1. REVIEW PROBLEMS FOR TEST 1 PART 3
CALCULUS III
REVIEW PROBLEMS FOR TEST 1 PART 3
SECTIONS/PROBLEMS: 12.2 (7), 12.3 (5, 23, 26, 64), 12.4 (3), 12.5 (5,26)
13.2 (42), 13.4 (5), 13.3 (5), 16.2 (34), 14.1 (10, 15), 14.3 (61), 14.4 (11)
Dr. Vladimir Visnjic

2. SECTION 12.2
a + b = 2c ⇒ 𝐜= 𝐚+ 𝐛 𝟐 b – a = 2d ⇒ 𝐝= 𝐛−𝐚 𝟐

3. SECTION 12.3
5. Find 𝐚∙𝐛 where 𝐚=<4, 1, 1 4 > and 𝐛=<6, −3−8>.
𝐚∙𝐛=4∙6−3− 1 4 ∙8=24−3−2=19
23. Determine whether the given vectors are orthogonal, parallel, or neither.
(a) 𝐚=<9, 3>, 𝐛=<−2, 6>
𝐚∙𝐛=9∙ −2 +3∙6=−18+18=0 The vectors are orthogonal.
(b) 𝐚=<4, 5, −2>, 𝐛=<3,−1, 5>
𝐚∙𝐛=12−5−10=−3 The vectors are neither orthogonal nor parallel
(c) a = − 8i + 12j + 4k, b = 6i − 9j − 3k
The vectors are scalar multiples of each other: b =− 3 4 𝐚. They are parallel.
(d) a = 3i − j + 3k, b = 5i + 9j − 2k
𝐚∙𝐛=3∙5−9 −6=15−15=0 The vectors are orthogonal.

4. 26. Find the values of x such that the angle between the vectors <2, 1,−1>, and <1,𝑥,0> is π/4.
cos 𝜋 4 = <2, 1,−1>∙<1,𝑥,0> <2, 1,−1> |<1,𝑥,0>|
= 2+𝑥 𝑥 2
1= 2+𝑥 𝑥 2
Cross-multiply, square both sides and simplify: 𝑥 2 −4𝑥−1=0 The solutions of this equation are 𝑥= and 𝑥= 2−

5. 64. Show that if u + v and u – v are orthogonal, then the vectors u and v must have the same length.
If u + v and u – v are orthogonal, their dot product is zero:
𝐮+𝐯 ∙ 𝐮−𝐯 =0 Compute the dot product:
𝐮+𝐯 ∙ 𝐮−𝐯 =𝐮∙𝐮−𝐮∙𝐯+𝐯∙𝐮−𝐯∙𝐯= 𝐮 2 − 𝐯 2 Here we used the properties that the dot product commutes, 𝐮∙𝐯= 𝐯∙𝐮, and that 𝐮∙𝐮= 𝐮 2 .
𝐮 2 − 𝐯 2 =0
𝐮 2 = 𝐯 2
Since the magnitudes of the vectors cannot be negative, it follows that 𝐮 = 𝐯 .

6. SECTION 12.4
3. Find the cross product 𝐚×𝐛 and verify that it is orthogonal to both a and b. a = 2j – 4k, b = – i + 3j + k
𝐚×𝐛= 𝐢 𝐣 𝐤 0 − − =𝐢 2 − −𝐣 0 −4 −1 1 +𝐤 0 2 −1 3 =14𝐢+4𝐣+2𝐤 Verify that 𝐚×𝐛 is orthogonal to both a and b:
𝐚∙ 𝐚×𝐛 =2∙4−4∙2=0  𝐛∙ 𝐚×𝐛 =−14+3∙4+2=0 

7. 𝐫=<1, 0, 6>+𝑡<1, 3, 1>,
SECTION 12.5
5. Find the vector equation and parametric equations for the line through the point (1, 0, 6) and perpendicular to the plane 𝑥+3𝑦+𝑧= The normal vector of the plane 𝐧=<1, 3, 1>. Since the line is perpendicular to the plane, the vector v of the line can be set equal to n. With the point (1, 0, 6), the vector equation of the line is
𝐫=<1, 0, 6>+𝑡<1, 3, 1>,
and the parametric equations are
𝑥=1+𝑡, 𝑦=3𝑡, 𝑧=6+𝑡

8. SECTION 12.5
26. Find an equation of the plane through the point (2, 0, 1) and perpendicular to the line 𝑥=3𝑡, 𝑦=2−𝑡, 𝑧=3+4𝑡. The direction vector of the line 𝐯=<3, −1, 4>. Since the plane is perpendicular to the line, the normal vector n of the plane can be set equal to v. With the point (2, 0, 1), the plane equation is
3 𝑥−2 − 𝑦−0 +4 𝑧−1 =0,
or, in the form ax + by + cz = d,
3𝑥−𝑦+4𝑧=10.

9. 𝐫 𝑡 =(1+ 𝑡 2 2 ) 𝐢+ 𝑒 𝑡 𝐣+ 𝑡 𝑒 𝑡 − 𝑒 𝑡 +2 𝐤.
SECTION 13.2
42. Find r(t) if r’(t) = t i + 𝑒 𝑡 𝐣+𝑡 𝑒 𝑡 𝐤 and 𝐫 0 =𝐢+𝐣+𝐤. 𝐫 𝑡 = 𝐫’ 𝑡 𝑑𝑡= 𝑡 𝐢 + 𝑒 𝑡 𝐣+𝑡 𝑒 𝑡 𝐤 𝑑𝑡
= 𝑡 𝐢+ 𝑒 𝑡 𝐣+ 𝑡 𝑒 𝑡 − 𝑒 𝑡 𝐤+𝐂
where C is the integration constant which here can be determined from the given initial condition:
𝐫 0 =𝐣−𝐤+𝐂=𝐢+𝐣+𝐤.
From this, C = i + 2k and
𝐫 𝑡 =(1+ 𝑡 2 2 ) 𝐢+ 𝑒 𝑡 𝐣+ 𝑡 𝑒 𝑡 − 𝑒 𝑡 +2 𝐤.

10. 13.4
 5 . Find the velocity, acceleration, and speed of a particle with the position function
r(t) =3 cos t i +2 sin t 𝐣.
Sketch the path of the particle and draw the velocity and acceleration vectors for t = π/3. (Note: Make sure to draw these vectors at the point on the path for t = π/3 and not at the origin.)
 𝐯= 𝐫 ′ 𝑡 =−3 sin 𝑡 𝐢+2 cos 𝑡 𝐣
𝐚 t =𝐯 ′ t = 𝐫 ′′ (t) = −3 cos t i −2 sin t 𝐣 speed = 𝐯 𝑡 = 9 sin 2 𝑡+4 cos 2 𝑡 = 4+5 sin 2 𝑡
Sketching:
Path: 𝑥=3 cos 𝑡, 𝑦=2 sin 𝑡 are parametric equations of the ellipse with the semiaxis 3 along the x axis and 2 along the y axis. For t = π/3, the particle is at the point x = 3/2, y = Velocity: 𝐯( 𝜋 3)=− 𝐢+𝐣 Acceleration: 𝐚( 𝜋 3)=− 𝐢− 3 𝐣

11. 5 . Find the length of the curve 𝐫 𝑡 =𝐢+ 𝑡 2 𝐣+ 𝑡 3 𝐤, 0≤𝑡≤1.
SECTION 13.3
5 . Find the length of the curve 𝐫 𝑡 =𝐢+ 𝑡 2 𝐣+ 𝑡 3 𝐤, 0≤𝑡≤1.
𝐿= 𝑑𝑥 𝑑𝑡 𝑑𝑦 𝑑𝑡 𝑑𝑧 𝑑𝑡 𝑑𝑡
𝑑𝑥 𝑑𝑡 =0, 𝑑𝑦 𝑑𝑡 =2𝑡, 𝑑𝑧 𝑑𝑡 =3 𝑡 2
𝐿= 𝑡 𝑡 4 𝑑𝑡= 0 1 𝑡 4+9 𝑡 2 𝑑𝑡= 1 27 ( −8)
u-substitution: u = 𝑡 2 , du = 18 t dt: 𝑢 𝑑𝑢= 𝑢 | 13 4

12. = 0 𝜋 2 𝑘 𝑎 2 sin 𝑡 cos 𝑡 𝑎 2 sin 2 𝑡+ 𝑎 2 cos 2 𝑡 𝑑𝑡
SECTION 16.2
34. A thin wire has the shape of the first quadrant part of the circle with center the origin and radius a. If the density function is ρ(x, y) = kxy, find the mass of the wire.
Parametric equations of the first quadrant part of the circle with center the origin and radius a are
𝑥=𝑎 cos 𝑡, 𝑦=𝑎 sin 𝑡, 0≤𝑡≤ 𝜋 2
𝑚= 0 𝜋 2 𝜌(𝑥, 𝑦) 𝑑𝑥 𝑑𝑡 𝑑𝑦 𝑑𝑡 2 𝑑𝑡
= 0 𝜋 2 𝑘 𝑎 2 sin 𝑡 cos 𝑡 𝑎 2 sin 2 𝑡+ 𝑎 2 cos 2 𝑡 𝑑𝑡
= 𝑘 𝑎 𝜋 2 sin 𝑡 cos 𝑡 𝑑𝑡 = 𝑘 𝑎 3

15. SECTION 14.3
61. Verify that the conclusion of Clairaut’s Theorem holds for 𝑢= cos ( 𝑥 2 𝑦) , that is, 𝑢 𝑥𝑦 = 𝑢 𝑦𝑥
𝑢 𝑥 =−2 𝑥𝑦 sin (𝑥 2 𝑦) 𝑢 𝑥𝑦 = −2𝑥 sin ( 𝑥 2 𝑦) −2 𝑥 3 𝑦 cos ( 𝑥 2 𝑦)
𝑢 𝑦 = − 𝑥 2 sin ( 𝑥 2 𝑦) 𝑢 𝑦𝑥 =−2𝑥 sin 𝑥 2 𝑦 −2 𝑥 3 𝑦 cos 𝑥 2 𝑦

16. SECTION 14.4 11. Find the linearization L(x, y) of 𝑓 𝑥,𝑦 =1+𝑥 ln (𝑥𝑦−5) at the point (2, 3).
The first partial derivatives of f(x, y) are 𝑓 𝑥 𝑥,𝑦 = ln 𝑥𝑦−5 + 𝑥𝑦 𝑥𝑦− and 𝑓 𝑦 𝑥, 𝑦 = 𝑥 2 𝑥𝑦−5 At the point (2, 3) their values are 𝑓 𝑥 2,3 =6 and 𝑓 𝑦 2,3 =4. The linearization of f(x, y) at the point (2, 3) is
𝐿 𝑥,𝑦 =𝑓 2,3 + 𝑓 𝑥 2,3 𝑥−2 + 𝑓 𝑦 (2,3)(𝑦−3)
𝐿 𝑥,𝑦 =1+6 𝑥−2 +4(𝑦−3)
𝐿 𝑥,𝑦 =6𝑥+4𝑦−23