MHB What Does the Graph of This Vector Equation Look Like?

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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?
 
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ineedhelpnow said:
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 $e^t$? :confused:

If so, suppose you define $x=e^{2t}$ and $y=e^t$.
What kind of relationship can you deduce from that? (Wondering)
 
oops. yeah the y component is e^t not e^r

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$x=y^2$?
 
ineedhelpnow said:
oops. yeah the y component is e^t not e^r

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$x=y^2$?

Yep! :D

What would the graph look like? (Wondering)
 

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

Yeah right!

be-rational-get-real.png
 
dude come on... grow up (Giggle)

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or should i say get real...
 
  • #10
ineedhelpnow said:
https://www.physicsforums.com/attachments/3102

That's what I got,using matlab:

View attachment 3104

(Blush)
 

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  • #11
evinda said:
That's what I got,using matlab:

View attachment 3104

(Blush)

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

It does look like some standard graph, doesn't it? (Wondering)
 
  • #12
I like Serena said:
But... but... how can $y=e^t$ be negative?? :confused:

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

I tried it again.. (Wasntme)

View attachment 3105

Is it better now? (Blush)
 

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  • #13
evinda said:
I tried it again.. (Wasntme)

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

Is it better now? (Blush)

It's perfect! (Nod)
 
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