# Conics, Parametric Equations, & Polar Coordinates

• Sep 12th 2007, 06:30 AM
googoogaga
Conics, Parametric Equations, & Polar Coordinates
I have no clue how to do these problems! Could someone help please??
Thanks.

1) Sketch the curve represented by the following parametric equations ( indicate the orientation of the curve), and write the corresponding rectangular equation by eliminating the parameter.

a) x= 2t^(2), y= t^(4) + 1

b) x= t^(2)+ t, y= t^(2) - t

c) x= e^(-t), y= e^(2t) - 1`

d) x= 4 + 2 cos Theta

e) x= 4 sec Theta, y= 3 tan Theta

2) Use a graphing utility to graph the curve represented by the following parametric equation. Indicate the direction the curve. Identify any points at which the curve is not smooth.

a) Witch of Agnesi: x= 2 cot Theta, y= 2 sin^(2) theta
• Sep 12th 2007, 07:03 AM
topsquark
Quote:

Originally Posted by googoogaga
1) Sketch the curve represented by the following parametric equations ( indicate the orientation of the curve), and write the corresponding rectangular equation by eliminating the parameter.

a) x= 2t^(2), y= t^(4) + 1

Let's take this one for example.
To plot it simply choose a number of t values, say $-2 \leq t \leq 2$. (You may wish to use a calculator.) This will give you a series of (x, y) points, which you can connect with a smooth curve.

To get an equation for y you need to "eliminate the parameter" t from the equations:
$x = 2t^2 \implies t = \sqrt{\frac{x}{2}}$

Thus
$y = t^4 + 1 = \left ( \sqrt{\frac{x}{2}} \right )^4 + 1 = \frac{x^2}{4} + 1$

So this appears to be a parabola.

-Dan
• Sep 12th 2007, 11:55 AM
googoogaga
Conics, Parametric Equations E& Polar Coordinates
How would you solve exercices c and e and 2) a? There are some algebraic rules that I am not really remembering:(
• Sep 12th 2007, 12:45 PM
red_dog
c) $\displaystyle x=e^{-t}=\frac{1}{e^t}\Rightarrow e^t=\frac{1}{x}$.
$\displaystyle y=e^{2t}-1=(e^t)^2-1\Rightarrow y=\frac{1}{x^2}-1$.

e) $\displaystyle x=\frac{4}{\cos\theta}\Rightarrow x^2=\frac{16}{\cos^2\theta}=16(\tan^2\theta+1)$.
Then $\displaystyle x^2=16\left(\frac{y^2}{9}+16\right)\Rightarrow\fra c{x^2}{256}-\frac{y^2}{144}=1$
which is a hyperbola.