What Does the Graph of This Vector Equation Look Like?

[ineedhelpnow]

sketch the plane curve with the given vector equation

$r(t)=e^{2t}$i $+e^{r}$j

what does the graph of this look like. is there any way to get a simple equation out of the two?

 

[I like Serena]

sketch the plane curve with the given vector equation

$r(t)=e^{2t}$i $+e^{r}$j

what does the graph of this look like. is there any way to get a simple equation out of the two?

Can it be that the y-component should be $e^t$?

If so, suppose you define $x=e^{2t}$ and $y=e^t$.
What kind of relationship can you deduce from that? (Wondering)

 

[ineedhelpnow]

oops. yeah the y component is e^t not e^r

- - - Updated - - -

$x=y^2$?

 

[I like Serena]

oops. yeah the y component is e^t not e^r

- - - Updated - - -

$x=y^2$?

Yep! :D

What would the graph look like? (Wondering)

 

[ineedhelpnow]

View attachment 3102

 

[I like Serena]

https://www.physicsforums.com/attachments/3102

Hmm... what would be the meaning of that red line?
What kind of standard graph would it be anyway? (Wondering)

 

[ineedhelpnow]

red line is imaginary. ignore that part. the blue part represents x=y^2

 

[I like Serena]

red line is imaginary. ignore that part. the blue part represents x=y^2

Yeah right!

 

[ineedhelpnow]

dude come on... grow up (Giggle)

- - - Updated - - -

or should i say get real...

 

[evinda]

https://www.physicsforums.com/attachments/3102

That's what I got,using matlab:

View attachment 3104

(Blush)

 

[I like Serena]

That's what I got,using matlab:

View attachment 3104

(Blush)

But... but... how can $y=e^t$ be negative??

It does look like some standard graph, doesn't it? (Wondering)

 

[evinda]

But... but... how can $y=e^t$ be negative??

It does look like some standard graph, doesn't it? (Wondering)

I tried it again.. (Wasntme)

View attachment 3105

Is it better now? (Blush)

 

[I like Serena]

I tried it again.. (Wasntme)

https://www.physicsforums.com/attachments/3105

Is it better now? (Blush)

It's perfect! (Nod)