# Thread: [SOLVED] Calulus III: Parametric Equations

1. ## [SOLVED] Calulus III: Parametric Equations

Please, could someone please help me step by step with the following problems, I've tried to work them and it's driving me nuts to rewrite everything again and then to find out that I am making the same mistakes! My main problem is being able to differentiate correctly.
Thanks.

1) Find the arc length of the curve on the interval (0, 2pi) of the following:

Cylcoid arch: x= a (θ - sinθ), y= a (1-cosθ)

2) Find the arc length on the curve of the given interval (1, 2) of the following:

x= t , y= t^(5) /10 + 1 /6t^(3)

3) Find dy/dx and d^(2) y / dx^(2) (first derivative and 2nd derivative), and find the slope and concavity (if possible) at the given value of the parameter of the following:

a) x= 2 + sec θ, y= 1+ 2 tan θ When θ= pi/6b) x= cos θ, y= 3 sin θ When θ= 0

2. Originally Posted by googoogaga
Please, could someone please help me step by step with the following problems, I've tried to work them and it's driving me nuts to rewrite everything again and then to find out that I am making the same mistakes! My main problem is being able to differentiate correctly.
Thanks.
that's strange. the functions shown here are not really difficult to differentiate. maybe you're using the wrong formula

1) Find the arc length of the curve on the interval (0, 2pi) of the following:

Cylcoid arch: x= a (θ - sinθ), y= a (1-cosθ)
use: $\displaystyle \int_{0}^{2 \pi} \sqrt { \left( \frac {dx}{d \theta} \right)^2 + \left( \frac {dy}{d \theta} \right)^2}~d \theta$

2) Find the arc length on the curve of the given interval (1, 2) of the following:

x= t , y= t^(5) /10 + 1 /6t^(3)
use: $\displaystyle \int_{1}^{2} \sqrt { \left( \frac {dx}{dt} \right)^2 + \left( \frac {dy}{dt}\right)^2}~dt$

3) Find dy/dx and d^(2) y / dx^(2) (first derivative and 2nd derivative), and find the slope and concavity (if possible) at the given value of the parameter of the following:

a) x= 2 + sec θ, y= 1+ 2 tan θ When θ= pi/6b) x= cos θ, y= 3 sin θ When θ= 0
use: $\displaystyle \frac {dy}{dx} = \frac { \frac {dy}{d \theta}}{\frac {dx}{d \theta}}$ and $\displaystyle \frac {d^2 y}{dx^2} = \frac d{dx} \left( \frac {dy}{dx} \right) = \frac {\frac d{d \theta} \left( \frac {dy}{dx}\right)}{\frac {dx}{d \theta}}$

3. ## Calculus III: Parametric Equations

I did use those equations but my differentiation is not working for some reason, I could write the steps and everything that I did so you could show me where I went wrong, but I write so much that it would take me a a lifetime to write it because I don't have that software you use to answer the problems. Please I would appreciate it if anyone show me how to do the problems step by step especially the first two pleeaase. Thanks

4. Originally Posted by googoogaga
I did use those equations but my differentiation is not working for some reason, I could write the steps and everything that I did so you could show me where I went wrong, but I write so much that it would take me a a lifetime to write it because I don't have that software you use to answer the problems. Please I would appreciate it if anyone show me how to do the problems step by step especially the first two pleeaase. Thanks
There is another way to approach this, though the Math gets a bit nasty. You could remove the parameters $\displaystyle r \text{ and } \theta$ and do the problem in the usual fashion:
1) $\displaystyle x = a( \theta - sin(\theta)), ~ y = a(1 - cos(\theta)$

So
$\displaystyle \theta = cos^{-1} \left ( a - \frac{y}{a} \right )$

So
$\displaystyle x = a \left [ cos^{-1} \left ( a - \frac{y}{a} \right ) - sin \left ( cos^{-1} \left ( a - \frac{y}{a} \right ) \right ) \right ]$

Now use the inverse function theorem to get $\displaystyle \frac{dy}{dx}$.

(Hey, I said it was going to be nasty! )

-Dan

5. Is there any other way to approach this problem , because I am not seeing this method clearly, I am really confused, please I would like to see the whole problem worked out I just can't see it.

6. Originally Posted by googoogaga
Is there any other way to approach this problem , because I am not seeing this method clearly, I am really confused, please I would like to see the whole problem worked out I just can't see it.
Originally Posted by Jhevon
use: $\displaystyle \int_{0}^{2 \pi} \sqrt { \left( \frac {dx}{d \theta} \right)^2 + \left( \frac {dy}{d \theta} \right)^2}~d \theta$
I'll work the first one using Jhevon's method:
$\displaystyle x = a(\theta - sin(\theta)) \text{ and }y = a~cos(\theta)$

So
$\displaystyle \frac{dx}{d\theta} = a - a~cos(\theta)$
and
$\displaystyle \frac{dy}{d\theta} = -a~sin(\theta)$

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

$\displaystyle = \int_{0}^{2 \pi} \sqrt {(a - a~cos(\theta) )^2 + ( -a~sin(\theta))^2}~d \theta$

$\displaystyle = \int_{0}^{2 \pi} \sqrt {a^2 - 2a^2~cos(\theta) + a^2~cos^2(\theta) + a^2~sin^2(\theta)}~d \theta$

$\displaystyle = \int_{0}^{2 \pi} \sqrt {2a^2 - 2a^2~cos(\theta)}~d \theta$

I'll leave the rest to you.

-Dan

7. Hello, googoogaga!

The Calculus is not hard, but the Algebra is mind-boggling . . .

1) Find the arc length of the curve on the interval (0, 2pi) of:
. . Cylcoid arch: .$\displaystyle \begin{array}{ccc}x &= & a(\theta - \sin\theta) \\ y & = & a(1 - \cos\theta) \end{array}$
We have the formula: .$\displaystyle L \;=\;\int^b_a\sqrt{\left(\frac{dx}{d\theta}\right) ^2 + \left(\frac{dy}{d\theta}\right)^2}\,d\theta$

We have: .$\displaystyle \begin{array}{ccc}\frac{dx}{d\theta} & = & a(1 - \cos\theta) \\ \frac{dy}{d\theta} & = & a\sin\theta \end{array}$

Then: .$\displaystyle \left(\frac{dx}{d\theta}\right)^2 + \left(\frac{dy}{d\theta}\right)^2 \;=\;a^2(1-\cos\theta)^2 + a^2\sin^2\theta$

. . $\displaystyle = \;a^2 - 2a^2\cos\theta + \underbrace{a^2\cos^2\!\theta + a^2\sin^2\!\theta}_{a^2(\cos^2\!\theta+\sin^2\!\th eta)} \;=\;a^2 - 2a^2\cos\theta + a^2 \;=\;2a^2 - 2a^2\cos\theta$

. . $\displaystyle = \;2a^2(1 - \cos\theta) \;=\;4a^2\left(\frac{1-\cos\theta}{2}\right) \;=\;4a^2\sin^2\!\left(\frac{\theta}{2}\right)$

Hence: .$\displaystyle \sqrt{\left(\frac{dx}{d\theta}\right)^2 + \left(\frac{dy}{d\theta}\right)^2} \;=\;\sqrt{4a^2\sin^2\!\left(\frac{\theta}{2}\righ t)} \;=\;2a\sin\left(\frac{\theta}{2}\right)$

The problem becomes: .$\displaystyle L \;=\;2a\int^{2\pi}_0\sin\left(\frac{\theta}{2}\rig ht)\,d\theta$

. . Can you finish it now?

.

8. Hello, googoogaga!

With most Arc Length problems, a lot of algebra is expected
. . and it usually comes out something "nice".

2) Find the arc length on the interval (1, 2) of .$\displaystyle \begin{Bmatrix}x & = & t \\y& = &\frac{t^5}{10} + \frac{1}{6t^3}\end{Bmatrix}$

We have: .$\displaystyle \begin{Bmatrix}x \: = \: t & \Rightarrow & \frac{dx}{dt}\:=\:1 \\ \\ y \:= \:\frac{1}{10}t^5 + \frac{1}{6}t^{-3} & \Rightarrow & \frac{dy}{dt} \:=\: \frac{1}{2}t^4 - \frac{1}{2}t^{-4}\end{Bmatrix}$

Then: .$\displaystyle \left(\frac{dx}{dt}\right)^2 + \left(\frac{dy}{dt}\right)^2 \;=\;(1)^2 + \left(\frac{1}{2}t^4 - \frac{1}{2}t^{-4}\right)^2 \;=\;1 + \frac{1}{4}t^8 - \frac{1}{2} + \frac{1}{4}t^{-8}$

. . $\displaystyle = \;\frac{1}{4}t^8 + \frac{1}{2} + \frac{1}{4}t^{-8} \;=\; \left(\frac{1}{2}t^4 + \frac{1}{2}t^{-4}\right)^2$

Hence: .$\displaystyle \sqrt{\left(\frac{dx}{dt}\right)^2 + \left(\frac{dy}{dt}\right)^2} \;=\;\sqrt{\left(\frac{1}{2}t^4 + \frac{1}{2}t^{-4}\right)^2} \;=\;\frac{1}{2}t^4 + \frac{1}{2}t^{-4} \;=\;\frac{1}{2}\left(t^4 + t^{-4}\right)$

And your integral becomes: .$\displaystyle L \;=\;\frac{1}{2}\int^2_1\left(t^4 + t^{-4}\right)\,dt$

9. Hello, googoogaga!

Second derivatives of parametric is always a tricky ordeal . . .

3) Find $\displaystyle \frac{dy}{dx}$ and $\displaystyle \frac{d^2\!y}{dx^2}$
Find the slope and concavity (if possible) at the given value of the parameter.

$\displaystyle a)\;\begin{array}{ccc}x & = & 2 + \sec\theta \\y &=& 1+ 2\tan\theta\end{array}$ . at .$\displaystyle \theta = \frac{\pi}{6}$

We have: .$\displaystyle \begin{Bmatrix}\frac{dx}{d\theta} & = & \sec\theta\tan\theta \\ \\ \frac{dy}{dx} & = & 2\sec^2\theta\end{Bmatrix}$

Then: .$\displaystyle \frac{dy}{dx}\;=\;\frac{\frac{dy}{d\theta}}{\frac{ dx}{d\theta}} \;=\;\frac{2\sec^2\theta}{\sec\theta\tan\theta} \;=\;2\,\frac{\sec\theta}{\tan\theta} \;=\;2\,\frac{\frac{1}{\cos\theta}}{\frac{\sin\the ta}{\cos\theta}} \;=\;\frac{2}{\sin\theta}\quad\Rightarrow\quad\box ed{\frac{dy}{dx} \;=\;2\csc\theta}$

At $\displaystyle \theta = \frac{\pi}{6}\!:\;\;\frac{dy}{dx}\;=\;2\csc\frac{\ pi}{6} \;=\;2(2)\quad\Rightarrow\quad\boxed{ \text{slope} \:=\:4}$

Formula: .$\displaystyle \frac{d^2\!y}{dx^2} \;=\;\frac{\frac{d}{d\theta}\!\left(\frac{dy}{dx}\ right)}{\frac{dx}{d\theta}} \;=\;\frac{-2\csc\theta\cot\theta}{\sec\theta\tan\theta}$

At $\displaystyle \theta = \frac{\pi}{6}\!:\;\;\frac{d^2\!y}{dx^2} \;=\;\frac{-2\csc\frac{\pi}{6}\cot\frac{\pi}{6}}{\sec\frac{\pi }{6}\tan\frac{\pi}{6}} \;=\;\frac{-2(2)\sqrt{3})}{\frac{2}{\sqrt{3}}\!\cdot\!\frac{1} {\sqrt{3}}}\quad\Rightarrow\quad\boxed{\text{conca vity} \:=\:-6\sqrt{3}}$

10. Thank you dear sirs for clearing the pathway for me, these differentiations are indeed much easier once the algebraic hardaches have been solved.