Problem 57
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.
Step-by-Step Solution
Verified Answer
Direction vector is \( \cos(5) \mathbf{i} - \sin(5) \mathbf{j} + 7 \mathbf{k}. \)
1Step 1: Find when the particle is 12 meters above the ground
We need to solve for when the \[ t^2 - 3t + 2 = 12. \]This simplifies to:\[ t^2 - 3t + 2 - 12 = 0, \]which further simplifies to:\[ t^2 - 3t - 10 = 0. \]
2Step 2: Solve the quadratic equation
To solve the quadratic equation \( t^2 - 3t - 10 = 0 \), use the quadratic formula:\[ t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}, \]where \( a = 1 \), \( b = -3 \), and \( c = -10 \). Then\[ t = \frac{3 \pm \sqrt{(-3)^2 - 4 \cdot 1 \cdot (-10)}}{2 \cdot 1}. \]This results in\[ t = \frac{3 \pm \sqrt{9 + 40}}{2}, \]\[ t = \frac{3 \pm \sqrt{49}}{2}. \]Therefore, \( \sqrt{49} = 7 \), and\[ t = \frac{3 + 7}{2} \quad \text{or} \quad t = \frac{3 - 7}{2}. \]Thus, \( t = 5 \) or \( t = -2 \). Since \( t \geq 0 \), \( t = 5 \).
3Step 3: Determine the tangent vector at \( t = 5 \)
Find the derivative of \( \mathbf{r}(t) = \sin t \mathbf{i} + \cos t \mathbf{j} + (t^2 - 3t + 2) \mathbf{k} \) with respect to \( t \):\[ \mathbf{r}'(t) = \cos t \mathbf{i} - \sin t \mathbf{j} + (2t - 3) \mathbf{k}. \]Substitute \( t = 5 \) to find the tangent vector:\[ \mathbf{r}'(5) = \cos(5) \mathbf{i} - \sin(5) \mathbf{j} + (2\times5 - 3) \mathbf{k}. \]
4Step 4: Calculate the components of the tangent vector
Calculate the components at \( t = 5 \):- \( \mathbf{i} \) component: \( \cos(5) \)- \( \mathbf{j} \) component: \( -\sin(5) \)- \( \mathbf{k} \) component: \( 2 \cdot 5 - 3 = 7 \)Thus, the tangent vector at \( t = 5 \) is:\[ \cos(5) \mathbf{i} - \sin(5) \mathbf{j} + 7 \mathbf{k}. \]
5Step 5: Finalize the direction vector
The direction vector for the line tangent to the helix when the particle is 12 meters above the ground is \( \cos(5) \mathbf{i} - \sin(5) \mathbf{j} + 7 \mathbf{k} \).
Key Concepts
Parametric EquationsTangent VectorsQuadratic EquationsParticle Motion
Parametric Equations
Parametric equations are fundamental in describing motion, especially in cases that don't conform to single-variable functions. In this context, we have the motion of a particle along a helical path, dictated by the parametric equation \( \mathbf{r}(t) = \sin t \mathbf{i} + \cos t \mathbf{j} + (t^2 - 3t + 2) \mathbf{k} \). This equation involves three parameters that individually describe movement in the horizontal plane (\( \mathbf{i} \) and \( \mathbf{j} \) directions) and vertical movement (\( \mathbf{k} \) direction). By using parametric equations, we can track the position of the particle in all three spatial dimensions simultaneously.
Parametric equations provide several advantages:
Parametric equations provide several advantages:
- They allow for the representation of complex 3D paths that would be difficult to analyze using standard Cartesian equations.
- They offer flexibility to independently vary each component over time, which is crucial for accurately modeling motion.
Tangent Vectors
Tangent vectors play a vital role in understanding the instantaneous direction of motion of a particle traveling along a path, such as a helix. In our exercise, we derive the tangent vector by differentiating the parametric equation \( \mathbf{r}(t) \) with respect to time \( t \).
To find the tangent vector, we compute the derivative:
Understanding tangent vectors is important as they:
To find the tangent vector, we compute the derivative:
- \( \mathbf{r}'(t) = \cos t \mathbf{i} - \sin t \mathbf{j} + (2t - 3) \mathbf{k} \)
Understanding tangent vectors is important as they:
- Help in determining the path's direction at a specific point.
- Allow for finding equations for lines tangent to curves, critical in geometry and physics.
Quadratic Equations
Quadratic equations frequently appear in problems involving motion, like those in kinematic studies, where variables change with time. Here, a quadratic equation helps solve for the time \( t \) when the height of the particle is 12 meters. The equation
\( t^2 - 3t + 2 = 12 \)
Is transformed into a standard quadratic form:
Quadratic equations are cornerstone elements in mathematical analysis:
\( t^2 - 3t + 2 = 12 \)
Is transformed into a standard quadratic form:
- \( t^2 - 3t - 10 = 0 \)
- \( t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \)
- \( t = 5 \)
Quadratic equations are cornerstone elements in mathematical analysis:
- They emerge naturally in problems involving acceleration and forces.
- Provide solutions through straightforward and logical calculations, as demonstrated in solving for time \( t \).
Particle Motion
Motion of particles, like our particle traveling along a helix, is vital in physics and engineering. Understanding particle motion involves analyzing how position changes over time, using models like the parametric equations described.
In our problem, we examine the trajectory of the particle, especially focusing on when it leaves the helical path to follow a tangent line. This requires understanding both the spatial position and the velocity direction at the departure point. Key aspects include:
In our problem, we examine the trajectory of the particle, especially focusing on when it leaves the helical path to follow a tangent line. This requires understanding both the spatial position and the velocity direction at the departure point. Key aspects include:
- Identifying when a particle reaches a specific height (12 meters in this case), utilizing components of motion equations.
- Calculating and understanding the tangent vector, which gives insight into the immediate directionality of motion
- Analyzing trajectories in mechanics and aerodynamics.
- Designing paths for moving components in machinery and robotics.
Other exercises in this chapter
Problem 56
Find the angle between a main diagonal of a cube and one of its faces.
View solution Problem 57
Find the smallest angle between the main diagonals of a rectangular box 4 feet by 6 feet by 10 feet.
View solution Problem 58
Find the angles formed by the diagonals of a cube.
View solution Problem 58
An object moves along the curve \(y=\sin 2 x\). Without doing any calculating, decide where \(a_{N}=0\).
View solution