Problem 41 - 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 \(t=t_{1} .\) $$ \mathbf{r}(t)=3 t \mathbf{i}+3 t^{2} \mathbf{j} ; t_{1}=\frac{1}{3} $$

Step-by-Step Solution

Answer

At \( t = \frac{1}{3} \), \( a_{T} = \frac{12}{\sqrt{13}} \) and \( a_{N} = \frac{6}{\sqrt{13}} \).

Step 1: Compute the Velocity Vector

To find the tangential and normal components, we first need the velocity vector. The velocity vector \( \mathbf{v}(t) \) is the derivative of the position vector \( \mathbf{r}(t) \). Compute \( \mathbf{v}(t) \):\[ \mathbf{v}(t) = \frac{d}{dt}(3t \mathbf{i} + 3t^2 \mathbf{j}) = 3 \mathbf{i} + 6t \mathbf{j}. \]

Step 2: Compute the Acceleration Vector

Now, compute the acceleration vector \( \mathbf{a}(t) \). This is the derivative of the velocity vector \( \mathbf{v}(t) \). Find \( \mathbf{a}(t) \):\[ \mathbf{a}(t) = \frac{d}{dt}(3 \mathbf{i} + 6t \mathbf{j}) = 6 \mathbf{j}. \]

Step 3: Calculate the Tangential Component of Acceleration

The tangential component \( a_{T} \) is given by the projection of \( \mathbf{a}(t) \) onto \( \mathbf{v}(t) \). It can be calculated using the formula \( a_{T} = \frac{\mathbf{a} \cdot \mathbf{v}}{\|\mathbf{v}\|} \). First, compute \( \mathbf{a} \cdot \mathbf{v} \) and \( \|\mathbf{v}\| \): \[ \mathbf{a} \cdot \mathbf{v} = 0 \cdot 3 + 6 \cdot 6t = 36t. \]\[ \|\mathbf{v}\| = \sqrt{3^2 + (6t)^2} = \sqrt{9 + 36t^2}. \]So, \( a_{T} = \frac{36t}{\sqrt{9 + 36t^2}} \).

Step 4: Calculate the Normal Component of Acceleration

The normal component \( a_{N} \) can be found using the formula \( a_{N} = \frac{\sqrt{\|\mathbf{a}\|^2 - (a_{T})^2}}{\|\mathbf{v}\|} \). First, compute \( \|\mathbf{a}\| \):\[ \|\mathbf{a}\| = \sqrt{0^2 + 6^2} = 6. \]Next, evaluate \( a_{N} = \frac{\sqrt{\|\mathbf{a}\|^2 - (a_{T})^2}}{\|\mathbf{v}\|} \) which simplifies to \( a_{N} = \frac{\sqrt{36 - \left( \frac{36t}{\sqrt{9 + 36t^2}} \right)^2}}{\sqrt{9 + 36t^2}} \).

Step 5: Evaluate at t = t_1

Substitute \( t = \frac{1}{3} \) into the expressions for \( a_{T} \) and \( a_{N} \):For \( a_{T} \):\[ a_{T} = \frac{36 \times \frac{1}{3}}{\sqrt{9 + 36 \times \left(\frac{1}{3}\right)^2}} = \frac{12}{\sqrt{13}}. \]For \( a_{N} \): Plug \( t = \frac{1}{3} \) into the expression and simplify:\[ a_{N} = \frac{\sqrt{36 - \left( \frac{12}{\sqrt{13}} \right)^2}}{\sqrt{13}} = \frac{6\sqrt{1}}{\sqrt{13}} = \frac{6}{\sqrt{13}}. \]

Key Concepts

Velocity Vector

The velocity vector is a core component in understanding motion. It represents how fast an object is moving and in which direction. When we talk about a velocity vector, we are looking at the rate of change of the position vector with respect to time. In mathematical terms, if we have a position vector \( \mathbf{r}(t) = 3t \mathbf{i} + 3t^2 \mathbf{j} \), the velocity vector \( \mathbf{v}(t) \) is the first derivative of this vector.

• Taking the derivative of \( \mathbf{r}(t) = 3t \mathbf{i} + 3t^2 \mathbf{j} \) gives \( \mathbf{v}(t) = 3 \mathbf{i} + 6t \mathbf{j} \).
• Here, \( 3 \mathbf{i} \) indicates a constant motion in the x-direction, while \( 6t \mathbf{j} \) shows that motion in the y-direction is speeding up over time.

These components help us understand both the direction and the change in speed of the object over time.

Acceleration Vector

An acceleration vector indicates how the velocity of an object changes over time. It is essentially the derivative of the velocity vector. When we differentiate the velocity vector \( \mathbf{v}(t) = 3 \mathbf{i} + 6t \mathbf{j} \), we derive the acceleration vector \( \mathbf{a}(t) \).

• In our example, differentiating leads to \( \mathbf{a}(t) = 0 \mathbf{i} + 6 \mathbf{j} \).
• This shows there is no acceleration in the x-direction, while the acceleration in the y-direction is constant at \( 6 \mathbf{j} \).

Understanding the acceleration vector helps us gauge the change in velocity, and therefore, predict future motion. It is a critical component in analyzing the dynamics of moving objects.

Projection of Acceleration

The projection of acceleration is where we begin to break down the acceleration vector into components that line up with the motion trajectory. The tangential component \( a_T \), is essentially how much of the acceleration is acting in the same direction as the velocity vector.

• This is computed using the formula: \( a_T = \frac{\mathbf{a} \cdot \mathbf{v}}{\|\mathbf{v}\|} \).
• The dot product \( \mathbf{a} \cdot \mathbf{v} \) measures how aligned the acceleration vector is with the velocity vector.
• In many physics problems, the tangential acceleration gives insight into how the speed changes over time.

Finding \( a_T \) gives us the „push” or „pull” this acceleration effectively puts onto the speed of the object.

Vector Magnitude

In the analysis of dynamics, the concept of vector magnitude is crucial to understand the quantity, or "size", of vectors. For any vector \( \mathbf{v} = v_x \mathbf{i} + v_y \mathbf{j} \), its magnitude \( \|\mathbf{v}\| \) is calculated using the Pythagorean theorem: \( \|\mathbf{v}\| = \sqrt{v_x^2 + v_y^2} \).

• In the problem, the velocity vector's magnitude is \( \|\mathbf{v}\| = \sqrt{3^2 + (6t)^2} = \sqrt{9 + 36t^2} \).
• Magnitude helps determine the "speed" similar to how speed is the magnitude of velocity.

Understanding vector magnitudes allows us to translate vector quantities into more intuitive scalar answers, such as the actual speed from a velocity vector or the actual strength from a force vector.