1. ## Parametric equations

1) Let c be given parametrically by x= cost and y=sint with t[0,pi]. Rotate this curve about the x axis. Set up integral from surface area.

2)The curve c is defined by the parametric equation x=t^3 + 3t and y= t^2 e^-t. determine any values of t where the derivative is horizontal.

3) Consider the curve C given parametrically by x= tant and y= sect with t[-pi/2, pi/2]. eliminate the parameter to get a familiar equation.

Do as many as you can please. I really need these

2. Originally Posted by tiga killa
1) Let c be given parametrically by x= cost and y=sint with t[0,pi]. Rotate this curve about the x axis. Set up integral from surface area.

$x = \cos t$

$y =\sin t$

$x^2+y^2=\sin^2 t+\cos^2 t=1$

This is a circle with centre (0, 0) and radius 1 unit. With the limits, [0, pi] . See graph.

$x = \cos 0=1, \;y = \sin 0 =0$ so, point (1,0)

$x = \cos \pi=-1, \;y = \sin \pi =0$ so, point (-1,0)

$\frac{dx}{dt}=- \sin t$

$\Rightarrow dx = - \sin t \;dt$

$\frac{dy}{dt}=-\cos t$

$\frac{dy}{dx}=\frac{\cos t}{-\sin t}$

$
1+\left(\frac{dy}{dx}\right)^2= 1+\left(\frac{\cos t}{-\sin t}\right)^2
$

$1+\left(\frac{dy}{dx}\right)^2=\frac{1}{\sin^2 t}$

Surface Area

$S = 2\pi \int\limits_0^\pi {y\sqrt {1 + \left( {\frac{{dy}}
{{dx}}} \right)^2 } } .dx \hfill \\$

$= 2\pi \int\limits_0^\pi {\sin t\sqrt {\left( {\frac{1}
{{\sin ^2 t}}} \right)} } .\left( { - \sin t{\text{ }}dt} \right) \hfill \\$

$= - 2\pi \int\limits_0^\pi {\sin t{\text{ }}dt} \hfill \\$

$= 4\pi \hfill \\$

3. Originally Posted by tiga killa
1)

3) Consider the curve C given parametrically by x= tant and y= sect with t[-pi/2, pi/2]. eliminate the parameter to get a familiar equation.
$x^2 = \tan ^2 t \hfill \\$

$y^2 = \sec ^2 t \hfill \\$

$y^2 - x^2 = \sec ^2 t - \tan ^2 t = 1 \hfill \\$

$y^2 - x^2 = 1 \hfill \\$

Now , you recognize, what type of curve it is? Hyperbola?

4. Hello, tiga killa!

2) The curve $C$ is defined by: . $\begin{array}{ccc}x&=&t^3 + 3t \\ \\[-4mm] y&=& t^2\, e^{-t}\end{array}$

Determine any values of $t$ where the derivative is horizontal.
The derivative is: . $\frac{dy}{dx} \:=\:\frac{\frac{dy}{dt}}{\frac{dx}{dt}}$

The tangent is horizontal when $\frac{dy}{dx} \,=\,0$
. . That is, when $\frac{dy}{dt} = 0$

We have: . $\frac{dy}{dt} \;=\;-t^2e^{-t} + 2te^{-t} \:=\:0 \quad\Rightarrow\quad -t\cdot e^{-t}\cdot(t - 2) \:=\:0$

. . Hence: . $\begin{array}{ccccccc}-t &=& 0 & \Rightarrow & t\:=\:0 \\
e^{-t} &=& 0 & \Rightarrow & \text{no solution} \\
t-2 &=& 0 & \Rightarrow & t \:=\: 2 \end{array}$

Therefore, horizontal tangents when $t \;=\;0,\:2$

5. Originally Posted by tiga killa
2)The curve c is defined by the parametric equation x=t^3 + 3t and y= t^2 e^-t. determine any values of t where the derivative is horizontal.
${\text{For, derivative to be horizontal, slope = 0, }}\frac{{dy}}
{{dx}} = 0 \hfill \\
$

finish it.