Problem 45 - 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 \(t=t_{1} .\) $$ \mathbf{r}(t)=a \cosh t \mathbf{i}+a \sinh t \mathbf{j} ; t_{1}=\ln 3 $$

Step-by-Step Solution

Answer

At \( t = \ln 3 \), \( a_T = 8a \) and \( a_N = 2a \).

Step 1: Calculate the velocity vector

First, we need the velocity vector \( \mathbf{v}(t) \). To find this, we differentiate the position vector \( \mathbf{r}(t) = a \cosh t \mathbf{i} + a \sinh t \mathbf{j} \) with respect to time \( t \). This gives us:\[ \mathbf{v}(t) = \frac{d}{dt}(a \cosh t \mathbf{i} + a \sinh t \mathbf{j}) = a \sinh t \mathbf{i} + a \cosh t \mathbf{j} \].

Step 2: Calculate the acceleration vector

Next, we find the acceleration vector \( \mathbf{a}(t) \) by differentiating the velocity vector:\[ \mathbf{a}(t) = \frac{d}{dt}(a \sinh t \mathbf{i} + a \cosh t \mathbf{j}) = a \cosh t \mathbf{i} + a \sinh t \mathbf{j} \].

Step 3: Determine the tangential component of acceleration

The tangential component of acceleration \( a_{T} \) is given by the derivative of the speed magnitude. The speed \( v \) is:\[ v = \sqrt{(a \sinh t)^2 + (a \cosh t)^2} = a \sqrt{\sinh^2 t + \cosh^2 t} = a \cosh(2t) \].Then, \[ a_T = \frac{dv}{dt} = \frac{d}{dt}(a \cosh(2t)) = 2a \sinh(2t) \].

Step 4: Determine the normal component of acceleration

The normal component \( a_{N} \) is found using the formula:\[ a_N = \sqrt{\|\mathbf{a}(t)\|^2 - a_T^2} \], where \( \|\mathbf{a}(t)\| = a \cosh(2t) \). Thus,\[ a_N = \sqrt{(a \cosh(2t))^2 - (2a \sinh(2t))^2} = \sqrt{4a^2} = 2a. \]

Step 5: Evaluate components at \( t = t_1 = \ln 3 \)

Substitute \( t = \ln 3 \) into \( a_T \) and \( a_N \):For \( a_T \):\[ a_T(\ln 3) = 2a \sinh(2 \ln 3) = 2a \sinh(\ln 9) = 2a \cdot \frac{81-1}{18} = 8a. \]For \( a_N \):\( a_N \) does not depend on \( t \), so \( a_N = 2a \).

Key Concepts

Tangential Acceleration

When analyzing motion, tangential acceleration is a vital concept to grasp. It represents how quickly an object speeds up or slows down along its path of motion. In other words, it measures the rate of change of speed along the direction of the velocity vector. It is calculated as the derivative of the speed or the magnitude of the velocity. In this case, the tangential acceleration, denoted as \( a_T \), was derived from differentiating the speed function, where the speed is calculated using the hyperbolic identities, as shown in the exercise. Interestingly, the speed function derived from the velocity in this context involves hyperbolic trigonometric functions. Differentiating the speed function \( v = a \, \text{cosh}(2t) \) results in the expression for tangential acceleration, revealing how speed changes over time:

Normal Acceleration

Normal acceleration is another critical component in understanding acceleration fully. While tangential acceleration deals with changes in speed, normal acceleration, often called centripetal acceleration, is concerned with the changes in direction of the velocity vector. It acts perpendicular to the velocity direction. The normal acceleration component, denoted as \( a_N \), is determined by isolating and comparing the total acceleration with the tangential component. In this exercise, it is concise:

• The magnitude of normal acceleration, \( a_N \), is found using the Pythagorean relation between the total acceleration and the tangential component.

Interestingly, in this specific motion, the calculation simplifies to a constant value \( 2a \), which means the directional changes have a steadiness in speed.

Hyperbolic Functions

Hyperbolic functions like \( \text{cosh}(t) \) and \( \text{sinh}(t) \) play a crucial role in this motion analysis. These functions are quite similar to regular trigonometric functions, yet they have distinct properties and arise frequently in problems involving hyperbolic shapes and equations.In the given problem, the position vector is represented by

• \( \mathbf{r}(t) = a \, \text{cosh}(t) \, \mathbf{i} + a \, \text{sinh}(t) \, \mathbf{j} \),

which inherently involves these hyperbolic functions. The intriguing feature of hyperbolic functions is their connection to exponential functions, which can simplify calculations significantly in many cases. This smooth breakdown allows them to be essential in expressing both velocity and acceleration in concise forms, as demonstrated in the derivation of both \( a_T \) and \( a_N \).

Differentiation

Differentiation is at the heart of understanding rates of change in mathematics, and it is foundational in deriving both tangential and normal components of acceleration. In our exercise, the procedure begins with differentiating the position vector to obtain the velocity vector, which illustrates the object's movement direction.

• The position vector \( \mathbf{r}(t) \) differentiates into the velocity vector \( \mathbf{v}(t) \).

This step involves applying derivative rules to the individual components - mainly dealing with hyperbolic functions in this scenario.Following that, the velocity vector is once more differentiated to yield the acceleration vector, revealing the dynamic changes in the movement. This sequence essentially unravels key insights into how the speed and direction of a moving object vary with time.