Toughest Math P3 doubts! Here.

[Thought blocker]

Thought blocker are you taking A2 Maths this session too?

The first time m gonna give A2

 

[I Believe]

but why we taking 1m^2 ?

you know area of square is 1m x 1m, so 1 m^2, right?

 

[Thought blocker]

you know area of square is 1m x 1m, so 1 m^2, right?

FRom where did square come from?
I am sorry I being stupid, but seriously I am not getting it.

 

[I Believe]

FRom where did square come from?
I am sorry I being stupid, but seriously I am not getting it.

Since it's a rectangular base, base area is square, and we are given that each side measures 1 m
the formula for area of square is x^2, (that's where the square comes from)
so 1m^2 is the base area

 

[Thought blocker]

Since it's a rectangular base, base area is square, and we are given that each side measures 1 m
the formula for area of square is x^2, (that's where the square comes from)
so 1m^2 is the base area

OH YEAH Thank you! ^_^

 

[Thought blocker]

I Believe can u please explain that thing that u said that we did with understanding.
"You know it's the same method that they do like dv/dt =dv/dh x dh/dt, only that we did it with understanding instead of using formula"

 

[I Believe]

I Believe can u please explain that thing that u said that we did with understanding.
"You know it's the same method that they do like dv/dt =dv/dh x dh/dt, only that we did it with understanding instead of using formula"

 

[Thought blocker]

View attachment 61272 View attachment 61273

from where u got this?

 

[I Believe]

from where u got this?

I have solved topical past papers by Redspot though I'm recommended never to use them... so I've stopped using them now.
Unsolved papers and markschemes is the best way if you want an A
solved papers are for average students only, and they will get you no more than Bs

 

[Thought blocker]

I have solved topical past papers by Redspot though I'm recommended never to use them... so I've stopped using them now.
Unsolved papers and markschemes is the best way if you want an A
solved papers are for average students only, and they will get you no more than Bs

True!
Now i get this, thank you!

 

[I Believe]

I can't seem to proceed after taking the derivative which is : -2tanx(e^-2x) + (sec^2)x (e^-2x)

 

[Thought blocker]

View attachment 61274
I can't seem to proceed after taking the derivative which is : -2tanx(e^-2x) + (sec^2)x (e^-2x)

use the identity sec^2(x) = 1 + tan^2(x)
then take e^-(2x) common, and then u will get a quadratic expression in terms of tan(x)
Then make that quadratic equation in completing square thing a = -1 b = 1

 

[I Believe]

use the identity sec^2(x) = 1 + tan^2(x)
then take e^-(2x) common, and then u will get a quadratic expression in terms of tan(x)
Then make that quadratic equation in completing square thing a = -1 b = 1

Got it but
I'm getting [x+(-tanx)]^2
that means a =1 but b=-1
are you getting the same?

 

[I Believe]

use the identity sec^2(x) = 1 + tan^2(x)
then take e^-(2x) common, and then u will get a quadratic expression in terms of tan(x)
Then make that quadratic equation in completing square thing a = -1 b = 1

Got it but
I'm getting [x+(-tanx)]^2
that means a =1 but b=-1
are you getting the same?

 

[Thought blocker]

Got it but
I'm getting [x+(-tanx)]^2
that means a =1 but b=-1
are you getting the same?

No
e^-(2x)[-1 + tan(x)]^2

 

[Thought blocker]

Got it but
I'm getting [x+(-tanx)]^2
that means a =1 but b=-1
are you getting the same?

how did u get that x?

 

[I Believe]

how did u get that x?

you know to solve quadratic (x^2 +bx +c) by completing the square, it's (x +b/2)^2 that simply gives us [x+(-tanx)]^2 or (x-tanx)^2 isn't it?

 

[I Believe]

No
e^-(2x)[-1 + tan(x)]^2

can you please show me the solution?

 

[Thought blocker]

can you please show me the solution?

tan^2(x) - 2tan(x) + 1
let u = tan(x)
u^2 - 2u + 1 which is equivalent to (u - 2/2)^2 -(-2/2)^2 + 1 = (u -1)^2
therefore (tanx -1)^2 thus we want in form of (a + btanx)^2 thus (-1 + tanx)^2 hence a = -1 and b =1

 

[I Believe]

tan^2(x) - 2tan(x) + 1
let u = tan(x)
u^2 - 2u + 1 which is equivalent to (u - 2/2)^2 -(-2/2)^2 + 1 = (u -1)^2
therefore (tanx -1)^2 thus we want in form of (a + btanx)^2 thus (-1 + tanx)^2 hence a = -1 and b =1

get it now
thanks