Math 265Homework#3

12.7.12) Change from rectangular to cylindrical coordinates the point (3, 4, 5).

Proof. r2 = x2 + y2 = 32 + 42 = 25 so r = 5: tan θ = 43 and the point (3, 4)

is in the first quadrant of the xy-plane, so θ = tan−1( 43 ) + 2π ≈ 0.93 +

2π; z = 5. Thus, one set of cylindrical coordinates is (5, tan−1( 43 , 5)) ≈

(5, 0.93, 5). �

12.7.20) Change from rectangular to spherical coordinates the point (0,√

3, 1).

Proof. ρ =√

x2 + y2 + z2 =√

0 + 3 + 1 = 2, cos φ = zρ = 1

2 ⇒ φ = π3 ,

and cos θ = xρ sin φ = 0

2 sin π3

= 0 ⇒ θ = π2 (since y > 0). Thus the spherical

coordinates are (2, π2 , π

3 ). �

12.7.46) Identify the surface whose equation is ρ2(sin2 φ− 4 cos2 φ) = 1.

Proof. Since ρ2(sin2 φ − 4 cos2 φ) = 1, ρ2(sin2 φ − 4 cos2 φ) + ρ2 cos2 φ −ρ2 cos2 φ = 1,or ρ2(sin2 φ + cos2 φ − 5 cos2 φ) = 1, or ρ2(1 − 5 cos2 φ) = 1.But ρ2 = x2 + y2 + z2 and z2 = ρ2 cos2 φ, so we can rewrite the equationof the surface as x2 + y2 + z2 − 5z2 = 1 or x2 + y2 − 4z2 = 1. Thus thesurface is a hyperboloid of one sheet with axis the z-axis. �

12.7.62) Sketch the solid described by the inequalities 0 ≤ φ ≤ π3 , ρ ≤ 2.

Proof. ρ = 2 ⇔ x2 + y2 + z2 = 4, which is a sphere of radius 2, centered atthe origin. Hence ρ ≤ 2 is the sphere and its interior. 0 ≤ φ ≤ π

3 restrictsthe solid to that section of this ball that lies above the cone φ = π

3 .

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13.1.2) Find the domain of the vector function r(t) = t−2t+2 i + sin tj + ln(9− t2)k

Proof. The component functions t−2t+2 , sin t, and ln(9 − t2) are all defined

when t 6= 2 and 9 − t2 > 0 ⇒ −3 < t < 3, so the domain of r(t) is(−3,−2) ∪ (−2, 3). �

13.1.14) Sketch the curve with the vector equation r(t) = sin t i+sin t j+√

2 cos t k.Indicate with an arrow the direction in which t increases.

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Proof. The parametric equations give x2+y2+z2 = 2 sin2 t+2 cos2 t = 2, sothe curve lies on the sphere of radius

√2, and center (0, 0, 0). Furthermore

x = y = sin t, so the curve is the intersection of this sphere with the planex = y, that is, the curve is the circle of radius

√2, center (0, 0, 0) in the

plane x = y.

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13.1.34) Find a vector function that represents the curve of intersection of the cylin-der x2 + y2 = 4 and the surface z = xy.

Proof. The projection of the curve C of intersection onto the xy-planeis the circle x2 + y2 = 4, z = 0. Then we can write x = 2 cos t, y =2 sin t, 0 ≤ t ≤ 2π. Since C also lies on the surface z = xy, we havez = xy = (2 cos t)(2 sin t) = 4 cos t sin t, or 2 sin(2t). Then the parametricequations for C are x = 2 cos t, y = 2 sin t, z = 2 sin(2t), 0 ≤ t ≤ 2π, and thecorresponding vector function is r(t) = 2 cos t i + 2 sin t j + 2 sin(2t) k, 0 ≤t ≤ 2π. �

13.2.14) Find the derivative of the vector functionr(t) = at cos 3t i + b sin3 t j + c cos3 t k.

Proof. r′(t) = [at(−3 sin 3t)+a cos 3t]i+b·2 sin2 t cos tj+c·3 cos2 t(− sin t)k =(a cos 3t− 3at sin 3t)i + 3b sin2 t cos tj− 3c cos2 t sin tk. �

13.2.24) Find parametric equations for the tangent line to the curve with the para-metric equations x = t2 − 1, y = t2 + 1 z = t + 1 at the point (−1, 1, 1).

Proof. The vector equation for the curve is r(t) = 〈t2 − 1, t2 + 1, t + 1〉, sor′(t) = 〈2t, 2t, 1〉. The point (−1, 1, 1) corresponds to t = 0, so the tangentvector there is r′(0) = 〈0, 0, 1〉. Thus, the tangent line is parallel to thevector 〈0, 0, 1〉 and parametric equations are x = −1 + 0 · t = −1, y =1 + 0 · t = 1, z = 1 + 1t = 1 + t. �

13.2.29) Determine whether the curve is smooth.a) r(t) = 〈t3, t4, t5〉b) r(t) = 〈t3 + t, t4, t5〉c) r(t) = 〈cos3 t, sin3 t〉

Proof. a) r(t) = 〈t3, t4, t5〉 ⇒ r′(t) = 〈3t2, 4t3, 5t4〉, and since r′(0) =〈0, 0, 0〉 = 0, the curve is not smooth.b) r(t) = 〈t3 + t, t4, t5〉 ⇒ r′(t) = 〈3t2 + 1, 4t3, 5t4〉.r′(t) is continuous

3

since its component functions are continuous. Also, r′(t) 6= 0, as the y-andz−components are 0 only for t = 0, but r′(0) = 〈1, 0, 0〉 6= 0. Thus, thecurve is smooth.c) r(t) = 〈cos3 t, sin3 t〉 ⇒ r′(t) = 〈−3 cos2 t sin t, 3 sin2 t cos t〉. SInce r′(0) =〈−3 cos2 0 sin 0, 3 sin2 0 cos 0〉 = 〈0, 0〉 = 0, the curve is not smooth. �

13.2.40) Find r(t) if r′(t) = sin t i− cos t j + 2t k and r(0) = i + j + 2 k.

Proof. r′(t) = sin t i−cos t j+2t k ⇒ r(t) = (− cos t) i−(sin t) j+t2 k+C.But i + j + 2k = r(0) = −i + (0)j + (0)k + C. Thus C = 2i + j + 2k andr(t) = (2− cos t)i + (1− sin t)j + (2 + t2)k. �

13.3.6) Find the length of the curve r(t) = 12t i + 8t3/2 j + 3t2 k, 0 ≤ t ≤ 1.

Proof. r′(t) = 12i+12√

tj+6tk ⇒ |r′(t)| =√

144 + 144t + 36t2 =√

36(t + 2)2 =6|t + 2| = 6(t + 2) for 0 ≤ t ≤ 1. Then

L =∫ 1

0

6(t + 2) dt = [3t2 + 12t]10 = 15.

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13.3.12) Reparametrize the curve

r(t) =( 2

t2 + 1− 1

)i +

2t

t2 + 1j

with respect to arc length measured from the point (0, 1) in the directionof increasing t. Express the reparametrization in its simplest form. Whatcan you conclude about the curve?

Proof. r′(t) = −4t(t2+1)2 i + −2t2+2

(t2+1)2 j,

ds

dt= |r′(t)| =

√[ −4t

(t2 + 1)2]2

+[−2t2 + 2(t2 + 1)2

]2

=

√4t4 + 8t2 + 4

(t2 + 1)4

=

√4(t2 + 1)2

(t2 + 1)4=

√4

(t2 + 1)2=

2t2 + 1

Since the initial point (1, 0) corresponds to t = 0, the arc length function

s(t) =∫ t

0

|r′(t)| du =∫ t

0

2u2 + 1

du = 2 arctan t.

Then arctan t = 12s ⇒ t = tan 1

2s. Substituting, we have

r(t(s)) =[ 2tan2( 1

2s) + 1− 1

]i +

2 tan( 12s)

tan2( 12s) + 1

j =1− tan2( 1

2s)1 + tan2( 1

2s)i +

2 tan( 12s)

sec2( 12s)

j

=1− tan2( 1

2s)sec2( 1

2 )si + 2 tan(

12s) cos2(

12s)j

[cos2(12s)− sin2(

12s)]i + 2 sin(

12s) cos(

12s)j = cos si + sin sj

With this parametrization, we recognize the function as representing theunit circle. Note here that the curve approaches, but does not include, thepoint (−1, 0), since cos s = −1 for s = π + 2kπ (k an integer) but thent = tan( 1

2s) is undefined. �

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13.3.20) Find the curvature of r(t) = 〈et cos t, et sin t, t〉 at the point (1, 0, 0).

Proof. r′(t) = 〈et cos t − et sin t, et cos t + et sin t, 1〉. The point (1, 0, 0)corresponds to t = 0, and r′(0) = 〈1, 1, 1〉 ⇒ |r′(0)| =

√12 + 12 + 12 =

√3.

r′′(t) = 〈et cos t− et sin t− et cos t− et sin t, et cos t− et sin t + et cos t + et sin t, 0〉= 〈−2et sin t, 2et cos t, 0〉 ⇒ r′′(0) = 〈0, 2, 0〉.

r′(0)×r′′(0) = 〈−2, 0, 2〉. |r′(0)×r′′(0)| =√

(−2)2 + 02 + 22 =√

8 = 2√

2.Then κ(0) = |r′(0)×r′′(0)|

|r′(0)|3 = 2√

2(√

3)3= 2

√2

3√

3or 2

√6

9 . �

13.3.45) At what point on the curve x = t3, y = 3t, z = t4 is the normal planeparallel to the plane 6x + 6y − 8z = 1?

Proof. The tangent vector is normal to the normal plane, and the vector〈6, 6, 8〉 is normal to the given plane. But T(t)||r′(t) and 〈6, 6, 8〉||〈3, 3,−4〉,so we need to find t such that r′(t)||〈3, 3,−4〉. r(t) = 〈t3, 3t, t4〉 ⇒ r′(t) =〈3t2, 3, 4t3〉||〈3, 3,−4〉 when t = −1. So the planes are parallel at the pointr(−1) = (−1,−3, 1). �