Length of Parametric curve

In summary, to estimate the length of the curve with parametric equations x = t^2, y = t^3, z = t^4, 0<=t<=3, we first find the magnitude of the velocity vector ||v|| using the formula sqrt((dx/dt)^2 + (dy/dt)^2 + (dz/dt)^2) and then take the integral of ||v|| from 0 to 3. This results in an estimate of 86.628, rounded to three decimal places. However, there may be some uncertainty in using this method as it is an approximation.
  • #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
 
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  • #2
I didn't get the same expression for ||v|| as you did. Where did 4t^2 and 9t^4 come from?
 
  • #3
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
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
ya that's a typo sorry

x = t2
 
  • #6
Well, so could you get the answer after that?
 
  • #7
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
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
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 I am 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
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
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
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
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
joemama69 said:
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.
 

What is the length of a parametric curve?

The length of a parametric curve is the distance traveled from the starting point to the end point along the curve. It is also known as the arc length.

How is the length of a parametric curve calculated?

The length of a parametric curve can be calculated using integration. The formula is L = ∫√(dx/dt)^2 + (dy/dt)^2 + (dz/dt)^2 dt, where dx/dt, dy/dt, and dz/dt are the derivatives of the parametric equations.

What is the significance of the length of a parametric curve?

The length of a parametric curve is important in many applications, such as physics, engineering, and computer graphics. It can be used to calculate the distance traveled by an object, the speed of an object, and the curvature of a curve.

Can the length of a parametric curve be negative?

No, the length of a parametric curve cannot be negative. It represents a physical distance and therefore must be a positive value. In some cases, the length may be zero if the starting and ending points are the same.

What factors can affect the length of a parametric curve?

The length of a parametric curve can be affected by the parametric equations used, the starting and ending points, and the curvature of the curve. It can also be affected by the scale or unit of measurement used to measure the distance along the curve.

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