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[OK Lor]

Hi,


If y = eˣ / cos x, show that d²y/dx² = 2 tanx dy/dx + 2y.

Thanks.

[iFruit]

OK Lor:

Hi,


If y = eˣ / cos x, show that d²y/dx² = 2 tanx dy/dx + 2y.

Thanks.

y = eˣ / cos x = eˣ sec x

dy/dx = eˣ sec x + eˣ sec x tan x = eˣ sec x (1+ tan x) = y (1+ tan x)

d²y/dx² = dy/dx (1 + tan x) + y ( sec² x) = dy/dx (1 + tan x) + y ( 1 + tan² x) = y(1+ tan x)(1+tan x) + y (1 + tan² x) = 2y (1 + tan x + tan² x) ----(1)

2 tanx dy/dx + 2y = 2 tanx . y (1+ tan x) + 2y = 2y ( 1 + tan x + tan² x)---- (2)

From (1) and (2), d²y/dx² = 2 tanx dy/dx + 2y.

HTH

[OK Lor]

iFruit:

OK Lor:

Hi,


If y = eˣ / cos x, show that d²y/dx² = 2 tanx dy/dx + 2y.

Thanks.

y = eˣ / cos x = eˣ sec x

dy/dx = eˣ sec x + eˣ sec x tan x = eˣ sec x (1+ tan x) = y (1+ tan x)

d²y/dx² = dy/dx (1 + tan x) + y ( sec² x) = dy/dx (1 + tan x) + y ( 1 + tan² x) = y(1+ tan x)(1+tan x) + y (1 + tan² x) = 2y (1 + tan x + tan² x) ----(1)

2 tanx dy/dx + 2y = 2 tanx . y (1+ tan x) + 2y = 2y ( 1 + tan x + tan² x)---- (2)

From (1) and (2), d²y/dx² = 2 tanx dy/dx + 2y.

HTH

Hi iFruit,

Thanks for your help. Was stuck and coundn't continue from (1) to get the answer without (2), which it was just one more step to go, the reverse of (2) :lol:

[iFruit]

OK Lor:

iFruit:

[quote=\"OK Lor\"]Hi,


If y = eˣ / cos x, show that d²y/dx² = 2 tanx dy/dx + 2y.

Thanks.

y = eˣ / cos x = eˣ sec x

dy/dx = eˣ sec x + eˣ sec x tan x = eˣ sec x (1+ tan x) = y (1+ tan x)

d²y/dx² = dy/dx (1 + tan x) + y ( sec² x) = dy/dx (1 + tan x) + y ( 1 + tan² x) = y(1+ tan x)(1+tan x) + y (1 + tan² x) = 2y (1 + tan x + tan² x) ----(1)

2 tanx dy/dx + 2y = 2 tanx . y (1+ tan x) + 2y = 2y ( 1 + tan x + tan² x)---- (2)

From (1) and (2), d²y/dx² = 2 tanx dy/dx + 2y.

HTH

Hi iFruit,

Thanks for your help. Was stuck and coundn't continue from (1) to get the answer without (2), which it was just one more step to go, the reverse of (2) :lol:[/quote]Hi OK Lor,

Sure. Yeah, sometimes it is easier to simplify both sides to a common third.

[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)