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

The plane should contain r(pi/4) = <-√(2), 0, 3>. You should find another normal at pi/4; your current normal is the same as the tangent.
  • #1
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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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  • #2
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.
 
  • #3
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?
 
  • #4
Right.
1) lhs of ⇒ is correct, but r(pi/4) seems wrong.
2) Form is correct, but seems the numbers are wrong.
 

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

What is the definition of a tangent plane?

A tangent plane is a plane that touches a curve at a specific point and has the same slope as the curve at that point.

How is the tangent plane calculated?

The tangent plane is calculated by finding the derivative of the curve at the given point and using that derivative to determine the slope of the tangent line. This slope is then used to find the equation of the tangent plane.

What is the normal vector of a curve?

The normal vector of a curve is a vector that is perpendicular to the tangent plane at a specific point on the curve. It helps to determine the direction in which the curve is curving at that point.

What is an osculating plane?

An osculating plane is a plane that touches a curve at a specific point and has the same curvature as the curve at that point. It is also known as a tangent plane of order 2.

How is the osculating plane determined?

The osculating plane is determined by finding the second derivative of the curve at the given point and using that to calculate the curvature of the curve at that point. This curvature is then used to find the equation of the osculating plane.

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