91Ó°ÊÓ

When working with a curve, such as the elliptical helix given by the position vector \( \mathbf{c}(t) = (4 \cos t, \sin t, t) \), a tangent line provides the straightest path that touches the curve at a specific point. To find this tangent line at \( t = \pi/4 \), we first compute the derivative of the position vector, resulting in \( \mathbf{c}'(t) = (-4 \sin t, \cos t, 1) \). This derivative, evaluated at \( t = \pi/4 \), determines the direction the tangent line points in, giving \( \mathbf{c}'(\pi/4) = (-2\sqrt{2}, \frac{\sqrt{2}}{2}, 1) \).

Next, we find the position on the curve at \( t = \pi/4 \), which is \( \mathbf{c}(\pi/4) = (2\sqrt{2}, \frac{\sqrt{2}}{2}, \pi/4) \). With both the direction vector and position point, we construct the equation of the tangent line as follows:

\[ \mathbf{r}(t) = (2\sqrt{2}, \frac{\sqrt{2}}{2}, \frac{\pi}{4}) + s(-2\sqrt{2}, \frac{\sqrt{2}}{2}, 1), \] where \(s\) is a parameter. This signifies that the tangent line extends infinitely in both directions along this line.

The force exerted on a particle moving along a path depends on its acceleration, derived from the second derivative of the position vector. For our particle moving along the elliptical helix, we first compute the acceleration by differentiating the velocity vector (or the first derivative of the position vector) again. This gives us \( \mathbf{c}''(t) = (-4\cos t, -\sin t, 0) \).

To find the force at a specific time, such as \( t = \pi/4 \), we evaluate this second derivative at that time: \( \mathbf{c}''(\pi/4) = (-2\sqrt{2}, -\frac{\sqrt{2}}{2}, 0) \).

Using Newton's second law of motion, the force \( \mathbf{F} \) is calculated as \( m\mathbf{a} \), with \( m \) representing the mass of the particle. Therefore, the force acting on the particle at \( t = \pi/4 \) is given by:

\[ \mathbf{F}(t=\pi/4) = m(-2\sqrt{2}, -\frac{\sqrt{2}}{2}, 0) \].
This expression indicates how factors such as the particle's mass and its acceleration affect the resulting force.

The arc length of a curve is a measure of the distance along the curve between two points. For the elliptical helix defined by \( \mathbf{c}(t) = (4 \cos t, \sin t, t) \), we want to find the length from \( t = 0 \) to \( t = \pi/4 \). To do this, we use the arc length formula:

\[ S = \int_{a}^{b} \left| \mathbf{c}'(t) \right| \, dt \]
with \( a = 0 \) and \( b = \pi/4 \). Here, \( \mathbf{c}'(t) \) is the derivative of the position vector.

First, compute the magnitude of the derivative:

\[ \left| \mathbf{c}'(t) \right| = \sqrt{(-4\sin t)^2 + (\cos t)^2 + 1^2} = \sqrt{16\sin^2 t + \cos^2 t + 1} \].

Substitute this expression into the integral to set up the arc length formula:

\[ S = \int_{0}^{\pi/4} \sqrt{16\sin^2 t + \cos^2 t + 1} \, dt \].

This integral provides the arc length, allowing us to measure the total distance traveled along the curve over the specified interval.

An elliptical helix is a fascinating three-dimensional curve that extends the concept of a standard helix. For the given problem, the helix is parameterized by

\( \mathbf{c}(t) = (4 \cos t, \sin t, t) \).

This representation combines both circular motion, in the \( x \)-\( y \) plane, and linear motion along the \( z \)-axis.

• The term \( 4 \cos t \) describes an ellipse rather than a circle, due to the scaling by 4 in the \( x \)-coordinate, stretching the circle horizontally into an ellipse.
• The \( \sin t \) stays as the \( y \)-coordinate term, maintaining the vertical component of the ellipse.
• The third coordinate \( t \) indicates that while the particle follows the elliptical path in the \( x \)-\( y \) plane, it simultaneously ascends in the \( z \)-direction, creating the helical aspect.

This distinctive pattern results in a smooth and graceful curve, widely applicable in areas such as electrical engineering and structural design, where the blend of rotational symmetry and elevation is advantageous.