Question

when solving parametric equations, exactly when do we need to eliminate the t?

Answer 1

do we need to eliminate over here?

Answer 2

I don't think you ever "need to" eliminate the parameter.
You can simply choose to do so when it's convenient.

In this case, we have a squared t in each equation,
it might be fairly difficult to eliminate the parameter.

Answer 3

\[\large\rm \frac{dy}{dx}=\frac{(dy/dt)}{(dx/dt)}\]So in this problem, it's probably easier to just take your derivatives separately, and then plug them in.\[\large\rm y=t^2-t+1\qquad\to\qquad \frac{dy}{dt}=2t-1\]So that's what would go in our numerator, ya?\[\large\rm \frac{dy}{dx}=\frac{2t-1}{(dx/dt)}\]

Answer 4

yes. \[\frac{ 2t-1 }{ 2t }\]

Answer 5

we're not allowed to eliminate 2t together, right? what would be the next step?

Answer 6

Woops, your denominator doesn't look quite right.

Answer 7

You got 0 for the derivative of 1/t? O_o whuuuu

Answer 8

\[\frac{ 0 \times v - 1 \times 1 }{ t^2} = \left(\begin{matrix}-1 \\ t^2\end{matrix}\right)\]

Answer 9

\[\large\rm \frac{dy}{dx}=\frac{2t-1}{-\frac1{t^2}+2t}\]Mmm ok that looks better.

Answer 10

\[\frac{ 2t-1 }{ \frac{ -1 }{ t^2 } }\]

Answer 11

where'd you get the 2t?

Answer 12

Where did you get the 2t from the first time?

Answer 13

Ohhh I see.

Answer 14

It's from the original equation I'm sorry

Answer 15

So we have a derivative function.. that we need to evaluate at (x,y)=(2,1).

This next step might seem a little confusing at first.
We don't have any x or y in our derivative function! So what can we do?

Well, we can plug our (x,y) into our original system to find the corresponding t value.

Answer 16

\[\large\rm x=\frac1t+t^2\qquad\qquad\qquad y=t^2-t+1\]Plugging in (2,1) gives us,\[\large\rm 2=\frac1t+t^2\qquad\qquad\qquad 1=t^2-t+1\]

Answer 17

Oh? But that still does not make sense to me

Answer 18

Well let's look at the second equation.
Maybe we can come up with some values for t.\[\large\rm 1=t^2-t+1\]Subtracting 1 from each side,\[\large\rm 0=t^2-t\]From here, maybe factor a t out of each term,\[\large\rm 0=t(t-1)\]

Answer 19

Do you see the t values?
There should be two of them.

Answer 20

1 and 0

Answer 21

\[t+t^2=2\]

Answer 22

t = 1

Answer 23

Only 1? Does it have to satisfy both equations?

Answer 24

\[\large\rm 2=\frac1t+t^2\]Well it turns out that t=0 does not satisfy this other equation.
Plugging in t=0 would give us this undefined value 1/0.

So that's great! It gives us our one and only value for t, t=1.

Yes, it has to satisfy both equations.

Answer 25

So we've determined that the point (x,y) = (2,1) is the time when t=1.
So that's when we'll evaluate our derivative function, at t=1.

Answer 26

The entire answer to this problem seems to be 1.

Answer 27

I plugged in 1 for dy/dx and got 1.

Answer 28

\[\large\rm \frac{dy}{dx}=\frac{2t-1}{-\frac1{t^2}+2t}\]Umm let's see..\[\large\rm \frac{dy}{dx}=\frac{2(1)-1}{-\frac1{1^2}+2(1)}=\frac11\]Mmmm yes good job!

Answer 29

Thank you for helping :)

Answer 30

np