mathematic problems and the solution

PROBLEMS (by Anis Septiana Sari)
1.If |z = xy + x + y , x = r + s + t , y = rst

Determine
∂z/∂s |r=1,s=-1,t=2

2.A lawyer, every day, go for work from his house to his office. In average, the trip takes 24 minutes with standard deviation of 3,8 minutes. Assume that the time of the trip spread normally. Count the opportunity that 2 of three next trips to the office will spent more than 30 minutes.

3.Prove that 1 ≤ ∫ from 0 until 1 √(1+x4) dx ≤ 6/5
Clue : 1 ≤ √(1+x4) ≤ 1+x4

4.Prove if a|(b-1) so a|(b4-1)

5.How to find the distance between two parallel planes?



SOLUTION (by Endah Ayu Wardani)

1.To find the solution of problem number 1, there is a theorm that can helps us, it is theorm of chain rule :
For example x=x(s,t) and y=y(s,t) have a first differential at (s,t) and for example z=f(x,y) can be differentiated at (x(s,t),y(s,t)) so z=f(x(s,t),y(s,t)) has a first partial differential that be given by :

i.∂z/∂s = ∂z/ ∂x . ∂x/∂s + ∂z/∂y . ∂y/∂s

ii.∂z/∂t = ∂z/∂x . ∂x/∂s + ∂z/∂y . ∂y/∂t


From that theorm, we know that the way to solve it is
∂z/∂s = ∂z/∂x . ∂x/∂s + ∂z/∂y . ∂y/ ∂s ………Eq.1


Find each first differential:
∂z/∂x = y+1 ∂x/∂s = 1 ∂z/∂y =x+1 ∂y/∂s=rt


Subtitude them to Eq. 1
∂z/∂s = (y+1)(1) + (x+1)(rt)

∂z/∂s = y + 1 + rtx + rt

Subtitude x = r + s + t and y = rst
∂z/∂s = rst + 1 + rt(r+s+t) + rt

∂z/∂s = 2rst + r2t + rt2 + rt + 1

When r=1, s=-1, and t=2
∂z /∂s | r=1,s=-1,t=2 = 2(1)(-1)(2) + (1)^2(2) + (1)(2)^2 + (1)(2) + 1 = -4 + 2 + 4 + 2 + 1 = 5


2.The opportunity cannot be found in a way. First, find the opportunity when the trip spent more than 30 minutes.

For example X is the normal random variable with µ (median) = 24 minutes and σ(standard deviation) = 3,8 minutes.
P(X>30) = P(Z>z1)
z1 =(30 - µ)/σ = (30 – 24)/3,8 = 1,58
P(X>30) = P(Z>1,58) = 1 – P(Z<1,58) = 1- P(ZSo, the opportunity of the trip spent more than 30 minutes is 0,0571

Then, we bring it to binomial deviation.
• The success opportunity is p = 0,0571
• The failure opportunity is q = 1 – 0,0571 = 0,9249

The opportunity when twice success from three times repetition :
b( 2 ; 3 ; 0,0571 ) = 3C2 . (0,0571)2 . (0,9429)3-2 = 3 . (0,0033) . (0,9429) = 0,0092

So, the opportunity that twice of three times next trips spent more than 30 minutes is 0,0092.


3.From the clue, we know that 1≤√(1+x4)≤1+x4. Then, integral all of them in limit [0,1] toward x, become :
∫01 1 dx ≤ ∫0 until1 √(1+x4) dx ≤ ∫01 (1+x4) dx
<=> 1 ≤ ∫ from 0 until 1 √(1+x4) dx ≤ [x+(1/5) x5]01
<=> 1 ≤ ∫ from 0 until 1 √(1+x4) dx ≤ (1+(1/5)-0)
<=> 1 ≤ ∫ from 0 until 1 √(1+x4) dx ≤ 6/5 PROVED!!


4.There is a theorm in numeric theory contain :
If a|b => a|mb for every m is round number
From that theorm, we can solve it when (b4-1) is m(b-1)
•b^4-1 = (b^2+1)(b^2-1) ; b^2-1 = (b+1)(b-1)
<=> b^4-1 = (b^2+1)(b+1)(b-1)
Assume that (b2+1)(b+1) = m so
<=> b^4-1 = m(b-1)
a|(b-1) => a|m(b-1), when m=(b^2+1)(b+1) so a|(b^4-1) PROVED!!


5.For example, plane-α // plane-β
The steps to find the distance between plane-α and plane-β :

1)Investigate plane-α as same as plane-β or not, if they are same, the distance is zero (0).
2)When plane-α is different with plane-β, the ways to find the distance are :

2.1 Choose a point in plane-α, for example A(xA,yA,zA)
2.2 Find the normal vectors of both planes, because they are parallel, the normal vectors are same. For example the vector is k.
2.3 Find the perpendicular plane-α passing A. For example line g.

g = x = xA + ρk ; ρ is parameter
y yA
z zA
2.4 Find the passage point of line g with plane-β by equating the right side of equation g with the right side of plane-β equation. We will get system of linear equation, solve it and get the value of parameters and then subtitude them to equation g or equation of plane-β, so we get the passage point, for example point B(xB,yB,zB).
2.5 Find the distance between A and B.
Distance(A,B) = √((xA-xB)^2 + (yA-yB)^2 + (zA-zB)^2)
The distance between A and B is the distance of plane-α and plane-β.