1. ## Parametric equations

So I don't seem to understand how to find a parametric equation from a vector function or to find an equation in x and y whose graph is the path of the particle.

Ok first. Let's just say we have v (t) = (sin t)i + t^2j + e^tk .

I thought all you do is x = sin t etc... but that's not the case apparently.

I thought you did something similar for my latter inquiry... but that's not the case apparently.

2. Originally Posted by Oneiromancy
So I don't seem to understand how to find a parametric equation from a vector function or to find an equation in x and y whose graph is the path of the particle.

Ok first. Let's just say we have v (t) = (sin t)i + t^2j + e^tk .

I thought all you do is x = sin t etc... but that's not the case apparently.

I thought you did something similar for my latter inquiry... but that's not the case apparently.
That is how I have always done it

Let $\displaystyle \bold{r(t)}=\left\langle{f(t),g(t),h(t)}\right\ran gle$

Then we parametrize it by

$\displaystyle x=f(t)$
$\displaystyle y=g(t)$
$\displaystyle z=h(t)$

3. Originally Posted by Oneiromancy
So I don't seem to understand how to find a parametric equation from a vector function or to find an equation in x and y whose graph is the path of the particle.

Ok first. Let's just say we have v (t) = (sin t)i + t^2j + e^tk .

I thought all you do is x = sin t etc... but that's not the case apparently.

I thought you did something similar for my latter inquiry... but that's not the case apparently.
It is the case.

Post the exact question you're working on.

4. Ok I have 2 problems that I can't do.

1) Find an equation in x and y whose graph is the path of the particle. Then find the particle's velocity and acceleration vectors at the given value of t.

r(t) = (t + 1)i + (t^2 - 1)j , t = 1

The answer is y = x^2 - 2x

I know how to get v and a.

2) Find parametric equations for the line that is tangent to the given curve at the given parameter value t_0 = 0

r(t) = (sin t)i + (t^2 - cos t)j + (e^t)k , t_0 = 0

The answer should be x = t , y = -1 , z = 1 + t

I thought all you do is find dr/dt then just plug in 0 but that didn't work.

These problems seem way too easy. I know I'm simply doing it completely wrong.

5. Originally Posted by Oneiromancy
]
1) Find an equation in x and y whose graph is the path of the particle. Then find the particle's velocity and acceleration vectors at the given value of t.

r(t) = (t + 1)i + (t^2 - 1)j , t = 1

The answer is y = x^2 - 2x

I know how to get v and a.
So
$\displaystyle x = t + 1$
and
$\displaystyle y = t^2 - 1$

So from the x equation we know that
$\displaystyle t = x - 1$

Inserting this into the y equation:
$\displaystyle y = (x - 1)^2 - 1$

Now simplify.

-Dan

6. Originally Posted by Oneiromancy
Ok I have 2 problems that I can't do.

1) Find an equation in x and y whose graph is the path of the particle. Then find the particle's velocity and acceleration vectors at the given value of t.

r(t) = (t + 1)i + (t^2 - 1)j , t = 1

The answer is y = x^2 - 2x

I know how to get v and a.

2) Find parametric equations for the line that is tangent to the given curve at the given parameter value t_0 = 0

r(t) = (sin t)i + (t^2 - cos t)j + (e^t)k , t_0 = 0

The answer should be x = t , y = -1 , z = 1 + t

I thought all you do is find dr/dt then just plug in 0 but that didn't work.

These problems seem way too easy. I know I'm simply doing it completely wrong.
1) x = t + 1 => t = x - 1 .... (1)
y = t^2 - 1 .... (2)

Substitute (1) into (2): y = (x - 1)^2 - 1 = x^2 - 2x + 1 - 1 = x^2 - 2x.

2) dr/dt at t = 0 gives a vector v in the direction of the line. And a point on the line is (0, -1, 1). So a vector equation of the line is:

r = 0 i - 1 j + 1 k + t v = -j + k + t v. (You can put in what v is).

Therefore a parametric equation is x = ...., y = ....., z = ......

7. Got it now, thank you.