Inverse Trig: Help! (1 Viewer)

[foram]

The velocity of a particle is given by
dx/dt = x^2 + 4 cm/s

At the time t=0, x= -2

Find the displacement after pi/4 seconds.

ANS: 2cm

___________________________________________
My working so far:

dx/dt = x^2 + 4

dt/dx = 1/(x^2 + 4)
t= 1/2 tan^-1 (x/2) + C

Sub t=0, x= -2

1/2 tan^-1 (-1) + C = 0
tan^-1 (-1) + C = 0
-tan^-1 (1) + C = 0
C = tan^-1 (1)

Therefore: t=1/2 tan^-1 (x/2) + tan^-1 (1)

At t= pi/4

1/2 tan^-1 (x/2) + tan^-1 (1) = pi/4
1/2 tan^-1 (x/2) = pi/4 - tan^-1 (1)
tan^-1 (x/2) = pi/2 - 2tan^-1(1)
x/2 = tan (pi/2 - 2tan^-1(1))
x = 2 tan (pi/2 - 2tan^-1(1))

By Calculator, x=0 ?! Why doesn't x=2?!

What did I do wrong? How do i find x? Can sombody help me please?

 

[vds700]

you evaluated c wrong...

(1/2) arctan (-1) + c = 0
(1/2) (-pi/4) + c = 0
c = pi/8

sub that in and u should get x = 2

 

[foram]

you evaluated c wrong...

(1/2) arctan (-1) + c = 0
(1/2) (-pi/4) + c = 0
c = pi/8

sub that in and u should get x = 2

Oh, I made a rather careless mistake. lol at myself. Thankyou very much.

 

[foram]

I'm having trouble with this question. Can somebody show me how it is done?


Express cos@ in terms of cos (@/2) and hence show that

tan^-1 [sqrt.[(1-x)/(1+x)]] = 0.5 cos^-1 (x)

Hence find d/dx [ tan^-1 sqrt.((1-x)/(1+x)) ]

 

[lyounamu]

I'm having trouble with this question. Can somebody show me how it is done?


Express cos@ in terms of cos (@/2) and hence show that

tan^-1 [sqrt.[(1-x)/(1+x)]] = 0.5 cos^-1 (x)

Hence find d/dx [ tan^-1 sqrt.((1-x)/(1+x)) ]

cos @ = 2cos^2 (@/2) - 1

 

[foram]

cos @ = 2cos^2 (@/2) - 1

yea i got up to that bit, but i couldn't do the 'hence show' part.

 

[lyounamu]

yea i got up to that bit, but i couldn't do the 'hence show' part.

Sorry didn't finish the question, yet. I accidentally clicked the button.

I will post up answer in few sec

EDIT: I am going around a same circle over and over again. Aahhh!!! Haven't learn this yet so I don't fully understand. It will take some time. Meanwhile, can anyone help??

 

[vds700]

hmmm this is a tricky question, i tried simplifying the left hand side and i got.

LHS = arctan [tan^2(0.5arccosx)]. I cant work out how to simplify this any further

 

[Mark576]

Just a moment I will post up my solution soon.

EDIT: http://img504.imageshack.us/my.php?image=mathssolutionae8wv6.jpg

 

[3unitz]

cos (@/2) = sqrt[0.5(cos@ + 1)] -----(1)
sin (@/2) = sin@ / [2 sqrt(0.5cos@ + 0.5] -----(2)

(1) / (2):
tan (@/2) = sin@ / (cos@ + 1)

sub, @ = cos^-1(x)
tan[0.5 cos^-1(x)] = sin[cos^-1(x)] / (x + 1)
= sqrt(1 - x^2) / (x + 1)
= sqrt[(1 - x) / (1 + x)]

tan^-1 [sqrt[(1 - x) / (1 + x)]] = 0.5 cos^-1(x)]

 

[foram]

Thanks! That was pretty smart working out. I understand now.

 

[lyounamu]

Damn. I wish I knew how to work that out.

Better study now...