Consider the following curve: x = sin(4t), y = −cos(4t), z = 16t Find the equation of the osculating plane of the given curve at the point (0, 1, 4𝜋).

To find the equation of the osculating plane of the curve \( x = \sin(4t) \), \( y = -\cos(4t) \), \( z = 16t \) at the point \( (0, 1, 4\pi) \), follow these steps:

1. **Determine the value of \( t \) that corresponds to the point \( (0, 1, 4\pi) \):**

   Given that \( z = 16t \), we solve for \( t \):
   \[
   16t = 4\pi \implies t = \frac{\pi}{4}
   \]

2. **Compute the derivatives of the parametric equations to find the tangent vector:**

   The parametric equations are:
   \[
   x = \sin(4t), \quad y = -\cos(4t), \quad z = 16t
   \]

   First derivatives:
   \[
   \frac{dx}{dt} = 4\cos(4t), \quad \frac{dy}{dt} = 4\sin(4t), \quad \frac{dz}{dt} = 16
   \]

   At \( t = \frac{\pi}{4} \):
   \[
   \frac{dx}{dt}\bigg|_{t = \frac{\pi}{4}} = 4\cos\left(4 \cdot \frac{\pi}{4}\right) = 4\cos(\pi) = 4(-1) = -4
   \]
   \[
   \frac{dy}{dt}\bigg|_{t = \frac{\pi}{4}} = 4\sin\left(4 \cdot \frac{\pi}{4}\right) = 4\sin(\pi) = 4(0) = 0
   \]
   \[
   \frac{dz}{dt}\bigg|_{t = \frac{\pi}{4}} = 16
   \]

   So the tangent vector \( \mathbf{T} \) at \( t = \frac{\pi}{4} \) is:
   \[
   \mathbf{T} = (-4, 0, 16)
   \]

3. **Compute the second derivatives to find the normal vector:**

   Second derivatives:
   \[
   \frac{d^2x}{dt^2} = -16\sin(4t), \quad \frac{d^2y}{dt^2} = 16\cos(4t), \quad \frac{d^2z}{dt^2} = 0
   \]

   At \( t = \frac{\pi}{4} \):
   \[
   \frac{d^2x}{dt^2}\bigg|_{t = \frac{\pi}{4}} = -16\sin(\pi) = 0
   \]
   \[
   \frac{d^2y}{dt^2}\bigg|_{t = \frac{\pi}{4}} = 16\cos(\pi) = 16(-1) = -16
   \]
   \[
   \frac{d^2z}{dt^2}\bigg|_{t = \frac{\pi}{4}} = 0
   \]

   So the second derivative vector \( \mathbf{N} \) at \( t = \frac{\pi}{4} \) is:
   \[
   \mathbf{N} = (0, -16, 0)
   \]

4. **Find the binormal vector \( \mathbf{B} \) by taking the cross product of \( \mathbf{T} \) and \( \mathbf{N} \):**

   \[
   \mathbf{B} = \mathbf{T} \times \mathbf{N} = \begin{vmatrix}
   \mathbf{i} & \mathbf{j} & \mathbf{k} \\
   -4 & 0 & 16 \\
   0 & -16 & 0
   \end{vmatrix} = (0 \cdot 0 - 16 \cdot (-16)) \mathbf{i} - (-4 \cdot 0 - 16 \cdot 0) \mathbf{j} + (-4 \cdot (-16) - 0 \cdot 0) \mathbf{k}
   \]
   \[
   \mathbf{B} = (256, 0, 64)
   \]

5. **Determine the normal vector of the osculating plane:**

   The normal vector of the osculating plane is \( \mathbf{B} \):
   \[
   \mathbf{B} = (256, 0, 64)
   \]

6. **Write the equation of the plane:**

   The equation of a plane with normal vector \( \mathbf{B} = (A, B, C) \) and passing through the point \( (x_0, y_0, z_0) \) is:
   \[
   A(x - x_0) + B(y - y_0) + C(z - z_0) = 0
   \]

   Plugging in the values:
   \[
   256(x - 0) + 0(y - 1) + 64(z - 4\pi) = 0
   \]
   \[
   256x + 64(z - 4\pi) = 0
   \]
   \[
   256x + 64z - 256\pi = 0
   \]
   \[
   4x + z = 4\pi
   \]

   So the equation of the osculating plane at the point \( (0, 1, 4\pi) \) is:
   \[
   4x + z = 4\pi
   \]

Would you like any further details or have any questions about this solution? Here are five more questions you might consider:

1. How do you find the curvature of a parametric curve?
2. What is the difference between the osculating plane and the normal plane?
3. How do you determine the torsion of a space curve?
4. Can you explain how to find the Frenet-Serret frame for a curve?
5. What is the significance of the binormal vector in the context of space curves?

**Tip:** When working with parametric curves, it's often helpful to compute the derivatives and cross products step-by-step to avoid mistakes and better understand the geometry of the curve.

Learn how to find the equation of the osculating plane for the parametric curve x = sin(4t), y = -cos(4t), z = 16t at the point (0, 1, 4π). Includes step-by-step calculation of derivatives, tangent vector, normal vector, and the final plane equation. Suitable for undergraduate students studying vector calculus and parametric curves.

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