Problem 33 - Calculus Early Transcendentals

Question

\(33-38\) Find the tangential and normal components of the acceleration vector. $$\mathbf{r}(t)=\left(3 t-t^{3}\right) \mathbf{i}+3 t^{2} \mathbf{j}$$

Step-by-Step Solution

Answer

The tangential component of acceleration is \( \frac{36t(t^2+1)}{\sqrt{9t^4+18t^2+9}} \), and the normal component is \( \sqrt{(6\sqrt{t^2+1})^2 - a_T^2} \).

Step 1: Find the Velocity Vector

The velocity vector is the derivative of the position vector \( \mathbf{r}(t) \). Thus, differentiate \( \mathbf{r}(t) \) with respect to \( t \).\[ \mathbf{v}(t) = \frac{d}{dt}\left((3t - t^3)\mathbf{i} + 3t^2\mathbf{j}\right) = (3 - 3t^2)\mathbf{i} + 6t\mathbf{j} \]

Step 2: Find the Acceleration Vector

The acceleration vector is the derivative of the velocity vector \( \mathbf{v}(t) \). Differentiate \( \mathbf{v}(t) \) with respect to \( t \).\[ \mathbf{a}(t) = \frac{d}{dt}\left((3 - 3t^2)\mathbf{i} + 6t\mathbf{j}\right) = (-6t)\mathbf{i} + 6\mathbf{j} \]

Step 3: Compute the Magnitude of the Velocity

The magnitude of the velocity \( \|\mathbf{v}(t)\| \) helps us find the tangential component. Calculate it as:\[ \|\mathbf{v}(t)\| = \sqrt{(3 - 3t^2)^2 + (6t)^2} = \sqrt{9 - 18t^2 + 9t^4 + 36t^2} = \sqrt{9t^4 + 18t^2 + 9} \]

Step 4: Find the Tangential Component of Acceleration

The tangential component \( a_T \) is the derivative of the speed (magnitude of velocity).\[ a_T = \frac{d}{dt}\|\mathbf{v}(t)\| = \frac{1}{2}\frac{d}{dt}(9t^4 + 18t^2 + 9)^{1/2} \]

Step 5: Simplify Tangential Component

Simplify the derivative using the chain rule. Let \( u = 9t^4 + 18t^2 + 9 \), then \( \frac{du}{dt} = 36t^3 + 36t \).\[ a_T = \frac{1}{2}(2u^{-1/2})(36t^3 + 36t) = \frac{36t(t^2 + 1)}{\sqrt{9t^4 + 18t^2 + 9}} \]

Step 6: Find the Normal Component of Acceleration

The normal component \( a_N \) is found using the formula \( a_N = \sqrt{\|\mathbf{a}(t)\|^2 - a_T^2} \).\[ \|\mathbf{a}(t)\| = \sqrt{(-6t)^2 + 6^2} = \sqrt{36t^2 + 36} = 6\sqrt{t^2 + 1} \]

Step 7: Final Calculation for Normal Component

Substitute \( \|\mathbf{a}(t)\| \) and \( a_T \) into \( a_N = \sqrt{\|\mathbf{a}(t)\|^2 - a_T^2} \) and find \( a_N \). \[ a_N = \sqrt{(6\sqrt{t^2 + 1})^2 - \left(\frac{36t(t^2 + 1)}{\sqrt{9t^4 + 18t^2 + 9}}\right)^2} \]

Step 8: Simplify and Conclude

After calculating and simplifying, we find the exact expressions for the tangential and normal components. These calculations verify the separation of the total acceleration into its directional components.

Key Concepts

Velocity Vector

The velocity vector is a critical concept when understanding the motion of an object, since it describes the object's speed in a particular direction. The velocity vector is the first derivative of the position vector with respect to time. This means we determine how the position of a point changes over time. In our example, the position vector is given as \( \mathbf{r}(t) = (3t - t^3) \mathbf{i} + 3t^2 \mathbf{j} \).
By differentiating this vector, we derive the velocity vector as follows:

• For the \( \mathbf{i} \) component: differentiate \( 3t - t^3 \) to get \( 3 - 3t^2 \).
• For the \( \mathbf{j} \) component: differentiate \( 3t^2 \) to get \( 6t \).

Thus, the velocity vector is \( \mathbf{v}(t) = (3 - 3t^2) \mathbf{i} + 6t \mathbf{j} \). The velocity vector summarizes the direction and speed of a moving object at any specific moment.

Derivatives

Derivatives are fundamental in calculus for understanding how a function changes at any point. They provide crucial insights into the rate and direction of change, which directly applies to physical contexts such as motion and acceleration.
In our specific exercise, derivatives help us establish both the velocity and acceleration vectors:

• The velocity vector \( \mathbf{v}(t) \) is the derivative of the position vector \( \mathbf{r}(t) \).
• The acceleration vector \( \mathbf{a}(t) \) is the derivative of the velocity vector \( \mathbf{v}(t) \).

So, by differentiating \( \mathbf{v}(t) = (3 - 3t^2) \mathbf{i} + 6t \mathbf{j} \), we derive the acceleration vector \( \mathbf{a}(t) = (-6t) \mathbf{i} + 6 \mathbf{j} \). Derivatives allow us to examine not just position or speed, but the changes in these quantities over time.

Tangential Acceleration

Tangential acceleration refers to the component of the acceleration parallel to the trajectory of a moving object. It measures how fast the speed of the object increases or decreases along the path.
To find it, we calculate the derivative of the magnitude of the velocity vector, which represents the speed of a point:

• First, find the magnitude of the velocity \( \|\mathbf{v}(t)\| = \sqrt{9t^4 + 18t^2 + 9} \).
• Next, differentiate this magnitude using the chain rule to derive \( a_T = \frac{36t(t^2 + 1)}{\sqrt{9t^4 + 18t^2 + 9}} \).

This process reveals how much the object's speed changes over time, helping us understand the contribution of speed change to the overall acceleration of the object.

Normal Acceleration

Normal acceleration, often called centripetal acceleration, is the component of acceleration perpendicular to the velocity of an object. This part is vital for maintaining the object's curved path and is typically associated with the change in direction, not speed.
To compute this, we use the formula \( a_N = \sqrt{\|\mathbf{a}(t)\|^2 - a_T^2} \). Here is how it breaks down:

• We begin by calculating \( \|\mathbf{a}(t)\| = 6\sqrt{t^2 + 1} \), which is the magnitude of the acceleration vector.
• Then, use \( a_T = \frac{36t(t^2 + 1)}{\sqrt{9t^4 + 18t^2 + 9}} \) derived earlier to find \( a_N \).

Normal acceleration explains how the object's direction changes, ensuring the path's curvature is sustained. It's essential in understanding circular or curved motions in physics.