# Thread: length of curve

1. ## length of curve

find the length of the curve C from t=0 to t pi/2 where x(t)=2sin^3t and y(t)=2cos^3t. set up the correct integral and simplify the integrand.

i think im supposed to use the formula L= sqrt[(dx/dt)^2 + (dy/dt)^2]

for x' i got 36sin^4tcos^2t
y' = -36sin^2tcos^4t

how am i supposed to add(or subtract) them so that it =6sintcost ? that was the answer given for the part inside the integrand. so before rooting im supposed to get 36sin^2tcos^2t but how?

2. Originally Posted by jeph
find the length of the curve C from t=0 to t pi/2 where x(t)=2sin^3t and y(t)=2cos^3t. set up the correct integral and simplify the integrand.

i think im supposed to use the formula L= sqrt[(dx/dt)^2 + (dy/dt)^2]

for x' i got 36sin^4tcos^2t
y' = -36sin^2tcos^4t

how am i supposed to add(or subtract) them so that it =6sintcost ? that was the answer given for the part inside the integrand. so before rooting im supposed to get 36sin^2tcos^2t but how?
$L = \int_0^{\pi /2} dt \, \sqrt{ \left ( \frac{dx}{dt} \right ) ^2 + \left ( \frac{dy}{dt} \right )^2 }$

So
$\frac{dx}{dt} = 6sin^2(3t)cos(3t)$
$\frac{dy}{dt} = -6cos^2(3t)sin(3t)$

Thus
$L = \int_0^{\pi /2} dt \, \sqrt{ \left ( 6sin^2(3t)cos(3t) \right ) ^2 + \left ( -6cos^2(3t)sin(3t) \right ) ^2 }$

$L = \int_0^{\pi /2} dt \, \sqrt{ 36sin^4(3t)cos^2(3t) + 36cos^4(3t)sin^2(3t) }$

$L = 6 \int_0^{\pi /2} dt \, \sqrt{ sin^2(3t)cos^2(3t) (sin^2(3t) + cos^2(3t) ) }$

$L = 6 \int_0^{\pi /2} dt \, sin(3t)cos(3t)$

Can you take it from here?

-Dan

3. i think i might have screwed up on my math typing again...i wanted to say x=2sin^3(t) and y=2cos^3(t). just t not 3t so "sin and cos cubed t" o.o if not then i dont know how to get 3t...

whats the technique to deferentiate those trigs with cubes. can you go
2cos^3(t)
2(cos(t))^3 u=cos(t)
6(u^2)u'
-6cos^2(t)sin(t)

or does the sign switch both times you differentiate? it becomes negative then postive again? or the sign only switches when you differentiate the last cos(t)?

4. Originally Posted by jeph
find the length of the curve C from t=0 to t pi/2 where x(t)=2sin^3t and y(t)=2cos^3t. set up the correct integral and simplify the integrand.

i think im supposed to use the formula L= sqrt[(dx/dt)^2 + (dy/dt)^2]

...
Hello,

I assume that your functions read like:
$x(t)=2\sin^3(t)$ and $y(t)=2\cos^3(t)$

Then

$\frac{dx}{dt} = 6sin^2(t)cos(t)$ and
$\frac{dy}{dt} = -6cos^2(t)sin(t)$

If you square both sides of these equations you'll get:

$\left(\frac{dx}{dt} \right)^2= 36sin^4(t)cos^2(t)$
$\left( \frac{dy}{dt} \right)^2 = 36cos^4(t)sin^2(t)$

Now calculate the integral:

$L=\int_{0}^{\frac{\pi}{2}} \sqrt{36sin^4(t)cos^2(t) + 36cos^4(t)sin^2(t)} \cdot dt$

$= \int_{0}^{\frac{\pi}{2}} \sqrt{36sin^2(t)cos^2(t)\underbrace{\left( \sin^2(t)+\cos^2(t) \right)}} _{= 1}\cdot dt$

$= 6 \cdot \int_{0}^{\frac{\pi}{2}} |{sin(t)cos(t)} | \cdot dt$

I'll leave the rest for you .

(For confirmation only: I've got L = 3)

5. Originally Posted by jeph
...or the sign only switches when you differentiate the last cos(t)?
Hello,

have a look here:

$x(t) = 2 \sin^3(t)$

$\frac{dx}{dt}= 2 \cdot 3 \cdot \sin^2(t) \cdot(-\cos(t))$

6. Originally Posted by earboth
Hello,

have a look here:

$x(t) = 2 \sin^3(t)$

$\frac{dx}{dt}= 2 \cdot 3 \cdot \sin^2(t) \cdot(-\cos(t))$
shouldnt it be the dy/dt that becomes negative? you differentiate sin(t) right? that just becomes cos(t) not -cos(t)? o.o

7. Originally Posted by jeph
shouldnt it be the dy/dt that becomes negative? you differentiate sin(t) right? that just becomes cos(t) not -cos(t)? o.o
Hello,

of course you are right - but I hope that you caught the method even though I did an ugly mistake. My apologies.

8. Originally Posted by topsquark
$L = \int_0^{\pi /2} dt \, \sqrt{ \left ( \frac{dx}{dt} \right ) ^2 + \left ( \frac{dy}{dt} \right )^2 }$

So
$\frac{dx}{dt} = 6sin^2(3t)cos(3t)$
$\frac{dy}{dt} = -6cos^2(3t)sin(3t)$

Thus
$L = \int_0^{\pi /2} dt \, \sqrt{ \left ( 6sin^2(3t)cos(3t) \right ) ^2 + \left ( -6cos^2(3t)sin(3t) \right ) ^2 }$

$L = \int_0^{\pi /2} dt \, \sqrt{ 36sin^4(3t)cos^2(3t) + 36cos^4(3t)sin^2(3t) }$

$L = 6 \int_0^{\pi /2} dt \, \sqrt{ sin^2(3t)cos^2(3t) (sin^2(3t) + cos^2(3t) ) }$

$L = 6 \int_0^{\pi /2} dt \, sin(3t)cos(3t)$

Can you take it from here?

-Dan
Dan, is there a real reason why you always put $dt$ at the begining of the integral rather than the end? or is it just another one of your personality quirks?

9. Originally Posted by Jhevon
Dan, is there a real reason why you always put $dt$ at the begining of the integral rather than the end? or is it just another one of your personality quirks?
It's a personality quirk. It's also fairly standard in the Physics books I tend to use. There are advantages and disadvantages to its use, but for this level of work I don't see any particular advantage/disadvantage.

-Dan