Mathematics: Post your doubts here!

syed1995

syed1995

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PhyZac or anyone else..

Know of any place (online or via calculator) where one can get inverse of phi? without referring to the normal tables I mean...
 
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syed1995 said:
naw.. I was talking about physics and chemistry.. they are 1 hour each for theory.. It takes 1 hour to solve them.. so one can solve like 3 or 4 papers in a day.. but with maths P1 I can't even solve 2 papers a day :\

And no one can solve the paper in 45 minutes! .. takes me about 1hr 30-40 minutes to solve a complete paper..
Click to expand...
Honestly for P1 1hr 45 is plenty.
 
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PhyZac

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syed1995 said:
PhyZac or anyone else..

Know of any place (online or via calculator) where one can get inverse of phi? without referring to the normal tables I mean...
Click to expand...
Wait, you mean phi of statistic.

Well, i dont think so there is a way, and if there is, in exam the only way is table. (what i think)

EDIT: check this
 
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iKhaled

iKhaled

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PhyZac said:
Okay, from first part you get that dy/dx = -sint/cost

Now they want the equation of tangent when t is t!!

Equation is in form

(y-y1) = m (x-x1)

M is the gradient, and when t is t, the gradient is -sint/cost ( the dy/dx)

the y1 is asin^3 (t)

and the x1 is acos^3(t)

Sub those value, and do some algebra and then you get the answer! [P.S, i am sure you know how to do this step, if you get any difficulty ask]
Click to expand...
thanks man that really helped!
 
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minie23 said:
http://papers.xtremepapers.com/CIE/...S Level/Mathematics (9709)/9709_s11_qp_31.pdf
Please help for no. 7 (i) :(
Click to expand...
OKaay..see..!

First we will do the limits.
in the first one it is 2 and 0

so, the sub is x = t^2 + 1
when t is 2, x is 5
when t is 0, x is 1

So now new limit are 5 and 1

Now to main thing!

First thing I do is find the dy/dx of substitution.

not of this >> x = t^2 + 1

make t the subject first.

t^2 = x - 1
t = (x - 1)^1/2 [power 1/2 is square root]

okay now, take dy/dx, but to be precise dt/dx

dt/dx = 1/2 ( x - 1)^(-1/2)

Now next thing!

We now should substitute the t's of formula.

since x=t^2 +1

then this
4t^3 ln(t^2 + 1)

is
4t^3 lnx

One more t is left! what is t alone?

x = t^2 +1
t = (x - 1)^1/2

so sub this and get

4((x-1)^1/2)^3 lnx

4(x-1)^3/2 ln x

Now add the differential to the top equation

4(x-1)^3/2 ln x 1/2 ( x - 1)^(-1/2)

4/2 (x-1) ^(3/2 -1/2) lnx

2 (x-1) ^1 ln x

2x - 2 ln x , with limits 5 to 1
 
minie23

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PhyZac said:
OKaay..see..!

First we will do the limits.
in the first one it is 2 and 0

so, the sub is x = t^2 + 1
when t is 2, x is 5
when t is 0, x is 1

So now new limit are 5 and 1

Now to main thing!

First thing I do is find the dy/dx of substitution.

not of this >> x = t^2 + 1

make t the subject first.

t^2 = x - 1
t = (x - 1)^1/2 [power 1/2 is square root]

okay now, take dy/dx, but to be precise dt/dx

dt/dx = 1/2 ( x - 1)^(-1/2)

Now next thing!

We now should substitute the t's of formula.

since x=t^2 +1

then this
4t^3 ln(t^2 + 1)

is
4t^3 lnx

One more t is left! what is t alone?

x = t^2 +1
t = (x - 1)^1/2

so sub this and get

4((x-1)^1/2)^3 lnx

4(x-1)^3/2 ln x

Now add the differential to the top equation

4(x-1)^3/2 ln x 1/2 ( x - 1)^(-1/2)

4/2 (x-1) ^(3/2 -1/2) lnx

2 (x-1) ^1 ln x

2x - 2 ln x , with limits 5 to 1
Click to expand...

Thank you :)
 
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