MathQA tutor - Ask your A-level Maths questions here!

[Mathducator]

@atutor2001


Primary school mathematics is harder in a sense: you are like fighting with bare hands (e.g. you normally don’t use algebra, but bar model diagrams). Secondary and JC mathematics is like fighting with knives and guns – it’s easier to ‘kill’, but you have to learn how to use the weapons first.

[OK Lor]

Hi,


Please evaluate
http://www.postimage.org/

Thanks.

[mathqa]

OK Lor:

Hi,


Please evaluate
http://www.postimage.org/

Thanks.

Let u=tan(x). Rewrite it in term of u, it would become an integral of an elementary function in form of

http://lh6.ggpht.com/_nr85VD4DdiA/TPjKaT7U5HI/AAAAAAAAAE0/jSSE2KNWb9E/s800/elementary-function-integral.png\">

[OK Lor]

mathqa:

OK Lor:

Hi,


Please evaluate
http://www.postimage.org/

Thanks.

Let u=tan(x). Rewrite it in term of u, it would become an integral of an elementary function in form of

http://lh6.ggpht.com/_nr85VD4DdiA/TPjKaT7U5HI/AAAAAAAAAE0/jSSE2KNWb9E/s800/elementary-function-integral.png\">

Hi mathqa, thanks .

[OK Lor]

Hi,

Pls help on (ii):
A curve has parametric equation x = 2t – 1, y = 1/(t² + 1).
(i) Prove that the equation of the tangent at the point with parameter t is (t² + 1)² y + tx = 3t² - t + 1.
(ii) The tangent at point where t = 3 meets the curve again at the point where t = q. Find the value of q.
Ans: -4/3

Thanks.

[FrekiWang]

OK Lor:

Hi,

Pls help on (ii):
A curve has parametric equation x = 2t – 1, y = 1/(t² + 1).
(i) Prove that the equation of the tangent at the point with parameter t is (t² + 1)² y + tx = 3t² - t + 1.
(ii) The tangent at point where t = 3 meets the curve again at the point where t = q. Find the value of q.
Ans: -4/3

Thanks.

Assuming (i) has been proven.

tangent at t=3 has equation
[(3^2+1)^2]y+3x=3*3^2-3+1
simplified, we have 100y+3x=25

to find the point of intersection between the curve and the straight line, we need to solve
100y+3x=25 and x=2t-1, y=1/(t^2+1)

By subsititution,
100/(t^2+1)+3(2t-1)=25
100+(6t-3)(t^2+1)=25(t^2+1)
100+6t^3-3t^2+6t-3=25t^2+25
6t^3-28t^2+6t+72=0
3t^3-14t^2+3t+36=0

since we know t=3 is a repeated root as a tangent, (t-3)^2 must be a factor, (or t^2-6t+9 must be a factor)

after doing a long division, we have
(3t+4)(t-3)^2=0

the other root is t=-4/3 (when 6t+8=0)

[equink]

I’ve been trying to figure this out but I still don’t understand.

Why is dx/dy proven to be 1/(dy/dx)? I know that dx/dy cannot be treated as a fraction because it simply isn’t a fraction. Please help,thanks so much!

[albertong]

[quote]
equink posted: I've been trying to figure this out but I still don't understand.
Why is dx/dy proven to be 1/(dy/dx)? I know that dx/dy cannot be treated as a fraction because it simply isn't a fraction. Please help,thanks so much![/quote]@mathstuition88 answered:

Good question! A rigorous proof of that would require analysis (a math course in university), but one can prove it briefly using chain rule,

(dy/dx)(dx/dy)=dy/dy=1, hence dx/dy=1/(dy/dx).

Hope this helps!

Post and get your questions answered here: http://fbl.me/Tt

[Studentx]

deleted.

[student101]

Hi , Any expert in perm&comb in H2 Maths can explain to me clearly how the below questions red circled highlighted parts can get. I did not figure out how that step get . Please help

http://i59.tinypic.com/vhe2h2.jpg\">