Unit Tangent, Normal & Binormal Vectors - Calculus 3

1. For the helix $\mathbf{r}(t) = \langle a\cos t, a\sin t, bt \rangle$ where $a > 0$ and $b \neq 0$, the binormal vector $\mathbf{B}(t)$ has the property that:
   Answer: $\mathbf{B}(t)$ rotates around the $z$-axis with period $2\pi$ For this helix, r'(t) = ⟨-a sin t, a cos t, b⟩ and ||r'(t)|| = √(a² + b²). So T(t) = ⟨-a sin t, a cos t, b⟩/√(a² + b²). Then T'(t) = ⟨-a cos t, -a sin t, 0⟩/√(a² + b²), and N(t) = ⟨-cos t, -sin t, 0⟩. Computing B(t) = T(t) × N(t) gives B(t) = ⟨b sin t, -b cos t, a⟩/√(a² + b²). This shows B(t) rotates around the z-axis with period 2π. Choice A is wrong because B(t) is not constant. Choice C is wrong because B(t) has a non-zero z-component. Choice D is incomplete since all unit vectors have constant magnitude; the key insight is the specific rotational behavior.
2. Let $\mathbf{r}(t) = \langle 2\cos t, 2\sin t, 3t \rangle$ be a vector-valued function. If $\mathbf{T}(t)$ is the unit tangent vector and $\mathbf{N}(t)$ is the principal unit normal vector, what is $\mathbf{T}(\pi/2) \cdot \mathbf{N}(\pi/2)$?
   Answer: $0$ The unit tangent vector T(t) and principal unit normal vector N(t) are always orthogonal by definition. Therefore, their dot product is always 0, regardless of the specific value of t. First, we find T(t) = r'(t)/||r'(t)|| where r'(t) = ⟨-2sin t, 2cos t, 3⟩ and ||r'(t)|| = √(4sin²t + 4cos²t + 9) = √13. So T(t) = ⟨-2sin t, 2cos t, 3⟩/√13. Then N(t) = T'(t)/||T'(t)|| where T'(t) = ⟨-2cos t, -2sin t, 0⟩/√13. Since T(t) and N(t) are orthogonal by construction, T(t) · N(t) = 0. Choice B represents 2/√13 which might come from incorrectly computing a component of T. Choice C represents 3/√13 which is the z-component of T. Choice D represents 1, which might come from confusing this with T·T.
3. Consider a space curve where the unit tangent vector is $\mathbf{T}(t) = \langle \cos t, \sin t, 0 \rangle$ and the principal unit normal vector is $\mathbf{N}(t) = \langle -\sin t, \cos t, 0 \rangle$. Which of the following statements about the binormal vector $\mathbf{B}(t)$ is correct?
   Answer: $\mathbf{B}(t) = \langle 0, 0, 1 \rangle$ and the torsion $\tau = 0$ When analyzing space curves, the binormal vector $\mathbf{B}(t)$ is always the cross product of the unit tangent and principal normal vectors: $\mathbf{B}(t) = \mathbf{T}(t) \times \mathbf{N}(t)$. This vector completes the moving coordinate frame (called the Frenet frame) that travels along the curve. Let's compute the binormal vector using the given information: $\mathbf{B}(t) = \mathbf{T}(t) \times \mathbf{N}(t) = \langle \cos t, \sin t, 0 \rangle \times \langle -\sin t, \cos t, 0 \rangle$ Using the cross product formula: $\mathbf{B}(t) = \langle (0)(0) - (0)(\cos t), (0)(-\sin t) - (\cos t)(0), (\cos t)(\cos t) - (\sin t)(-\sin t) \rangle$ $= \langle 0, 0, \cos^2 t + \sin^2 t \rangle = \langle 0, 0, 1 \rangle$ Since $\mathbf{B}(t)$ is constant (doesn't vary with $t$), the torsion $\tau = 0$. Torsion measures how much a curve twists out of its osculating plane, and it equals $-\frac{d\mathbf{B}}{dt} \cdot \mathbf{N}$. When $\mathbf{B}$ is constant, $\frac{d\mathbf{B}}{dt} = \mathbf{0}$, so $\tau = 0$. Option A is wrong because $\mathbf{B}(t)$ is constant, not varying. Option B has the wrong sign for the binormal vector. Option C correctly identifies $\mathbf{B}(t)$ but incorrectly claims non-zero torsion. Study tip: Remember that constant binormal vector always means zero torsion. When both $\mathbf{T}$ and $\mathbf{N}$ lie in the $xy$-plane ($z$-components are zero), $\mathbf{B}$ will point along the $z$-axis and typically be constant.
4. For the curve $\mathbf{r}(t) = \langle e^t \cos t, e^t \sin t, e^t \rangle$, the angle between the unit tangent vector $\mathbf{T}(t)$ and the binormal vector $\mathbf{B}(t)$ is:
   Answer: $\pi/2$ radians (they are perpendicular) By definition, the unit tangent vector T(t), principal unit normal vector N(t), and binormal vector B(t) form an orthonormal basis called the Frenet frame. This means T(t) ⊥ N(t), N(t) ⊥ B(t), and T(t) ⊥ B(t). Therefore, the angle between T(t) and B(t) is always π/2 radians, regardless of the specific curve. This is a fundamental property of the TNB frame. Choices A, B, and D represent common misconceptions about the relationships between these vectors.
5. Consider the curve $\mathbf{r}(t) = \langle t^2, t^3, t \rangle$. At the point where $t = 1$, if $\mathbf{B}$ is the binormal vector and $\kappa$ is the curvature, which expression correctly relates the binormal vector to the tangent and normal vectors?
   Answer: $\mathbf{B} = \mathbf{T} \times \mathbf{N}$ and $\mathbf{B}$ is perpendicular to both $\mathbf{T}$ and $\mathbf{N}$ The binormal vector is defined as B = T × N, where T is the unit tangent vector and N is the principal unit normal vector. Since B is the cross product of two orthogonal unit vectors, it is perpendicular to both T and N, and ||B|| = 1. Choice A is incorrect because B does not point in the direction of r''(1); the normal vector N points toward the center of curvature. Choice B has the wrong order in the cross product (N × T = -B). Choice D is incorrect because ||B|| = 1 always, not κ (the curvature).

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