How To Solve Applied Mathematics Problems Pdf

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1

VECTOR ALGEBRA

One of the fundamental objectives of Applied Mathematics is to use mathematics to aid the understanding of the nature of the physical world. To this end it is necessary to develop a mathematical structure which is capable of dealing with events in three-dimensional space.

The problem of constructing an algebra for a triplet of numbers, representing a point in space, was resolved by the American Josiah Willard Gibbs (1839-1903). Further developments were made by the English mathematical physicist Oliver Heaviside (1850-1925) in 1891.

A controversy arose over the relative merits of vectors and quaternions, 2 × 2 matrices, introduced by the Irish mathematician William Rowen Hamilton (1805-1865), which satisfy the rules of an associative algebra. A strong supporter of quaternions was the Scotsman Peter Guthrie Tait (1831-1901) who was professor of mathematics at Queen's College Belfast from 1854 to 1860.

However the non-associative, non-commutative vector algebra developed by Gibbs and Heaviside proved to be the most practical system.

by using bold type, for example a, and scalar and vector products by · and × respectively.

We start this chapter with the application of vector methods to problems in geometry.

1.01 Prove the theorem of Apollonius, that if D is the mid-point of the side BC of a triangle ABC then AB² + AC² = 2 (AD² + BD²).

• Solution

The solution of any applied mathematics problem depends on knowing which are the relevant fundamental formulae.

In this problem we use the triangle law. For the triangle ABC Hence

In the next three problems we use the following important result:

•Theorem.

are two vectors and X is a point in the line AB such that

then

1.02 Show that the position vector of the centroid G of a triangle ABC is given by

where O is an arbitrary origin.

• Solution

Let D, E, F be the mid-points of the sides BC, CA, AB respectively of a triangle ABC.

where X1 divides AD in the ratio 2:1. Hence

Similarly

where X2 divides BE in the ratio 2:1, and

where X3 divides CF and so the points X1, X2, X3 coincide. The point of coincidence is called the centroid G and

1.03 Show that the four lines joining the centroids of the faces of a tetrahedron ABCD to the opposite vertices are concurrent.

• Solution

If G is the centroid of the face ABC we have from the previous problem that

where X divides DG in the ratio 3:1. Similarly X lies on the other lines joining the centroids to the opposite vertices and divides them in the same ratio. The point of concurrence is given by

1.04 If E, F, G, H are the mid-points of the edges AB, BC, CD, DA respectively of a tetrahedron ABCD, show that EG and FH meet at their mid-points and that this is the same as the point of concurrence of the lines joining the vertices to the centroids of the opposite faces given in the previous problem.

Further show that EFGH is a parallelogram.

• Solution

Hence

where X1 is the mid-point of EG. Similarly

where X2 is the mid-point of FH. Hence EG and FH meet at their mid-points X where

It follows that EFGH is a parallelogram since its opposite sides are equal in length and are parallel.

1.05 Any five points may be taken in 10 ways, as one pair of points and a set of three points. H is the mid-point of the line joining the pair and K is the centroid of the triangle formed by the other three points. Show that the ten lines HK are concurrent.

• Solution

This is a problem which would require some considerable skill to establish using ordinary geometrical methods but which comes out easily using (1.2).

Let the five points be A1, A2, A3, A4, A5. Further let H12 be the mid-point of the line A1A2 and K345 be the centroid of the triangle A3A4A5. Take any origin O. Then

Hence

is the position vector of a point G12,345 on the line H12K345 with 2H12G12,345 = 3K345G12,345. Similarly it is the position vector of a point G on each of the other lines HK. Hence the lines are concurrent at G.

Next we consider the application of vector methods to some trigonometry problems.

1.06 If a, b, c are vectors determining the sides BC, CA, AB respectively of a triangle ABC, show that a + b + c =0. Hence prove the cosine law a² = b² + c² − 2bc cos A where A is the angle opposite to the side BC of length a.

• Solution

it follows at once that a + b + c = 0. Hence a = − (b + c) and so

using the definition of the scalar product b.c =bccos A.

1.07 If a + b + c = 0 show that a × b = b × c = c × a and hence prove the sine law

where A, B, C are the internal angles of a triangle opposite to the sides of lengths a, b, c respectively.

• Solution

Since a = −(b + c) we have that a × b = −(b + c) × b = −c × b = b × c using b × b = 0 and b × c = c × b. Similarly a × b = c × a and thus we get a × b = b × c = c × a. Now, using the definition of the vector product a × b = absinC is a unit vector perpendicular to a and b, we obtain ab sin C = bcsin A = ca sin B and the sine law follows.

1.08 Show that

and hence show that

cos(A + B) = cos A cos B − sin A sin B

• Solution

Using the cyclic interchange property of the scalar triple product

and the formula for the vector triple product

we see that

(a × b).(b × c) = [b × (b × c)].a = (b.c b − b² c).a = a.b b.c − b² a.c

which proves the vector formula (1.4). Now taking A to be the angle between a and b, and taking B to be the angle between b and c, and further assuming that a, b, c are coplanar, we obtain sin A sin B = cos A cos B − cos (A + B) which yields the trigonometric formula for the cosine of the sum of two angles.

1.09 Solve the simultaneous vector equations r + c × s = a, s + c × r = b.

• Solution

Since a scalar triple product vanishes if two of its vectors are the same, we see that c.r = c.a and c.s = c.b. Now from the first vector equation c × r + c × (c × s) = c × a so that c × r + c.s c − c² s = c × a. It follows that b − s + c.b c − c² s = c × a and so we get

Hence

1.10 Show that the volume of a tetrahedron whose vertices have position vectors a, b, c, d is given by

• Solution

The volume of a tetrahedron ABCD is given by

Hence

and the result follows.

1.11 If a, b, c, d are four vectors such that a + b + c + d = 0 show that the magnitude of each vector is proportional to the volume of the parallelepiped determined by the unit vectors of the other three vectors.

• Solution

Let a , b = b , c = c , d = d so that we have

a + b + c + d = 0.

. Then we get a ) + b ) = 0. It follows that

Similarly

and

Hence

.( , , .

1.12 If p, q, r are three linearly independent vectors and

a = p + q, b = q + r,c = r + p

, , = −1 if and only if a, b, c are parallel to the same plane and that in this case

a b -1c.

• Solution

If a, b, c are parallel to the same plane we may write a b c are scalars. Then

p + q (q + (r + p)

so that

)p )q + )r = 0

Since p, q and r + = −1.

= −1 then

a = p + q = p + (b − r) = b + p − r = b + p −1r = b + −1c.

1.13 Two straight lines are represented by r = a + û and r = b + where û .(a × û .(b × û) and find the point of intersection.

• Solution

The lines intersect if we have a+ û = b+ , that is if a. ) = b.(û .(a × û) = .(b × û). At the point of intersection a.(b ) + û.(b ) =0

so that

Hence the point of intersection is given by the position vector

1.14 Show that the perpendicular distance of the origin from a plane passing through three non-collinear points A, B, C with position vectors a, b, c is given by

• Solution

A vector which is perpendicular to the plane is (a − b) × (a − c) = a × b + b × c + c × a and so a unit vector in this direction is

The perpendicular distance of the origin from the plane is therefore

We conclude this chapter on vectors with the following interesting result on vector triple products.

1.15 Show that a × (b × c) + b × (c × a) + c × (a × b) = 0.

• Solution

Using the formula for the vector triple product a × (b × c) = a.c b − a.b c we get a × (b × c) + b × (c × a) + c × (a × b) = a.c b − a.b c + b.a c − b.c a + c.b a − c.a b = 0

2

KINEMATICS

This chapter is concerned with problems on the motion of points or particles in space, without taking account of the dynamics of the motion.

The first pair of problems we shall discuss are three-dimensional.

2.01 The position vectors of the vertices A1, A2, A3 of a triangle A1A2A3 are r1 r2, r3 is the vector area of the triangle. If the time rates of change of the position vectors are given by

find d /dt and the constant n.

If the position vectors of A1, A2, A3 are a1, a2, a3 respectively at time t as a function of the time t.

• Solution

The vector area of a triangle with sides determined by the vectors a and b Hence we have

and so

giving

Therefore

2.02 Four points A1, A2, A3, A4 with position vectors r1, r2, r3, r4 are moving with velocities v1, v2, v3, v4 respectively. Find the time rate of change of the volume of the tetrahedron A1A2A3A4.

• Solution

The volume of a tetrahedron having three concurrent edges determined by the vectors a, b, c Hence the volume of the tetrahedron A1A2A3A4 is

It follows that

where we have denoted differentiation with respect to time t 1 = v1, 2 = v2 3 = v3.

Hence

which may be rewritten, on expanding the cross products, in the simpler form

2-1 Radial and transverse resolutes of velocity and acceleration

The next few problems are two-dimensional and use the radial and transverse resolutes formulae for the velocity v and the acceleration a:

are unit vectors in the radial and transverse directions respectively and r are circular polar coordinates. Here we have denoted differentiation with respect to time t = dr/dt = d²r/dt².

2.03 If

i j, i j

where i and j . = 0 and

• Solution

= 0 and

2.04 A point P moves along a circle of radius a so that its acceleration is directed towards a point O of its circumference. Show that the acceleration towards O is 8a²h²/r⁵ where h is the areal constant.

• Solution

The equation of a circle of radius a is r = 2a = −2a = −2a cos ² −2a . Since the acceleration is purely radial, the transverse component of acceleration must vanish. Now