2 are joined ata point P to form a curve C, then we will say that C1 and C2 make a smooth transitionat P if the curvature of C is continuous at P .

The accompanying figure shows the arc of a circle of radius r with center at (0, r). Findthe value of a so that there is a smooth transition from the circle to the parabola y = ax2

at the point where x = 0.

Solution: The curvature for a circle of radius r is 1/r. For the parabola y = ax2,dy/dx = 2ax and d2y/dx2 = 2a. So, using the result of Exercise 17(b),

κ(x) =|d2y/dx2|

[1 + (dy/dx)2]3/2=

|2a|[1 + 4a2x2]3/2

,

and κ(0) = |2a| = 2a (since a > 0 for the parabola in the figure). Equating the curvaturefor the circle and parabola results in 1/r = 2a or a = 1/2r.

(61) In Exercises 61-64, we assume that s is an arc length parameter for a smooth vector-valuedfunction r(s) in 3-space and that dT/ds and dN/ds exist at each point on the curve. Thisimplies that dB/ds exists as well, since B = T×N.

Show thatdT

ds= κ(s)N(s)

and use this result to obtain the formula in (10).

Solution: We know that T(s) = r′(s), N(s) = r′′(s)/||r′′(s)||, and κ(s) = ||r′′(s)||.Therefore,

dT

ds= T′(s) = r′′(s) = ||r′′(s)|| r′′(s)

||r′′(s)||= κ(s)N(s).

2

(62) (a) Show that dB/ds is perpendicular to B(s).

Solution: B is unit vector, so B(s)•B(s) = 1. Differentiating both sides with respect tos gives

dB

ds•B + B•dB

ds= 0,

or

2B•dBds

= 0.

Thus, B•dBds = 0, implying that dB/ds is perpendicular to B(s).

(b) Show that dB/ds is perpendicular to T(s).[Hint: Use the fact that B(s) is perpendicularto both T(s) and N(s), and differentiate B•T with respect to s.]

Solution: Since B and T are orthogonal, B•T = 0. Differentiating this equation withrespect to s gives

dB

ds•T + B•dT

ds= 0.

Since dT/ds = κN (from Exercise 61) and B is orthogonal to N, the second term on theleft is 0, implying that

dB

ds•T = 0.

Hence, dB/ds is perpendicular to T(s).

(c) Use the results in (a) and (b) to show that dB/ds is a scalar multiple of N(s). Thenegative of this scalar is called the torsion of r(s) and is denoted by τ(s). Thus,

dB

ds= −τ(s)N(s).

Solution: Since N = B × T, N is orthogonal to both B and T, as is dB/ds from theresults in (a) and (b). Since B and T are orthogonal unit vectors that define a plane, bothN and dB/ds must be orthogonal to this plane. Hence, dB/ds must be a scalar multipleof N.

(d) Show that τ(s) = 0 for all s if the graph of r(s) lies in a plane. [Note: For reasons thatwe cannot discuss here, the torsion is related to the “twisting” properties of the curve, andτ(s) is regarded as a numerical measure of this tendency for the curve to twist out of theosculating plane.]

Solution: If the graph of r(s) lies in a plane, then both T and N must lie in this planesince they are the unit tangent and unit “inward” normal vectors respectively. (Note thatif T lies in a plane with normal vector n then T•n = 0. Differentiating gives T′•n = 0,which implies that N•n = 0. Therefore N also lies in the same plane.) Since T and N liein the plane containing r, B = T×N must always be normal to this plane. Since the unitvector B has constant magnitude and the direction of the vector is always perpendicularto the plane of r, B must be constant, implying that dB/ds = 0. Since dB

ds = −τ(s)N(s)and N(s) is never zero, τ(s) must be 0 for all s.

3

(63) Let κ be the curvature of c and τ the torsion (defined in Exercise 62). By differentiatingN = B×T with respect to s, show that dN/ds = −κT + τB.

Solution: Consider dN/ds = (dB/ds) × T + B × (dT/ds). Since dB/ds = −τN anddT/ds = κN, we obtain dN/ds = −τN×T+B×κN. Since N×T = −B, B×N = −T,and a scalar can be factored out of a cross product,

dN/ds = τB− κT = −κT + τB.

(64) The following derivatives, known as the Frenet-Serret formulas, are fundamental in thetheory of curves in 3-space:

dTds = κN [Exercise 61]dNds = −κT + τB [Exercise 63]dBds = −τN [Exercise 62(c)]

Use the first two Frenet-Serret formulas and the fact that r′(s) = T if r = r(s) to showthat

τ =[r′(s)× r′′(s)]•r′′′(s)

||r′′(s)||2and B =

r′(s)× r′′(s)

||r′′(s)||

Solution: The equation for B is easy, and we leave it to you. To find τ , dot both sidesof the second Frenet-Serret formula with B:

B•dNds

= B•(−κT + τB) = −κB•T + τB•B.

Since B is a unit vector orthogonal to T, B•B = 1 and B•T = 0, which implies that theabove equation simplifies to

B•dNds

= τ.

Since N = r′′(s)/||r′′(s)|| = r′′(s)/κ(s), we can differentiate with respect to s to obtain

dN

ds=

κ(s)r′′′(s)− r′′(s)κ′(s)

(κ(s))2

=r′′′(s)

κ(s)− r′′(s)

κ(s)· κ′(s)

κ(s)

=r′′′(s)

κ(s)−N · κ

′(s)

κ(s).

Therefore,

B•dNds

=B•r′′′(s)κ(s)

−B•N · κ′(s)

κ(s)

=B•r′′′(s)||r′′(s)||

because B•N = 0 and κ(s) = ||r′′(s)||. Finally, since B = r′(s)×r′′(s)||r′′(s)|| , we can write

τ = B•dNds

=B•r′′′(s)||r′′(s)||

=

r′(s)×r′′(s)||r′′(s)|| •r

′′′(s)

||r′′(s)||=

[r′(s)× r′′(s)]•r′′′(s)||r′′(s)||2

,

as desired.

4


(9) As illustrated in the accompanying figure, suppose that the equations of motion of a particlemoving along an elliptic path are x = a cosωt, y = b sinωt.

(a) Show that the acceleration is directed toward the origin.

Solution: Let the parametric equations of the particle motion be represented by r(t) =x(t)i + y(t)j = a cosωti + b sinωtj, so that v(t) = r′(t) = ω(−a sinωti + b cosωtj) anda(t) = r′′(t) = −ω2(a cosωti + b sinωtj) = −ω2r(t). Since the acceleration vector isproportional to the negative of the position vector, the acceleration vector always pointstoward the origin.

(b) Show that the magnitude of the acceleration is proportional to the distance from theparticle to the origin.

Solution: The magnitude of the acceleration is given by ||a(t)|| = ||−ω2r(t)|| = ω2||r(t)||,which shows that it is proportional to the distance from the particle to the origin.

(12) Suppose that the motion of a particle is described by the position vector r = (t− t2)i− t2j.Find the minimum speed of the particle and the location when it has this speed.

Solution: r′(t) = (1− 2t)i− 2tj and speed of the particle is

v(t) = ||r′(t)|| =√

(1− 2t)2 + 4t2 =√

8t2 − 4t+ 1.

The expression 8t2 − 4t + 1 is minimal at t = 1/4. So the minimum speed occurs whent = 1/4, which corresponds to the point (x, y) = ( 3

16 ,−116 ). The speed at this point

v(1/4) = 1/√

2.

(19) What can you say about the trajectory of a particle that moves in 2-space or 3-space withzero acceleration? Justify your answer.

Solution: If a(t) = 0 for all t, then v(t) =∫a(t) dt = C1 for some constant vector C1.

So the velocity is constant. Integrating again gives r(t) =∫v(t) dt = C1t + C2 for some

constant C2. Therefore the trajectory of the particle must lie on a line.

(29) The position vector of two particles are given. Show that the particles move along thesame path but the speed of the first is constant and the speed of the second is not.

r1 = 2 cos 3ti + 2 sin 3tjr2 = 2 cos(t2)i + 2 sin(t2)j (t ≥ 0)

Solution: If x(t) and y(t) denote the components of the vector functions, then for bothr1 and r2 we have (x(t))2 + (y(t))2 = 4. So the path of both particles is the circle of radius2 centered at the origin. The speed of r1 is given by ||r′1(t)|| = ||−6 sin 3ti+ 6 cos 3tj|| = 6,so it is constant. The speed of r2 is given by ||r′2(t)|| = || − 2(2t) sin t2i + 2(2t) cos t2j|| =2(2t) = 4t, so it is not constant.

1

(48) Suppose that a particle moves with nonzero acceleration along the curve y = f(x). Usepart (b) of Exercise 17 in Section 14.5 to show that the acceleration vector is tangent tothe curve at each point where f ′′(x) = 0.

Solution: According to Theorem 14.6.2, the acceleration of a particle can be expressedas a = (d2s/dt2)T + κ(ds/dt)2N, where s is an arc length parameter for the curve, κ isthe curvature, T is the unit tangent vector, and N is the unit normal vector. The resultof Exercise 17(b) in Section 14.5 states that the curvature for a curve y = f(x) can beexpressed as:

κ(x) =|d2y/dx2|

[1 + (dy/dx)2]3/2.

So κ = 0 when f ′′(x) = d2y/dx2 = 0. Thus, a = (d2s/dt2)T, implying that the (nonzero)acceleration vector must be tangent to the curve.

(62) A ball rolls off a table 4 ft high while moving at a constant speed of 5 ft/s.

(a) How long does it take for the ball to hit the floor after it leaves the table?

Solution: Choose a coordinate system such that the origin lies at the edge of the table, thepositive x-axis points horizontally in the initial direction of the ball, and the positive y-axispoints upward. The parametric equations for the motion of the ball are x(t) = (v0 cosα)tand y(t) = s0 + (v0 sinα)t − 1

2gt2, where s0 = 0 is the initial height, v0 = ||v0|| = 5 ft/s

is the initial speed, α = 0 is initial angle of release, g is the acceleration due to gravity(32ft/s2), and t is the time in seconds. So y(t) = s0 + (v0 sinα)t− 1

2gt2 = −16t2, and the

ball hits the floor when y(t) = −16t2 = −4, or t = 1/2.

(b) At what speed does the ball hit the floor?

Solution: The components of the velocity of the ball are given by vx = v0 cosα andvy = v0 sinα − gt. So at t = 1/2, vx = 5 ft/s and vy = −16 ft/s. The final speed isvf =

√vx2 + vy2 =

√281 ft/s.

(c) If a ball were dropped from rest at table height just as the rolling ball leaves the table,which ball would hit the ground first? Justify your answer.

Solution: If the ball were dropped from rest, v0 = 0 and α = −π/2, so the parametricequations of motion would be x = 0 and y = −16t2. At the ground, y = −4 = −16t2,implying that the ball would hit the ground at t = 1/2 s, which is the same as before.Therefore, both balls would strike simultaneously. This makes sense: the time it takes forthe ball to hit the floor is completely dependent on the vertical component y(t), whichitself does not depend on the initial horizontal velocity.

(67) A shell is fired from ground level at an elevation angle of α and a muzzle speed of v0.

(a) Show that the maximum height reached by the shell is

maximum height =(v0 sinα)2

2g.

Solution: Choose the (obvious) coordinate system such that the origin is the point fromwhich the shell is fired, the ground is y = 0, and the positive x-axis corresponds to zeroangle of elevation. The parametric equations for the motion of the shell are x = (v0 cosα)t

2

and y = s0 + (v0 sinα)t− 12gt

2, where s0 = 0 is the initial height, and g is the accelerationdue to gravity. The shell is at its maximum height when dy/dt = v0 sinα− gt = 0, whichoccurs when t = v0 sinα/g. So the maximum height is:

y(v0 sinα/g) = (v0 sinα)(v0 sinα/g)− 1

2g(v0 sinα/g)2 =

(v0 sinα)2

2g.

(b) The horizontal range R of the shell is the horizontal distance traveled when the shellreturns to ground level. Show that R = (v0

2 sin 2α)/g. For what elevation angle will therange be maximum? What is the maximum range?

Solution: When the shell returns back to the ground, y = (v0 sinα)t− 12gt

2 = 0, implyingthat either t = 0 or t = 2v0 sinα/g. Since t = 0 corresponds to the initial release of theshell, t = 2v0 sinα/g is the desired time for the shell to return to the ground. Therefore,the horizontal range is:

R = x(2v0 sinα/g) = (v0 cosα)t = (v0 cosα)(2v0 sinα/g) = (v02 sin 2α)/g.

(We’ve made use of the identity sin 2α = 2 sinα cosα.) The maximum range occurs whensin 2α = 1, or α = π/4, and this maximum range is v0

2/g.

3

ohsx xm521 multivariable differential calculus: homework ... · homework solutions x14.3 (17)...

Documents