Line Integral along a Parabola

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SUMMARY

The discussion focuses on evaluating a line integral along a parabolic segment defined by the parametric equation r(t) = for the interval 1 ≤ t ≤ 2. The integral evaluated is ∫_C x ds, which transforms into ∫_1^2 t √(1 + 4t²) dt. The user successfully applies the substitution u = 1 + 4t², leading to the integral being simplified to (1/8) ∫_{u=5}^{u=17} √u du, which is correctly evaluated to yield (17^(3/2) - 5^(3/2))/12.

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  • Understanding of parametric equations in calculus
  • Knowledge of line integrals and their evaluation
  • Familiarity with substitution techniques in integration
  • Proficiency in evaluating definite integrals
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  • Study the properties of line integrals in vector calculus
  • Learn about parametric curves and their applications
  • Explore advanced integration techniques, including trigonometric and hyperbolic substitutions
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Students studying calculus, particularly those focusing on line integrals and parametric equations, as well as educators looking for examples to illustrate these concepts.

Char. Limit
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Homework Statement


a. Find a parametric equation to describe a parabola from the point (1,1) to the point (2,4).

b. Evaluate the line integral \int_C x ds along the parabolic segment in part a.

Homework Equations



\int_C x ds = \int_{t1}^{t2} x(t) |r'(t)| dt

The Attempt at a Solution



Well, for part a, the parametric equation r(t)=<t,t2>, 1≤t≤2, seemed to suffice. So I used that.

For part b, first I found r'(t) and got \sqrt{1+4 t^2}. Since x(t) = t, I then plugged this into my equation to get my new integral...

\int_1^2 t \sqrt{1+4 t^2} dt

Then I used the transform u=1+4t2, du = 8t dt to transform my integral to...

\frac{1}{8} \int_{t=1}^{t=2} \sqrt{u} du = \frac{1}{8} \int_{u=5}^{u=17} \sqrt{u} du

Which I then evaluated to get...

\frac{1}{8} \frac{2}{3} \left[ u^{\frac{3}{2}} \right]_5^{17}

Which seems to equal...

\frac{17^{\frac{3}{2}} - 5^{\frac{3}{2}}}{12}

My question is... is this right?
 
Physics news on Phys.org
Looks fine to me.
 

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