Problem 42 - 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 \(t=t_{1} .\) $$ \mathbf{r}(t)=t^{2} \mathbf{i}+t \mathbf{j} ; t_{1}=1 $$

Step-by-Step Solution

Answer

At \(t=1\), \(a_T = \frac{4}{\sqrt{5}}\) and \(a_N = \frac{2}{\sqrt{5}}\).

Step 1: Find the Velocity Vector

First, find the derivative of the position vector \(\mathbf{r}(t)=t^{2} \mathbf{i}+t \mathbf{j}\) with respect to time \(t\) to get the velocity vector \(\mathbf{v}(t)\). This gives \(\mathbf{v}(t)=\frac{d}{dt}(t^{2}) \mathbf{i} + \frac{d}{dt}(t) \mathbf{j} = 2t \mathbf{i} + \mathbf{j}\).

Step 2: Find the Acceleration Vector

Differentiate the velocity vector \(\mathbf{v}(t)=2t \mathbf{i} + \mathbf{j}\) with respect to time \(t\) to obtain the acceleration vector \(\mathbf{a}(t)\). This results in \(\mathbf{a}(t)=\frac{d}{dt}(2t) \mathbf{i} + \frac{d}{dt}(1) \mathbf{j} = 2 \mathbf{i}\).

Step 3: Compute the Magnitude of the Velocity Vector

Compute the magnitude of the velocity vector, \(|\mathbf{v}(t)|\), which is \(\sqrt{(2t)^{2}+(1)^{2}} = \sqrt{4t^2 + 1}\).

Step 4: Calculate the Tangential Component of Acceleration \(a_T\)

The tangential component \(a_{T}\) is given by the formula \(a_{T} = \frac{d}{dt}|\mathbf{v}(t)|\). First, compute \(\frac{d}{dt}|\mathbf{v}(t)|\): \(|\mathbf{v}(t)|=\sqrt{4t^2+1}\). Using the chain rule, \(\frac{d}{dt}|\mathbf{v}(t)| = \frac{1}{2}(4t^2+1)^{-1/2} \cdot (8t)\). This simplifies to \(\frac{4t}{\sqrt{4t^{2}+1}}\). So, \(a_T = \frac{4t}{\sqrt{4t^{2}+1}}\).

Step 5: Calculate the Normal Component of Acceleration \(a_N\)

The normal component \(a_{N}\) is given by \(a_{N} = \sqrt{\mathbf{a}(t) \cdot \mathbf{a}(t) - a_{T}^{2}}\). First, calculate \(\mathbf{a}(t) \cdot \mathbf{a}(t) = 2^2 = 4\). Therefore, \(a_{N}= \sqrt{4 - \left(\frac{4t}{\sqrt{4t^2+1}}\right)^2}\). This simplifies to \(a_{N}\) at time \(t\).

Step 6: Evaluate at \(t = 1\)

Substitute \(t = 1\) into the expressions for \(a_T\) and \(a_N\). This yields \(a_T = \frac{4 \cdot 1}{\sqrt{4 \cdot 1^2 + 1}} = \frac{4}{\sqrt{5}}\) and \(a_N = \sqrt{4 - \left(\frac{4 \cdot 1}{\sqrt{5}}\right)^2} = \sqrt{4 - \frac{16}{5}} = \sqrt{\frac{4 \cdot 5 - 16}{5}} = \sqrt{\frac{4}{5}} = \frac{2}{\sqrt{5}}\).

Key Concepts

Tangential and Normal Components

When an object moves along a curved path, its acceleration can be separated into tangential and normal components. These components explain how the object's speed and direction change at a given moment.

• Tangential Component (\(a_T\)): This represents the rate of change in speed along the path. It shows how fast the object is speeding up or slowing down.
• Normal Component (\(a_N\)): This captures the change in direction of the velocity and points towards the center of the curvature. It relates to how sharply the object is turning.

These concepts allow us to analyze changes in motion efficiently by separating speed and direction changes.

Acceleration Vector

The acceleration vector \( \mathbf{a}(t) \) represents the rate of change of the velocity vector with respect to time. It not only tells us about the change in speed but also how the direction of an object’s path changes. Differentiating the velocity vector will yield this acceleration vector. For the function \( \mathbf{v}(t) = 2t \mathbf{i} + \mathbf{j} \), the corresponding acceleration vector is found by performing component-wise differentiation:

• The differentiation of \(2t\) with respect to \(t\) is \(2\).
• The \(\mathbf{j}\) component is constant, so its derivative is \(0\).

Thus, our acceleration vector becomes \( \mathbf{a}(t) = 2 \mathbf{i} \).

Velocity Vector

The velocity vector \( \mathbf{v}(t) \) describes the direction and speed of an object at any given time. It is the first derivative of the position vector with respect to time. For example, the derivative of the position vector \( \mathbf{r}(t) = t^2 \mathbf{i} + t \mathbf{j} \) results in:

• The derivative of \(t^2\) is \(2t\).
• The derivative of \(t\) is \(1\).

Putting these together gives the velocity vector \( \mathbf{v}(t) = 2t \mathbf{i} + \mathbf{j} \).
This vector helps in understanding how the object’s position changes over time with both magnitude and direction.

Differentiation

Differentiation is a fundamental process in calculus used to find the rate at which a function is changing at any given point. In the context of vector calculus, differentiation is used to determine both the velocity and acceleration vectors by following simple rules:

• The derivative of \(x^n\) with respect to \(x\) is \(nx^{n-1}\).
• For constant terms, the derivative is \(0\).

By applying these rules, we can calculate derivatives of each component of a vector separately to find important results like velocity and acceleration.

Magnitude of a Vector

The magnitude of a vector, often termed as its length or norm, gives an idea of the size of the vector. For a vector \( \mathbf{v} = a \mathbf{i} + b \mathbf{j} \), the magnitude is calculated using the formula:\[|\mathbf{v}| = \sqrt{a^2 + b^2}\]This formula involves squaring each component, summing them, and then taking the square root of the result.
For the velocity vector \( \mathbf{v}(t) = 2t \mathbf{i} + \mathbf{j} \), the magnitude is calculated as:\[|\mathbf{v}(t)| = \sqrt{(2t)^2 + 1^2} = \sqrt{4t^2 + 1}\]This measure helps to quantify how fast the object is moving through space, regardless of its direction.