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J. M.'s missing motivation
J. M.'s missing motivation
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This is yet another case where the Weierstrass substitutionWeierstrass substitution is useful:

With[{m = Quantity[6, "Kilograms"], θ = 38 °,
      v1 = Quantity[13, ("Meters")/("Seconds")] {-1, 0},
      t = Quantity[3, "Seconds"]},
     v2 = Quantity[25, ("Meters")/("Seconds")] AngleVector[θ];
     L1 = m v1; L2 = m v2; ΔL = L2 - L1;
     F = Norm[ΔL]/t // UnitSimplify // N;
     i = t F AngleVector[ϕ] /. ϕ -> 2 ArcTan[ff] // TrigExpand;
     2 ArcTan[ff] /. First[NSolve[UnitSimplify /@ L1 + i == L2, ff]]]

   0.4399220723025391

This is yet another case where the Weierstrass substitution is useful:

With[{m = Quantity[6, "Kilograms"], θ = 38 °,
      v1 = Quantity[13, ("Meters")/("Seconds")] {-1, 0},
      t = Quantity[3, "Seconds"]},
     v2 = Quantity[25, ("Meters")/("Seconds")] AngleVector[θ];
     L1 = m v1; L2 = m v2; ΔL = L2 - L1;
     F = Norm[ΔL]/t // UnitSimplify // N;
     i = t F AngleVector[ϕ] /. ϕ -> 2 ArcTan[ff] // TrigExpand;
     2 ArcTan[ff] /. First[NSolve[UnitSimplify /@ L1 + i == L2, ff]]]

   0.4399220723025391

This is yet another case where the Weierstrass substitution is useful:

With[{m = Quantity[6, "Kilograms"], θ = 38 °,
      v1 = Quantity[13, ("Meters")/("Seconds")] {-1, 0},
      t = Quantity[3, "Seconds"]},
     v2 = Quantity[25, ("Meters")/("Seconds")] AngleVector[θ];
     L1 = m v1; L2 = m v2; ΔL = L2 - L1;
     F = Norm[ΔL]/t // UnitSimplify // N;
     i = t F AngleVector[ϕ] /. ϕ -> 2 ArcTan[ff] // TrigExpand;
     2 ArcTan[ff] /. First[NSolve[UnitSimplify /@ L1 + i == L2, ff]]]

   0.4399220723025391
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Bob Hanlon
Bob Hanlon
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This is yet another case where the Weierstrass substitutionWeierstrass substitution is useful:

With[{m = Quantity[6, "Kilograms"], θ = 38 °,
      v1 = Quantity[13, ("Meters")/("Seconds")] {-1, 0},
      t = Quantity[3, "Seconds"]},
     v2 = Quantity[25, ("Meters")/("Seconds")] AngleVector[θ];
     L1 = m v1; L2 = m v2; ΔL = L2 - L1;
     F = Norm[ΔL]/t // UnitSimplify // N;
     i = t F AngleVector[ϕ] /. ϕ -> 2 ArcTan[ff] // TrigExpand;
     2 ArcTan[ff] /. First[NSolve[UnitSimplify /@ L1 + i == L2, ff]]]

   0.4399220723025391

This is yet another case where the Weierstrass substitution is useful:

With[{m = Quantity[6, "Kilograms"], θ = 38 °,
      v1 = Quantity[13, ("Meters")/("Seconds")] {-1, 0},
      t = Quantity[3, "Seconds"]},
     v2 = Quantity[25, ("Meters")/("Seconds")] AngleVector[θ];
     L1 = m v1; L2 = m v2; ΔL = L2 - L1;
     F = Norm[ΔL]/t // UnitSimplify // N;
     i = t F AngleVector[ϕ] /. ϕ -> 2 ArcTan[ff] // TrigExpand;
     2 ArcTan[ff] /. First[NSolve[UnitSimplify /@ L1 + i == L2, ff]]]

   0.4399220723025391

This is yet another case where the Weierstrass substitution is useful:

With[{m = Quantity[6, "Kilograms"], θ = 38 °,
      v1 = Quantity[13, ("Meters")/("Seconds")] {-1, 0},
      t = Quantity[3, "Seconds"]},
     v2 = Quantity[25, ("Meters")/("Seconds")] AngleVector[θ];
     L1 = m v1; L2 = m v2; ΔL = L2 - L1;
     F = Norm[ΔL]/t // UnitSimplify // N;
     i = t F AngleVector[ϕ] /. ϕ -> 2 ArcTan[ff] // TrigExpand;
     2 ArcTan[ff] /. First[NSolve[UnitSimplify /@ L1 + i == L2, ff]]]

   0.4399220723025391
Source Link
J. M.'s missing motivation
J. M.'s missing motivation
  • 126.7k
  • 11
  • 413
  • 593

This is yet another case where the Weierstrass substitution is useful:

With[{m = Quantity[6, "Kilograms"], θ = 38 °,
      v1 = Quantity[13, ("Meters")/("Seconds")] {-1, 0},
      t = Quantity[3, "Seconds"]},
     v2 = Quantity[25, ("Meters")/("Seconds")] AngleVector[θ];
     L1 = m v1; L2 = m v2; ΔL = L2 - L1;
     F = Norm[ΔL]/t // UnitSimplify // N;
     i = t F AngleVector[ϕ] /. ϕ -> 2 ArcTan[ff] // TrigExpand;
     2 ArcTan[ff] /. First[NSolve[UnitSimplify /@ L1 + i == L2, ff]]]

   0.4399220723025391