Question

Consider the motion of a particle along a helix given by \(\mathbf{r}(t)=\sin t \mathbf{i}+\cos t \mathbf{j}+\left(t^{2}-3 t+2\right) \mathbf{k}\), where the \(\mathbf{k}\) component measures the height in meters above the ground and \(t \geq 0 .\) If the particle leaves the helix and moves along the line tangent to the helix when it is 12 meters above the ground, give the direction vector for the line.

Short answer

The direction vector is \( \cos 5 \mathbf{i} - \sin 5 \mathbf{j} + 7 \mathbf{k} \).

Step-by-step solution

Identify the Component for the Specific Condition

To solve the exercise, you need to understand that a direction vector for the line tangent to the helix can be found using the derivative of the position vector \( \mathbf{r}(t) \). First, note the condition that the particle is 12 meters above the ground. The height is given by the \( \mathbf{k} \) component of \( \mathbf{r}(t) \) which is \( t^2 - 3t + 2 \). Set this equal to 12 to find the specific time \( t \). Thus, solve:\[ t^2 - 3t + 2 = 12 \]

Solve the Equation for Specific Time

Solve the quadratic equation \( t^2 - 3t + 2 = 12 \), which simplifies to:\[ t^2 - 3t - 10 = 0 \]Use the quadratic formula: \[ t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \] where \( a = 1 \), \( b = -3 \), and \( c = -10 \). Calculate the discriminant:\[ b^2 - 4ac = (-3)^2 - 4(1)(-10) = 9 + 40 = 49 \]So, \( t = \frac{3 \pm 7}{2} \). Calculate the values of \( t \):\[ t = \frac{10}{2} = 5 \quad \text{or} \quad t = \frac{-4}{2} = -2 \]Since \( t \geq 0 \), we discard \( t = -2 \) and accept \( t = 5 \).

Calculate the Tangent Vector

To find the direction vector, calculate the derivative \( \mathbf{r}'(t) \) at \( t = 5 \). The derivative is:\[ \mathbf{r}'(t) = \frac{d}{dt}(\sin t)\mathbf{i} + \frac{d}{dt}(\cos t)\mathbf{j} + \frac{d}{dt}(t^2 - 3t + 2)\mathbf{k} \]Thus, \[ \mathbf{r}'(t) = \cos t \mathbf{i} - \sin t \mathbf{j} + (2t - 3) \mathbf{k} \].Substitute \( t = 5 \):\[ \mathbf{r}'(5) = \cos 5 \mathbf{i} - \sin 5 \mathbf{j} + (10 - 3) \mathbf{k} = \cos 5 \mathbf{i} - \sin 5 \mathbf{j} + 7 \mathbf{k} \].

Direction Vector for the Line

The direction vector for the tangent line at \( t = 5 \) is:\[ \mathbf{d} = \cos 5 \mathbf{i} - \sin 5 \mathbf{j} + 7 \mathbf{k} \],as derived from \( \mathbf{r}'(5) \). This vector indicates the direction in which the particle will move away from the helix.

Key concepts

Helical Motion

Helical motion describes the path of a particle moving along a three-dimensional spiral, much like a spring or a corkscrew. The helical path is represented by a position vector function. For example, in our problem, the helical motion is given by the vector \( \mathbf{r}(t) = \sin t \mathbf{i} + \cos t \mathbf{j} + (t^2 - 3t + 2) \mathbf{k} \). This equation can be broken down into components:

• The \( \mathbf{i} \) component, \( \sin t \), indicates the motion in the x-direction.
• The \( \mathbf{j} \) component, \( \cos t \), indicates the motion in the y-direction.
• The \( \mathbf{k} \) component, \( t^2 - 3t + 2 \), measures the height above the ground, i.e., the motion in the z-direction.

The \'helical\' nature comes from the combination of circular motion in the xy-plane with the vertical motion in the z-direction.

Direction Vector

When analyzing the motion of a particle, particularly when it decides to leave its path (like our particle departing the helix), the direction vector is crucial. It points in the direction of motion at that specific moment. To find this, we need the tangent vector derived from the position vector function.
In our case, the direction vector is \[ \mathbf{d} = \cos 5 \mathbf{i} - \sin 5 \mathbf{j} + 7 \mathbf{k} \]This vector is derived from calculating the derivative of the position vector, \( \mathbf{r}(t) \), to yield \( \mathbf{r}'(t) \). At \( t=5 \), we substitute back into this derivative to compute the specific vector the particle follows at that height.

Differentiation

Differentiation is a key mathematical tool we use to find how a function changes with respect to one of its variables. In the context of motion, it's used to calculate the velocity or the slope of a curve at any point. For a vector function \( \mathbf{r}(t) = f(t)\mathbf{i} + g(t)\mathbf{j} + h(t)\mathbf{k} \), the derivative \( \mathbf{r}'(t) \) gives the rate of change of the position vector.
To find the direction vector, we differentiate each component:

• The derivative of \( \sin t \) is \( \cos t \).
• The derivative of \( \cos t \) is \( -\sin t \).
• The derivative of \( t^2 - 3t + 2 \) is \( 2t - 3 \).

At \( t = 5 \), substituting back gives us the direction vector which shows the tangent direction.

Quadratic Equation

A quadratic equation is a polynomial equation of the second degree, often written in the standard form \( ax^2 + bx + c = 0 \). Solving quadratic equations is essential to find specific conditions, like when our helical particle is exactly 12 meters above the ground.
For this problem:

• Given \( t^2 - 3t + 2 \) (height) = 12, we set up the equation \( t^2 - 3t - 10 = 0 \).
• We use the quadratic formula \( t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \) to solve for \( t \).

Plugging in values \( a=1 \), \( b=-3 \), \( c=-10 \), we find the discriminant and solutions. The valid solution, considering \( t \geq 0 \), is \( t = 5 \). This time value is crucial for determining when the particle reaches the desired height on the helix.