12.1 Three-Dimensional Coordinate Systems

1. Points in space. First choose a fixed point O (the origin) and three directed lines throughO that are perpendicular to each other, called the coordinate axes and labeled the x-axis,y-axis, and z-axis.

Usually x- and y-axes are horizontal and the z-axis is vertical. The direction of the z-axis isdetermined by the right-hand rule .

Right-hand rule. If you curl the fingers of your right hand around the z-axis in the directionof a 90◦ counterclockwise rotation from the positive x-axis to the positive y-axis, then yourthumb points in the positive direction of the z-axis.

The three coordinate axes determine the three coordinate planes. The xy-plane is theplane that contains the x- and y-axes; the yz-plane contains the y- and z-axes; the xz-planecontains the x- and z-axes. These three coordinate planes divide space into eight parts,called octants. The first octant is determined by the positive axes.

2. Coordinate. Now if P is any point in space, let a be the (directed) distance from theyz-plane to P, let b be the distance from the xz-plane to P, and let c be the distance from thexy-plane to P. We represent the point by the ordered triple (a,b,c) of real numbers and wecall a, b, and c the coordinates of P; a is the x-coordinate, b is the y-coordinate, and c is thez-coordinate.

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Exercise 1. Sketch the points

(a) (0,5,2)

(b) (4,0,-1)

(c) (2,4,6)

(d) (1,-1,2)

3. Projection. The point P(a,b,c) determines a rectangular box. If we drop a perpendicularfrom P to the xy-plane, we get a point Q with coordinates (a,b,0) called the projection of Ponto the xy-plane. Similarly, R(0,b,c) and S(a,0,c) are the projections of P onto the yz-planeand xz-plane, respectively.

Exercise 2. Find the projections of P onto the xy-, yz-, and xz-plane.

(a) P(-4,3,-5)

(b) P(3,-2,-6)

4. Cartesian product. The Cartesian product R×R×R = {(x, y, z), x, y, z ∈ R} is the set ofall ordered triples of real numbers and is denoted by R3 . We have given a one-to-onecorrespondence between points P in space and ordered triples (a,b,c) in R3. It is called athree dimensional rectangular coordinate system. Notice that, in terms of coordinates, thefirst octant can be described as the set of points whose coordinates are all positive.

In three-dimensional analytic geometry, an equation in x, y, and z represents a surface inR3.

Exercise 3. What surface in R3 are represented by the following equations?

(a) x=3

(b) x = -2

(c) y = 3

(d) y = -2

(e) z = 3

(f) z = -2

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Exercise 4. What surface in R3 are represented by the following equations?

(a) x2 + y2 = 1 and z = 3(b) x2 + y2 = 1(c) y2 + z2 = 16 and x = 3(d) y2 + z2 = 16

Exercise 5. What surface in R3 are represented by the following equations?

(a) y = x(b) z = x

Exercise 6. What surface in R3 are represented by the following equations?

(a) x ≥−3(b) y < 8(c) z2 = 1

5. Distance Formula in Three Dimensions. The distance |P1P2|between the points P1(x1, y1, z1)and P2(x2, y2, z2) is

|P1P2| =√

(x1 −x2)2 + (y1 − y2)2 + (z1 − z2)2 (1)

Exercise 7. Find the distance between the following two points

(a) (2,-1,7) and (1,-3,5)

(b) (3,-2,-3) and (7,0,1)

6. Equation of a Sphere. An equation of a sphere with center C(h,k,l) and radius r is

(x −h)2 + (y −k)2 + (z − l )2 = r 2 (2)

An equation of a sphere with center C(0,0,0) and radius r is

x2 + y2 + z2 = r 2 (3)

Exercise 8. Show that x2 + y2 + z2 +4x −6y +2z +6 = 0 is the equation of a sphere and findits center and radius.

Exercise 9. Show that x2+ y2+z2−2x −4y +8z = 15 is the equation of a sphere and find itscenter and radius.

Exercise 10. Show that x2 + y2 + z2 +8x −6y +2z +17 = 0 is the equation of a sphere andfind its center and radius.

Exercise 11. Show that 2x2 +2y2 +2z2 = 8x −24z +1 is the equation of a sphere and findits center and radius.

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12.1 Problem Set

1. Given three points A(−4,0,−1), B(3,1,−5) and C (2,4,6)(a) Sketch A, B, and C on a single set of coordinate axes.

(b) What are the projections of A on the xy-, yz-, and xz-planes?

(c) What are the projections of B on the xy-, yz-, and xz-planes?

(d) What are the projections of C on the xy-, yz-, and xz-planes?

(e) Find the distance from each point to the xy-plane. Which of the points A, B, and C isclosest to the xy-plane?

(f) Find the distance from each point to the xz-plane. Which of the points A, B, and C isclosest to the xz-plane?

(g) Find the distance from each point to the yz-plane. Which of the points A, B, and C isclosest to the yz-plane?

(h) Find the distance between A and B.

(i) Find the distance between B and C.

(j) Find the distance between A and C.

(k) Three points A, B and C make a triangle. Is it a right triangle? Is it an isosceles triangle?

2. Determine whether the points lie on straight line.

(a) A(2,4,2), B(3,7,-2), C(1,3,3)

(b) A(0,-5,5), E(1,-2,4), F(3,4,2)

3. Find an equation of the sphere with the following given center and radius.

(a) Center (-3,2,5), radius 4.

(b) Center (2,-6,4), radius 5.

4. Show that the equation 3x2 +3y2 +3z2 = 10+6y +12z is a sphere and find its center andradius.

5. Describe the region of R3 by the equations or inequalities

(a) x=5

(b) 0 ≤ z ≤ 6(c) x2 + y2 + z2 = 4(d) x2 + y2 + z2 ≤ 4(e) x2 + y2 + z2 ≤ 4, x ≥ 0

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12.2 Vectors

1. Vector. The term vector indicates a quantity that has both magnitude and direction. Avector is often represented by an arrow or a directed line segment. The length of the arrowrepresents the magnitude of the vector and the arrow points in the direction of the vector.

Notation: v or~v

2. Suppose a particle moves along a line segment from point A to point B. The correspondingdisplacement vector v has initial point A (the tail) and terminal point B(the tip) and weindicate this by writing v = ~AB .Vector u = ~C D has the same length and the same direction as v even though it is in adifferent position. We say that u and v are equivalent (or equal) and we write u = v. Thezero vector, denoted by 0, has length 0. It is the only vector with no specific direction.

3. Combining Vectors. Suppose a particle moves from A to B, so its displacement vector is~AB . Then the particle changes direction and moves from B to C , with displacement vector~BC . The combined effect of these displacements is that the particle has moved from A to

C. The resulting displacement vector ~AC is called the sum of ~AB and ~BC and we write

~AC = ~AB + ~BC (4)

Definition of Vector Addition. If u and v are vectors positioned so the initial point of u isat the terminal point of v, then the sum u+v is the vector from the initial point of v to theterminal point of u.

Figure 1: Triangle Law

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4. Definition of Scalar Multiplication. If c is a scalar and v is a vector, then the scalar multi-ple cv is the vector whose length is c times the length of v and whose direction is the sameas v if c>0 and is opposite to v if c or v =< a,b,c > (6)

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6. Vector. Given the points A(x1, y2, z1) and B(x2, y2, z2), the vector v with representation ~ABis

v =< x2 −x1, y2 − y1, z2 − z1 > (7)

Exercise 12. Find the vector represented by the directed line segment with initial pointA(2,−3,4) and terminal point B(−2,1,1).

7. Length of Vector. The magnitude or length of the vector v is the length of any of itsrepresentations and is denoted by the symbol |v| or ||v||.The length of the two dimensional vector v =< x, y > is

|v| =√

x2 + y2 (8)

The length of the three dimensional vector v =< x, y, z > is

|v| =√

x2 + y2 + z2 (9)

8. Vector Algebra. If u =< a1, a2 > and v =< b1,b2 >, then

(a) u+v =< a1 +b1, a2 +b2 >(b) u−v =< a1 −b1, a2 −b2 >(c) cu =< ca1,ca2 >

If u =< a1, a2, a3 > and v =< b1,b2,b3 >, then

(a) u+v =< a1 +b1, a2 +b2, a3 +b3 >(b) u−v =< a1 −b1, a2 −b2, a3 −b3 >(c) cu =< ca1,ca2,ca3 >

Exercise 13. If a =< 4,0,3 > and b =, find(a) |a| and |b|(b) a+b(c) a−b(d) −3b(e) 2a+5b

9. Properties of Vectors. If a, b, and c are vectors and c and d are scalars, then

(a) a+b = b + a

(b) a + (b + c) = (a + b) + c

(c) a+0 = a

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(d) a+ (-a) = 0

(e) c(a+b) = ca + cb

(f) (c+d)a = ca + cb

(g) (cd)a = c(da)

(h) 1 a = a

10. Standard Basis Vector.

In two dimensions, i =< 1,0 > and j =< 0,1 > are the standard basis vectors.In three dimensions, i =< 1,0,0 > and j =< 0,1,0 > and k =< 0,0,1 > are the standard basisvectors.

If a =< a1, a2 >, then

a = a1 < 1,0 >+a2 < 0,1 >= a1i+a2j (10)

If a =< a1, a2, a3 >, then

a = a1 < 1,0,0 >+a2 < 0,1,0 >+a2 < 0,0,1 >= a1i+a2j+a3k (11)

Exercise 14. Express the following vectors in terms of i, j, and k

(a) a =

(b) b =

Exercise 15. If a = i+2j−3k and b = 4i+7k, express the vector 2a+3b in terms of i, j, and k11. Unit Vector. A unit vector is a vector whose length is 1.

i, j, and k are all unit vectors.

If a 6= 0, the the unit vector that has the same direction as a is

u = a|a| (12)

Exercise 16. Find the unit vector in the direction of the vector 2i− j - 2 k

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12.2 Problem Set

1. Find a vector a with representation given by the directed line segment ~AB . Draw ~AB andthe equivalent representation starting at the origin.

(a) A(-1,1), B(3,2)

(b) A(-4,-1),B(1,2)

(c) A(0,3,1),B(2,3,-1)

(d) A(4,0,-2),B(4,2,1)

2. Find a + b, 2a + 3 b, |a|, and |a - b|

(a) a = , b =

(b) a = i+2j -3k, b = -2 i - j + 5 k

(c) a = 4 i + j, b = i - 2 j

3. Find a unit vector that has the same direction as the given vector.

(a) -3 i + 7 j

(b)

(c)

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12.3 The Dot Product

1. Dot Product. If a =< a1, a2 > and b =< b1,b2 >, then the dot product of a and b is thenumber a · b given by

a ·b =< a1, a2 > · < b1,b2 >= a1b1 +a2b2 (13)

If a =< a1, a2, a3 > and b =< b1,b2,b3 >, then the dot product of a and b is the number a ·b given by

a ·b =< a1, a2, a3 > · < b1,b2,b3 >= a1b1 +a2b2 +a3b3 (14)

Thus, to find the dot product of a and b, we multiply corresponding components andadd. The result is not a vector. It is a real number, that is, a scalar. For this reason, the dotproduct is sometimes called the scalar product (or inner product).

Exercise 17. Find the following dot product

(a) · (b) · (c) ( i + 2 j - 3 k) · (2 j - k)

2. Properties of the Dot Product. If a, b, and c are vectors and c is a scalar, then

(a) a ·a = |a|2(b) a ·b = b ·a(c) a · (b+c) = a ·b+a ·c(d) (ca) ·b = c(a ·b) = a · c(b)(e) 0 ·a = 0

3. Angle Between Vectors. The dot product a ·b can be given a geometric interpretationin terms of the angle θ between a and b, which is defined to be the angle between therepresentations of a and b that start at the origin, where 0 ≤ θ ≤π . In other words, θ is theangle between the line segments ~O A and ~OB . Note that if a and b are parallel vectors, thenθ = 0 or θ =π.

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If θ is the angle between the vectors a and b, then

a ·b = |a||b|cosθ (15)

θ 0 π6π4

π3

π2

2π3

3π4

5π6 π

3π2 2π

sinθ 0 121p2

p3

2 1p

32

1p2

12 0 -1 0

cosθ 1p

32

1p2

12 0 -

12 -

1p2

-p

32 -1 0 1

Exercise 18. Find the angle between the vectors a =< 2,2,−1 > and b =< 5,−3,2 >

4. Perpendicular or Orthogonal. Two nonzero vectors a and b are called perpendicular ororthogonal if the angle between them is θ = π2 .Two nonzero vectors a and b are called perpendicular or orthogonal if and only if a ·b = 0.Exercise 19. Show that 2 i + 2 j - k is perpendicular to 5 i - 4 j + 2 k.

5. If two nonzero vectors a and b are in exactly the same direction, then θ = 0 and a·b = |a| |b|.6. If two nonzero vectors a and b are in exactly the opposite direction, then θ = π and

a ·b =−|a| |b|.7. Direction Angles, Direction Cosines. The direction angles of a nonzero vector a are the

angles α, β, and γ (in the interval[0,π]) that a makes with the positive x-, y-, and z-axes.

cosα, cosβ and cosγ are direction cosines of the vector a.

cosα= a · i|a| |i| =a1|a|

cosβ= a · j|a| |j| =a2|a|

cosγ= a ·k|a| |k| =a3|a| (16)

Some properties

cos2α+cos2β+cos2γ= 1 (17)a = |a| < cosα,cosβ,cosγ> (18)1

|a|a =< cosα,cosβ,cosγ> (19)

Exercise 20. Find the direction cosine of the vector a =< 1,2,3 >.

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dirangl.gifMedia File (image/gif)

8. Vector Projection. a and x are two vectors with the same initial point A. If C is the foot ofthe perpendicular from B to the line containting ~AD , then the vector with representation~AC is called the vector projection of a onto x and is denoted by projxa (you can think of it

as a shadow of a).

9. Scalar Projection. The scalar projection of a onto x is defined to be the signed magnitudeof the vector projection, denoted by compxa.

compxa = |a|cosθ (20)

where θ is the angle between a and x.

compxa is negative if π/2 < θ ≤π.

Scalar projection of b onto a: compab =a ·b|a| (21)

Vector projection of b onto a: projab = (a ·b|a| )

a

|a| =a ·b|a|2 a (22)

Exercise 21. Find the scalar projection and vector projection of b = onto a =

Projection in Physics. The constant force is a vector F = ~PR pointing in some direction. Ifthe force moves the object from P to Q, then the displacement vector is D = ~PQ. The workdone by this force is the product of the component of the force along D and the distancemoved:

W = (|Fcosθ|)|D| = F ·D (23)

Exercise 22. A wagon is pulled a distance of 100 m along a horizontal path by a constantforce of 70 N. The handle of the wagon is held at an angle of 35◦ above the horizontal. Findthe work done by the force.

Exercise 23. A force is given by a vector F = 3i+4j+5 k and moves a particle from the pointP(2,1,0) to the point Q(4,6,2). Find the work done.

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12.3 Problem Set

1. Find a ·b(a) a = , b =

(b) a = , b =

(c) a = 2 i+ j, b = i - j + k

(d) a = 3i + 2 j - k, b = 4 i + 5 k

(e) |a| = 6, |b| = 5, the angle between a and b is 2π/3.

2. Find the angle between the vector

(a) a = , b =

(b) a = , b =

(c) a = i + 2 j - 2 k, b = 4 i -3k

3. Determine whether the given vectors are orthogonal, parallel, or neither.

(a) a = , b =

(b) a = , b =

(c) a = -i + 2 j + 5 k, b = 3 i + 4 j - k

4. Find the value of x such that the angle between the vectors and is π/4.

5. Find the direction cosine of the vector

(a)

(b)

(c)

6. Find the scalar and vector projection of b onto a

(a) a = , b =

(b) a = , b =

7. Find the work done by the force F =< 8,−6,9 > that moves an object from the point (0,10,8)to the point (6,12,20) along a straight line. The distance is measured in meters and theforce in newtons.

8. A tow truck drags a stalled car along a road. The chain makes an angle of 30◦ with the roadand the tension in the chain is 1500 N. How much work is done by the truck in pulling thecar 1 km?

9. The following inequality is called Cauchy-Schwarz Inequality. Prove this inequality.

|a ·b| ≤ |a| |b| (24)

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12.4 Cross Product

1. Cross Product. If a=< a1, a2, a3 > and b = < b1,b2,b3 >, then the cross product of a and bis the vector

a×b =< a2b3 −a3b2, a3b1 −a1b3, a1b2 −a2b1 > (25)

Notice that the cross product of two vectors a and b, unlike the dot product, is a vector. Forthis reason it is also called the vector product. Note that is defined only when a and b arethree-dimensional vectors.

2. Determinant. A determinant of order 2 is defined by∣∣∣∣a bc d∣∣∣∣= ad −bc (26)

Exercise 24. Find the determinant of ∣∣∣∣ 2 1−6 4∣∣∣∣

A determinant of order 3 is defined by∣∣∣∣∣∣a1 a2 a3b1 b2 b3c1 c2 c3

∣∣∣∣∣∣= a1∣∣∣∣b2 b3c2 c3

∣∣∣∣−a2 ∣∣∣∣b1 b3c1 c3∣∣∣∣+a3 ∣∣∣∣b1 b2c1 c2

∣∣∣∣ (27)Exercise 25. Find the determinant of ∣∣∣∣∣∣

1 2 −13 0 1−5 4 2

∣∣∣∣∣∣3. Cross Product. If a =< a1, a2, a3 >= a1i+ a2j+ a3k and b =< b1,b2,b3 >= b1i+b2j+b3k,

then

a×b =∣∣∣∣∣∣

i j ka1 a2 a3b1 b2 b3

∣∣∣∣∣∣ (28)=

∣∣∣∣a2 a3b2 b3∣∣∣∣ i− ∣∣∣∣a1 a3b1 b3

∣∣∣∣ j+ ∣∣∣∣a1 a2b1 b2∣∣∣∣k (29)

Exercise 26. If a =< 1,3,4 > and b =< 2,7,−5 >, find a×b.Exercise 27. Show that a×a = 0 for any vector a.

4. The vector a×b is orthogonal to both a and b.

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(a×b) ·a = 0(a×b) ·b = 0

5. Right hand rule. If a and b are represented by directed line segments with the same initialpoint , then the cross product a×b points in a direction perpendicular to the plane througha and b . It turns out that the direction of a×b is given by the right-hand rule: If the fingersof your right hand curl in the direction of a rotation (through an angle less than 180◦ ) froma to b , then your thumb points in the direction of a×b.

If θ is the angle between a and b (so 0 ≤ θ ≤π), then

|a×b| = |a| |b|sinθ (30)

6. a×b is the vector that is perpendicular to both a and b, whose orientation is determinedby the right-hand rule, and whose length is |a| |b|sinθ.

7. Two nonzero vectors a and b are parallel if and only if

a×b = 0 (31)

8. The length of the cross product a×b is equal to the area of the parallelogram determinedby a and b.

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Exercise 28. Find a vector perpendicular to the plane that passes through the points P(1,4,6),Q(-2,5,-1) and R(1,-1,1).

Exercise 29. Find the area of the triangle with vertices P(1,4,6), Q(-2,5,-1) and R(1,-1,1) .

9. Cross Product of Standard Basis Vectors.

i× j = k j×k = i k× i = j (32)j× i =−k k× j =−i i×k =−j (33)

10. The cross product is not commutative. And the associative law for multiplication does notusually holds.

11. Laws of Algebra for Cross Product. If a, b and c are vectors and c is a scalar, then

(a) a×b =−b×a(b) (ca)×b = c(a×b) = a× (cb)(c) a× (b+c) = a×b+a×c(d) (a+b)×c = a×c+b×c(e) a · (b×c) = (a×b) ·c(f) a× (b×c) = (a ·c)b− (a ·b)c

12. Triple Products. a · (b×c) is called the scalar triple product of the vectors a, b and c. Thescalar triple product is a determinant

a · (b×c) =∣∣∣∣∣∣a1 a2 a3b1 b2 b3c1 c2 c3

∣∣∣∣∣∣ (34)The volume of the parallelepiped determined by the vectors a, b, and c is the magnitude oftheir scalar triple product:

V = |a · (b×c)| (35)

The area of the base parallelogram is A = |b×c|. If ψ is the angle between a and b×c, thenthe helght h of the parallelepiped is h = |a||cosψ|.

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Exercise 30. Use the scalar triple product to show that the vector a =< 1,4,−7 >, b =<2,−1,4 > and c =< 0,9,18 > are coplanar (volume = 0).

13. Vector Triple Product. a× (b×c) is called the vector triple product.14. Torque. If we consider a force F acting on a rigid body at a point given by a position vector

r. The torque τ (relative to the origin) is defined to be the cross product of the position andforce vectors

τ= r×F (36)

and measures the tendency of the body to rotate about the origin. The direction of thetorque vector indicates the axis of rotation. The direction of the torque vector indicatesthe axis of rotation.The magnitude of the torque vector is

|τ| = |r×F| = |r| |F|sinθ (37)

where θ is the angle between the position and force vectors.

The only component of F that can cause a rotation is the one perpendicular to r , that is,|F|sinθ . The magnitude of the torque is equal to the area of the parallelogram determinedby r and F.

Exercise 31. A bolt is tightened by applying a 40-N force to a 0.25-m wrench. Find themagnitude of the torque about the center of the bolt.

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12.4 Problem Set

1. Find the cross product a×b and verify that it is orthogonal to both a and b.(a) a =< 6,0,−2 > and b =< 0,8,0 >.(b) a = i+3j−2k and b =−i+5k.

2. If a = i−2k and b = j+k, find a×b. Sketch a, b and a×b as vectors starting at the origin.3. Using the properties of cross product, find

(a) (i× j)×k(b) (j−k)× (k− i)(c) (i+ j)× (i− j)

4. If a = , b = and c = , show that a× (b×c) 6= (a×b)×c.5. Find two unit vectors orthogonal to both and .

6. Find the area of the parallelogram with vertices A(-2,1), B(0,4), C(4,2) and D(2,-1).

7. Find the area of the parallelogram with vertices A(1,2,3), B(1,3,6), C(3,8,6), and D(3,7,3).

8. Find a nonzero vector orthogonal to the plane through the points P, Q, and R. Find the areaof triangle PQR.

(a) P(1,0,1), Q(-2,1,3), R(4,2,5)

(b) P(0,0,-3), Q(4,2,0), R(3,3,1)

9. Find the volume of the parallelepiped with adjacent edges PQ, PR, and PS.

(a) P(-2,1,0), Q(2,3,2), R(1,4,-1), S(3,6,1)

(b) P(3,0,1), Q(-1,2,5), R(5,1,-1), S(0,4,2)

10. Use the scalar triple product to verify that the vectors u = i+5j−2k, v = 3i− j and w =5i+9j−4k are coplanar.

11. Use the scalar triple product to determine whether the points A(1,3,2), B(3,-1,6), C(5,2,0),D(3,6,-4) lie in the same plane.

12. A bicycle pedal is pushed by a foot with a 60-N force. The shaft of the pedal is 18 cm long.Find the magnitude of the torque about the center.

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12.5 Equations of Lines and Planes

1. A line L in the xy-plane is determined when a point on the line and the direction of the line(its slope or angle of inclination) are given. The equation of the line can then be writtenusing the point-slope form.

A line L in three-dimensional space is determined when we know a point P0(x0, y0, z0) on Land the direction of L. In three dimensions the direction of a line is conveniently describedby a vector, so we let v be a vector parallel to L. Let P (x, y, z) be an arbitrary point on L andlet r0 and r be the position vectors of P0 and P (that is, they have representations ~OP0 and~OP ). If a is the vector with representation ~P0P , then the Triangle Law for vector addition

gives r = r0 +a . But, since a and v are parallel vectors, there is a scalar t such that a = tv.

Thus

r = r0 + tv (38)

which is a vector equation of L. Each value of the parameter t gives the position vector rof a point on L.

In other words, as t varies, the line is traced out by the tip of the vector r . Positive valuesof t correspond to points on L that lie on one side of P0, whereas negative values of tcorrespond to points that lie on the other side of P0.

If the vector v that gives the direction of the line is written in component form as v =<a,b,c > , then we have tv =< t a, tb, tc >. We can also write r =< x, y, z > and r0 =<x0, y0, z0 >, so the vector equation become

< x, y, z >=< x0 + t a, y0 + tb, z0 + tc > (39)

That is

x = x0 + t a y = y0 + tb z = z0 + tc (t ∈R) (40)

These equations are called parametric equations of the line through the point P0(x0,0 , z0)and parallel to the vector v = . Each value of the parameter t gives a point P(x,y,z)on L.

19

Exercise 32. Find a vector equation and parametric equations for the line that passesthrough the point (5,1,3) and is parallel to the vector i+4j−2k. Find two other points on theline.

2. In general, if a vector v = is used to describe the direction of a line L, then thenumbers a, b, and c are called direction numbers of L.

3. Symmetric Equations.

x −x0a

= y − y0b

= z − z0c

(a 6= 0,b 6= 0,c 6= 0) (41)

are called symmetric equations of L.

If a = 0, then the equations of L is

x = x0 y − y0b

= z − z0c

(42)

This means that L lies in the vertical plane x = x0.Exercise 33. Find parametric equations and symmetric equations of the line that passesthrough the points A(2,4,-3) and B(3,-1,1). At what point does this line intersect the xy-plane?

4. The vector equation of a line through the tip of the vector r0 in the direction of a vector v isr = r0 + tv.If the line also passes through the tip of r1, then the line segment from r0 to r1 is given bythe vector equation

r = (1− t )r0 + tr1 (43)

Exercise 34. Show that the lines L1 and L2 with parametric equations

x = 1+ t y =−2+3t z = 4− tx = 2s y = 3+ s z =−3+4s

are skew lines; that is, they do not intersect and are not parallel (and therefore do not line inthe same plane.)

5. Plane. A plane in space is determined by a point P0(x0, y0, z0) in the plane and a vectorn that is orthogonal to the plane. This orthogonal vector n is called a normal vector. LetP(x,y,z) be an arbitrary point in the plane, and let r0 and r be the position vectors of P0 andP. Then the vector r− r0 is represented by ~P0P . The normal vector n is orthogonal to everyvector in the given plane. In particular, n is orthogonal to r− r0 and so we have

n · (r− r0) = 0 (44)

which can be rewritten as

n · r = n · r0 (45)

20

Those two equations are called a vector equation of the plane.

To obtain a scalar equation for the plane, we write n = , r=, and r0 =<x0, y0, z0 > . Then the vector equation becomes

< a,b,c > · < x −x0, y − y0, z − z0 >= 0

or

a(x −x0)+b(y − y0)+ c(z − z0) = 0 (46)

This is the scalar equation of the plane through P0(x0, y0, z0) with normal vector n = .

Exercise 35. Find an equation of the plane through the point P(2,4,-1) with normal vectorn = . Find the intercepts and sketch the plane.

6. The linear equation is

ax +by + cz +d = 0 where d =−(ax0 +by0 + cz0) (47)

Exercise 36. Find an equation of the plane that passes through the points P(1,3,2), Q(3,-1,6)and R(5,2,0).

Exercise 37. Find the point at which the line with parametric equations x=2+3t, y=-4t,z=5+t intersects the plane 4x+5y-2z = 18

Exercise 38. Find the angle between the planes x+y+z = 1 and x-2y+3z=1.

Exercise 39. Find the symmetric equations for the line of intersection L of these two planesx+y+z = 1 and x-2y+3z=1.

7. Two planes are parallel if their normal vectors are parallel.

If two planes are not parallel, then they intersect in a straight line and the angle betweenthe two planes is defined as the acute angle between their normal vectors.

21

8. Distance from a point to the plane The distance D from a point P1(x1, y1, z1) to the planeax +by + cz +d = 0 is

D = |ax1 +by1 + cz1 +d |pa2 +b2 + c2

(48)

Exercise 40. Find the distance between the parallel planes 10x+2y-2z=5 and 5x+y-z=1.

Exercise 41. Find the distance between the following two lines

x = 1+ t y =−2+3t z = 4− tx = 2s y = 3+ s z =−3+4s

22

12.5 Problem Set

1. Find a vector equation and parametric equations for the line

(a) the line through the point (6,-5,2) and parallel to the vector < 1,3,−23 >.(b) the line through the point (0,-14,-10) and parallel to the line x =−1+2t , y = 6−3t ,

z = 3+9t .(c) the line through the point (1,0,6) and perpendicular to the plane x+3y+z = 5.

2. Find parametric equations and symmetric equations for the line

(a) the line through the origin and the point (4,3,-1).

(b) the line through the points (-8,1,4) and (3,-2,4).

(c) the line through (2,1,0) and perpendicular to both i+ j and j+k.(d) the line through (1,-1,1) and parallel to the line x +2 = 12 y = z −3(e) the line of intersection of the planes x +2y +3z = 1 and x − y + z = 1

3. Is the line through (-4,6,1) and (-2,0,-3) parallel to the line through (10,18,4) and (5,3,14) ?

4. Is the line through (2,4,0) and (1,1,1) perpendicular to the line through (2,3,4) and (3,-1,-8)?

5. Find symmetric equations for the line that passes through the point (1,-5,6) and is parallelto the vector . Find the points in which this required line intersects the coordinateplanes.

6. Find parametric equations for the line through (2,4,6) that is perpendicular to the planex-y+3z = 7. In what points does this line intersect the coordinate planes?

7. Find a vector equation for the line segment from (2,-1,4) to (4,6,1).

8. Find a parametric equation for the line segment from (10,3,1) to (5,6,-3).

9. Determine whether the lines are parallel, skew, or intersecting. If they intersect, find thepoint of intersection.

(a) L1 : x = 3+2t , y = 4− t , z = 1+3tL2 : x = 1+4s, y = 3−2s, z = 4+5s

(b) L1 : x = 5−12t , y = 3+9t , z = 1−3tL2 : x = 3+8s, y =−6s, z = 7+2s

(c) L1 :x−2

1 =y−3−2 = z−1−3

L2 :x−3

1 =y+4

3 = z−2−7(d) L1 :

x1 =

y−1−1 = z−23

L2 :x−2

2 =y−3−2 = z7

10. Find an equation of the plane

23

(a) The plane through the origin and perpendicular to the vector

(b) The plane through the point (5,3,5) and with normal vector 2 i + j - k

(c) The plane through the point (2,0,1) and perpendicular to the line x=3t, y=2-t, z=3+4t

(d) The plane through the point (2,4,6) and parallel to the plane z=x+y

(e) The plane that contains the line x=1+t, y=2-t, and z=4-3t is parallel to the plane5x+2y+z=1

(f) The plane through the points (3,-1,2), (8,2,4), and (-1,-2,-3)

(g) The plane that passes through the point (1,2,3) and contains the line x=3t, y=1+t,z=2-t.

(h) The plane that passes through the point (1,-1,1) and contains the line with symmetricequations x=2y=3z.

(i) The plane that passes through the points (0,-2,5) and (-1,3,1) and is perpendicular tothe plane 2z=5x+4y

11. Find the point at which the line intersects the given plane.

(a) x=3-t, y=2+t, z=5t; x-y+2z=9

(b) x-1+2t, y=4t, z=2-3t; x+2y-z+1=0

12. Find direction numbers for the line of intersection of the planes x+y+z=1 and x+z=0

13. Find the cosine of the angle between the planes x+y+z=0 and x+2y+3z=1

14. Determine whether the planes are parallel, perpendicular, or neither. If neither, find theangle between them.

(a) 2z=4y-x, 3x-12y+6z = 1

(b) x+y+z=1, x-y+z = 1

(c) x=4y-2z, 8y=1+2x+4z

15. Find symmetric equations for the line of intersection of the planes.

(a) 5x-2y-2z=1, 4x+y+z=6

(b) z=2x-y-5, z=4x+3y-5

16. Find the distance from the point to the given line

(a) (4,1,-2); x=1+t, y=3-2t, z=4-3t.

(b) (0,1,3); x=2t, y=6-2t, z=3+t

17. Find the distance from the point to the given plane

(a) (1,-2,4); 3x+2y+6z=5

(b) (-6,3,5); x-2y-4z=8

24

18. Find the distance between the given parallel planes.

(a) 2x-3y+z = 4, 4x-6y+2z = 3

(b) 6z = 4y-2x, 9z=1-3x+6y

25

12.6 Cylinders and Quadric Surfaces

1. Cylinders. A cylinder is a surface that consists of all lines (called rulings) that are parallelto a given line and pass through a given plane curve.

Exercise 42. Sketch the graph of the surface z = x2.Notice that the equation of the graph, z = x2 , does not involve y. This means that anyvertical plane with equation y=k (parallel to the xz-plane) intersects the graph in a curvewith equation z = x2. So these vertical traces are parabolas. Figure 1 shows how the graphis formed by taking the parabola z = x2 in the xz-plane and moving it in the direction ofthe y-axis. The graph is a surface, called a parabolic cylinder, made up of infinitely manyshifted copies of the same parabola. Here the rulings of the cylinder are parallel to the y-axis.

Exercise 43. Identify and sketch the surfaces.

(a) x2 + y2 = 1(b) y2 + z2 = 1

2. Quadric Surfaces. A quadric surface is the graph of a second-degree equation in threevariables x, y, and z. The most general such equation is

Ax2 +B y2 +C z2 +Dx y +E y z +F xz +Gx +H y + I z + J = 0 (49)

where A, B, C,...,J are constants. But by translation and rotation it can be brought into oneof the two standard forms,

Ax2 +B y2 +C z2 + J = 0 or Ax2 +B y2 + I z = 0 (50)

3. Summary

26

Exercise 44. Use traces to sketch the quadric surface with equation

x2 + y2

9+ z

2

4= 1

.

Note: It is called an ellipsoid because all of its traces are ellipses.

27

Exercise 45. Use traces to sketch the surface z = 4x2 + y2.Note: Because of the elliptical and parabolic traces, this quadric surface is called an ellipticparaboloid.

Exercise 46. Sketch the surface z = y2 −x2.Note: it is called hyperbolic paraboloid.

28

Exercise 47. Sketch the surface x2

4 + y2 − z2

4 = 1.

Exercise 48. Identify and sketch the surface 4x2 − y2 +2z2 +4 = 0.

Exercise 49. Identify and sketch the surface x2 +2z2 −6x − y +10 = 0

29

12.6 Problem Set

1. Describe and sketch the surface.

(a) x2 + z2 = 1(b) 4x2 + y2 = 4(c) z = 1− y2(d) y = z2(e) x y = 1

2. Use traces to sketch and identify the surface

(a) x = y2 +4z2(b) 9x2 − y2 + z2 = 0(c) x2 = y2 +4z2(d) 25x2 +4y2 + z2 = 100(e) 4x2 −16y2 + z2 = 16

3. Match the equation with its graph.

30

Vector Functions and Space Curves

1. Vector Function. A vector-valued function, or vector function, is simply a function whosedomain is a set of real numbers and whose range is a set of vectors. We are most interestedin vector functions r whose values are three-dimensional vectors.

r(t ) =< f (t ), g (t ),h(t ) >= f (t )i+ g (t )j+h(t )k (51)

where f (t ), g (t ) and h(t ) are the components of the vector r(t ). f, g, and h are real-valuedfunctions called the component functions of r.

The domain of r consists of all values of t for which the expression for r(t) is defined.

Exercise 50. If r(t ) =< t 3, ln(3− t ),pt >, find the component functions and domain of r.

2. Limit of Vector Function. If r(t ) =< f (t ), g (t ),h(t ) >, then

limt→a r(t ) =< limt→a f (t ), limt→a g (t ), limt→a h(t ) > (52)

provided the limits of the component functions exist.

Exercise 51. Fine limt→a r(t ) where r(t ) = (1+ t 3)i+ te−t j+ sin tt k

3. Continuity of Vector Function. A vector function r is continuous at a if

limt→a r(t ) = r(a) (53)

4. There is a close connection between continuous vector functions and space curves. Sup-pose that f, g, and h are continuous real-valued functions on an interval I. Then the set Cof all points (x,y,z) in space, where

x = f (t ) y = g (t ) z = h(t ) (54)

and t varies throughout the interval I, is called a space curve. The equations are calledparametric equations of C and t is called a parameter. We can think of C as being tracedout by a moving particle whose position at time t is (f(t), g(t), h(t)). If we now consider thevector function r(t) = , then r(t) is the position vector of the point P(f(t), g(t),h(t)) on C. Thus any continuous vector function r defines a space curve C that is traced outby the tip of the moving vector r(t).

Exercise 52. Describe the curve defined by the vector function.

r(t ) =< 1+ t ,2+5t ,−1+6t > (55)

Exercise 53. Sketch the curve whose vector equation is

r(t ) =< cos t , sin t , t >

31

Note: it is called a helix.

Exercise 54. Find a vector equation and parametric equations for the line segment thatjoins the point P(1,3,-2) to the point Q(2,-1,3)

Exercise 55. Find a vector function that represents the curve of intersection of the cylinderx2 + y2 = 1 and the plane y + z = 2.

32

13.1 Problem Set

1. Find the domain of the vector function.

(a) r(t ) =(b) r(t ) = t−2t+2 i+ sin t j+ ln(9− t 2)k

2. Find the limit .

(a) limt→0(e−3t i+ t 2

sin2 tj+cos(2t )k)

(b) limt→1( t 2−t

t−1 i+p

t +8j+ sinπtln t k)

3. Sketch the curve with the given vector equation. Indicate with an arrow the direction inwhich t increases.

(a) r(t ) =< sin t , t >(b) r(t ) =< t 3, t 2 >(c) r(t ) =< t ,2− t ,2t >(d) r(t ) =< sinπt , t ,cosπt >(e) r(t ) =< t 2, t ,2 >

4. Find a vector equation and parametric equations for the line segment that joins P to Q.

(a) P(2,0,0), Q(6,2,-2)

(b) P(-1,2,-2), Q(-3,5,1)

(c) P(a,b,c), Q(u,v,w)

33

13.2 Derivatives and Integrals of Vector Functions

1. The derivative of a vector function r is defined in much the same way as for real-valuedfunctions:

dr

d t= r′(t ) = lim

h→0r(t +h)− r(t )

h(56)

if this limit exists.

If the points P and Q have position vectors r(t) and r(t+h), then ~PQ represents the vectorr(t+h)−r(t ), which can therefore be regarded as a secant vector. If h > 0, the scalar multiple1h (r(t +h)− r(t)) has the same direction as r(t +h)− r(t) . As h → 0, it appears that thisvector approaches a vector that lies on the tangent line. For this reason, the vector iscalled the tangent vector to the curve defined by r at the point P, provided that r′(t ) existsand r′(t) = 0. The tangent line to C at P is defined to be the line through P parallel to thetangent vector r′(t ) . We will also have occasion to consider the unit tangent vector, whichis

T(t ) = r′(t )

|r′(t )| (57)

2. Derivative of Vector Functions. If r(t) =< f (t), g (t),h(t) >= f (t)i+ g (t)j+h(t)k, where f,g and h are differentiable functions, then

r′(t ) =< f ′(t ), g ′(t ),h′(t ) >= f ′(t )i+ g ′(t )j+h′(t )k (58)

Exercise 56. Find the derivative of r(t ) =< 1+ t 3, t3−t , sin2t >. Find the unit tangent vectorat the point where t=0.

Exercise 57. For the curve r(t ) =, find r′(t ) and sketch the position vector r(1)and the tangent vector r′(1).

Exercise 58. Find parametric equations for the tangent line to the helix with parametric

34

equations

x = 2cos t y = sin t z = t

at the point (0,1, π2 ).

3. Second Derivative. The second derivative of r(t ) is the derivative of r′, that is, r′′ = (r′)′.4. Suppose u and v are differentiable vector functions, c is a scalar, and f is a real-valued

function. Then

(a) dd t [u(t )+v(t )] = u′(t )+v′(t )(b) dd t [cu(t )] = cu′(t )(c) dd t [ f (t )u(t )] = f ′(t )u(t )+ f (t )u′(t )(d) dd t [u(t ) ·v(t )] = u′(t ) ·v(t )+u(t ) ·v′(t )(e) dd t [u(t )×v(t )] = u′(t )×v(t )+u(t )×v′(t )(f) dd t [u( f (t ))] = f ′(t )u′( f (t ))

Exercise 59. Show that if |r(t )| = c (a constant), then r′(t ) is orthogonal to r(t ) for all t.

5. Integral. ∫ ba

r(t )d t =<∫ b

af (t )d t ,

∫ ba

g (t )d t ,∫ b

ah(t )d t > (59)

Exercise 60. If r(t ) =< 2cos t , sin t ,2t >, find ∫ r(t )d t and ∫ π/20 r(t )d t

35

13.2 Problem Set

1. Sketch the plan curve with the given vector equation. Find r′(t ). Sketch the position vectorr(t ) and the tangent vector r′(t ) for the given value of t.

(a) r(t ) =< t −2, t 2 +1 >, t =−1.(b) r(t ) =< t 2, t 3 >, t = 1

2. Find the derivative of the vector function.

(a) r(t ) =< t sin t 2, t cos2t >(b) r(t ) =< tan t , sec t , 1

t 2>

(c) r(t ) = t i+ j+2ptk(d) r(t ) = a+ tb+ t 2c.(e) r(t ) = ta× (b+ tc)

3. Find the unit tangent vector T(t ) at the point with the given value of the parameter t.

(a) r(t ) =< te−t ,2arctan t ,2e t >, t = 0.(b) r(t ) =< t 3 +3t , t 2 +1,3t +4 >, t = 1

4. If r(t ) =< t , t 2, t 3 >, find r′(t ),T(1),r′′(t ), and r′(t )× r′′(t )5. If r(t ) =< e2t ,e−2t , te2t >, find r′(t ),T(0),r′′(t ),r′′(0), and r′(t ) · r′′(t ).6. Find a vector equation for the tangent line to the curve of intersection of the cylinders

x2 + y2 = 25 and y2 + z2 = 20 at the point (3,4,2).7. Find parametric equations for the tangent line to the curve with the given parametric

equations at the specified point.

(a) x = t , y = e−t , z = 2t − t 2, (0,1,0)(b) x = t cos t , y = t , z = t sin t , (−π,π,0).

8. Evaluate the integral.

(a)∫ 2

0 (t i− t 3j+3t 5k)d t(b)

∫ 10 (

41+t 2 j+ 2t1+t 2 k)d t

9. Find r(t ) if r′(t ) =< 2t ,3t 2,pt > and r(1) =< 1,1,0 >.10. Find r(t ) if r′(t ) =< t ,e t , te t > and r(0) =< 1,1,1 >.11. If u(t ) =< sin t ,cos t , t > and v(t ) =< t ,cos t , sin t >, find

d

d t[u(t ) ·v(t )] and d

d t[u(t )×v(t )]

36

13.3 Arc Length and Curvature

1. Length of a Plane Curve The length of a plane curve with parametric equations x = f (t ),y = g (t ), a ≤ t ≤ b, f ′ and g ′ are continuous, is

L =∫ b

a

√[ f ′(t )]2 + [g ′(t )]2d t =

∫ ba

√[d x

d t]2 + [d y

d t]2d t (60)

Suppose that the curve has the vector equation r(t) =< f (t), g (t),h(t) >, a ≤ t ≤ b, or,equivalently, the parametric equations x = f (t), y = g (t), z = h(t), where f’, g’, and h’ arecontinuous.

If the curve is traversed exactly once as t increases from a to b, then it can be shown that itslength is

L =∫ b

a

√[ f ′(t )]2 + [g ′(t )]2 + [h′(t )]2d t =

∫ ba

√[d x

d t]2 + [d y

d t]2 + [d z

d t]2d t (61)

In conclusion, the length is

L =∫ b

a|r′(t )|d t (62)

Exercise 61. Find the length of the arc of the circular helix with vector equation r(t) =cos t i+ sin t j+ tk from the point (1,0,0) to the point (1,0,2π)

2. arc length function. Suppose that C is a curve given by a vector function

r(t ) =< f (t ), g (t ),h(t ) >, a ≤ t ≤ b (63)

where r′ is continuous and C is traversed exactly once as t increases from a to b. Its arc

37

length function s is

s(t ) =∫ t

a|r′(u)|du (64)

=∫ t

a

√[ f ′(u)]2 + [g ′(u)]2 + [h′(u)]2d t =

∫ ta

√[

d x

du]2 + [ d y

du]2 + [ d z

du]2du (65)

where s(t) is the length of the part of C between r(a) and r(t ).

By FTC,

d s(t )

d t= |r′(t )|

3. It is often useful to parametrize a curve with respect to arc length because arc lengtharises naturally from the shape of the curve and does not depend on a particular coordinatesystem. If a curve r(t) is already given in terms of a parameter t and s(t) is the arc lengthfunction,then we may be able to solve for t as a function of s: t = t(s). Then the curve canbe reparametrized in terms of s by substituting for t: r = r(t (s)) .Exercise 62. Reparametrize the helix r(t) = cos t i+ sin t j+ tk with respect to arc lengthmeasured from from the point (1,0,0) in the direction of increasing t.

4. Smooth. A parametrization r(t) is called smooth on an interval I if r′ is continuous andr′(t ) 6= 0 on I. A curve is called smooth if it has a smooth parametrization. A smooth curvehas no sharp corners or cusps; when the tangent vector turns, it does so continuously.

If C is a smooth curve defined by the vector function r, recall that the unit tangent vector isgiven by

T(t ) = r′(t )

|r′(t )|and indicates the direction of the curve. T(t ) changes direction very slowly when C is fairlystraight, but it changes direction more quickly when C bends or twists more sharply.

38

5. Curvature. The curvature of at a given point is a measure of how quickly the curve changesdirection at that point. Specifically, we define it to be the magnitude of the rate of changeof the unit tangent vector with respect to arc length.

The curvature of a curve is

κ= |dTd s

| = |dT/d td s/d t

| = |T′(t )|

|r′(t )| (66)

= |r′(t )× r′′(t )||r′(t )|3 (67)

where T is the unit tangent vector.

Exercise 63. Show that the curvature of a circle of radius a is 1a

Exercise 64. Find the curvature of the twisted cubic r(t ) =< t , t 2, t 3 > at a general point andat (0,0,0).

6. For the special case of a plane curve with equation y = f (x), we choose x as the parameterand write r(x) = xi+ f (x)j. Then r′(x) = i+ f ′(x)j and r′′(x) = f ′′(x)j. Since i× j = k andj× j = 0, it follows that r′(x)× r′′(x) = f ′′(x)k. We also have |r′(x)| =√1+ [ f ′(x)]2 and so

κ(x) = | f′′(x)|

[1+ ( f ′(x))2]3/2 (68)

Exercise 65. Find the curvature of the parabola y = x2 at the points (0,0), (1,1) and (2,4).

7. Normal Vector and Binormal Vectors. At a given point on a smooth space curve r(t) ,there are many vectors that are orthogonal to the unit tangent vector T(t) . We singleout one by observing that, because |T(t)| = 1 for all t, we have T(t) ·T′(t) = 0, so T′(t) isorthogonal to T(t) . At any point where κ 6= 0, we can define the principal unit normalvector (or simply unit normal) as

N(t) = T′(t )

|T′(t )| (69)

The vector B(t) = T(t)×N(t) is called the binormal vector. It is perpendicular to both Tand N and is also a unit vector.

39

Exercise 66. Find the unit normal and binormal vectors for the circular helix r(t ) = cos t i+sin t j+ tk.

8. The plane determined by the normal and binormal vectors N and B at a point P on a curveC is called the normal plane of C at P. It consists of all lines that are orthogonal to thetangent vector T. The plane determined by the vectors N and T is called the osculatingplane of C at P. The name comes from the Latin osculum, meaning "kiss." It is the plane thatcomes closest to containing the part of the curve near P. (For a plane curve, the osculatingplane is simply the plane that contains the curve.)

The circle that lies in the osculating plane of C at P, has the same tangent as C at P, lieson the concave side of C (toward which N points), and has radius ρ = 1κ (the reciprocal ofthe curvature) is called the osculating circle (or the circle of curvature) of C at P. It is thecircle that best describes how C behaves near P; it shares the same tangent, normal, andcurvature at P.

Exercise 67. Find the equations of the normal plane and osculating plane of the helixr(t ) = cos t i+ sin t j+ tk at the point P (0,1, π2 ).

Exercise 68. Find and graph the osculating circle of the parabola y = x2 at the origin.

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13.3 Problem Set

1. Find the length of the curve.

(a) r(t ) =< t ,3cos t ,3sin t >, −5 ≤ t ≤ 5.(b) r(t ) =< 2t , t 2, 13 t 3 >, 0 ≤ t ≤ 1.

2. Reparametrize the curve with respect to arc length measured from the point where t=0 inthe direction of increasing t.

(a) r(t ) =< 2t ,1−3t ,5+4t >(b) r(t ) =< e2t cos2t ,2,e2t sin2t >.

3. Find the unit tangent and unit normal vectors T(t ) and N(t ). Find the curvature.

(a) r(t ) =< t ,3cos t ,3sin t >(b) r(t ) =< t 2, sin t − t cos t ,cos t + t sin t >, t>0(c) r(t ) =< t , 12 t 2, t 2 >

4. Find the curvature

(a) r(t ) =< t 3, t 2 >(b) r(t ) =< t , t 2,e t >

5. Find the curvature of r(t ) =< t 2, ln t , t ln t > at (1,0,0).6. Find the curvature of r(t ) =< t , t 2, t 3 > at (1,1,1).7. Find the curvature.

(a) y = x4(b) y = xex

8. Find the curvature.

(a) x = t 2, y = t 3(b) x = e t cos t and y = e t sin t .

9. Find T, N and B at the given point.

(a) r(t ) =< t 2, 23 t 3, t >, (1, 23 ,1)(b) r(t ) =< cos t , sin t , lncos t >, (1,0,0)

10. Find equations of the normal plane and osculating plane of the curve at the given point.

(a) x = 2sin3t , y = t , z = 2cos3t , (0,π,−2)(b) x = t , y = t 2, z = t 3, (1,1,1)

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13.4 Motion in Space: Velocity and Acceleration

1. Velocity Vector. Suppose a particle moves through space so that its position vector at timet is r(t ). For small values of h, the vector

r(t +h)− r(t )h

(70)

approximates the direction of the particle moving along the curve r(t). Its magnitudemeasures the size of the displacement vector per unit time.

The vector above gives the average velocity over a time interval of length h and its limit isthe velocity vector v(t ) at time t:

v(t ) = limh→0

r(t +h)− r(t )h

= r′(t ) (71)

Thus the velocity vector is also the tangent vector and points in the direction of the tangentline.

2. Speed. The speed of the particle at time is the magnitude of the velocity vector,that is

|v(t )| = |r′(t )| = d sd t

= rate of change of distance with respect time

3. Acceleration. As in the case of one-dimensional motion, the acceleration of the particleis defined as the derivative of the velocity

a(t ) = v′(t ) = r′′(t ) (72)

Exercise 69. The position vector of an object moving in a plane is given by r(t) = t 3i+ t 2j.Find its velocity, speed, and acceleration when t=1 and illustrate geometrically.

Exercise 70. Find the velocity, acceleration, and speed of a particle with position vectorr(t ) =< t 2,e t , te t >.Exercise 71. A moving particle starts at an initial position r(0) =< 1,0,0 > with initialvelocity v(0) =< 1,−1,1 >. Its acceleration is a(t ) =< 4t ,6t ,1 >. Find its velocity and positionat time t.

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4. In general, vector integrals allow us to recover velocity when acceleration is known andposition when velocity is known:

v(t ) = v(t0)+∫ t

t0a(u)du r(t ) = r(t0)+

∫ tt0

v(u)du (73)

5. Newton’s Second Law of Motion. If the force that acts on a particle is known, then theacceleration can be found from Newton’s Second Law of Motion. The vector version ofthis law states that if, at any time t, a force F(t ) acts on an object of mass m producing anacceleration a(t), then

F(t ) = ma(t ) (74)

Exercise 72. An object with mass that moves in a circular path with constant angular ωspeed has position vector r(t ) = a cosωt i+a sinωt j. Find the force acting on the object andshow that it is directed toward the origin.

Exercise 73. A projectile is fired with angle of elevation α and initial velocity vo . Assumingthat air resistance is negligible and the only external force is due to gravity, find the positionfunction r(t ) of the projectile. What value of α maximizes the range (the horizontal distancetraveled)?

Exercise 74. A projectile is fired with muzzle speed 150m/s and angle of elevation 45◦ froma position 10 m above ground level. Where does the projectile hit the ground, and with whatspeed?

6. Tangential and Normal Components of Acceleration. When we study the motion of aparticle, it is often useful to resolve the acceleration into two components, one in thedirection of the tangent and the other in the direction of the normal. If u = |v| is the speedof particle, then

T(t ) = r′(t )

|r′(t )| =v(t )

|v′(t )| =v

u

⇒ v = uTa = v′ = (uT)′ = u′T+uT′

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Also

κ= |T|′

|r′| =|T′|u

⇒|T′| = κu

and

N = T′

|T′|Then

⇒ T′ = |T′|N = κuN

and

a = u′T+κu2N (75)

Write aT and aN for the tangential and normal components of acceleration, then

a = aT T+aN N (76)where aT = u′ and aN = κu2 (77)

7. The first thing to notice is that the binormal vector B is absent.

No matter how an object moves through space, its acceleration always lies in the plane ofT and N (the osculating plane). (Recall that T gives the direction of motion and N points inthe direction the curve is turning.)

Next we notice that the tangential component of acceleration is u′, the rate of change ofspeed, and the normal component of acceleration is κu2, the curvature times the squareof the speed.

This makes sense if we think of a passenger in a car — a sharp turn in a road means a largevalue of the curvature , so the component of the acceleration perpendicular to the motionis large and the passenger is thrown against a car door. High speed around the turn has thesame effect; in fact, if you double your speed, aN is increased by a factor of 4.

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8.

v ·a = uT · (u′T+κu2N)= uu′

as T ·T = 1 and T ·N = 0, therefore

aT = u′ = v ·au

= r′(t ) · r′′(t )|r′(t )| (78)

aN = κu2 = |r′(t )× r′′(t )||r′(t )|3 |r

′(t )|2 = |r′(t )× r′′(t )||r′(t )| (79)

Exercise 75. A particle moves with position function r(t ) =< t 2, t 2, t 3 > . Find the tangentialand normal components of acceleration.

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13.4 Problem Set

1. The figure shows the path of a particle that moves with position vector r(t) at time t.

(a) Draw a vector that represents the average velocity of the particle over the time interval2 ≤ t ≤ 2.5.

(b) Draw a vector that represents the average velocity of the particle over the time interval1.5 ≤ t ≤ 2.

(c) Draw an approximation to the velocity vector v(2)

2. Find the velocity, acceleration, and speed of a particle with the given position function.

(a) r(t ) =, t = 2(b) r(t ) =< 3cos t ,2sin t >, t =π/3(c) r(t ) =< e t e2t >, t = 0(d) r(t ) =< t , t 2,2 >, t = 1

3. Find the velocity, acceleration, and speed of a particle with the given position function.

(a) r(t ) =< t 2 + t , t 2 − t , t 3 >(b) r(t ) =< 2cos t ,3t ,2sin t >

4. Find the velocity and position vectors of a particle that has the given acceleration and thegiven initial velocity and position.

(a) a(t ) =< 1,2,0 >, v(0) =< 0,0,1 >, r(0) =< 1,0,0 >(b) a(t ) =< 2,6t ,12t 2 >, v(0) =< 1,0,0 >, r(0) =< 0,1,−1 >

5. Find the tangential and normal components of the acceleration vector.

(a) r(t ) =< 1+ t , t 2 −2t >(b) r(t ) =< t , t 2,3t >

6. What force is required so that a particle of mass m has the position function r(t) =<t 3, t 2, t 3 >?

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7. A force with magnitude 20 N acts directly upward from the xy-plane on an object withmass 4 kg. The object starts at the origin with initial velocity v(0) = i− j. Find its positionfunction and its speed at time t.

8. A projectile is fired with an initial speed of 200 m/s and angle of elevation 60c i r c. Find (a)the range of the projectile, (b) the maximum height reached.

9. A gun is fired with angle of elevation 30◦. What is the muzzle speed if the maximum heightof the shell is 500 m?

10. A gun has muzzle speed 150 m/s. Find two angles of elevation that can be used to hit atarget 800 m away.

11. A ball is thrown eastward into the air from the origin (in the direction of the positive x-axis).The initial velocity is v(0) =< 50,0,80 >, with speed measured in feet per second. Thespin of the ball results in a southward acceleration of 4 ft/s2, so the acceleration vector isa =< 0,−4,−32 >. Where does the ball land and with what speed?

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