Mat 267 Engineering Calculus IIIUpdated on 01/19/2011 Dr. Firoz

Chapter 10 Vectors and the Geometry of Space

Section 10.1 Three-Dimensional Coordinate Systems We will be using the right hand rule. Please review from your textbook at page number

517.

Suppose 1 1 1 1( , , )P x y z and 2 2 2 2( , , )P x y z are the given points. The distance D is defined as

2 2 2

1 2 1 2 1 2( ) ( ) ( )D x x y y z z

Important equations in 3-D to remember:

1. ax by cz d represents a plane (compare with straight line in 2 D as ax by c )

2. x a is a surface parallel to yz-plane (compare with a vertical straight line or a straight

line parallel to y axis in 2 D as x a )

3. y b is a surface parallel to zx-plane (compare with a horizontal straight line or a

straight line parallel to x axis in 2 D as y b )

4. z c is a surface parallel to xy-plane

5. y x is a vertical plane that intersects xy-plane in the line y x (compare with a

straight line through the origin in 2 D as y x )

6. 2 2 2 2( ) ( ) ( )x a y b z c d is a sphere center at ( , , )a b c and radius d (compare

with a circle center at (a, b) and radius c in 2 D as 2 2 2( ) ( )x a y b c )

7. 2 2 2 2( ) ( ) ( )x a y b z c d , z c is a hemisphere center at ( , , )a b c

Examples:

1. Find the equation of a sphere center at (1,2, 1) and radius 1.

Solution: 2 2 2( 1) ( 2) ( 1) 1x y z

2. Determine whether the points lie on a straight line

a) (5,1,3), (7,9 1), (1, 15,11)A B C

Solution: Check that 2 21, 6 21, 4 21AB BC AC and AB AC BC . The

points are on a line.

b) (0,3, 4), (1,2, 2), (3,0,1)K L C .

Check that the points are not on the same line.

3. Find the center and radius of the sphere given by 2 2 2 6 4 2 11x y z x y z

Mat 267 Engineering Calculus IIIUpdated on 01/19/2011 Dr. Firoz

Solution: Complete the square as 2 2 2 2( 3) ( 2) ( 1) 5x y z and then center is at

(3, 2,1) and radius 5.

4. Describe in words the region of 3 represented by the equations or inequalities

a) 5y b) 5x c) 4x d) 0y

e) 0 6z f) y z g) 2 2 2 1x y z h) 0xyz

i) 2 2 2 2 3x y z z j) 2 2 1x y k) 2 2 9x z

5. Find the volume of the solid that lies inside both of the spheres 2 2 2 4 2 4 5 0x y z x y z and 2 2 2 4 0x y z

6. Find the lengths of the medians of the triangle with vertices (1,2,3), ( 2,0,5)A B

and (4,1,5)C

Section 10.2 Vectors

Parallelogram law: If we place two vectors ,u v so that they start at a same point, then

u v lies along the diagonal of the parallelogram with ,u v vectors as sides.

For two vectors , , and , ,u a b c v x y z the vector represented and defined by

, ,a AB x a y b z c and , ,a BA a x b y c z

The length or magnitude of a vector: 2 2 2( ) ( ) ( )a x a y b z c

Examples:

1. Given 3, 1 , 5,3a b . Find , 2 , 3a b a b a b and

, 2 , 3a b a b a b

2. Find a vector that has same direction as the vector 2,4,5 and magnitude 6.

Solution: find unit vector 1

2,4,53 5

u , the vector we are looking for is

26 2,4,5

5w u

Section 10.3 The Dot Product of Vectors

For two vectors , , , , ,u a b c v x y z the dot product is defined as

u v ax by cz

If is the angle between the vectors ,u v , then cosu v u v

Case 1. 0 vectors are parallel in the same direction and u v u v

Case 2. vectors are parallel in the opposite direction and u v u v

Mat 267 Engineering Calculus IIIUpdated on 01/19/2011 Dr. Firoz

Case 3. 2

vectors are perpendicular (Orthogonal) and 0u v

Direction angles and direction cosine: If , , are the angles of a vector with the

coordinate axes x, y and z respectively then those are called the direction angles. And

cos , cos , cos are direction cosines and 2 2 2cos cos cos 1

Question: Show that 2 2 2cos cos cos 1

Projections: Prove that scalar projection of b onto a is a

a bcomp b

a and

Vector projection of b onto a is a

a b aproj b

a a

(See your text book page # 534)

Examples:

1. Find the dot product between two given vectors.

a) 1

,4 and 8, 32

u v b) 4 3 and 2 3 4u i k v i j k

2. Given that 04,and 10, 120u v find u v

3. Find the angle between the vectors 1

,4 and 8, 32

u v

4. Show that 2,6, 4 and 3, 9,6u v are parallel

5. Find a unit vector that is orthogonal to both andi j i k

Solution: Suppose , ,a a b c is the unit vector. Now 2 2 2 1a b c and

, , 1,1,0 0a b c and , , 1,0,1 0a b c . Solve for a, b, and c for the unit

vector 1

, , 1, 1, 13

a a b c .

6. Find direction cosines and direction angles of the vector 2,3, 6a

Solution: 2 2

cos arccos(2 / 7)74 9 36

,

3 3cos arccos(3/ 7)

74 9 36

and

6 6cos arccos( 6 / 7)

74 9 36

7. If two direction angles are given / 4, / 3 , find .

8. Determine the scalar and vector projection of 4,1b onto 1,2a

Mat 267 Engineering Calculus IIIUpdated on 01/19/2011 Dr. Firoz

9. A crate is hauled 8 cm up a ramp under a constant force of 300 N applied at an

angle of 30 degrees to the ramp. Find the work done.

Solution: Work done W = 0cos 300(8)cos30 2078.46F D F D Nm J

Nm stands for Newton-meter and 1 Nm = 1 Joules, which is the unit of work.

10. A force is given by a vector F = <3, 4, 5> and moves a particle from the point

P(2,1, 0) to the point Q(3, 2, 1). Find the work done.

Solution: 3,4,5 1,1,1F D units = 12 units of work, where D is the distance

vector from P to Q.

11. A woman exerts a horizontal force of 25 lb on a crate as she pushes it up a ramp

that is 20 ft long and inclined at an angle of 15 degrees above the horizontal. Find

the work done on the box.

Answer: About 483 ft-lb

12. A clothesline is tied between two poles 8m apart. The line is quite taut and has

negligible sag. When a wet shirt wet shirt with a mass of 0.8 kg is hung at the

middle of the line, the midpoint is pulled down 8 cm. Find the tension in each half

of the clothesline.

Solution. Draw a picture of the problem. Let 1 2,T T be the tensions having equal

vertical components (b) and opposite horizontal components (a). We then have the

following

1 2 1 2, 2T ai bj T ai bj T T bj

From your picture find that 50a b . And ( ) 0.8(9.8) 7.81W mg j j . From

equilibrium vectors we have 1 2 0 3.92T T W b and 50 196a b . Now write

the vectors as tension

Section 10.4 The Cross Product of Vectors

For two vectors , , , , ,u a b c v d e f the cross product is defined as

( ) ( ) ( )

i j k

u v a b c i bf ec j af dc k ae bd

d e f

You need to review determinants from any algebra book.

Theorem: The vector u v is orthogonal to both u and v

Theorem: If is the angle between two vectors then sin , 0u v u v

Corollary: Two nonzero vectors u and v are parallel if 0u v

Mat 267 Engineering Calculus IIIUpdated on 01/19/2011 Dr. Firoz

Note: The length of the cross product u v is equal to the area of the parallelogram

determined by the vectors.

Scalar Triple Product (STP): For three vectors u , v and w the scalar triple product is

defined as ( )u v w

Volume of a parallelepiped is given by ( )V u v w , which is the magnitude of a scalar

triple product.

Note: If ( ) 0u v w the vectors are coplanar

Properties:

1. ( ) ( ) ( )a b c c a b b c a

2. u v v u

3. 0i i , i j k , k i j and so on

4. ( ) ( ) ) ( )a b c a c b a b c

Examples:

1. Show that 1,4, 7 , 2, 1,4a b and 0, 9,18c are coplanar

Solution: One needs to verify that ( ) 0a b c

2. Find a vector perpendicular to the plane that that passes thru

(1,4,6), ( 2,5, 1)P Q and (1, 1,1)R

Solution: Find 40, 15,15PQ PR

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

Solution: Area 5 82

1/ 22

A PQ PR

4. Find a b for 1,4, 7 , 2, 1,4a b and show that the cross product is

orthogonal to both a and b

5. Find two unit vectors orthogonal to both 1,1,1 and 2,0,1

6. Show that ( ) 0a b a

7. Show that ( ) ( )a c b c

a b c da d b d

Solution:

( ) ( ) ( ) ( )

( ) [( ) ( ) ]

( )( ) ( )( )

a b c d a b v a b v

a b c d a b d c b c d

a c b d b c a d

Mat 267 Engineering Calculus IIIUpdated on 01/19/2011 Dr. Firoz

Section 10.5 Equations of lines and planes

Lines: For a given direction vectors , ,v a b c , the vectors , ,r x y z , and

0 0 0 0, ,r x y z lies on the line L

0

0 0 0

0 0 0

, , , , , ,

, ,

r r tv

x y z x y z t a b c

x at y bt z ct

The parametric equation of the line L is 0 0 0, ,x x at y y bt z z ct

Also we can write the symmetric form (eliminating t)

0 0 0x x y y z z

a b c

where a, b, and c are called direction numbers or direction ratios.

The line segment from 0 0 0 0, ,r x y z to 1 1 1 1, ,r x y z is the vector given by

0 1( ) (1 ) , 0 1r t t r tr t

For two points 0 0 0 0( , , )p x y z and 1 1 1 1( , , )p x y z on L has the symmetric equation

0 0 0

1 0 1 0 1 0

x x y y z z

x x y y z z

Planes: Vector equation of a plane is defined as 0( ) 0n r r , where n is the unit

normal vector to the plane, containing r , and 0r .

A plane passing thru a point 0 0 0 0( , , )p x y z with normal vector , ,n a b c has scalar

equation 0 0 0( ) ( ) ( ) 0a x x b y y c z z which also can be written as

0ax by cz d . The distance of a point 1 1 1 1( , , )p x y z from the plane

0ax by cz d is defined as 1 1 1

2 2 2

ax by cz dD

a b c

Angle between two planes 0ax by cz d and 1 1 1 1 0a x b y c z d is given by

1 2

1 2

cosn n

n n

, where 1 , ,n a b c and 2 1 1 1, ,n a b c

Distance of a point 0 0 0 0( , , )p x y z from a line through the given points 1 1 1 1( , , )p x y z

and 2 2 2 2( , , )p x y z is given by the formula

2 1 2 1 2 1 1 0 1 0 1 0

2 1 2 1 2 1

, , , ,

| , , |

x x y y z z x x y y z zD

x x y y z z

Examples:

Mat 267 Engineering Calculus IIIUpdated on 01/19/2011 Dr. Firoz

A. Lines

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

(5, 1, 3) and is parallel to 1,4, 2v

Vector equation

, , 5,1,3 1,4, 2

5 ,1 4 ,3 2 (5 ) (1 4 ) (3 2 )

x y z t

t t t t i t j t k

Parametric equation 5 , 1 4 , 3 2x t y t z t

2. Find a symmetric equation and parametric equations for the line that passes thru

(2, 4, 3) and (3, 1, 1)

The symmetric equation

0 0 0

1 0 1 0 1 0

2 4 3

1 3 2

x x y y z z

x x y y z z

x y z

The parametric form is 2 , 4 3 , 3 2x t y t z t

3. In example 2, find intersection of the line with xy-plane.

On the xy-plane z = 0. Then 2 , 4 3 , 0 3 2 3/ 2x t y t z t t

We have the point (7/2, -1/2, 0)

4. Find the distance of the point (3, 4, -1) from the straight line passing through the

points (3, -2, -3) and (5, -3, 0). Answer: 40

B. Planes

5. Find an equation of a plane through (2, 4, -1) with a normal vector 2,3,4n

The plane has equation

0( ) 0 2,3,4 2, 4, 1 0 2( 2) 3( 4) 4( 1) 0n r r x y z x y z

6. Find the equation of a plane thru P(1, 3, 2), Q(3, -1, 6) and R(5, 2, 0)

Derive vectors 2, 4,4 , 4, 1, 2 ,andPQ PR n PQ PR . Now you can

consider the point P(1, 3, 2) and the normal vector to find your plane

12( 1) 20( 3) 14( 2) 0x y z

Mat 267 Engineering Calculus IIIUpdated on 01/19/2011 Dr. Firoz

7. Find the angle between two given planes 1x y z and 2 3 1x y z .

Notice that we have 1 1,1,1n and 2 1, 2,3n . Now find 1 2

1 2

cosn n

n n

8. Find the symmetric equations of the line of intersection L of two planes

1x y z and 2 3 1x y z .

Suppose 1n and

2n are the normal vectors to the given planes. Then 1 1,1,1n and

2 1, 2,3n . The line L has direction vector 1 2 5, 2, 3v n n . Let us find a

point common to both the planes letting z = 0, which could be (1, 0, 0). Thus we have the

equation of L in symmetric form, 1

5 2 3

x y z

Section 10.6 Cylinders and quadric surfaces

A cylinder is a surface that consists of all lines (called rulings) that are parallel to a given

line and passes thru a given plane curve.

A quadric surface is the graph of a second-degree equation in three variables x, y and z.

The most general equation of a quadric surface is 2 2 2 0Ax By Cz Dxy Eyz Fzx x Hy Iz J

General forms:

Look at page number 557 on your text for the diagrams

1. 2 2 2

2 2 21

x y z

a b c is an ellipsoid. For a b c , the ellipsoid is a sphere

2. 2 2

2 2

z x y

c a b is an elliptic paraboloid. For a b it is circular paraboloid.

3. 2 2

2 2

z x y

c a b is a hyperbolic paraboloid.

4. 2 2 2

2 2 21

x y z

a b c is a hyperboloid of one sheet

5. 2 2 2

2 2 21

x y z

a b c is a hyperboloid of two sheets

6. 2 2 2

2 2 20

x y z

a b c is a cone

Selected examples:

1. Since z is missing in 2 2 1x y , we consider 2 2 1x y with z = k, is a hyperbola

on the z = k plane. The surface is hyperbolic cylinder.

Mat 267 Engineering Calculus IIIUpdated on 01/19/2011 Dr. Firoz

2. Find the traces of 2 24y x z in the planes x = k, y = k, and z = k.

When x = k: 2 24y k z is a parabola,

y = k: 2 24k x z is a circle

and z = k: 2 24y x k is also a parabola

Thus the surface is a circular paraboloid with axis in the y axis and vertex (0, 0, 0)

3. 2 2 2

2 2 29 4 1 11/ 9 1/ 4 1

x y zx y z is an ellipsoid with intercepts

( 1/ 3,0,0), (0, 1/ 2,0), (0,0, 1)

4. Reduce the equation 2 24 16 4 20 0y z x y z to one of the standard forms and

classify the surface and make a rough sketch.

Solution: We find the form 2 2( 2) ( 2)

4 1 4

x y z is an elliptic paraboloid vertex at

(0, 2, 2) and axis is the horizontal line y = 2, z = 2.

Section 10.7 Vector Functions and Space Curves

A vector valued function or simply a vector function is a function whose domain is a set

of real numbers and whose range is a set of vectors.

Representation of a vector function: The vector ( )r t along with components

( ), ( ), ( )f t g t h t , which are simple functions of t, represented as

( ) ( ), ( ), ( ) ( ) ( ) ( )r t f t g t h t f t i g t j h t k

Important properties:

1. lim ( ) lim ( ), lim ( ), lim ( )t a t a t a t a

r t f t g t h t

, provided each limit exists

2. The parametric representations ( ), ( ), ( )x f t y g t z h t is a space curve on I, t is a

parameter.

Examples:

7. Find the domain of 3( ) , ln(2 ),r t t t t

Solution: For 3t , t is in I, set of real numbers. For ln(2 )t , t < 2 and for t , 0t . Thus

the domain is [0, 2)

Mat 267 Engineering Calculus IIIUpdated on 01/19/2011 Dr. Firoz

8. Determine 0

lim ( )t

r t

where 3 1 sin( ) 1 , ,

1

tr t t

t t

Solution: 0

lim ( ) 1, 1,1t

r t

9. Describe the curve defined by ( ) 1 ,2 5 , 1 6r t t t t

Solution: The parametric form of the curve is 1 , 2 5 , 1 6x t y t z t , which

represents a straight line thru (1, 2, -1) in the direction of the vector 1,5,6

10. Sketch the curve ( ) cos ,sin ,r t t t t

Solution: The parametric form of the curve is cos , sin ,x t y t z t , which

represents a curve that spirals around a circular cylinder with level curves 2 2 1x y .

The curve is known as circular helix. See the figure at page # 560 in you text.

11. Find the vector function that represents the curve of intersection of the cylinder 2 2 1x y and the plane 2y z

Solution: Consider cos , sinx t y t , then 2 sinz t , 0 2t .

Now ( ) cos ,sin ,2 sin cos sin (2 sin )r t t t t i t j t k t

12. Find the vector equation of a line thru the points P(1, 0, 1) and Q(2, 3, 1)

Solution: Use the formula thru two given points as 0 1( ) (1 ) , 0 1r t t r tr t

We have ( ) (1 ) 1,0,1 2,3,1 , 0 1r t t t t

13. Identify the curve ( ) cos10 , sin10 ,t t tr t e t e t e

The curve is a spiral around a cone whose level curves are circles 2 2 2tx y e .

Observe that 2 2 2x y z . Use the variable t for the parameter.

14. Find a vector that represents the curve of intersection of the paraboloid 2 22 4x y z and the cylinder 24y x

Solution: Consider x t and find y and z in terms of t as 2 2 2 2 2 2 44 4 , 2 4(4 ) 2 64y x t z t t t t and write a vector

2 2 4( ) , 4 , 2 64t t t t t r

15. Consider the paraboloid 2 2z x y . The plane 5 3 9 0x y z cuts the

paraboloid, its intersection being a curve. Find "the natural" parametrization of

this curve.

Mat 267 Engineering Calculus IIIUpdated on 01/19/2011 Dr. Firoz

Solution: We write 2 25 3 9 0x y x y by substituting 2 2z x y

Now write completing square as 2 2( 5/ 2) ( 3/ 2) 35/ 2x y . Finally consider

5/ 2 35/ 2 cos 35/ 2 cos 5/ 2x x and

3/ 2 35/ 2 sin 35/ 2 sin 3/ 2y y and

2 2( 35/ 2 cos 5/ 2) ( 35/ 2 sin 3/ 2)z

16. Find the parametric equations for the tangent line to the curve

at the point (242, 10, 81). Use the variable t for your parameter.

Solution: First find t: from 5242 1t or 210 1t

4 4 35 5(3 ) 405, 2 6, 4 108dx dy dz

t t tdt dt dt

The parametric equation of the tangent line as follows:

242 405 , 10 6 , 81 108x t y t z t

Section 10.8 The Arc Length and Curvature

The arc length of a curve ( ) , ,r t f g h from tot a t b is defined by

2 2 2( ) ( ) ( ) ( )

b b

a a

L f g h dt r t dt .

Suppose ( )s t is the length of the part of C between r(a) and r(t) then

2 2 2( ) ( ) ( ) ( ) ( )

t t

a a

s t f g h du r u du

Curvature: If C is a smooth curve defined by ( ) , ,r t f g h , then ( ) 0r t , the unit

vector ( )

( )( )

r tT t

r t

. The curvature of a curve C at a given point is a measure of how

quickly the curve changes direction at that point. The curvature is defined as dT

ds ,

where T is the unit tangent vector and s is the arc length function. Remember that

dT dT ds

dt ds dt , then

/ ( )

/ ( )

dT dT dt T t

ds ds dt r t

For a function defined as ( )y f x , 2 3

( )[1 ( ) ]

yx

y

Mat 267 Engineering Calculus IIIUpdated on 01/19/2011 Dr. Firoz

Theorem: The curvature of the curve given by the vector function ( ) , ,r t f g h is

3

( ) ( )

( )

r t r t

r t

The Normal and Binormal: The vectors N and B defined below are called the normal

and Binormal vectors,

( )

( ) and ( ) ( ) ( )( )

T tN t B t T t N t

T t

The plane determined by the normal and binormal vectors N and B at a point P on the

curve C is called the normal plane of C at P. It consists all lines that are orthogonal to

the tangent vector T.

The plane determined by the vectors T and N is called the osculating plane of C at P.

The circle that lies in the osculating plane of C at P has the same tangent as C at P lies on

the concave side of C towards N and has radius 1

, is called the osculating circle.

Examples:

1. Find the length of the arc of the circular helix with vector equation

( ) cos sinr t i t j t kt from the point (1, 0, 0) to (1,0,2 )

Solution:

2 2

0 0

( ) 2 2 2L r t dt dt

Where ( ) sin ,cos ,1 1 1 2r t t t

2. Find the arclength of the curve ,

Solution:

1 1 1

18 18 9 9 9 9

0 0 0

( ) 9 2 9 ( )t t t tL r t dt e e dt e e dt e e

3. Reparametrize the helix ( ) cos sinr t i t j t kt with respect to arc length s

measured from (1, 0, 0) in the direction of increasing t.

Solution: We have ( ) 2ds

r tdt

from example 1. Now

0

2 22

ts

s dt t t and ( ( )) cos( / 2) sin( / 2) ( / 2)r t s i s j s k s

Mat 267 Engineering Calculus IIIUpdated on 01/19/2011 Dr. Firoz

4. Show that the curvature of a circle of radius r is 1/r.

Solution: Consider ( ) cos , sinu t r t r t a circle of radius r. Now ( )

( )

T t

u t

and

( ) sin , cos( ) sin ,cos

( )

u t r t r tT t t t

u t r

, ( ) cos , sin , ( ) 1T t t t T t

The curvature of the circle of radius r is ( ) 1

( )

T t

u t r

5. Find the unit normal vector and binormal vector for the circular helix

( ) cos ,sin ,r t t t t

Solution: ( ) sin ,cos ,1r t t t , ( ) 1

( ) sin ,cos ,1( ) 2

r tT t t t

r t

And ( )

( ) cos , sin ,0( )

T tN t t t

T t

,

1 1( ) sin cos 1 sin , cos ,1

2 2cos sin 0

i j k

B t T N t t t t

t t

6. Find the equation of a normal plane and a osculating plane for the circular helix

( ) cos ,sin ,r t t t t at the point (0,1, / 2)P

Solution: The normal plane has normal vector ( / 2) 1,0,1r .

The equation of the plane thru P is ( 0) 0( 1) 1( / 2) 0 / 2x y z z x .

The osculating plane at P contains the vectors T and N. So from example 4, its

normal vector is 1 1

sin , cos ,1 , ( / 2) 1,0,12 2

B T N t t B . The

equation of the osculating plane is 1 1

( 0) ( / 2) 0 / 22 2

x z z x

7. Find the curvature of the parabola 2 1y x at the point (1, 2)

Solution: The curvature is 2 3/ 22 3

2( )

(1 4 )[1 ( ) ]

yx

xy

.

At x = 1, 3/ 2

2(1) 0.18

(1 4)

8. Find the osculating circle of 2y x at (0, 0)

Mat 267 Engineering Calculus IIIUpdated on 01/19/2011 Dr. Firoz

Solution: 2 3/ 22 3

2( ) 2, 0

(1 4 )[1 ( ) ]

yx x

xy

. The radius of the osculating

circle is 1/ 2 , center at (0, ½). The osculating circle has the equation 2 2( 1/ 2) 1/ 4x y

9. Find curvature of ( ) cos , sin ,t tr t e t e t t at t = 0.

Solution: 3

( ) ( )( )

( )

r t r tt

r t

( ) cos sin , sin cos ,1t t t tr t e t e t e t e t , (0) 1,1,1r

( ) 2 sin ,2 cos ,0t tr t e t e t , (0) 0,2,0r

Check that (0) (0) 2,0,2r r

Then 3

(0) (0) 2 2(0)

3 3(0)

r r

r

10. Given ( ) , sin , cost t tr t e e t e t . Find N, T, B at (1, 0, 1)

Solution: ( ) ( )

( ) , ( ) and ( ) ( ) ( )( ) ( )

r t T tT t N t B t T t N t

r t T t

( ) 1,sin cos ,cos sintr t e t t t t

When 1 0te t , (0) 1,1,1r , (0) 3r and

( )( ) 1,sin cos ,cos sin

( )

tr tT t e t t t t

r t

( ) 0,cos sin , sin cosT t t t t t

(0) 1 (0) 1(0) 1,1,1 , (0) 0,1, 1

(0) (0)3 2

r TT N

r T

1(0) (0) (0) 2,1,1

6B T N

Section 10.9 Motion in Space: Velocity and Acceleration

Suppose a particle moves thru the space so that its position vector at time t is r(t). The

quantity

( ) ( )r t h r t

h

approximates the direction of the particle moving along the curve r(t). Its magnitude

measures the size of the displacement vector per unit time. Thus the quantity we have is a

vector that gives the average velocity over a time interval of length h and its limit is the

velocity vector v(t) at time t defined as

Mat 267 Engineering Calculus IIIUpdated on 01/19/2011 Dr. Firoz

0

( ) ( )( ) lim ( )

h

r t h r tv t r t

h

The speed of the particle at time t is the magnitude of the velocity, that is

( ) ( )

speed ( ) ( )r t h r t

v t r t Vh

As in the case of one-dimensional motion, the acceleration of the particle is defined as the

derivative of the velocity:

( ) ( ) ( )a t v t r t

Angular velocity ω versus linear velocity v:

The angular velocity is a ratio of the total angular measurement through which a particle

rotates in a given unit of time. If we use to stand for angular velocity, we have the

angular velocity d

dt

and linear velocity v r where r is the radius.

Two dimensions

In two dimensions the angular velocity is a single number which has no direction. A

single number which has no direction is either a scalar or a pseudo scalar, the difference

being that a scalar does not change its sign when the x and y axes are exchanged (or

inverted), while a pseudo scalar does. The angle as well as the angular velocity is a

pseudo scalar. The positive direction of rotation is taken, by convention, to be in the

direction towards the y axis from the x axis. If the axes are inverted, but the sense of a

rotation does not, then the sign of the angle of rotation, and therefore the angular velocity

as well, will change.

It is important to note that the pseudo scalar angular velocity of a particle depends upon

the choice of the origin and upon the orientation of the coordinate axes.

Three dimensions

In three dimensions, the angular velocity becomes a bit more complicated. The angular

velocity in this case is generally thought of as a vector, or more precisely, a pseudo

vector. It now has not only a magnitude, but a direction as well. The magnitude is the

Mat 267 Engineering Calculus IIIUpdated on 01/19/2011 Dr. Firoz

angular speed, and the direction describes the axis of rotation. The right hand rule

indicates the positive direction of the angular velocity pseudo vector, namely:

If you curl the fingers of your right hand to follow the direction of the rotation,

then the direction of the angular velocity vector is indicated by your right thumb.

Examples:

1. Consider the Earth which rotates on its axis once every 24 hours. The angular

velocity of the Earth’s rotation is 2

/ /24 12

rad h rad h

2. A ceiling fan rotates 30 times per minute, the angular velocity is

30(2 )/ min 60 / min

1rad rad

3. Consider the Earth which rotates on its axis once every 24 hours. The radius of

the Earth is approximately 4000 miles. The velocity of Earth’s rotation is

(4000) 1047.212

v r

miles/h

Force: 2

2vF T m m r

r

Components of the force: along x axis is cos( )x F t and along y axis is sin( )y F t

Centripetal force: When a force acts in the direction opposite to the radius vector r(t)

and therefore points toward the origin is called a centripetal force and in this case

cos( )x F t and sin( )y F t

Example: A body of mass 4 kg moves in a (counterclockwise) circular path of radius 3

meters, making one revolution every 6 seconds. You may assume the circle is in the xy-

plane, and so you may ignore the third component.

Mat 267 Engineering Calculus IIIUpdated on 01/19/2011 Dr. Firoz

A. The centripetal force acting on the body has cos( ) 13.16cos( )x F t t and

sin( ) 13.16sin( )y F t t where 2 / 6 /3

B. The magnitude of that force

222 2

4 3 13.166

vF T m m r Newton N

r

Projectiles: A projectile is any object that is cast, fired, flung, heaved, hurled, pitched,

tossed, or thrown. The path of a projectile is called its trajectory. Some examples of

projectiles include …

a baseball that has been pitched, batted, or thrown

a bullet the instant it exits the barrel of a gun or rifle

a bus driven off an uncompleted bridge

a moving airplane in the air with its engines and wings disabled

A projectile is any object with an initial non-zero, horizontal velocity whose acceleration

is due to gravity alone.

The Trajectory of a Simple Projectile is a Parabola

If the a particle is thrown at an initial velocity 0v and angle the particle travels at time

t horizontally 0 cos( )x v t and vertically 2

0

1sin( )

2y v t gt , where the gravitational

force 29.8 /g m s .

Note: When the particle hits the ground 0 cos( )x v t is called the range and in this case

2

0

10 sin( )

2y v t gt

Examples:

1. A ball is thrown at an angle of 45 degrees to the ground, and lands 90 meters

away. Find the initial speed of the ball.

Mat 267 Engineering Calculus IIIUpdated on 01/19/2011 Dr. Firoz

We have 2 200

21sin( ) 0 4.9 0

2 2

vy v t gt t t and

00

2cos( ) 90

2

vx v t t . Solving the equation one can find that

0 29.7v m/s

2. A projectile is fired from ground level with an initial speed of 350 m/sec and an

angle of elevation of 30 degrees. Find the range of the projectile, the maximum

height and the speed with which the projectile hits the ground.

Solution: 2 2

0

1 1(350)sin( ) 0 4.9 0,35.7

2 2y v t gt t t t . The range of

the projectile is 0 cos( ) 10824x v t meters. The maximum height of the

projectile is at 35.7

17.852

t s s (time for half of the range) and max

2

0

1sin( ) 1562.5

2y v t gt m . The vector equation of the projectile is

2

0 0( ) cos , sin 4.9r t v t v t t . Thus we have the magnitude of velocity of

the projectile

2 2 2 2| ( ) | | ( ) | ( ) ( ) (175 3) (175 9.8(35.71)) 349.98v t r t x y m/s

Tangent and normal components of acceleration: We have the tangent vector

( ) ( ) ( )

( )( ) ( ) V

r t v t v tT t v VT

r t v t

Differentiating velocity vector we have acceleration va v T VT

The curvature is then defined as V

T TT V

r

The unit normal vector is T T

N T N VT V

Thus we have 2

T Na v V T VT V T V N a T a N , where ( ) ( )

= V =( )

T

r t r ta

r t

is

the tangential component and 2( ) ( )

( )N

r t r ta V

r t

is the normal component of the

acceleration.

Examples:

Mat 267 Engineering Calculus IIIUpdated on 01/19/2011 Dr. Firoz

1. Find the velocity, acceleration and speed of a particle with position function 2 3 2( ) 1, , 1r t t t t

Solution: Velocity 2( ) ( ) 2 ,3 ,2v t r t t t t

Acceleration ( ) ( ) ( ) 2,6 ,2a t v t r t t

Speed 2 2 2 2( ) (2 ) (3 ) (2 )V r t t t t

2. Find the velocity, acceleration and speed of a particle with position function 2 3 2( ) i j kr t t t t

Solution: Velocity 2( ) ( ) 2 i 3 j 2 kv t r t t t t

Acceleration ( ) ( ) ( ) 2i 6 j 2ka t v t r t t

Speed 2 2 2 2( ) (2 ) (3 ) (2 )V r t t t t

3. Given the acceleration vector ( ) i 2j 2 ka t t , find velocity vector and position

vector when (0) 0, (0) i kv r

Solution: ( ) ( ) i 2j 2 kdv

a t v t tdt

. Now integrating we have velocity

2 2

1( ) i j k+cv t t t t , using condition 1(0) 0,c 0v .

Again, 2 2( ) ( ) i j kdr

v t r t t t tdt

, integrating we have position

2 3 3

2

1 1 1( ) i j k+c

2 3 3r t t t t , using initial condition (0) i kr we find 2c i k

4. Find the tangential and normal components of the acceleration vector 2( ) i j 3 kr t t t t

Solution: tangential component ( ) ( )

= V =( )

T

r t r ta

r t

and the normal component is

2( ) ( )

( )N

r t r ta V

r t

. You can find them now.