Problem 43 - 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 \(t=t_{1} .\) $$ \mathbf{r}(t)=(2 t+1) \mathbf{i}+\left(t^{2}-2\right) \mathbf{j} ; t_{1}=-1 $$

Step-by-Step Solution

Answer

\(a_T = -\sqrt{2}\), \(a_N = \sqrt{2}\) at \(t = -1\).

Step 1: Position and Velocity Vectors

First, we find the velocity vector by differentiating the position vector with respect to time. Given \( \mathbf{r}(t) = (2t + 1) \mathbf{i} + (t^2 - 2) \mathbf{j} \), the velocity \( \mathbf{v}(t) \) is the derivative:\[\mathbf{v}(t) = \frac{d}{dt}\left[(2t + 1) \mathbf{i} + (t^2 - 2) \mathbf{j} \right] = 2 \mathbf{i} + 2t \mathbf{j}.\]

Step 2: Acceleration Vector

We find the acceleration vector by differentiating the velocity vector:\[\mathbf{a}(t) = \frac{d}{dt} \left[ 2 \mathbf{i} + 2t \mathbf{j} \right] = 0 \mathbf{i} + 2 \mathbf{j} = 2 \mathbf{j}.\]

Step 3: Speed Calculation

Calculate the speed \( v \) as the magnitude of the velocity vector:\[v = \| \mathbf{v}(t) \| = \sqrt{(2)^2 + (2t)^2} = \sqrt{4 + 4t^2} = 2\sqrt{1 + t^2}.\]

Step 4: Tangential Component of Acceleration

The tangential component of the acceleration is found by taking the derivative of the speed with respect to time:\[a_T = \frac{dv}{dt} = \frac{d}{dt} \left[ 2\sqrt{1 + t^2} \right] = 2 \cdot \frac{1}{2}\left( 1 + t^2 \right)^{-\frac{1}{2}} \cdot 2t = \frac{2t}{\sqrt{1 + t^2}}.\]

Step 5: Normal Component of Acceleration

Use the formula \( a_N = \sqrt{ \| \mathbf{a}(t) \|^2 - a_T^2 } \):\[\| \mathbf{a}(t) \| = \sqrt{0^2 + 2^2} = 2.\]Then,\[a_N = \sqrt{ 2^2 - \left( \frac{2t}{\sqrt{1 + t^2}} \right)^2 } = \sqrt{4 - \frac{4t^2}{1 + t^2}} = \sqrt{4 \left( \frac{1}{1 + t^2} \right) } = \frac{2}{\sqrt{1 + t^2}}.\]

Step 6: Evaluate Components at \( t = t_1 = -1 \)

Replace \( t = -1 \) into \( a_T \) and \( a_N \):\[a_T = \frac{2(-1)}{\sqrt{1 + (-1)^2}} = \frac{-2}{\sqrt{2}} = -\sqrt{2},\]\[a_N = \frac{2}{\sqrt{1 + (-1)^2}} = \frac{2}{\sqrt{2}} = \sqrt{2}.\]

Key Concepts

Acceleration Components

In calculus, the acceleration vector plays a crucial role in understanding how a particle moves in space. It tells us how the velocity changes over time. To fully understand its behavior, it's divided into two components: tangential and normal.

• **Tangential Component**: This part of acceleration ( \( a_T \)) shows how fast the speed of a moving object is increasing or decreasing. It's essentially the rate of change of the speed.
• **Normal Component**: The normal component ( \( a_N \)) is perpendicular to the velocity vector and reflects how quickly the direction of the velocity is changing.

By analyzing both these components, you can better predict an object's future position and path. Calculating the tangential component involves differentiating the magnitude of the velocity vector, while the normal component requires evaluating the change in directional velocity.

Tangential and Normal Components

For any curve, the acceleration can be broken down into tangential and normal components. This helps in understanding both the speed and the path curvature at any given point. Here's how you can make these calculations:- **Tangential Component ( \( a_T \))**: It represents the acceleration along the path of the object. - To find it, take the derivative of the speed, which is the magnitude of the velocity vector. For example, given the speed \( v = 2\sqrt{1 + t^2} \), the tangential component will be determined by \[ a_T = \frac{d}{dt} \left[2\sqrt{1 + t^2}\right] = \frac{2t}{\sqrt{1 + t^2}}. \]- **Normal Component ( \( a_N \))**: This component measures the change in direction of the velocity vector. - It is calculated using the formula: \[ a_N = \sqrt{ \| \mathbf{a}(t) \|^2 - a_T^2 }. \] Here, it relies on knowing \( |\mathbf{a}(t)| = 2 \) and then computing the normal component, which results in \[ a_N = \frac{2}{\sqrt{1 + t^2}}. \]

Velocity and Acceleration Vectors

Velocity and acceleration vectors provide a deeper insight into the motion of objects, essential for calculus problems. Let's break them down:- **Velocity Vector**: This vector, represented as \( \mathbf{v}(t) \), shows both the speed and direction of motion at any point in time. - It is found by differentiating the position vector with respect to time. In our case, the velocity vector ended up being \[ \mathbf{v}(t) = 2 \mathbf{i} + 2t \mathbf{j}. \]- **Acceleration Vector**: This vector shows how the velocity changes over time and affects the trajectory and speed. - Through differentiation of the velocity vector, we found the acceleration vector: \[ \mathbf{a}(t) = 2 \mathbf{j}. \] By understanding the velocity and acceleration vectors, students can deduce how an object's path and speed evolve.Both vectors are fundamental in analyzing and predicting real-world phenomena, underlining their importance in calculus application.