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Length of Parametric curve

  • Thread starter joemama69
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  • #1
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Homework Statement



Estimate the length of the curve with parametric equations x = t^3, y = t^3, z = t^4, 0<=t<=3. Round to the 3 decimals place.

Homework Equations





The Attempt at a Solution



virst i found the magnitide.. ||v|| = (4t^2 + 9t^4 + 16t^6)^1/2

i tool the integral from 0 to 3 of ||v||dt and i got 86.628 did i do this correctly
 

Answers and Replies

  • #2
Defennder
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I didn't get the same expression for ||v|| as you did. Where did 4t^2 and 9t^4 come from?
 
  • #3
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well according to my book distance traveled is the integral from a to b of ||v(t)|dt
where ||v|| = sqrt((dx/dt)^2 + (dy/dt)^2 + (dz/dt)^2)

(dx/dt)^2 = (2t)^2 = 4t^2
(dy/dt)^2 = 3t^2)^2 = 9t^4
(dz/dt)^2 = 4t^3)^2 = 16t^6
 
  • #4
Cyosis
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The components x and y are the same, then how can the derivative with respect to the same variable differ? [tex]x=t^3 \Rightarrow \frac{dt^3}{dt}=?[/tex]?

Did you type out the wrong problem by any chance, should it be x=t^2?
 
  • #5
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ya thats a typo sorry

x = t2
 
  • #6
Defennder
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Well, so could you get the answer after that?
 
  • #7
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x = t^2, y = t^3, z = t^4, 0<=t<=3. Round to the 3 decimals place.

where ||v|| = sqrt((dx/dt)^2 + (dy/dt)^2 + (dz/dt)^2)

(dx/dt)^2 = (2t)^2 = 4t^2
(dy/dt)^2 = 3t^2)^2 = 9t^4
(dz/dt)^2 = 4t^3)^2 = 16t^6

first i found the magnitide.. ||v|| = (4t^2 + 9t^4 + 16t^6)^1/2

i took the integral from 0 to 3 of ||v||dt and i got 86.628 did i do this correctly
 
  • #8
Defennder
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It would help if you told us how you got that. The closed form expression of the integral looks very messy, and since the question said to "estimate" rather than give an exact answer some form of approximation must be involved.
 
  • #9
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according to my book, length of a curve is

Length of C = [tex]\int\sqrt{(dx/dt)^2 + (dy/dt)^2 + (dz/dt)^2}[/tex] dt from a to b

x = t^2, y = t^3, z = t^4, 0 <t<3

Length of C = [tex]\int\sqrt{(2t)^2 + (3t^2)^2 + (4t^3)^2}[/tex] dt from 0 to 3

Length of C = [tex]\int\sqrt{(4t)^2 + (9t^4) + (16t^6)}[/tex] dt from 0 to 3

i pluged it into my calculator and got 86.628

the problem says to round to three decimals. So im not really sure how i am suppose to approximate something that percise. But all the same, is there somethign wrong with what i did
 
  • #10
Defennder
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You mean all you did was enter that definite integral into your calculator and have it work the answer out for you? I'm not sure if that calculator you're using would be allowed in your exams, but right up to the point where you got the definite integral everything seems to be all right.
 
  • #11
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ya we are allowed to use them,

but are u saying my calc gave me the wrong answer or just that ur not sure if i should be using it or not
 
  • #12
Defennder
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I'm not saying that your calculator is wrong. It's probably correct if you entered the integrand properly. It's just that doing so may not be what the question had in mind.
 
  • #13
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Ya i was thinking maybe that is why it is worded to estimate it.

But i assume that the integral can be intigrated and would give a deffinite answer anyways.
 
  • #14
Redbelly98
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x = t^2, y = t^3, z = t^4, 0 <t<3

Length of C = [tex]\int\sqrt{(2t)^2 + (3t^2)^2 + (4t^3)^2}[/tex] dt from 0 to 3

Length of C = [tex]\int\sqrt{(4t)^2 + (9t^4) + (16t^6)}[/tex] dt from 0 to 3
Note, first term should be (4t2)

i pluged it into my calculator and got 86.628

the problem says to round to three decimals.
I get the same answer.
 

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