Finding Tangent, Normal & Osculating Planes of r(t) at t=π/4

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SUMMARY

The discussion focuses on finding the tangent line, normal plane, and osculating plane for the curve defined by r(t) = -2sin(t) i + 2cos(t) j + 3 k at t = π/4. The correct representation for the tangent line is l_T = r(π/4) + k * T^(π/4), where T^(t) is the unit tangent vector. The normal plane is defined by the equation <–√(2)/2, –√(2), 0> ⋅ = 0, while the osculating plane is represented by <0, 0, 1> ⋅ = 0. The unit normal vector N^(t) is derived from the derivative of the tangent vector.

PREREQUISITES
  • Understanding of vector calculus and parametric equations
  • Familiarity with the concepts of tangent, normal, and osculating planes
  • Knowledge of derivatives and unit vectors
  • Proficiency in using vector notation and operations
NEXT STEPS
  • Study the derivation of unit tangent vectors in vector calculus
  • Learn about the properties and applications of osculating planes
  • Explore the concept of curvature and its relation to tangent and normal vectors
  • Practice solving problems involving parametric curves and their associated planes
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Students studying calculus, particularly those focusing on vector calculus and geometry, as well as educators looking for detailed explanations of tangent and normal planes in three-dimensional space.

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Homework Statement


Find the equations of the tangent line, normal plane and osculating plane to the curve

r(t) = -2sin(t) i + 2cos(t) j + 3 k

at the point corresponding to t = π/4.

Homework Equations


T[/B]^(t) = r'(t) // ||r'(t)||
u = a i + b j + c k, ||u|| = √(a^2 + b^2 + c^2)
N^(t) = T^(t) / ||T^(t)||
B^ = T^ × N^

The osculating plane is plane formed by unit normal and unit tangent vectors where the unit binormal vector is the vector normal to the osculating plane.

The normal plane is formed by the unit normal and unit binormal vectors and the unit tangent vector is the vector normal to the normal plane.

The Attempt at a Solution


N[/B]^(t) = sin(t) i - cos(t) j

B
(t) = k (technically independent of t)

This was a question on a quiz I received, and I'm studying for an exam now, and I have the following three questions.:
  1. Would the tangent line be represented by P= a ⋅ T^(π/4) + [r(t) - r(π/4)] (where a is some scalar)?
  2. Would the normal plane be represented by T^(pi/4) ⋅ (r(t) -r(π/4)) = 0?
  3. Would the osculating plane be represented by B^(π/4) ⋅ (r(t) - r(π/4)) = 0?

Any input would be GREATLY appreciated!
 
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I think your equation defining the unit normal is wrong -it should have primes(derivatives) rather than hats on the rhs.
Then:
1) Close but, The tangent line should have a single parameter; is it a or t? Try to picture: you start at r(π/4) and add an arbitrary amount (a) along the tangent vector.
2) A plane should have 2 degrees of freedom; you are parametrizing it by t which gives only 1. Let r(t) be free, you are already constraining the plane to be normal to the tangent, and to pass through r(π/4).
3) Yes, but , same problem as in 2.
 
Sorry for the very-delayed response. (I have so much work that it takes time to get ahead of it so that I can do other things such as answer forum posts.)

Okay, so the equation for the unit normal should be N^(t) = [d/dt T^(t)] / || d/dt [T^(t)] ||, right?

1) So, would the tangent line be correctly represented by l_T = r(pi/4) + k * T^(π/4) ⇒ l_T = <–2,0,3> + k * <–√(2)/2, –√(2)/2, 0>? (where k is some scalar)

2) So, would the normal plane be correctly represented by <–√(2)/2, –√(2), 0> ⋅ <x – –2, y – 0, z – 3> = 0?

3) So, would the osculating plane be correctly represented by <0, 0, 1> ⋅ <x - -2, y – 0, z – 3> = 0?
 
Right.
1) lhs of ⇒ is correct, but r(pi/4) seems wrong.
2) Form is correct, but seems the numbers are wrong.
 

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