Midterm Exam I, Calculus III, Sample A

Midterm Exam I, Calculus III, Sample A

1. (10 points) Show that the 4 points P1 = (0, 0, 0), P2 = (2, 3, 0), P3 = (1, -1, 1), P4 = (1, 4, -1)

are coplanar (they lie on the same plane), and find the equation of the plane that contains

them.

---

---

Solution: u = P1P2 = 2, 3, 0 , v = P1P3 = 1, -1, 1 , w = 1, 4, -1 , and the scalar triple

product is equal to

2 u ? (v ? w) = 1

1

3 -1

4

0 1 -1

=2

-1 4

1 -1

-3

1 1

1 -1

+0

1 1

-1 4

= 0,

so the volume of the parallelepiped determined by u, v, and w is equal to 0. This means that these vectors are on the same plane. So, P1, P2, P3, and P4 are coplanar.

2. (10 points) Find the equation of the plane that is equidistant from the points A = (3, 2, 1)

and B = (-3, -2, -1) (that is, every point on the plane has the same distance from the two

given points).

Solution:

The

midpoint

of

the

points

A

and

B

is

the point --

C

=

1 2

3, 2, 1)+(-3, -2, -1)

=

(0, 0, 0). A normal vector to the plane is given by AB = 3, 2, 1 - -3, -2, -1 = 6, 4, 2 .

So, the equation of the plane is 6(x - 0) + 4(y - 0) + 2(z - 0) = 0, that is, 3x + 2y + z = 0.

3. (6 points) Find the vector projection of b onto a if a = 4, 2, 0 and b = 1, 1, 1 . Solution: Since |a|2 = 42 + 22 = 20, the vector projection of b onto a is is equal to

a ? b 4, 2, 0 ? 1, 1, 1 6

3

projab = |a|2 a =

20

= 4, 2, 0 = 2, 1, 0 .

20

5

4. (12 points) Consider the curve r(t) = 2 cos ti + sin tj + sin tk.

(a) (8 points) Find the unit tangent vector function T(t) and the unit normal vector function N(t).

(b) (4 points) Compute the curvature .

Solution:

(a) r (t) = - 2 sin t i + cos tj + cos tk and |r (t)| =

2 sin2 t + cos2 t + cos2 t =

2. So, the unit tangent vector T(t) is is equal to

r (t)

2

2

T(t) =

= - sin ti + cos tj + cos tk.

|r (t)|

2

2

Since T (t) = - cos ti -

2 2

sin

tj

-

2 2

sin tk

and

|T

(t)|

=

cos2

t

+

1 2

sin2

t

+

1 2

sin2

t

=

1,

the normal vector is equal to

2

2

N(t) = cos ti - sin tj - sin tk.

2

2

1

(b)

One

can

use

the

formula

(t) =

|T (t)| |r (t)|

and

|r (t)| = 2

form

part

(a)

and

|T (t)|

=1

form part (b) to get (t) = 1 =

2

2 2

.

One

can

also

use

the

formula

(t)

=

|r

(t)?r (t)| |r (t)|3

.

Since

r (t) = - 2 cos t i - sin t j - sin t k, |r (t)| = 2, and

i

j

k

r (t) ? r (t) = -2 sin t cos t cos t = - 2j + 2k = 0, - 2, 2 .

- 2 cos t - sin t - sin t

Then

|r (t) ? r

(t)| = 2

and

(t) =

2 ( 2)3

=

2 2

.

5. (10 points) Find the length of the curve with parametric equation:

r(t) = et, et sin t, et cos t ,

between the points (1, 0, 1) and (e2, 0, e2). Solution:

r (t) = et, et sin t + et cos t, et cos t - et sin t

and

|r (t)| = e2t + e2t(sin t + cos t)2 + e2t(sin t - cos t)2 = et 3.

Note that r(0) = 1, 0, 1 and r(2) = e2, 0, e2 . So, the length of the curve is equal to

2

2

length = |r (t)| dt =

3et dt = 3 e2 - 1).

0

0

6. (12 points) A spaceship is traveling with acceleration a(t) = et, t, sin 2t .

At t = 0, the spaceship was at the origin, r(0) = 0, 0, 0 , and had initial velocity v(0) = 1, 0, 0 . Find the position of the spaceship at t = .

Solution: The velocity is equal to

t

t

v(t) = v(0) + a(s) ds = 1, 0, 0 + es, s, sin 2s ds

0

0

t

t

t

= 1, 0, 0 + es ds, s ds, sin 2s ds

0

0

0

=

1, 0, 0

+

et - 1, t2 , 1 (1 - cos 2s)

=

et,

t2 ,

1 (1

-

cos 2t)

.

22

22

Since r(t) = r(0) +

t 0

v(s)

ds,

r(t) =

t

es,

s2 ,

1 (1 - cos 2s)

0

22

ds = =

t

es ds,

t s2

1

ds,

t

(1 - cos 2s) ds

0

02

20

et

-

1,

t3 ,

t

-

1

sin 2t

.

62 4

So, r() =

e

-

1,

3 6

,

2

.

2

7. (10 points) Write the equation of the tangent line to the curve with parametric equation

r(t) = t, 1, t4 ,

at the point (1, 1, 1). Solution: r (t) = 1 , 0, 4t3 . At (1, 1, 1), t = 1 and r (1) = 1/2, 0, 4 . Thus the

2t

parametric equations of the tangent line are

x = t/2 + 1, y = 1, z = 4t + 1.

8. (12 points) Using cylindrical coordinates, find the parametric equations of the curve that is the intersection of the cylinder x2 + y2 = 4 and the cone z = x2 + y2. (This problem refers to the material not covered before midterm 1.)

9. (6 points) Let f (x, y) = sin(x2 + y2) + arcsin(y2). Calculate:

2f .

xy

(This problem refers to the material not covered before midterm 1.)

10. (12 points) (This problem refers to the material not covered before midterm 1.) Show that the following limit does not exist:

7x2y(x - y)

lim

(x,y)(0,0)

x4 + y4

Justify your answer. (This problem refers to the material not covered before midterm 1.)

3

Midterm Exam I, Calculus III, Sample B

1. (6 Points) Find the center and radius of the following sphere x2 + y2 + z2 - 6x + 4z - 3 = 0. Completing the squares:

0 = x2 + y2 + z2 - 6x + 4z - 3 = (x2 - 6x + 9) + y2 + (z2 + 4z + 4) - 3 - 9 - 4 = (x - 3)2 + y2 + (z + 2)2 - 16.

So, the equation of the sphere is (x - 2)2 + y2 + (z - (-2))2 = 42, the center is (3, 0, -2) and radius 4.

2. (6 Points) Write each combination of vectors as a single vector

-- -- (a) AB + BC

- -- (b) AC - BC

-- -- -- (c) AD + DB + BA.

Solution: -- -- -

(a) AB + BC = AC,

- -- -- (b) AC - BC = AB,

(c)

-- AD

+

-- DB

+

-- BA

=

-0 .

3. (6 Points) Find the cosine of the angle between the vectors a = 1, 2, 3 and b = 2, 0, -1 . Solution: If is the angle between a and b, then

a ? b 1, 2, 3 ? 2, 0, -1

1

1

cos =

=

= - = - .

|a||b| 1, 2, 3 || 2, 0, -1 |

14 5

70

4. (6 Points) Find the vector projection of v onto u if u = 2i - k and v = 2i + 3j. Solution: The vector projection of v onto u is equal to

u ? v (2i - k) ? (2i + 3j)

4

projuv = |u|2 u =

|2i - k|2

2i - k = 2i - k . 5

4

5. (7 Points) Find the area the triangle with vertices P = (2, 1, 7), Q = (1, 1, 5), R = (2, -1, 1).

--

-

Solution: Since P Q = -1, 0, -2 and P R = 0, -2, -6 , and

-- -

i jk

P Q ? P R = -1 0 -2 = -4i - 6j + 2k,

0 -2 -6

the area of the triangle P QR is equal to

1 -- - 1 |P Q ? P R| =

(-4)2 + (-6)2 + 22 =

56 = 14.

2

2

2

6. (5 Points) Show that the line

x = 3 + t, y = 1 + 2t, z = 1 - 2t

is parallel to the plane

2x + 3y + 4z = 5.

Solution: A line is parallel to the plane if it is perpendicular to a normal vector to the plane. A normal vector to the plane is given by 2, 3, 4 and the direction of the line is given by the vector 1, 2, -2 . Compute the dot product of these vectors:

1, 2, -2 ? 2, 3, 4 = 2 + 6 - 8 = 0.

So, the line is parallel to the plane. 7. (6 Points) Find a vector parallel to the line of intersection for the two planes

x + 2y + 3z = 0 and x - 3y + 2z = 0.

Solution: A vector which gives the direction of the line of intersection of these planes is perpendicular to normal vectors to the planes. A norma vector to the first plane 1, 2, 3 and a normal vector to the second is is given by 1, -3, 2 . Then the vector

i jk 1, 2, 3 ? 1, -3, 2 = 1 2 3 = 13, 1, -5

1 -3 2

is parallel to the line of intersection of these planes. 8. (6 Points) Find cosine of the angle between intersection planes

2x + y + z = 0 and 3x - y + 2z = 0.

Solution: The angle between planes is equal to the angle between normal vectors of the

two planes. Since 2, 1, 1 is a normal vector to the plane and 3, -1, 2 is a normal vector to

the second plane,

cos = 2, 1, 1 ? 3, -1, 2 = 7

21 =.

| 2, 1, 1 || 3, -1, 2

6 14 6

5

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