You are Here: Home >< Maths

# help! intrinsic coordinates watch

Announcements
1. ive found the book to be completely useless when explaining these... can anyone help by solving these questions?

3. Measuring arc length from the origin O, find the intrinsic equation of the curve given parametically by x=2t+2sint and y=2-2cost. [ans: s=8sin(psi)]

6. For the curve with equation siny=e^x, where s is measured from the point (0, pi/2), show that e^s=tan(psi/2).

thanks!
2. I think it in P5 book:

The solution might be:

dx/dt = 2 + 2cos t
dy/dt = 2sin t

the arc length is:

s = INT ( sqrt [ (dx/dt)^2 + (dy/dt)^2 ] ) dt
= INT ( sqrt [ 4*( 1+ cos t)^2 + 4*sin ^2 t ]) dt
= INT ( 2* sqrt ( 2 + 2 cos t) ) dt
= 2*[ 2/3*( 2 + 2 cos t)^ 3/2 * ( 2t + 2sin t) ]

I don't know which is the limit of that? From 0 to Pi or what?
Can you tell me cearly?
3. (Original post by VCVT17)
= INT ( 2* sqrt ( 2 + 2 cos t) ) dt
= 2*[ 2/3*( 2 + 2 cos t)^ 3/2 * ( 2t + 2sin t) ]
i've never been taught to integrate like this...could you please share how; you came to this?

cheers

PK
4. Integrating by substituting:
Let sqrt ( 2 + 2 cos t) = a ^ 1/2
So integrating it become: 2/3*a^1/2
And then integrate a as it has 2t+ 2sin t

I will check again...
5. Sorry I did wrong

It must be:
= INT ( 2* sqrt ( 2 + 2 cos t) ) dt
= 2*INT(sqrt (4* cos ^2 t/2) ) dt
= 4* INT( cos t/2) dt
= 8* sin t/2

6. = INT ( 2* sqrt ( 2 + 2 cos t) ) dt
= 2*[ 2/3*( 2 + 2 cos t)^ 3/2 * ( 2t + 2sin t) ]

still,

how do you integrate s function of a function like above. I mean, i've been taught how to differentaite by teh chain rule. Is that a chain fule of integration or something?...bit lost...please enlighten me as to whether you can actually integrate stuff using stuff like what you did above?

thanks

PK
7. 3.
dx/dt = 2(1+cost)
dy/dt = 2sint
dy/dx = sint/(1+cost) = 2sin(t/2)cos(t/2)/(1+2cos²(t/2)-1) = sin(t/2)/cos(t/2) = tan(t/2) = tan(psi) => psi = (t/2)

(dx/dt)² + (dy/dt)² = 4+8cost+4cos²t + 4sin²t = 8(1+cost)
sqrt[(dx/dt)² + (dy/dt)²] = sqrt[8(1+cost)] = sqrt[8(1+2cos²(t/2)-1)] = 4cos(t/2)

s = ∫ sqrt[(dx/dt)² + (dy/dt)²] dt = ∫ 4cos(t/2) dt = 8sin(t/2) = 8sin(psi)

6.
siny = e^x => x = ln(siny)
dx/dy = coty = cot(psi) => psi = y

s = ∫ sqrt[1+(dx/dy)²] dy = ∫ sqrt[1+cot²y] dy = ∫ cosecy dy = ln|tan(y/2)| = ln|tan(psi/2)|
e^s = tan(psi/2)
8. (Original post by Phil23)
= INT ( 2* sqrt ( 2 + 2 cos t) ) dt
= 2*[ 2/3*( 2 + 2 cos t)^ 3/2 * ( 2t + 2sin t) ]

still,

how do you integrate s function of a function like above. I mean, i've been taught how to differentaite by teh chain rule. Is that a chain fule of integration or something?...bit lost...please enlighten me as to whether you can actually integrate stuff using stuff like what you did above?

thanks

PK
That's wrong. I think he applied the 'rule' for integrating (ax+b)^n to (2+2cost)^(0.5), but you can't do that since 2+2cost isn't linear.
9. (Original post by dvs)
That's wrong. I think he applied the 'rule' for integrating (ax+b)^n to (2+2cost)^(0.5), but you can't do that since 2+2cost isn't linear.
thanks for that...thought that method he employed didn't ring any bells

TSR Support Team

We have a brilliant team of more than 60 Support Team members looking after discussions on The Student Room, helping to make it a fun, safe and useful place to hang out.

This forum is supported by:
Updated: March 4, 2005
Today on TSR

10 quick tips

### University open days

• University of the West of England, Bristol
Wed, 23 Jan '19
• University of East London
Wed, 23 Jan '19
• University of Gloucestershire
School of Education Open Day Postgraduate
Wed, 23 Jan '19
Poll
Useful resources

### Maths Forum posting guidelines

Not sure where to post? Read the updated guidelines here

### How to use LaTex

Writing equations the easy way

### Study habits of A* students

Top tips from students who have already aced their exams