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#51 Re: Exercises » Question Bank : Age Group 16-17 : I » 2009-08-15 22:03:58
Hi ganesh
For the third problem:
w=x+2y+z^2 , x= cos(t) , y = sin(t) , z=t
dw/dt 
dw/dx)(dx/dt)+ (dw/dy)(dy/dt)+( dw/dz)(dz/dt)
=1(-sin (t) + 2(cos (t) + 2 z (1)
= -sin(t)+2 cos(t) + 2z
Best Regards
Riad Zaidan
#52 Re: Exercises » Question Bank : Age Group 16-17 : I » 2009-08-15 02:32:40
Hi ganesh
For the second problem:
Assume that the angle between the sides of fixed length is θ, so
A = ½(4)(5) sin θ=10 sin θ where A is the area of triangle at any time , so
dA/dt =10 cos θ (dθ/dt) when θ=pi/3 and dθ/dt = 0.06
dA/dt =10 cos (pi/3) * 0.06
=10 (½)(0.06)=0.3
Best Regards
Riad Zaidan
#53 Re: Exercises » Integration by Substitution » 2009-08-15 02:12:30
Hi Identity
This can be noticed from the derivative of ln(sec x + tan x) + C which is
sec x tan x + sec^2 x sec x ( tan x + sec x )
(d/dx) (ln(sec x + tan x) + C)=__________________ = ___________________ = sec x
secx + tan x sec x + tan x
Best Regards
Riad Zaidan
#54 Re: Exercises » Integration by Substitution » 2009-08-13 23:08:34
Hi Identity
sec x + tan x
∫ (1/cos x) dx=∫ sec x dx=∫ sec x ______________ dx
sec x + tan x
(sec x)^2 + sec x tan x
∫ ___________________ dx
sec x + tan x
= ln(sec x + tan x) + C since the numerator is the derivative of the denominator.
Best Regards
Riad Zaidan
#55 Re: Help Me ! » When can this be simplified? » 2009-08-13 20:11:47
Dear bobbym
thanks for your attention , and I am very pleased with you.
Best Wishes