1. ## Parametric Eqtns

Good afternoon Everyone,

I am at this again...

I have two questions with respect to parametric equations:

1) The problem was to find a parametric equation of a straight line sement from $\displaystyle P=(1,2)$ to $\displaystyle Q=(4,7)$.

I get the following:

$\displaystyle m=\frac{5}{3}$

$\displaystyle C(t) = (a+t , b+mt)$

$\displaystyle C(t) = (1+t , 2+\frac{5}{3}t)$

Is this all there is to this question?

2) The second question I have is deals with eliminating the parameter from:

$\displaystyle \left\{\begin{array}{cc}x=cos\theta\\y=\cos\theta+ \sin^2\theta\end{array}\right$

I am at a complete loss here, I don't follow what my book is trying to tell me at all. There was an entire paragraph on the subject of eliminating the parameter. The notes on this problem state that the equation relating x and y should not involve trig functions.

2. Let me throw up what I think I know about the second problem...

If I solve $\displaystyle x=\cos\theta$ for $\displaystyle \theta$ I come up with:

$\displaystyle \theta=\cos^{-1}x$

If we substitute $\displaystyle \cos^{-1}x$ for $\displaystyle \theta$ in our $\displaystyle y$ equation, we get:

$\displaystyle y=\cos(\cos^{-1}x)+\sin^2(\cos^{-1}x)$

$\displaystyle y=x+(-x-1)$

$\displaystyle y=-x+x+1$

What identity dictates that $\displaystyle \sin^2(\cos^{-1}x)=(-x-1)$ ???

3. 1) Your answer has t ranging from 0 to 3. It's certainly a valid parametrization (there are infinitely many, by the way). You might want to consider a parametrization that has t ranging from 0 to 1.

2) Try looking at $\displaystyle (y-x)+x^{2}.$ What does that give you?

4. I am not a fan of how my text approached parameterization (or the lack of an approach). In this instance to get a parameterization for t from 0 to 1 would I have the following:

$\displaystyle C(t) = (1+t , 2+\frac{5}{3}t)$

Simply multiply the constant for t by the facotr for which I want to decrease my range (in this case 3).

$\displaystyle C(t) = (1+3t , 2+5t)$

Suppose I wanted my t to range from 0-10...

$\displaystyle C(t) = (1+\frac{3}{10}t , 2+\frac{1}{2}t)$

Do I need to do something to indicate the range of t?

Does my answer for the second part make sense?

5. That's certainly a fine way to get your range to be from 0 to 1. You should always indicate the range of your t. Just say something like t is in the interval [0,1].

Your answer for the second part does not make a lot of sense to me. I think you're missing a squared term in there. Your final answer is y=1? Try this identity, and square the result. That'll get you where you need to go.

When I see parametric equations like that, I almost always try to use the Pythagorean theorem to simplify. Your method should end up with the same thing, however.

6. If $\displaystyle Pa,b)~\&~Qc,d)$ then the parametric equation of the line segment $\displaystyle \overline{PQ}$ is $\displaystyle \left( {a + t(c - a),b + t(b - d)} \right),\,0 \leqslant t \leqslant 1$

7. Oops... I seem to have omited a power of 2 in my answer. It should have read $\displaystyle y=-x^2+x+1$

I did try your $\displaystyle (y-x)+x^2$ method as well. This is how that worked out for me:

$\displaystyle y=\cos\theta+\sin^2\theta$

$\displaystyle x=\cos\theta$

$\displaystyle (\cos\theta+\sin^2\theta-\cos\theta)+\cos^2\theta$

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

So...

$\displaystyle y-x+x^2=1$

$\displaystyle y=-x^2+x+1$

Same answer as above... and slightly more intuitive. Or is this wrong?

If I look at the dentities you directed me to, I find that $\displaystyle \sin\cos^{-1}x=\sqrt{1-x^2}$ ...so, same thing.

8. Right. Looks like you're done with the second problem, at any rate!

9. Sweet! Thank you!

Last problem for the section actually.

Now, on to Arc Length, Surface Area, and Polar Coordinates...

10. Hello, MechEng!

Eliminate the parameter: .$\displaystyle \left\{\begin{array}{cc}x=cos\theta & [1] \\y=\cos\theta+\sin^2\theta & [2] \end{array}\right$

We have: .$\displaystyle \sin^2\!\theta \:=\:1-\cos^2\!\theta$

Substitute [1]: .$\displaystyle \sin^2\!\theta \;=\;1-x^2\;\;[3]$

Substitute [1] and [3] into [2]: . $\displaystyle y \;=\;x + 1 - x^2$