# Math Help - parametric equation

1. ## parametric equation

5)Consider the parametric equation

What is the length of the curve for to ?

How do you do this problem? I am studying for my final and I dont know how to!

2. Originally Posted by nhatie
5)Consider the parametric equation

What is the length of the curve for to ?

How do you do this problem? I am studying for my final and I dont know how to!

length of the curve

$\int_{0}^{\frac{9\pi}{10}} \sqrt{\left(\frac{dx}{d\theta}\right)^2 + \left(\frac{dy}{d\theta}\right)^2 } d\theta$

3. Originally Posted by Amer
length of the curve

$\int_{0}^{\frac{9\pi}{10}} \sqrt{\left(\frac{dx}{d\theta}\right)^2 + \left(\frac{dy}{d\theta}\right)^2 } d\theta$
So what I suppose to do next?

4. find

$\frac{dx}{d\theta} = 12(-\sin \theta + (\sin \theta + \theta \cos \theta ) = 12(\theta \cos \theta$

$\frac{dy}{d\theta} = 12(\cos \theta - ( \cos \theta - \theta \sin \theta )=12(\theta \sin \theta)$

$\left(\frac{dx}{d\theta}\right)^2 + \left(\frac{dy}{d\theta}\right)^2 = 144 \theta^2 \cos^2 \theta + 144 \theta^2 \sin^2 \theta = 144\; \theta^2 ( \sin ^2 \theta + \cos ^2 \theta)$

it is easy now hint $sin^2 a + cos ^2 a = 1$

5. Originally Posted by Amer
find

$\frac{dx}{d\theta} = 12(-\sin \theta + (\sin \theta + \theta \cos \theta ) = 12(\theta \cos \theta$

$\frac{dy}{d\theta} = 12(\cos \theta - ( \cos \theta - \theta \sin \theta )=12(\theta \sin \theta)$

$\left(\frac{dx}{d\theta}\right)^2 + \left(\frac{dy}{d\theta}\right)^2 = 144 \theta^2 \cos^2 \theta + 144 \theta^2 \sin^2 \theta = 144\; \theta^2 ( \sin ^2 \theta + \cos ^2 \theta)$

it is easy now hint $sin^2 a + cos ^2 a = 1$
i got int from 0 to 9pi/10 12sinx =>-12cosx | 0 to 9pi/10
is it rite?

6. $\int_{0}^{\frac{9\pi}{10}} \sqrt{144 \theta^2 (\sin ^2 \theta + \cos^2 \theta) } d\theta$

but $\sin ^2 \theta + \cos ^2 \theta = 1$

so

$\int_{0}^{\frac{9\pi}{10}} 12\theta d\theta = 12\left(\frac{\theta ^2 }{2}\right) \mid_{0}^{\frac{9\pi}{10}}$

$6 (9\pi/10)^2$

7. Originally Posted by Amer
$\int_{0}^{\frac{9\pi}{10}} \sqrt{144 \theta^2 (\sin ^2 \theta + \cos^2 \theta) } d\theta$

but $\sin ^2 \theta + \cos ^2 \theta = 1$

so

$\int_{0}^{\frac{9\pi}{10}} 12\theta d\theta = 12\left(\frac{\theta ^2 }{2}\right) \mid_{0}^{\frac{9\pi}{10}}$

$6 (9\pi/10)^2$
thank you so much!