Problem 85 - Calculus 3

Question

A particle travels along the path of a helix with the equation \(\mathbf{r}(t)=\cos (t) \mathbf{i}+\sin (t) \mathbf{j}+t \mathbf{k} .\) See the graph presented here: Find the following: Acceleration of the particle at any time

Step-by-Step Solution

Answer

The acceleration of the particle is \(-\cos(t) \mathbf{i} - \sin(t) \mathbf{j}\).

Step 1: Determine the Velocity Vector

The velocity of the particle is obtained by differentiating the position vector \(\mathbf{r}(t)\) with respect to time \(t\). The position vector is given as \(\mathbf{r}(t) = \cos(t) \mathbf{i} + \sin(t) \mathbf{j} + t \mathbf{k}\). Find the derivative:\[\mathbf{v}(t) = \frac{d\mathbf{r}}{dt} = -\sin(t) \mathbf{i} + \cos(t) \mathbf{j} + \mathbf{k}.\]

Step 2: Compute the Acceleration Vector

The acceleration of the particle is obtained by differentiating the velocity vector \(\mathbf{v}(t)\) with respect to time \(t\). Using the expression for velocity:\[\mathbf{a}(t) = \frac{d\mathbf{v}}{dt} = -\cos(t) \mathbf{i} - \sin(t) \mathbf{j} + 0 \mathbf{k} = -\cos(t) \mathbf{i} - \sin(t) \mathbf{j}.\]

Step 3: Verify the Components of Acceleration

Ensure that the acceleration vector derived has the correct form by checking the individual components. The \(\mathbf{i}\) component is \(-\cos(t)\), the \(\mathbf{j}\) component is \(-\sin(t)\), and the \(\mathbf{k}\) component is 0, consistent with the expected form of acceleration for circular motion in the xy-plane and constant linear motion along the z-axis.

Key Concepts

Velocity Vector

When a particle moves through space, especially along a curved path like a helix, we can describe its motion using vectors. A velocity vector is a crucial concept because it tells us the speed and direction of the particle at any given moment. To find this vector, we start with the position vector of the particle, \( \mathbf{r}(t) = \cos(t) \mathbf{i} + \sin(t) \mathbf{j} + t \mathbf{k} \).
By differentiating this position vector with respect to time, we get the velocity vector:

• The derivative of \( \cos(t) \) is \( -\sin(t) \), forming the \(\mathbf{i}\) component of \( \mathbf{v}(t) \).
• The derivative of \( \sin(t) \) is \( \cos(t) \), forming the \(\mathbf{j}\) component.
• The derivative of \( t \) is \( 1 \), forming the \(\mathbf{k}\) component.

This gives us:\[ \mathbf{v}(t) = -\sin(t) \mathbf{i} + \cos(t) \mathbf{j} + \mathbf{k} \].
This vector shows how the particle moves in a circular path in the xy-plane while it simultaneously moves upward along the z-axis.

Differentiation

Differentiation is a fundamental technique in calculus used to determine the rate at which a function is changing at any given point. When dealing with vectors, we apply differentiation component-wise. For example, differentiating our position vector, \( \mathbf{r}(t) = \cos(t) \mathbf{i} + \sin(t) \mathbf{j} + t \mathbf{k} \), means separately differentiating each of the functions \( \cos(t), \sin(t), \) and \( t \).
By doing this, we not only obtain the velocity vector, \( \mathbf{v}(t) = -\sin(t) \mathbf{i} + \cos(t) \mathbf{j} + \mathbf{k} \), but can continue to differentiate to find higher rates of change, such as acceleration.To find the acceleration vector, differentiate the velocity vector:

• \( -\sin(t) \) becomes \( -\cos(t) \) as the \(\mathbf{i}\) component.
• \( \cos(t) \) becomes \( -\sin(t) \) as the \(\mathbf{j}\) component.
• \( 1 \) becomes \( 0 \), the constant, as the \(\mathbf{k}\) component.

Thus, the differentiation of \( \mathbf{v}(t) \) leads to \( \mathbf{a}(t) = -\cos(t) \mathbf{i} - \sin(t) \mathbf{j} \). Understanding this process is key to analyzing changes in motion.

Helix Equation

A helix is a fascinating curve, often imagined as a spiral staircase or a slinky toy. In calculus, we describe a helix using parametric equations that depend on a parameter, usually time \( t \). The given helix has the equation:\[ \mathbf{r}(t) = \cos(t) \mathbf{i} + \sin(t) \mathbf{j} + t \mathbf{k} \].
This equation beautifully captures the complexity of a helix:

• The \( \cos(t) \) and \( \sin(t) \) terms represent circular motion in the xy-plane, hence a spiral.
• The \( t \mathbf{k} \) term allows for the linear rise along the z-axis, giving the spiral its height.

Each component of the equation is essential for the full 3D description.
Studying helix motion in calculus provides insights not just into particle movements, but also in understanding interesting mathematical structures in space, technology, and nature.