Expert: Josh - 4/28/2006

Question
-------------------------
Followup To
Question -
http://www.geocities.com/uttoransen/questions.doc
Answer -
I prefer not to open Microsoft and other embedded documents as they pose a risk to computer security.
If you have a question, please send it in text using the web form. Always attempt answering the question first yourself.

Regards
Q1) If  a,b,g are the roots of the equation

X3 – 7x2 + x + 5 = 0

Find the equation whose roots are

a2 + b2 ,  b2 + g2 , g2 + a2 .



q2) evaluate : cos-1x dx

q3) find all the fifth roots of (2+I)

q4) find the area bounded by the x – axis , the curve y = e2x and the ordinates x=2 and x=3


q6) find the equation of the line joining the points (-5,2,3) and (5,-2,3)


q7) find the equation of the sphere, which contains the circle
x2  + y2 + z2 = 9, 3x + 3y +3z =5

passes through the oregin,  

Answer
Q1) If a,b,g are the roots of the equation
We have k(x-a)(x-b)(x-g)=0, where k=1.
Expanding this, we get
(x^2-(a+b)x+ab)(x-g)=0
x^3-(a+b+g)x^2+(ab+ag+bg)x-abg=0
Make sure you familiarize with this technique.

Comparing the coefficients with x^3-7x^2+x+5=0
for x^2: (a+b+g)=7
for x^1: (ab+ag+bg)=1
for x^0: -abg=5.
You have three equations and three unknowns.
This enables you to solve for a,b and g.

Note: I'm not going to do the work for you. I'm only going to show you the way. You will have to find "a","b" and "g" yourself.

After determining the values of "a","b" and "g", finding the cubic polynomial with roots x1=a^2+b^2, x2=b^2+g^2, x3=a^2+g^2 is easy. The equation is simply y=(x-x1)(x-x2)(x-x3), where x1,x2 and x3 are the roots.

Q2) I'm going to solve this using a technique called integration by parts (IBP). This is not necessarily the most elegant solution, but it will do.

Part I: The idea behind integration by parts (IBP)

Consider a composite function f(x)g(x). For brevity, I'll write fg instead, making the dependence on x implicit.
Applying the chain rule of differentiation,
d/dx[fg] = f'g + g'f, where f'=df(x)/dx, g'=dg(x)/dx.
Integrating both sides w.r.t. x,
f(x)g(x) = integral f'(x)g(x) dx + integral g'(x)f(x) dx
========================================================
The integration by parts formula states that
integral f'(x)g(x) dx = f(x)g(x) - integral g'(x)f(x) dx ...[#1]
========================================================

Part II: Relating to your question
Observe that integral cos^(-1)(x) dx = integral 1*cos^(-1)(x) dx.
This is consistent with integral f'(x)g(x) dx, where f'=1, g=cos^(-1)(x). We have to find f(x) and g'(x) in order to apply the IBP formula.

Part III: Show that d/dx cos^(-1)(x)= -1/sqrt(1-x^2)

Proof: Let angle y=cos^(-1)(x) as shown in the following diagram. Angle subtended by ACB = y = cos^(-1)(x).

A
.
.
B..........C

Label |BC|=x, |AC|=1.
Pythagoras theorem gives |AB|=sqrt(1-x^2)
Since cos(y)=x, dx/dy=-sin(y) and sin(y)=sqrt(1-x^2) by construction, dy/dx = -1/sin(x) = -1/sqrt(1-x^2).
Hence, d/dx [cos^(-1)(x)]= -1/sqrt(1-x^2).

====================================================
Generalization: It can be shown that
d/dx [cos^(-1)k(x)] = k'(x)/sqrt(1-(k(x))^2) ...[#2]
====================================================

Continuing from Part II:
Let I = integral cos^(-1)(x) dx = integral f'(x)g(x) dx.
We identified f', g, f and g' as
f'(x)=1; f(x)=x
g(x)=cos^(-1)(x) and g'(x)=-1/sqrt(1-x^2) follows from [#2]

Applying IBP formula [#1],
I = f(x)g(x) - integral g'(x)f(x) dx
 = x cos^(-1)(x) - integral -x/sqrt(1-x^2) dx ...[#3]

Aside: In [#3], inside the integral, if we let k(x)=x^2, k'(x)=2x. Thus, we can write -x/sqrt(1-x^2)dx as 0.5k'(x)/sqrt(1-k(x)^2). This has the same form as [#2].

Thus, continuing from [#3]
I = x cos^(-1)(x) -0.5 integral k'(x)/sqrt(1-k(x)^2) dx
 = x cos^(-1)(x) -0.5 cos^(-1)(k(x)) + constant
 = x cos^(-1)(x) -0.5 cos^(-1)(x^2) + constant

Q3) Equivalent to finding the fifth roots of z=(2+i).
Write z=r*exp(i*theta), where r=sqrt(2^2+1)=sqrt(5), theta=tan^(-1)(1/2)=pi/6. [Note: pi radian = 180 degrees]
Applying De Moivre's theorem,
z^(1/5)= r^(1/5) exp(-i*(theta+2*pi*n)/5) for n=0,1,2,3,4.
You still need to wrap the angle back into the fundamental interval of from 0 to 2*pi as appropriate.

I assume you know how to convert from Euler's form into cartesian coordinates in the Argand plane [If not, z=x+iy=r*exp(i*theta), where x=r*cos(theta), y=r*sin(theta)].

Q4) Evaluate I=integral exp(2x) dx from x=2 to x=3.

Take note that d/dx exp(k(x))= k'(x)*exp(k(x)).
This is something fundamental that you need to remember.
Integrating both sides (switching left and right),
integral k'(x)*exp(k(x)) = exp(k(x))+constant

Here, let k(x)=2x, k'(x)=2. Thus,
I=(1/2)*integral k'(x)exp(k(x)) dx; x=2 to x=3
=(1/2) exp(k(x))
=(1/2) { [exp(2x)]|x=3 - [exp(2x)]|x=2 }
=(1/2) [exp(6)-exp(4)]

check that i've not made any mistake.

Q6) Let vector v=(x,y,z), v1=(x1,y1,z1) and v2=(x2,y2,z2); both v1 and v2 are known points in 3D.
Parametric equation is v = v1 + k (v2-v1) for 0<=k<=1