Mathematics: Post your doubts here!

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sagar65265

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tanmaydube said:
View attachment 23414




The answer to the first part is 1/x + 10/x^2 + 1/(10-x)


Cannot do part (ii) please help!
Click to expand...


Okay, the first thing to notice is that they've given you the rate of change of x with respect to t in terms of x; this is not a very easy form to work with,
it's like writing dy/dx in terms of y. What you can do now is take the reciprocal of each side; it is proved that

dx/dy = 1/(dy/dx)

so you can take the reciprocal on each side and write

dt/dx = 100/x^2(10-x)

so this is more useful, since this equation now tells us how t changes with respect to x in terms of x. Another thing about this new form is that it is now the same equation as the one they asked you to split up into partial fractions in part (i), so you can substitute the values you have obtained earlier:

dt/dx = 1/x + 10/x^2 + 1/(10-x)

now you have an integral that can be carried out more easily than the earlier one. Multiplying both sides by dx,

dt = 1/x + 10/x^2 + 1/(10-x) dx

This you can integrate to give

t = ln(x) - 10/x - ln(10-x) + k ------- (1)

This constant is still an issue, so substitute the given values :- t = 0 and x = 1 to obtain:

0 = 0 - 10 - ln(9) + k
10 + ln(9) = k

So you can plug this value of k back in your equation (1) to get

t = ln(x) - 10/x - ln(10-x) + 10 + ln(9)

And using the properties of logarithms ( ln(a) + ln(b) = ln(ab) and ln(a) - ln(b) = ln(a/b) )

t = ln((9 times x)/(10-x)) - 10/x + 10
t = ln(9x/(10-x)) - 10/x + 10

which is the answer in the marking scheme.
Hope this helped!

Note: Before multiplying both sides by dx, make sure that you have separated out the variables in the equation, i.e. all terms in t must be on the left side and all terms in x must be on the right side, because it may not be possible to integrate t in terms of x or vice versa; separate the terms to the correct side and integrate both sides to get a differential equation with the constant.

Good Luck!
 
iKhaled

iKhaled

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Beaconite007 said:
http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Mathematics (9709)/9709_s12_qp_42.pdf
Can anybody please help me with question 4 part ii)?
Click to expand...
in part i we got v= 40 + Tsin25 and x= Tcos25

we have been told that the system is in equilibrium so this means the frictional force (F) will be equal to horizontal component (x) of the tension and we know the equation F = μR
R is the contact force which is the force that the rod is acting on ring which is 40 + Tsin25

substitute these in our equation and we will get:

T cos25 = 0.4(40 + T sin25)
T cos 25 = 16 + 0.4T sin25
T cos25 - 0.4T sin25 = 16
T = 16/(cos 25-0.4sin25)
T= 21.7 N...thats it!

i hope u got it
 
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yubakkk

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Please answer this question
For a biased cubical dice the probability of any particular score between 1 and 6 (inclusive) being obtained is inversely proportional to that score. Find the probability of scoring a 1.​
 
syed1995

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yubakkk said:
Please answer this question​
For a biased cubical dice the probability of any particular score between 1 and 6 (inclusive) being obtained is inversely proportional to that score. Find the probability of scoring a 1.​
Click to expand...

I don't know .. I think this is what it will be... What's the answer?

1 ----- 2 ----- 3 ---- 4 ---- 5 ---- 6
x/1 + x/2 + x/3+ x/4 + x/5 + x/6 = 1

49x/20 = 1
x = 20/49


Probability of 1 = x/1
= 20/49 Answer
 
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syed1995 said:
I don't know .. I think this is what it will be... What's the answer?

1 ----- 2 ----- 3 ---- 4 ---- 5 ---- 6
x/1 + x/2 + x/3+ x/4 + x/5 + x/6 = 1

49x/20 = 1
x = 20/49


Probability of 1 = x/1
= 20/49 Answer
Click to expand...
what i thought at first thought but i dont know if it is rite :confused:
 
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Swechhya

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http://papers.xtremepapers.com/CIE/...AS Level/Mathematics (9709)/9709_w08_qp_3.pdf

question no. 9 please please please help :/

and

http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Mathematics (9709)/9709_s09_qp_3.pdf


question number 6

The parametric equations of a curve are
x = a cos^3 t, y = a sin^3 t,
where a is a positive constant and 0 < t < pie/2
(i) Express dy/dx in terms of t. [3]
(ii) Show that the equation of the tangent to the curve at the point with parameter t is
x sin t + y cos t = a sin t cos t.

I've done part one. How do you do part (ii) ?

please help! Thanks in advance :)
 
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can anyone help me by telling me how to find out angle between tangents and angle between tangents and x axis :/. for some reason i keep forgetting
 
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