Hey, guys greetings to the entire Nairaland family , I'm Benjamin .K a.k.a benbuks , a mathematician (calculus lover ) by passion & a statistician by profession .

Just thought of creating this interesting & educative thread for all those interested , its purely calculus (differential or integral with maybe ODE)

Questions will be posted , first to post correct answer (maybe + solution) gets 5pts , then person with the highest number of point at the end the day wins N200 airtime of any network , then that person will host the next contest . in similar manna on & on ,


questions/ contributions /criticisms. welcomed .

Let the questions roll in na

Oya

Am a programmer. Let's do it. Rep uniosun

I'm in, i hope they wont be too tough to solve sha! Lol. Let the game begin

hmmm.

Ok guys , the issues now is Timing ,

when do you think will be the best time ?. morning , evening or night .?




all vote now ..

Morning...when thr brain is still fresh

Let it begin.....

agentofchange1:


Ok guys , the issues now is Timing ,

when do you think will be the best time ?. morning , evening or night .?




all vote now ..

That's exactly the problem, anyway, I think evening time after the day's work would be fine by me

I and my friend bolkay will always win your questions.

We have a whatsapp group we created simply because of calculus.

Roll your question, bro.

Following...

Help me differenciate log3 base x. Thanks

ValentineMary:
Help me differenciate log3 base x. Thanks

y'=3/xlog3basex

Daughterzion:
Let the questions roll in na

Lolz , can't wait right ?.

don't worry Asap.

OK guys , the game starts. tomorrow by 7pm.










BTW: exercise your brains with this

$(3x+2)/(x^4 +4) dx

ValentineMary:
Help me differenciate log3 base x. Thanks

forward your questions to the math clinic not here .

.

agentofchange1:



OK guys , the game starts. tomorrow by 7pm.










BTW: exercise your brains with this

$(3x+2)/(x^4 +4) dx

this your question na tough one, but let me try
$(3x+2)/(x^4 +4) dx
=$(3(x^2)^1/2 +2)/((x^2)2 +4) dx
let x^2 = p
dx = dp to simply things, re-writing
$(3p^1/2 +2)/(p^2 +4) dp
using trig sub method
=$(3p^1/2 + 2)/2 d¤
=(1/2)$3p^1/2 + 2 d¤
the rest are easy but my problem is p^1/2 = tan¤, forget the constant (2^1/2).
How can I integrate the square root of tan¤

agentofchange1:


put. log 3 base x = y

=> x^y= 3


taking Napierien log of both sides

=> ylnx=ln3

put f(x,y)= ylnx-ln3=0

y'= dy/dx = - f_y / f_x

now , f_y= lnx

& f_x= y/x


thus we have y'= - lnx /(y/x)


simplify further & replace 'y' that's it

please, let me shine
y = log_x3.
Method 1:
y = log_e3/log_ex
for easy typing, permit me to change to
y = 1n3 / 1nx
y = (1n3)(1nx)^-1
using product rule (short cut)
dy/dx = (1n3)(-1)(1nx)^-2(1/x)
dy/dx = -1n3/x1nx1nx.
Method 2:
y = 1n3/1nx
quotien rule
dy/dx = (-1n3/x)/(1nx{1nx})
dy/dx = -1n3/x1nx1nx.
Method 3:
y = 1n3/1nx
implicit method
1nx (y) = 1n3
1nx (dy/dx) + y/x = 0
dy/dx = -y/x1nx
substitue y from the question
dy/dx = -1n3/x1nx1nx

Lemme also drop mine. Btw, its off head assumption, but its a lil bit realistic.
Assuming that the rate of outflow of water from a tank of capacity 2500 dm³ as a result of a hole at the base of the tank is directly proportional to the volume of water in it at each point in time, if it takes 10 hours for the water to drop from full capacity to 1%, find
A) how long it'll take to drop to 10%
B) the volume of water left after 1 hour
C) the volume of water left after 9 hours
In short in a tabular form, show the % of amount left for each of the first ten hours.
*the solution is in the thinking*

ohh so sorry guys , got really busy with school tinz ...but nonetheless the quiz/ game will hold by his grace .

BTW: any 1 is free to host it b4 I get ready. ,plz feel free to post & solve calculus questions here



greetings ..



1luv

#shalom..

@agentofchange please email me at themoringashop at g mail dot com and when do please include your nairaland name.

Thank you

agentofchange1:



OK guys , the game starts. tomorrow by 7pm.










BTW: exercise your brains with this

∫(3x+2)/(x^4 +4) dx

∫(3x + 2)dx/(x⁴ + 4)

To solve this, we will have to factor the denominator into linear and irreducible quadratic terms:::

So that::: (x⁴ + 4) => (x² - 2x + 2)(x² + 2x + 2)
Which brings us to => ∫(3x + 2)dx/(x² - 2x + 2)(x² + 2x + 2)

Breaking the expression (3x + 2)/(x² - 2x + 2)(x² + 2x + 2) using partial fractions =>
::: ∫{[(5 - x)/4(x² - 2x + 2)] + [(x + 1)/4(x² + 2x + 2)]}dx
::: 1/4∫(x + 1)dx/(x² + 2x + 2) + 1/4∫ (5 - x)dx/(x² - 2x + 2)

This is where we will apply a little trick to the expression ::: (x + 1)/(x² + 2x + 2) rewriting it as => {(2x + 2)/2(x² + 2x + 2)} - 2/(x² + 2x + 2). Then we continue...integrating term by term and factoring out constants

::: 1/4∫{[(2x + 2)/2(x² + 2x + 2)] - 2/(x² + 2x + 2)}dx + 1/4∫ (5 - x)dx/(x² - 2x + 2)
::: 1/8∫(2x + 2)dx/(x² + 2x + 2) - 1/2∫dx/(x² + 2x + 2) + 1/4∫ (5 - x)dx/(x² - 2x + 2)

To integrate => (2x + 2)dx/(x² + 2x + 2), Let's Put s = x² + 2x + 2, so that ds = (2x + 2)dx; substituting back will give us:::
::: 1/8∫ds/s - 1/2∫dx/(x² + 2x + 2) + 1/4∫ (5 - x)dx/(x² - 2x + 2)
::: 1/8ln(s) - 1/2∫dx/(x² + 2x + 2) + 1/4∫ (5 - x)dx/(x² - 2x + 2)

To integrate the next expression 1/(x² + 2x + 2); let's first complete the square of the denominator => x² + 2x + 2 = (x + 1)² + 1²

::: 1/8ln(s) - 1/2∫dx/[(x + 1)² + 1] + 1/4∫ (5 - x)dx/(x² - 2x + 2)
Then let's put t = x + 1, so that dt = dx; substituting back will give us =>
::: 1/8ln(s) - 1/2∫dt/(t² + 1) + 1/4∫ (5 - x)dx/(x² - 2x + 2)
::: 1/8ln(s) - 1/2 arcTan(t) + 1/4∫ (5 - x)dx/(x² - 2x + 2)

Then let's express::: (5 - x)/(x² - 2x + 2) as ==> 4/(x² - 2x + 2) - (2x - 2)/2(x² - 2x + 2)

::: 1/8ln(s) - 1/2 arcTan(t) + 1/4∫ {4/(x² - 2x + 2) - (2x - 2)/2(x² - 2x + 2)}dx
::: 1/8ln(s) - 1/2 arcTan(t) - 1/8∫(2x - 2)dx/(x² - 2x + 2) + ∫dx/(x² - 2x + 2)
::: Then we continue integrating term by term factoring out constants =>

To integrate (2x - 2)dx/(x² - 2x + 2), let's put u = x² - 2x + 2 so that du = (2x - 2)dx, substituting back will now yield :::

::: 1/8ln(s) - 1/2 arcTan(t) - 1/8∫du/u + ∫dx/(x² - 2x + 2)
::: 1/8ln(s) - 1/2 arcTan(t) - 1/8ln(u) + ∫dx/(x² - 2x + 2)

Then finally we take the integrand of dx/(x² - 2x + 2) by rewriting dx/{(x - 1)² + 1} ==>

::: 1/8ln(s) - 1/2 arcTan(t) - 1/8ln(u) + ∫ dx/{(x - 1)² + 1}
Put v = x - 1, so that dv = dx; substituting back yields =>
::: 1/8ln(s) - 1/2 arcTan(t) - 1/8ln(u) + ∫ dv/(v² + 1)
::: 1/8ln(s) - 1/2 arcTan(t) - 1/8ln(u) + arcTan(v) + Constant

Then finally, we can now replace s, t, u and v back into the answer.

Recall that:::

s = x² + 2x + 2
t = x + 1
u = x² - 2x + 2
v = x - 1

:. 1/8ln(s) - 1/2 arcTan(t) - 1/8ln(u) + arcTan(v) + Constant
::: 1/8ln(x² + 2x + 2) - 1/2arcTan(x + 1) - 1/8ln(x² - 2x + 2) + arcTan(x - 1) + Constant
::: 1/8ln(x² + 2x + 2) - 1/8ln(x² - 2x + 2) + arcTan(x - 1) - 1/2arcTan(x + 1) + Constant
::: 1/8ln(x² + 2x + 2) - 1/8ln(x² - 2x + 2) - arcTan(1- x) - 1/2arcTan(x + 1) + Constant

Nice thread! I will drop one question later!

Drum roll. Grrroll, grroll