Problem 50 - Calculus III

Question

Suppose that the position function for an object in three dimensions is given by the equation \(\mathbf{r}(t)=t \cos (t) \mathbf{i}+t \sin (t) \mathbf{j}+3 t \mathbf{k}\) Find the tangential and normal components of acceleration when \(t=1.5\).

Step-by-Step Solution

Answer

The tangential and normal components of acceleration at \( t=1.5 \) are calculated through specific formulas and moments when derived from the equations of motion provided.

Step 1: Find Velocity Vector

The velocity vector \( \mathbf{v}(t) \) is the derivative of the position vector \( \mathbf{r}(t) \) with respect to \( t \). We derive each component:\[\mathbf{v}(t) = \frac{d}{dt}[t\cos(t)]\mathbf{i} + \frac{d}{dt}[t\sin(t)]\mathbf{j} + \frac{d}{dt}[3t]\mathbf{k}\]Use the product rule for \( [t\cos(t)] \) and \( [t\sin(t)] \):\[\frac{d}{dt}[t\cos(t)] = \cos(t) - t\sin(t)\]\[\frac{d}{dt}[t\sin(t)] = \sin(t) + t\cos(t)\]Therefore, the velocity vector is:\[\mathbf{v}(t) = (\cos(t) - t\sin(t))\mathbf{i} + (\sin(t) + t\cos(t))\mathbf{j} + 3\mathbf{k}\]

Step 2: Find Acceleration Vector

The acceleration vector \( \mathbf{a}(t) \) is the derivative of the velocity vector \( \mathbf{v}(t) \) with respect to \( t \). Derive each component:\[\mathbf{a}(t) = \frac{d}{dt}[\cos(t) - t\sin(t)]\mathbf{i} + \frac{d}{dt}[\sin(t) + t\cos(t)]\mathbf{j} + \frac{d}{dt}[3]\mathbf{k}\]Derivatives:\[\frac{d}{dt}[\cos(t) - t\sin(t)] = -\sin(t) - (\sin(t) + t\cos(t)) = -2\sin(t) - t\cos(t)\]\[\frac{d}{dt}[\sin(t) + t\cos(t)] = \cos(t) - (\cos(t) - t\sin(t)) = 2t\sin(t)\]Therefore, the acceleration vector is:\[\mathbf{a}(t) = (-2\sin(t) - t\cos(t))\mathbf{i} + (2\cos(t) - t\sin(t))\mathbf{j} + 0\mathbf{k}\]

Step 3: Compute the Magnitude of Velocity

Find the magnitude of the velocity vector, \( \|\mathbf{v}(t)\| \):\[\|\mathbf{v}(t)\| = \sqrt{(\cos(t) - t\sin(t))^2 + (\sin(t) + t\cos(t))^2 + 3^2}\]Simplify:\[\|\mathbf{v}(1.5)\| = \sqrt{(\cos(1.5) - 1.5\sin(1.5))^2 + (\sin(1.5) + 1.5\cos(1.5))^2 + 9}\]

Step 4: Find Tangential Component of Acceleration

The tangential component \( a_t \) is calculated as:\[ a_t = \frac{\mathbf{a}(t) \cdot \mathbf{v}(t)}{\|\mathbf{v}(t)\|} \]Compute the dot product \( \mathbf{a}(t) \cdot \mathbf{v}(t) \):\[-2\sin(t) - t\cos(t) \cdot (\cos(t) - t\sin(t)) + 2t\cos(t) \sin(t) + 3\cdot0\]Evaluate at \( t=1.5 \).

Step 5: Find Normal Component of Acceleration

The normal component \( a_n \) can be found using:\[ a_n = \sqrt{\|\mathbf{a}(t)\|^2 - a_t^2} \]Firstly calculate \( \|\mathbf{a}(1.5)\| \):\[ \| \mathbf{a}(t) \| = \sqrt{(-2\sin(t) - t\cos(t))^2 + (2\cos(t) - t\sin(t))^2}\]Then substitute to find \( a_n \).

Key Concepts

Velocity Vector

The velocity vector is a crucial concept in understanding motion. It describes the rate of change of the position of an object with respect to time. Essentially, it tells us how fast and in what direction the object is moving at any given moment. To find the velocity vector, we take the derivative of the position function. For our exercise, the position function is given as \[ \mathbf{r}(t)=t \cos (t) \mathbf{i}+t \sin (t) \mathbf{j}+3 t \mathbf{k} \] The velocity vector \( \mathbf{v}(t) \) is then found by calculating the derivative of each of the components of the position function:

• \( \frac{d}{dt}[t\cos(t)] \mathbf{i} = \cos(t) - t\sin(t) \mathbf{i} \)
• \( \frac{d}{dt}[t\sin(t)] \mathbf{j} = \sin(t) + t\cos(t) \mathbf{j} \)
• \( \frac{d}{dt}[3t] \mathbf{k} = 3 \mathbf{k} \)

The resulting velocity vector becomes:\[\mathbf{v}(t) = (\cos(t) - t\sin(t))\mathbf{i} + (\sin(t) + t\cos(t))\mathbf{j} + 3\mathbf{k}\]Understanding the velocity vector helps us in analyzing various components of motion, paving the way for evaluating acceleration and other aspects of dynamics.

Acceleration Vector

The acceleration vector is derived from the velocity vector and it represents the rate of change of velocity with respect to time. This vector is crucial as it indicates how an object's speed or direction is changing, allowing us to understand more about the forces acting on it. In essence, it helps in understanding how quickly an object is speeding up or slowing down, or changing its path.To compute the acceleration vector \( \mathbf{a}(t) \), we differentiate each component of the velocity vector \( \mathbf{v}(t) \):

• \( \frac{d}{dt}[\cos(t) - t\sin(t)] = -2\sin(t) - t\cos(t) \mathbf{i} \)
• \( \frac{d}{dt}[\sin(t) + t\cos(t)] = 2t\sin(t) \mathbf{j} \)
• \( \frac{d}{dt}[3] \mathbf{k} = 0 \mathbf{k} \)

The complete acceleration vector is then:\[ \mathbf{a}(t) = (-2\sin(t) - t\cos(t))\mathbf{i} + (2t\sin(t))\mathbf{j} \]Understanding the acceleration vector is key to discerning changes in an object's motion which is vital in fields like physics and engineering.

Position Function

The position function is a fundamental concept in studying motion in physics. It provides the coordinates of a point object in space as a function of time. Essentially, it tells you exactly where an object is at any moment. For the exercise, the position function is given in three dimensions as:\[\mathbf{r}(t)=t \cos (t) \mathbf{i}+t \sin (t) \mathbf{j}+3 t \mathbf{k}\]Each component of this vector function tells us the position along the respective axis (\( \mathbf{i} \), \( \mathbf{j} \), and \( \mathbf{k} \)) at any time \( t \).

• The \( \mathbf{i} \) component \((t \cos (t))\) represents the position along the x-axis.
• The \( \mathbf{j} \) component \((t \sin (t))\) represents the position along the y-axis.
• The \( 3t \mathbf{k} \) indicates the straightforward progression along the z-axis.

Such functions are highly applicable in engineering and physics, allowing analysts to track the movement and position of objects over time.

Magnitude of Velocity

The magnitude of velocity, often referred to as speed, gives us the scalar measure of how fast an object is moving along its path, regardless of direction. It is obtained from the velocity vector and provides insight into the rate at which an object's position is changing over time.To compute the magnitude, we use the formula for the magnitude of a vector, namely:\[\|\mathbf{v}(t)\| = \sqrt{(\cos(t) - t\sin(t))^2 + (\sin(t) + t\cos(t))^2 + 3^2}\]When calculated at a specific time, such as \( t = 1.5 \), we substitute for \( t \) and simplify:\[\|\mathbf{v}(1.5)\| = \sqrt{(\cos(1.5) - 1.5\sin(1.5))^2 + (\sin(1.5) + 1.5\cos(1.5))^2 + 9}\]Understanding the magnitude of velocity helps in evaluating how the speed of an object changes over time. This understanding is crucial in applications ranging from vehicle dynamics to flight navigation.