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Vector calculus

  • Thread starter gordda
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  • #1
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i having so much trouble with this vector calculus question, someone please help.

At time t, a moving particle has position r(t)=((e^t) + (e^-t))/2 i + ((e^t) - (e^-t))/2 j.

a) find the cartesian equation of the path

I know all you have to do is let the i component to equal x(t) and then make t the subject, and then with the j component let it equal y and sub t into y. but i can't get t to be the subject in the equation.
someone please give it a go
 

Answers and Replies

  • #2
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i don't get it how can you get that equation from the original equation?
 
  • #3
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This is because the coeffcient of the x-coordinate and the y-coordinate is the same: [tex]\left( {\frac{{e^t + e^{ - t} }}{2}} \right)[/tex]
 
  • #4
HallsofIvy
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Oerg said:
This is because the coeffcient of the x-coordinate and the y-coordinate is the same: [tex]\left( {\frac{{e^t + e^{ - t} }}{2}} \right)[/tex]
No, that's incorrect (or the original poster has edited the post since you wrote this). The vector equation is
[tex]\left(\frac{e^t+ e^{-t}}{2}\right)i+ \left(\frac{e^t- e^{-t}}{2}\right)j[/tex]

The x and y components are not equal. It should be easy to see that the x-component is cosh(t) and the y-component is sinh(t). Now, use an identity: cosh2(t)- sinh2(t)= 1 so that x2-y2= 1, a hyperbola (in fact, that's why those called "hyperbolic" functions!).

If was not "easy" :biggrin: to see that [itex]\frac{e^t+ e^{-t}}{2}= cosh(t)[/itex] and that [itex]\frac{e^t- e^{-t}}{2}= sinh(t)[/itex], then it is a little harder: (et+ e-t)2= e2t+ 2+ e-2t. (You see how the et and e-t cancel when you mutiply them?)
Similarly, (et-et)2= e2t+ 2+ e-2t. Subtract those and you just get 4 (which is why you need "2" in the denominators).
 
  • #5
Hurkyl
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Halls already knew where this was going; that's how he figured out to do it that way. :smile:

If I had not already known where this was going, the first thing I would do is to recognize that you have two things of this form:

x = A + B
y = A - B

(for your actual problem, A = exp(t) / 2 and B = exp(-t) / 2)

You might recognize that it's easy to isolate A and B by adding and subtracting the equations respectively. If not, you might recognize this is a system of two linear equations in A and B, which can then be solved by A and B.

So, in any case, the first thing I would have done is exactly that:

x + y = 2A
x - y = 2B

so:

exp(t) = x + y
exp(-t) = x - y

And then, it should be clear how to get rid of t!
 

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