Solving for the mass using Parametric Equations (Density Forumula)

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SUMMARY

The mass of a spring defined by the parametric equations r(t) = <2cos(t), t, 2sin(t)> for 0 ≤ t ≤ 5π can be calculated using the density function ρ(x, y, z) = y. The correct formula for mass in this context is given by the integral ∫ ρ(x) dσ, where dσ represents the differential of arc length. For the provided parametric equations, dσ is calculated as √((df/dt)² + (dg/dt)² + (dh/dt)²) dt, which incorporates the derivatives of the parametric functions.

PREREQUISITES
  • Understanding of parametric equations in three dimensions
  • Knowledge of calculus, specifically double and single integrals
  • Familiarity with the concept of density functions in physics
  • Ability to compute derivatives of parametric functions
NEXT STEPS
  • Study the derivation of the differential of arc length for parametric curves
  • Learn how to apply density functions in three-dimensional integrals
  • Practice calculating mass using various density functions and parametric equations
  • Explore advanced topics in multivariable calculus, such as line integrals
USEFUL FOR

Students preparing for calculus exams, particularly those focusing on parametric equations and mass calculations, as well as educators seeking to enhance their teaching methods in multivariable calculus.

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Hey,

I have a calculus 3 test coming up that involves calculating the mass of an object using a density function. I know how to use the double integral form for x-y equations, but the practice test my professor gave me has a question using parametric equations and I don't know how to calculate the answer.

Find the mass of a spring in the shape of a helix defined by r(t) = <2cost,t,2sint>, for 0<=t<=5*pi, density function p(x,y,z)=y.



I know the mass is calculated using Int(Int(p(x,y)dx)dy) for a 2 variable system but don't know how to incorporate the t or z variables.
 
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You are mistaken in thinking that "the mass is calculated using Int(Int(p(x,y)dx)dy) for a 2 variable system". That is correct for a two-dimensional situation- a surface. What you have is a one dimensional figure in three dimensions. The mass is given by [itex]\int \rho(x) d\sigma[/itex] where [itex]d\sigma[/itex] is the "differential of arc length. If the figure is given by x= f(t), y= g(t), z= h(t), as you have here, then
[tex]d\sigma= \sqrt{\left(\frac{df}{dt}\right)^2+ \left(\frac{dg}{dt}\right)^2+ \left(\frac{dh}{dt}\right)^2}dt[/tex].
 

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