Problem 48 - Calculus Early Transcendentals: Pearson New International Edition

Question

find the tangential and normal components \(\left(a_{T}\right.\) and \(\left.a_{N}\right)\) of the acceleration vector at \(t .\) Then evaluate at \(t=t_{1} .\) $$ x=t, y=t^{2}, z=t^{3} ; t_{1}=2 $$

Step-by-Step Solution

Answer

\( a_T(2) \approx 3.3 \), \( a_N(2) \approx 6.6 \).

Step 1: Parametric Representation of the Curve

The position vector of the particle as a function of time is given by \( \mathbf{r}(t) = t \mathbf{i} + t^2 \mathbf{j} + t^3 \mathbf{k} \).

Step 2: Differentiate Position to Find Velocity

The velocity vector \( \mathbf{v}(t) \) is the first derivative of the position vector: \( \mathbf{v}(t) = \frac{d}{dt}(t \mathbf{i} + t^2 \mathbf{j} + t^3 \mathbf{k}) = \mathbf{i} + 2t \mathbf{j} + 3t^2 \mathbf{k} \).

Step 3: Differentiate Velocity to Find Acceleration

The acceleration vector \( \mathbf{a}(t) \) is the derivative of the velocity vector: \( \mathbf{a}(t)= \frac{d}{dt}(\mathbf{i} + 2t \mathbf{j} + 3t^2 \mathbf{k}) = 2 \mathbf{j} + 6t \mathbf{k} \).

Step 4: Calculate Speed as Magnitude of Velocity

The speed of the particle is the magnitude of the velocity vector: \( v(t) = \| \mathbf{v}(t) \| = \sqrt{(1)^2 + (2t)^2 + (3t^2)^2} = \sqrt{1 + 4t^2 + 9t^4} \).

Step 5: Determine Tangential Component of Acceleration

The tangential component of acceleration is given by \( a_T = \frac{d}{dt}(v(t)) \). Use the chain rule to differentiate: \( a_T = \frac{d}{dt}(\sqrt{1 + 4t^2 + 9t^4}) = \frac{\dot{v}(t)(1 + 4t^2 + 9t^4)}{2\sqrt{1 + 4t^2 + 9t^4}} = \frac{8t + 18t^3}{2\sqrt{1 + 4t^2 + 9t^4}} \).

Step 6: Evaluate Tangential Component at \( t = 2 \)

Plug \( t = 2 \) into the formula for \( a_T \) calculated in Step 5: \( a_T(2) = \frac{8(2) + 18(2)^3}{2\sqrt{1 + 4(2)^2 + 9(2)^4}} = \frac{16 + 144}{2\sqrt{393}} \).

Step 7: Compute Normal Component of Acceleration

The normal component of the acceleration is given by \( a_N = \sqrt{\| \mathbf{a}(t) \|^2 - a_T^2} \). Compute \( \| \mathbf{a}(t) \| = \sqrt{(2)^2 + (6t)^2} = \sqrt{4 + 36t^2} \).

Step 8: Evaluate Normal Component at \( t = 2 \)

Using \( \| \mathbf{a}(t) \| \, \), evaluate at \( t = 2 \): \( \| \mathbf{a}(2) \| = \sqrt{4 + 36(2)^2} = \sqrt{148} \). Then calculate \( a_N = \sqrt{\| \mathbf{a}(2) \|^2 - a_T(2)^2} = \sqrt{148 - \left( \frac{160}{2\sqrt{393}} \right)^2} \).

Step 9: Final Evaluation

Calculate \( a_T(2) \) and \( a_N(2) \) using their respective expressions from earlier steps.

Key Concepts

Parametric Curves

Parametric curves are a mathematical tool used to describe the position of a point in space as a function of a parameter, typically time. In this exercise, we describe a curve using parametric equations:

• The curve is defined by three equations: \(x = t\), \(y = t^2\), and \(z = t^3\).
• The parameter \(t\) usually represents time, and each equation expresses the respective coordinate in terms of \(t\).
• This type of representation is useful in physics and engineering to model the trajectory of particles moving in space.

Using parametric equations allows us to easily explore how the position of the particle changes over time.

Velocity and Acceleration

Velocity and acceleration are key concepts when analyzing motion using parametric curves. They describe how the position and speed of an object change over time.

• Velocity refers to the rate at which an object changes its position. It's the derivative of the position vector.
• The velocity vector for our curve is \(\mathbf{v}(t) = \mathbf{i} + 2t \mathbf{j} + 3t^2 \mathbf{k}\).
• Acceleration, on the other hand, is the rate at which the velocity of the object changes and is found by differentiating the velocity vector.
• For this problem, the acceleration vector is \(\mathbf{a}(t) = 2\mathbf{j} + 6t \mathbf{k}\).

Both velocity and acceleration are vector quantities, meaning they have both magnitude and direction.

Differentiation

Differentiation is a fundamental mathematical process used to find the rate of change of a function. In the context of motion along parametric curves:

• The position, velocity, and acceleration are all derived from the original position vector through differentiation.
• To find the velocity vector \(\mathbf{v}(t)\), we differentiate the position vector with respect to \(t\).
• Similarly, to determine the acceleration vector \(\mathbf{a}(t)\), we differentiate the velocity vector.
• Differentiation helps us understand how fast something is moving and how the speed changes over time.

This step is crucial in forming expressions for velocity and acceleration vectors.

Vector Magnitude

The magnitude of a vector is a measure of its length or size in space. For velocity and acceleration vectors, finding their magnitudes gives insights into speed and the rate of change of speed.

• The speed of a particle is the magnitude of its velocity vector, calculated as \( v(t) = \sqrt{1 + 4t^2 + 9t^4} \).
• Finding the magnitude involves taking the square root of the sum of the squares of the components of the vector.
• The magnitude of the acceleration vector \(\|\mathbf{a}(t)\| = \sqrt{4 + 36t^2} \) is similarly found.
• These magnitudes help determine the tangential and normal components of the acceleration, which describe acceleration in directions parallel and perpendicular to the velocity, respectively.

Understanding vector magnitude is central to analyzing the dynamics of motion along parametric curves.