1. ## Parametric equations

The curve C has parametric equations

$\displaystyle x=cos^3\theta$

$\displaystyle y=sin^3\theta$

$\displaystyle 0<=\theta.=\frac{\pi}{2}$

a) find the arc length of C.

b) The curve C is rotated through 4 right angles about the x-axis. find the area of the curve surface generated.

OK so I've learned everything now for FP3 it's just a case of putting it into practice. Also my notes are slightly hard to understand in some places lol.

part A i would think differentiate x, then differentiate y, then form dy/dx

getting

$\displaystyle \frac{dy}{dx}=\frac{-3cos^2\theta sin\theta}{-3sin^2\theta cos\theta}$

thats what i did initally but looking at my notes it looks more like differentiation from first principles ... and pythagoras ... so i'm a bit stumped

2. Use the arclength form for parametric equations:

s = integral {[ (dx/dt)^2 + (dy/dt)^2]^1/2 dt}

integrated from 0 to pi/2

3. Originally Posted by djmccabie
The curve C has parametric equations

$\displaystyle x=cos^3\theta$

$\displaystyle y=sin^3\theta$

$\displaystyle 0<=\theta.=\frac{\pi}{2}$

a) find the arc length of C.

b) The curve C is rotated through 4 right angles about the x-axis. find the area of the curve surface generated.

OK so I've learned everything now for FP3 it's just a case of putting it into practice. Also my notes are slightly hard to understand in some places lol.

part A i would think differentiate x, then differentiate y, then form dy/dx

getting

$\displaystyle \frac{dy}{dx}=\frac{3cos^2\theta}{-3sin^2\theta}$

thats what i did initally but looking at my notes it looks more like differentiation from first principles ... and pythagoras ... so i'm a bit stumped

When you have parametric equaitons the arc length is given by the fomula

$\displaystyle L=\int_{t_0}^{t_1}\sqrt{\left( \frac{dx}{dt} \right)^2+\left( \frac{dy}{dt} \right)^2}$

$\displaystyle \int_{0}^{\pi/2}\sqrt{(3cos^2(\theta)(-sin(\theta)))^2+(3\sin^2(\theta))\cos(\theta))^2}d \theta$

$\displaystyle 9\int\sqrt{\cos^4(\theta)\sin^2(\theta)+\sin^4(\th eta)\cos^2(\theta)}d\theta$

$\displaystyle 9\int\sqrt{\cos^4(\theta)\sin^2(\theta)+\sin^2(\th eta)(1-\cos^2(\theta))\cos^2(\theta)}d\theta$

$\displaystyle 9\int \sqrt{\sin^2(\theta)\cos^2(\theta)}d\theta=9\int |\sin(\theta)\cos(\theta)|d\theta$

You can drop the absolute value becuase they are both posative on $\displaystyle \theta=[0,\frac{\pi}{2}]$

4. Hello, djmccabie!

The curve $\displaystyle C$ has parametric equations: .$\displaystyle \begin{array}{ccc}x&=&\cos^3\!\theta \\ y&=&\sin^3\!\theta\end{array} \quad 0 \leq \theta \leq \tfrac{\pi}{2}$

a) Find the arc length of $\displaystyle C.$
There is an Arc Length formula for parametric functions: .$\displaystyle L \;=\;\int^b_a\sqrt{\left(\frac{dx}{d\theta}\right) ^2 + \left(\frac{dy}{d\theta}\right)^2}\,d\theta$

We have: .$\displaystyle \begin{array}{ccccccc}\dfrac{dx}{d\theta} &=& \text{-}3\cos^2\!\theta\sin\theta & \Rightarrow & \left(\dfrac{dx}{d\theta}\right)^2 &=& 9\sin^2\!\theta\cos^4\!\theta \\ \\[-4mm] \dfrac{dy}{d\theta} &=& 3\sin^2\!\theta\cos\theta & \Rightarrow & \left(\dfrac{dy}{d\theta}\right)^2 &=& 9\sin^4\!\theta\cos^2\!\theta \end{array}$

Then: .$\displaystyle \left(\frac{dx}{d\theta}\right)^2 + \left(\frac{dy}{d\theta}\right)^2 \;=\;9\sin^2\!\cos^4\!\theta + 9\sin^4\!\theta\cos^2\!\theta$

. . . . . . . . . . $\displaystyle =\;9\sin^2\!\theta\cos^2\!\theta\underbrace{(\cos^ 2\!\theta + \sin^2\!\theta)}_{\text{This is 1}} \;=\;9\sin^2\!\theta\cos^2\!\theta$

. . Hence: .$\displaystyle \sqrt{\left(\frac{dx}{d\theta}\right)^2 + \left(\frac{dy}{d\theta}\right)^2} \;=\;\sqrt{9\sin^2\!\theta\cos^2\!\theta} \;=\;3\sin\theta\cos\theta$

Therefore: .$\displaystyle L \;=\;3\int^{\frac{\pi}{2}}_0 \sin\theta\cos\theta\,d\theta \quad\hdots$ Got it?

5. thanks with help i now understand my notes