ositadima1: brachistochrone problem: Suppose that there is a
rollercoaster. There is point 1 and point 2 . Point 1 is at the higher place when compared to the point 2, so the rollercoaster rolls down from
point 1 to point 2. Assuming no friction, we want to
build a rollercoaster path from point 1 to point 2, so
that the rollercoaster will reach point 2 from point 1 in
shortest time. Find the curve dat would help us achieve this?

This question is tricky, emmm, my fellow engineers lets tackle this one.

We know that rollercoasters employs the use of wheels, so i think the curve which will allow us reach point 2 from point 1 in the shortest possible time is the path traced by a point at rim of the wheels (a point at the edge of the wheel) and this path produces a cycloid.

well i'm still doing some thinking and still trying to see how i can mathematically prove it with calculus of variations.

NOTICE;
there is speculation that this thread was created only for maths gurus, that is not true. rather, this thread is a meeting point for gurus who can spare time to encourage people who are still struggling with mathematics. You will agree with me that it is better for the best hands to solve your problems. so feel free everyone don't be intimidated, bring your questions however simple or difficult they are.
A CANDLE LOOSES NOTHING BY LIGHTING ANOTHER CANDLE

Abeg, i need help wit dis
2^(2x) = 3^(x+4)

2. Prove by inductn
11^(n) - 8^(n) is the multiple of J for all nEN

Richiez:
Isaac Newton was faced with a similar problem. guess wot gurus, he solved it the next day even without the internet or textbooks

Newton was a too much bruv....Morning people!

doubleDx: Newton was a too much bruv....Morning people!

morning General! You're welcome, hope u dreamt of maths

Richiez:
morning General! You're welcome, hope u dreamt of maths

Lol, when you start getting those dreams, it means your mind is battling a math problem and you'll eventually solve it. It happened to me a lot back in school days but not anymore Greetings General richiez

doubleDx:

Lol, when you start getting those dreams, it means your mind is battling a math problem and you'll eventually solve it. It happened to me a lot back in school days but not anymore Greetings General richiez

hahaha then i guess now maths dreams of you

yungryce: Abeg, i need help wit dis
2^(2x) = 3^(x+4)

2. Prove by inductn
11^(n) - 8^(n) is the multiple of J for all nEN

QUESTION 1

2^(2x) = 3^(x+4)
Taking log of both sides;
log[2^(2x)] = log[3^(x+4)]

2xlog2 = (x+4)log3
(x+4)/2x = log2/log3
x/2x + 4/2x = log2/log3
1/2 + 4/2x = 0.63093
4/2x = 0.63093 - 1/2
4/2x = 0.63093 - 0.5
4/2x = 0.13093
cross multiplying;
2x = 4/0.13093
2x = 30.55
x = 30.55/2
x = 15.275 (approx)

QUESTION 2
2. Prove by inductn
11^(n) - 8^(n) is the multiple of J for all nEN

i don't quite get you, by 'J' do you mean complex number?

yungryce: Abeg, i need help wit dis
2^(2x) = 3^(x+4)

2. Prove by inductn
11^(n) - 8^(n) is the multiple of J for all nEN

Richiez has solved your question. Check question 2 again, seems there is an error. You can also use this method if you want to leave your answer in log form.

2^(2x) = 3^(x+4)
Taking log of both sides yields=>

log2^(2x) = log3^(x + 4)
log4^x = log(3^x).(3^4) <= Just breaking with the laws of indices. Now, breaking the RHS with the laws of log yields=>
xlog4 = log3^x + log3^4, collecting like terms =>
xlog4 - xlog3 = 4log3
x (log4 - log3) = 4log3
:.x = 4log3/(log4 - log3)

Which is also the same as=>

x = 4(0.477121255)/(0.602059991 - 0.477121255)
x = 1.90848502/0.124938736
x = 15.275 (approx)

Richiez:

QUESTION 1

2^(2x) = 3^(x+4)
Taking log of both sides;
log[2^(2x)] = log[3^(x+4)]

2xlog2 = (x+4)log3
(x+4)/2x = log2/log3
x/2x + 4/2x = log2/log3
1/2 + 4/2x = 0.63093
4/2x = 0.63093 - 1/2
4/2x = 0.63093 - 0.5
4/2x = 0.13093
cross multiplying;
2x = 4/0.13093
2x = 30.55
x = 30.55/2
x = 15.275 (approx)

QUESTION 2
2. Prove by inductn
11^(n) - 8^(n) is the multiple of J for all nEN

i don't quite get you, by 'J' do you mean complex number?

i tink so. Dats hw d questn was given

yungryce: i tink so. Dats hw d questn was given

well then, i think it is a typographical error. if i am to guess, the' J' should be replaced by '3'

And i thought i was extremely good in maths. Lol. I dont stand a damn chance with these gurus here...una be winch?
Anyways am a maths 100l stdnt and can the o.p give me some words of advise?? Cos i wan bail from this course from next session ooo.

topmostg: And i thought i was extremely good in maths. Lol. I dont stand a damn chance with these gurus here...una be winch? Anyways am a maths 100l stdnt and can the o.p give me some words of advise?? Cos i wan bail from this course from next session ooo.

let's look at it dis way, if u and i run away from mathematics, then who will stay? mathematics is a straight4ward course...no wonder the saying 'mathematics does not lie' a solution to a math problem solved here in Nigeria will definitely be the same with that solved somewhere else. It is very easy to get a 100% in mathematics without having to write long story or do some detailed explanation or cram some long and unending definitions. math is cool so be as cool as it is and don't runaway. however, you wear the shoe and know where it pinches you most. so your decision is paramount. but know that we'l always do our best to help.

Richiez:

well then, i think it is a typographical error. if i am to guess, the' J' should be replaced by '3'

lets take it 2 be 3. Hw will go about it

yungryce: lets take it 2 be 3. Hw will go about it

N- represents natural numbers, i.e positive whole numbers such as; 1,2,3,4....
multiples of 3 are numbers dat divide 3 without remainder.
first, let us see if the statement ir true for n=1
11^1 - 8^1 =
11-8 = 3 ....statement is true
again, let us see if the statement is true for n=2
11^2 - 8^2 = 121 - 64 = 57 ....this is a multiple of 3 so statement is true

now, let's assume that it is true for n=k
11^k - 8^k ...is true

to assert this, we must show that it is also true for n=k+1 11^(k+1) - 8^(k+1) 11*(11^k) - 8*(8^k) which can further be splitted as;
3(11^k) + 8(11^k) - 8(8^k)
3(11^k) - 8(11^k - 8^k)
now looking at the above expression; the first part is multiplied by 3 and is clearly a multiple of 3.
the other part is a multiple of the expression n=k which in turn is a multiple of 3

PROVEN

Good day house. Please help with these questions. Thanks guys.

1. Evaluate:

∫{sec^3(x) - sin(x)cos^3(x)[1 - sec^2(x)} dx

2.(a) Find an equation for the gradient to the tangent of the curve:

y = x^4/2 + x^3/3 - 13x^2/2 + 12x - 4

(b) What is the value(s) of x for which the gradient to the tangent of the curve is 6?

Ghettoguru: Good day house. Please help with these questions. Thanks guys.

1. Evaluate:

∫{sec^3(x) - sin(x)cos^3(x)[1 - sec^2(x)} dx

2.(a) Find an equation for the gradient to the tangent of the curve:

y = x^4/2 + x^3/3 - 13x^2/2 + 12x - 4

(b) What is the value(s) of x for which the gradient to the tangent of the curve is 6?

Question 1

∫{sec^3(x) - sin(x)cos^3(x)[1 - sec^2(x)} dx

This can be splitted =>

= ∫sec^3 x dx - ∫sin x cos^3 x(1 - sec^2 x) dx
=∫sec^3 x dx + ∫sin x cos^3 x(sec^2 x - 1) dx
Now, since sec^2 x - 1 = tan^2 x = sin^2 x/cos^2 x, substituting yields =>

= ∫sec^3 x dx + ∫sin x cos^3 x sin^2 x dx/cos^2 x, cancelling out reduces the question to =>
= ∫sec^3 x dx + ∫sin ^3 x cos x dx

Now, let's integrate the function individually starting with ∫sec^3 x dx which can be rewritten as =>

∫sec^2 x sec x dx, using integration by parts=>
Let's put

u = sec x, differentiating yields du = secxtanx dx.
Again let's put dv = sec^2 x dx, integrating yields v = tan x. Now the formula for integration by parts is =>

∫udv = uv - ∫vdu, substituting yields=>
∫sec^3 x dx = sec x tan x - ∫tan x (sec x tan x)dx
∫sec^3 x dx = sec x tan x - ∫tan^2 x sec x dx
remember that tan^2 x = sec^2 x - 1, substituting yields=>
∫sec^3 x dx = sec x tan x - ∫sec x (sec^2 x - 1) dx
∫sec^3 x dx = sec x tan x - ∫sec^3 x dx + ∫sec x dx

Collecting like terms by taking the bolded part to the LHS yields =>
2∫sec^3 x dx = sec x tan x + ∫sec x dx
2∫sec^3 x dx = sec x tan x + ln|sec x + tan x| + C
:. ∫sec^3 x dx = (1/2)sec x tan x + (1/2)ln|sec x + tan x| + C

Having solved the first part, we can now go back to the main question which has been reduced to => ∫sec^3 x dx + ∫sin ^3 x cos x dx and subtitute the solution for ∫sec^3 x dx.

∫sec^3 x dx + ∫sin ^3 x cos x dx = (1/2)sec x tan x + (1/2)ln|sec x + tan x| + ∫sin ^3 x cos x dx

To integrate the second part, let put u = sin x, differentiating yields du = cos x dx.

:. ∫sin ^3 x cos x dx = ∫u^3 du
= 1/4 (u^4) + C. Substituting u = sin x
∫sin ^3 x cos x dx = (1/4) sin^4 x + C

:. ∫{sec^3(x) - sin(x)cos^3(x)[1 - sec^2(x)} dx = ∫sec^3 x dx + ∫sin ^3 x cos x dx =>

(1/2)sec x tan x + (1/2)ln|sec x + tan x| + (1/4)sin ^4 x + C

*Where C is the arbitrary constant*

Question 2 coming up later!

Ghettoguru: Good day house. Please help with these questions. Thanks guys.

1. Evaluate:

∫{sec^3(x) - sin(x)cos^3(x)[1 - sec^2(x)} dx

2.(a) Find an equation for the gradient to the tangent of the curve:

y = x^4/2 + x^3/3 - 13x^2/2 + 12x - 4

(b) What is the value(s) of x for which the gradient to the tangent of the curve is 6?

Question 2a

y = x^4/2 + x^3/3 - 13x^2/2 + 12x - 4

Since y is a function of x. Then the gradient of the tangent to the curve is simply => dy/dx which is the first derivative of the equation of the curve.

dy/dx = 4(x^3/2) + 3(x^2/3) - 2(13x/2) + 12
dy/dx = 2x^3 + x^2 - 13x + 12.
Hence the gradient of the tangent to the curve is simply=>
2x^3 + x^2 - 13x + 12

Question 2b. Follows that if the gradient to the tangent of the curve is 6, what is x?. So, I guess that's clear and simple.

Since the gradient = dy/dx = 2x^3 + x^2 - 13x + 12

Then it follows that=>

2x^3 + x^2 - 13x + 12 = 6
2x^3 + x^2 - 13x + 6 = 0
^ solving the above polynomial gives us the values of x at which the gradient equals 6.

So, I think (x - 2) is a factor of the polynomial. Using division of polynomials yields=>
(x - 2)(2x^2 + 5x - 3) = 0

:. x = 2 or 2x^2 + 5x - 3 = 0
2x^2 + 6x - x - 3 = 0
2x(x + 3) - 1(x + 3) = 0
:. (x + 3)(2x -1) = 0

So the values of x for which the gradient of the curve equals 6 are =>

x = 2, -3 and 1/2.

I hope I have helped you.

^^^

Oh I hail thee Bros. I doff my hat to you sir. Thanks a bunch, appreciate y'all.

doubleDx:

Question 1

∫{sec^3(x) - sin(x)cos^3(x)[1 - sec^2(x)} dx

This can be splitted =>

= ∫sec^3 x dx - ∫sin x cos^3 x(1 - sec^2 x) dx
=∫sec^3 x dx + ∫sin x cos^3 x(sec^2 x - 1) dx
Now, since sec^2 x - 1 = tan^2 x = sin^2 x/cos^2 x, substituting yields =>

= ∫sec^3 x dx + ∫sin x cos^3 x sin^2 x dx/cos^2 x, cancelling out reduces the question to =>
= ∫sec^3 x dx + ∫sin ^3 x cos x dx

Now, let's integrate the function individually starting with ∫sec^3 x dx which can be rewritten as =>

∫sec^2 x sec x dx, using integration by parts=>
Let's put

u = sec x, differentiating yields du = secxtanx dx.
Again let's put dv = sec^2 x dx, integrating yields v = tan x. Now the formula for integration by parts is =>

∫udv = uv - ∫vdu, substituting yields=>
∫sec^3 x dx = sec x tan x - ∫tan x (sec x tan x)dx
∫sec^3 x dx = sec x tan x - ∫tan^2 x sec x dx
remember that tan^2 x = sec^2 x - 1, substituting yields=>
∫sec^3 x dx = sec x tan x - ∫sec x (sec^2 x - 1) dx
∫sec^3 x dx = sec x tan x - ∫sec^3 x dx + ∫sec x dx

Collecting like terms by taking the bolded part to the LHS yields =>
2∫sec^3 x dx = sec x tan x + ∫sec x dx
2∫sec^3 x dx = sec x tan x + ln|sec x + tan x| + C
:. ∫sec^3 x dx = (1/2)sec x tan x + (1/2)ln|sec x + tan x| + C

Having solved the first part, we can now go back to the main question which has been reduced to => ∫sec^3 x dx + ∫sin ^3 x cos x dx and subtitute the solution for ∫sec^3 x dx.

∫sec^3 x dx + ∫sin ^3 x cos x dx = (1/2)sec x tan x + (1/2)ln|sec x + tan x| + ∫sin ^3 x cos x dx

To integrate the second part, let put u = sin x, differentiating yields du = cos x dx.

:. ∫sin ^3 x cos x dx = ∫u^3 du
= 1/4 (u^4) + C. Substituting u = sin x
∫sin ^3 x cos x dx = (1/4) sin^4 x + C

:. ∫{sec^3(x) - sin(x)cos^3(x)[1 - sec^2(x)} dx = ∫sec^3 x dx + ∫sin ^3 x cos x dx =>

(1/2)sec x tan x + (1/2)ln|sec x + tan x| + (1/4)sin ^4 x + C

*Where C is the arbitrary constant*

Question 2 coming up later!

Richiez:
N- represents natural numbers, i.e positive whole numbers such as; 1,2,3,4....
multiples of 3 are numbers dat divide 3 without remainder.
first, let us see if the statement ir true for n=1
11^1 - 8^1 =
11-8 = 3 ....statement is true
again, let us see if the statement is true for n=2
11^2 - 8^2 = 121 - 64 = 57 ....this is a multiple of 3 so statement is true

now, let's assume that it is true for n=k
11^k - 8^k ...is true

to assert this, we must show that it is also true for n=k+1 11^(k+1) - 8^(k+1) 11*(11^k) - 8*(8^k) which can further be splitted as;
3(11^k) + 8(11^k) - 8(8^k)
3(11^k) - 8(11^k - 8^k)
now looking at the above expression; the first part is multiplied by 3 and is clearly a multiple of 3.
the other part is a multiple of the expression n=k which in turn is a multiple of 3

PROVEN

guru. 10 gbosa 4 u. Abeg, e get anoda one.
Prove that for any two nos a & b, & for any nEN that
(a+b)^n = a^n + na^(n-1)b + [n(n-1)/2!]a^(n-2)b^(2) + ...+ [n!/(n-k)!k!]a^(n-k)b^k + ... + b^(k)

Wooooow!!! Am soooo thrilled by what am seeing here....is it that u solve and practice Maths everyday Cause this is really beyond me!
Keepit up guys!

@General DOUBLEDX, you really did a great job.

yungryce:
guru. 10 gbosa 4 u. Abeg, e get anoda one.
Prove that for any two nos a & b, & for any nEN that
(a+b)^n = a^n + na^(n-1)b + [n(n-1)/2!]a^(n-2)b^(2) + ...+ [n!/(n-k)!k!]a^(n-k)b^k + ... + b^(k)

thanks, u'r welcome
i'll check ur question later in the day.

SpicyMimi: Wooooow!!! Am soooo thrilled by what am seeing
here....is it that u solve and practice Maths everyday Cause this is
really beyond me!
Keepit up guys!

lol...not really the everyday thing, i think the passion for it contributes more...we are not geniuses oh we are still learning. thanks anyway

SpicyMimi: Wooooow!!! Am soooo thrilled by what am seeing here....is it that u solve and practice Maths everyday Cause this is really beyond me!
Keepit up guys!

^^^ Hi

doubleDx:

Lol, when you start getting those dreams, it means your mind is battling a math problem and you'll eventually solve it. It happened to me a lot back in school days but not anymore Greetings General richiez

oboy u 1 giv am nightmares.

@General Osita, i've been expecting you to shed more light on the solution to the brachistochrone problem, and please help me send soft copies of the advanced engineering maths textbooks to richobiagba@gmail.com
thanks in anticipation

Richiez: @General Osita, i've been expecting you to shed more light on the solution to the brachistochrone problem, and please help me send soft copies of the advanced engineering maths textbooks to richobiagba@gmail.com
thanks in anticipation

Ok, I will post the solution and the books here, that will be tomorrow.

ositadima1: Ok, I will post the solution and the books here, that will be tomorrow.

okay thanks

@ Richie nd Doubledx I must commend ur work here...u guys 're doing a great Job here..I enjoy reading stuffs on dis page...alot. m not a maths guru neither am I a learner..I read Economics & Statistics and I am professional in my field..but when I see guys like dis..I luv 2 b a learner.kip it up!

ControG: @ Richie nd Doubledx I must commend ur work here...u guys 're doing a great Job here..I enjoy reading stuffs on dis page...alot. m not a maths guru neither am I a learner..I read Economics & Statistics and I am professional in my field..but when I see guys like dis..I luv 2 b a learner.kip it up!

Thanks alot bro, we're just following the saying; 'a candle looses nothing by lighting another candle'

Ghettoguru: ^^^

Oh I hail thee Bros. I doff my hat to you sir. Thanks a bunch, appreciate y'all.

You are welcome bruv!

Richiez: @General DOUBLEDX, you really did a great job.

Thanks General richiez

ositadima1:

Ok, I will post the solution and the books here, that will be tomorrow.

Ok good Gen! I troway salute.