Multivariable Calculus Math 2080
Multivariable Calculus Math 2080 (Spring 2009) Exam 1 Solutions Instructions: There are 100 points on the exam and an additional extra credit problem worth 10 points. As always, a 100 counts as perfect. No technology allowed on this exam. Show your work. 1. (5 points each) Let a = h1, 1, 2i and b = h1, −1, −1i. Find each of the following. (a) a + 2b Solution: a + 2b = h3, −1, 0i
(b) kak and kbk Solution:
kak = and kbk =
√ √ 1+1+4= 6 √ √ 1+1+1= 3
(c) cos (θ) where θ is the angle between a and b. Solution: √ 2 −2 −2 h1, 1, 2i · h1, −1, −1i a·b √ √ = √ √ = √ =− = cos (θ) = kak kbk 3 6 3 6 3 3 2 2. (10 points) A circular racetrack has a radius of 50 feet. A car drives around the track at a constant speed of 20 feet per second. You are told that the vector parametrization of the motion of the car is of the form r (t) = R hcos (αt) , sin (αt)i for some constants R, α > 0. Find the constants R and α. Solution: Since 50 = kr (t)k = R, we get R = 50. Also, r0 (t) = 50α h− sin (αt) , cos (αt)i 2 and 20 = kr0 (t)k = 50α, so α = . 5 2 Summarizing, R = 50 and α = . 5
1
3. (3 points each) Each of the four plots is a contour map of a function f , showing the level curves of the form f (x, y) = c for 10 equally spaced values of c. (Each is shown on the square −2 ≤ x ≤ 2 , −2 ≤ y ≤ 2.)
2
2
2
2
1
1
1
1
0
0
0
0
-1
-1
-1
-1
-2
-2
-2
-1
0
1
2
-2
-2
-1
Plot A
0
1
-2
-2
2
Plot B
-1
0
Plot C
1
2
-2
-1
0
1
2
Plot D
Match these plots with the functions listed in each part below. If the function matches no plot, write NONE. (Each plot matches one function. Answers only are OK here.) (a) f (x, y) = x + y
Your answer: NONE
(b) f (x, y) = x − y
Your answer: B
(c) f (x, y) = (x − y)3
Your answer: C
(d) f (x, y) = (2x − y)3
Your answer: D
(e) f (x, y) = 2x − y
Your answer: A
xy exist? (x,y)→(0,0) x2 + y 2 Hint: What happens as (x, y) → (0, 0) along the x-axis? What happens as (x, y) → (0, 0) along the line y = x?
4. (5 points ) Does
lim
Solution: • Along the x-axis (y = 0 and x 6= 0),
x2
xy = 0. So if the limit exists, it must + y2
be 0.
• Along the line y = x (x 6= 0), 1 be . 2
x2 1 xy = = . So if the limit exists, it must 2 2 2 x +y 2x 2
1 Since the limit (which must be unique if it exists) cannot be both 0 and , the limit 2 does not exist.
2
5. In this problem P = (1, 0, 0), Q = (0, 2, 0), and R = (0, 0, 4) are points in space. (a) (8 points) Find the equation of the plane containing the points P , Q and R. Solution: The plane passes through the point P = (1, 0, 0) and a normal n to the plane is ⎡ ⎤ i j k −→ −→ n = P Q × P R = det ⎣ −1 2 0 ⎦ = 8i − (−4) j + 2k −1 0 4 = 8i + 4j + 2k = h8,4, 2i A point (x, y, z) is on the plane if and only if hx − 1, y, zi · h8,4, 2i = 0 or, expanding and rewriting, 8x + 4y + 2z = 8 or, dividing both sides by 2, 4x + 2y + z = 4. (b) (8 points) Shown is the triangle whose vertices are at the points P , Q and R. Find the area of this triangle.
z
x
Solution: We use the calculation from part (a). The desired area is ° 1° °−→ −→° 1 A = °P Q × P R° = kh8,4, 2ik = kh4,2, 1ik 2 2 √ √ = 16 + 4 + 1 = 21.
3
(c) (7 points) Consider the plane whose equation you found in part (a). Find the point S on this plane which is closest to the origin. Then find the distance between this point S and the origin. Hint: Think geometrically. Solution: Move along the line tn until the line intersects the plane. (Here −→ −→ n =P Q × P R was computed in part (a). This line will cross the plane at a right angle and this is exactly what we want.) Now, tn = h8,4, 2i is on the plane if 4 (8t) + 2 (4t) + 1 (2t) = 4 4 2 i.e. if 42t = 4 or t = = . The coordinates of the point S are 42 21 ¶ µµ ¶ µ ¶ µ ¶ ¶ µ 2 2 16 8 4 2 8, 4, 2 = , , . 21 21 21 21 21 21 Finally, the distance from the point to the plane is ktnk for this value of t, which is 4 4√ kh4,2, 1ik = 21 ≈ 0.87. 21 21 6. A particle moves in the x-y plane so that at time t, its position is given by the vector parametrization ® r (t) = t2 + t , t2 − 4t . The trajectory of the particle is shown over some time interval.
(a) (3 points) Draw an arrow or two on the curve indicating the direction of motion. (b) (5 points) The velocity and acceleration vectors are orthogonal at exactly one point on the curve. Find the coordinates of this point. Solution: v (t) = r0 (t) = h 2t + 1 ,2t − 4 i and a (t) = v0 (t) = h 2 , 2 i v (t) · a (t) = (2t + 1) 2 + (2t − 4) 2 = 2 (4t − 3) = 0
3 when t = . 4 µ ¶2 À µ ¶+ ¿ µ ¶ * µ ¶2 3 3 21 39 3 3 3 = = , − + , −4 r 4 4 4 4 4 16 16 so the point where this occurs is
µ 4
21 39 ,− 16 16
¶
(c) (5 points) Write, but do not evaluate, a definite integral whose value is the arc length of that part of the path of r (t) shown in the figure below.
y
x
Solution: The curve meets the x-axis when its y-coordinate = 0, i.e. 0 = t2 − 4t = t (t − 4) so t = 0 and t = 4. The arc length is s=
Z4 0
Z4 q kv (t)k dt = (2t + 1)2 + (2t − 4)2 dt 0
Additional Observations: Scientific WorkPlace evaluates this integral as µ ¶ µ ¶ √ √ 25 √ 3√ 3√ 13 √ 25 √ 13 √ 2 ln 2 + 97 − 2 ln 17 − 2 + 17 + 97 ≈ 22.5. 16 2 16 2 8 8 This makes sense since the curve meets the x-axis when x = 0 and when x = 42 + 4 = 20, so the length of the curve should be a bit more than 20.
5
(d) The rest of this problem concerns the behavior of the particle at the point P where it has its smallest y-coordinate. This point P is shown as a dot and labeled in the graph below.
y
P
x
Find i. (5 points) the time t when this occurs. Solution: The y-coordinate is y (t) = t2 − 4t. We want to minimize y (t). Since y 0 (t) = 2t − 4, there is one critical point at t = 2. Since y 00 (t) = 2 > 0, t = 2 is an absolute minimum for y. So y (t) is minimized when t = 2. ii. (3 points) the coordinates of the point P . Solution: ® r (2) = 22 + 2 , 22 − 4 (2) = h 6 , − 4 i so P = (6, − 4) .
iii. (3 points) the velocity vector V at P . Solution: V = v (2) = h 5 ,0 i iv. (2 points) the acceleration vector A at P . Solution: A = a (2) = h 2 ,2 i v. (2 points) the speed of the particle at P . Solution: The speed is kVk = 5. vi. (2 points) the tangential component aT of the acceleration vector A. Solution: Since A = a (2) = h 2 ,2 i and T = h 1 ,0 i, aT = A · T = 2. (You could have done this one correctly just by inspection.) vii. (2 points) the normal component aN of the acceleration vector A. Solution: There are many ways to do this. Here’s one way. q √ √ aN = kAk2 − |aT |2 = 8 − 4 = 4 = 2 6
7. (10 points Extra Credit) Only attempt this problem if you are satisfied with the rest of the exam. Extra credit problems are harder and partial credit is harder to earn! The figure on the left shows the circles whose polar equations are r = 1, r = 2 and r = 2 cos θ. In the figure on the right, only the part of this diagram in the first quadrant is shown and a region has been shaded. Find the area A of this shaded region. 2.0
y 1.5
1.0
x 0.5
0.0 0.0
0.5
1.0
1.5
2.0
π Solution: The curves r = 1 and r = 2 cos θ intersect when θ = . Let’s add the line 3 π θ = and change the colors within the region. 3 2.0
1.5
1.0
0.5
0.0 0.0
0.5
1.0
1.5
2.0
Now A = P urple + Y ellow where P urple is the area of the purple region and Y ellow is the area of the yellow region. We compute each separately and add the results. 1 P urple = 2
Zπ/3 Zπ/3 Zπ/3 ¢ ¡ 2 ¡ 2¢ 2 2 − (2 cos θ) dθ = 2 1 − cos θ dθ = 2 sin2 θ dθ 0
0
0
µ ¶¯π/3 Zπ/3 sin 2θ ¯¯ 1 1√ (1 − cos 2θ) dθ = θ − = 3 = π − ¯ 2 3 4 0 0
Also,
1 Y ellow = 2
Zπ/2 Zπ/2 ¡ 2 ¢ 1 3 1 dθ = π 2 − 12 dθ = 2 4
π/3
π/3
Adding these, we get 1 1 1√ 7 1√ A = P urple + Y ellow = π − 3+ π = π− 3. 3 4 4 12 4 7
(b) kak and kbk Solution:
kak = and kbk =
√ √ 1+1+4= 6 √ √ 1+1+1= 3
(c) cos (θ) where θ is the angle between a and b. Solution: √ 2 −2 −2 h1, 1, 2i · h1, −1, −1i a·b √ √ = √ √ = √ =− = cos (θ) = kak kbk 3 6 3 6 3 3 2 2. (10 points) A circular racetrack has a radius of 50 feet. A car drives around the track at a constant speed of 20 feet per second. You are told that the vector parametrization of the motion of the car is of the form r (t) = R hcos (αt) , sin (αt)i for some constants R, α > 0. Find the constants R and α. Solution: Since 50 = kr (t)k = R, we get R = 50. Also, r0 (t) = 50α h− sin (αt) , cos (αt)i 2 and 20 = kr0 (t)k = 50α, so α = . 5 2 Summarizing, R = 50 and α = . 5
1
3. (3 points each) Each of the four plots is a contour map of a function f , showing the level curves of the form f (x, y) = c for 10 equally spaced values of c. (Each is shown on the square −2 ≤ x ≤ 2 , −2 ≤ y ≤ 2.)
2
2
2
2
1
1
1
1
0
0
0
0
-1
-1
-1
-1
-2
-2
-2
-1
0
1
2
-2
-2
-1
Plot A
0
1
-2
-2
2
Plot B
-1
0
Plot C
1
2
-2
-1
0
1
2
Plot D
Match these plots with the functions listed in each part below. If the function matches no plot, write NONE. (Each plot matches one function. Answers only are OK here.) (a) f (x, y) = x + y
Your answer: NONE
(b) f (x, y) = x − y
Your answer: B
(c) f (x, y) = (x − y)3
Your answer: C
(d) f (x, y) = (2x − y)3
Your answer: D
(e) f (x, y) = 2x − y
Your answer: A
xy exist? (x,y)→(0,0) x2 + y 2 Hint: What happens as (x, y) → (0, 0) along the x-axis? What happens as (x, y) → (0, 0) along the line y = x?
4. (5 points ) Does
lim
Solution: • Along the x-axis (y = 0 and x 6= 0),
x2
xy = 0. So if the limit exists, it must + y2
be 0.
• Along the line y = x (x 6= 0), 1 be . 2
x2 1 xy = = . So if the limit exists, it must 2 2 2 x +y 2x 2
1 Since the limit (which must be unique if it exists) cannot be both 0 and , the limit 2 does not exist.
2
5. In this problem P = (1, 0, 0), Q = (0, 2, 0), and R = (0, 0, 4) are points in space. (a) (8 points) Find the equation of the plane containing the points P , Q and R. Solution: The plane passes through the point P = (1, 0, 0) and a normal n to the plane is ⎡ ⎤ i j k −→ −→ n = P Q × P R = det ⎣ −1 2 0 ⎦ = 8i − (−4) j + 2k −1 0 4 = 8i + 4j + 2k = h8,4, 2i A point (x, y, z) is on the plane if and only if hx − 1, y, zi · h8,4, 2i = 0 or, expanding and rewriting, 8x + 4y + 2z = 8 or, dividing both sides by 2, 4x + 2y + z = 4. (b) (8 points) Shown is the triangle whose vertices are at the points P , Q and R. Find the area of this triangle.
z
x
Solution: We use the calculation from part (a). The desired area is ° 1° °−→ −→° 1 A = °P Q × P R° = kh8,4, 2ik = kh4,2, 1ik 2 2 √ √ = 16 + 4 + 1 = 21.
3
(c) (7 points) Consider the plane whose equation you found in part (a). Find the point S on this plane which is closest to the origin. Then find the distance between this point S and the origin. Hint: Think geometrically. Solution: Move along the line tn until the line intersects the plane. (Here −→ −→ n =P Q × P R was computed in part (a). This line will cross the plane at a right angle and this is exactly what we want.) Now, tn = h8,4, 2i is on the plane if 4 (8t) + 2 (4t) + 1 (2t) = 4 4 2 i.e. if 42t = 4 or t = = . The coordinates of the point S are 42 21 ¶ µµ ¶ µ ¶ µ ¶ ¶ µ 2 2 16 8 4 2 8, 4, 2 = , , . 21 21 21 21 21 21 Finally, the distance from the point to the plane is ktnk for this value of t, which is 4 4√ kh4,2, 1ik = 21 ≈ 0.87. 21 21 6. A particle moves in the x-y plane so that at time t, its position is given by the vector parametrization ® r (t) = t2 + t , t2 − 4t . The trajectory of the particle is shown over some time interval.
(a) (3 points) Draw an arrow or two on the curve indicating the direction of motion. (b) (5 points) The velocity and acceleration vectors are orthogonal at exactly one point on the curve. Find the coordinates of this point. Solution: v (t) = r0 (t) = h 2t + 1 ,2t − 4 i and a (t) = v0 (t) = h 2 , 2 i v (t) · a (t) = (2t + 1) 2 + (2t − 4) 2 = 2 (4t − 3) = 0
3 when t = . 4 µ ¶2 À µ ¶+ ¿ µ ¶ * µ ¶2 3 3 21 39 3 3 3 = = , − + , −4 r 4 4 4 4 4 16 16 so the point where this occurs is
µ 4
21 39 ,− 16 16
¶
(c) (5 points) Write, but do not evaluate, a definite integral whose value is the arc length of that part of the path of r (t) shown in the figure below.
y
x
Solution: The curve meets the x-axis when its y-coordinate = 0, i.e. 0 = t2 − 4t = t (t − 4) so t = 0 and t = 4. The arc length is s=
Z4 0
Z4 q kv (t)k dt = (2t + 1)2 + (2t − 4)2 dt 0
Additional Observations: Scientific WorkPlace evaluates this integral as µ ¶ µ ¶ √ √ 25 √ 3√ 3√ 13 √ 25 √ 13 √ 2 ln 2 + 97 − 2 ln 17 − 2 + 17 + 97 ≈ 22.5. 16 2 16 2 8 8 This makes sense since the curve meets the x-axis when x = 0 and when x = 42 + 4 = 20, so the length of the curve should be a bit more than 20.
5
(d) The rest of this problem concerns the behavior of the particle at the point P where it has its smallest y-coordinate. This point P is shown as a dot and labeled in the graph below.
y
P
x
Find i. (5 points) the time t when this occurs. Solution: The y-coordinate is y (t) = t2 − 4t. We want to minimize y (t). Since y 0 (t) = 2t − 4, there is one critical point at t = 2. Since y 00 (t) = 2 > 0, t = 2 is an absolute minimum for y. So y (t) is minimized when t = 2. ii. (3 points) the coordinates of the point P . Solution: ® r (2) = 22 + 2 , 22 − 4 (2) = h 6 , − 4 i so P = (6, − 4) .
iii. (3 points) the velocity vector V at P . Solution: V = v (2) = h 5 ,0 i iv. (2 points) the acceleration vector A at P . Solution: A = a (2) = h 2 ,2 i v. (2 points) the speed of the particle at P . Solution: The speed is kVk = 5. vi. (2 points) the tangential component aT of the acceleration vector A. Solution: Since A = a (2) = h 2 ,2 i and T = h 1 ,0 i, aT = A · T = 2. (You could have done this one correctly just by inspection.) vii. (2 points) the normal component aN of the acceleration vector A. Solution: There are many ways to do this. Here’s one way. q √ √ aN = kAk2 − |aT |2 = 8 − 4 = 4 = 2 6
7. (10 points Extra Credit) Only attempt this problem if you are satisfied with the rest of the exam. Extra credit problems are harder and partial credit is harder to earn! The figure on the left shows the circles whose polar equations are r = 1, r = 2 and r = 2 cos θ. In the figure on the right, only the part of this diagram in the first quadrant is shown and a region has been shaded. Find the area A of this shaded region. 2.0
y 1.5
1.0
x 0.5
0.0 0.0
0.5
1.0
1.5
2.0
π Solution: The curves r = 1 and r = 2 cos θ intersect when θ = . Let’s add the line 3 π θ = and change the colors within the region. 3 2.0
1.5
1.0
0.5
0.0 0.0
0.5
1.0
1.5
2.0
Now A = P urple + Y ellow where P urple is the area of the purple region and Y ellow is the area of the yellow region. We compute each separately and add the results. 1 P urple = 2
Zπ/3 Zπ/3 Zπ/3 ¢ ¡ 2 ¡ 2¢ 2 2 − (2 cos θ) dθ = 2 1 − cos θ dθ = 2 sin2 θ dθ 0
0
0
µ ¶¯π/3 Zπ/3 sin 2θ ¯¯ 1 1√ (1 − cos 2θ) dθ = θ − = 3 = π − ¯ 2 3 4 0 0
Also,
1 Y ellow = 2
Zπ/2 Zπ/2 ¡ 2 ¢ 1 3 1 dθ = π 2 − 12 dθ = 2 4
π/3
π/3
Adding these, we get 1 1 1√ 7 1√ A = P urple + Y ellow = π − 3+ π = π− 3. 3 4 4 12 4 7
