Problem 49 - Calculus Early Transcendentals: Pearson New International Edition

Question

Find the tangential and normal components \(\left(a_{T}\right.\) and \(a_{N}\) ) of the acceleration vector at \(t\). Then evaluate at \(t=t_{1}\). \(x=e^{-t}, y=2 t, z=e^{t} ; t_{1}=0\)

Step-by-Step Solution

Answer

Evaluate \( a_{T} \) and \( a_{N} \) at \( t=0 \).

Step 1: Find the velocity vector

The velocity vector \( \mathbf{v}(t) \) is the derivative of the position vector \( \mathbf{r}(t) = \langle e^{-t}, 2t, e^{t} \rangle \). Compute the derivatives: \( \frac{dx}{dt} = -e^{-t} \), \( \frac{dy}{dt} = 2 \), and \( \frac{dz}{dt} = e^{t} \). Thus, \( \mathbf{v}(t) = \langle -e^{-t}, 2, e^{t} \rangle \).

Step 2: Find the acceleration vector

The acceleration vector \( \mathbf{a}(t) \) is the derivative of the velocity vector \( \mathbf{v}(t) = \langle -e^{-t}, 2, e^{t} \rangle \). Compute the derivatives: \( \frac{d(-e^{-t})}{dt} = e^{-t} \), \( \frac{d(2)}{dt} = 0 \), and \( \frac{d(e^{t})}{dt} = e^{t} \). Thus, \( \mathbf{a}(t) = \langle e^{-t}, 0, e^{t} \rangle \).

Step 3: Calculate the velocity magnitude

Find the magnitude of the velocity vector \( \mathbf{v}(t) = \langle -e^{-t}, 2, e^{t} \rangle \) as \( |\mathbf{v}(t)| = \sqrt{(-e^{-t})^2 + 2^2 + (e^{t})^2} = \sqrt{e^{-2t} + 4 + e^{2t}} \).

Step 4: Find the tangential component \(a_{T}\)

The tangential component of the acceleration is \( a_{T} = \frac{d}{dt}|\mathbf{v}(t)| \). First, find \( |\mathbf{v}(t)| \) from Step 3, then differentiate: \( \frac{d}{dt} \sqrt{e^{-2t} + 4 + e^{2t}} \). Calculate using the chain rule.

Step 5: Evaluate \( a_{T} \) at \( t=t_{1} \)

By evaluating the derivative from Step 4 at \( t=0 \), we find \( a_{T}(0) \).

Step 6: Find the normal component \(a_{N}\)

The normal component is given by \( a_{N} = \sqrt{|\mathbf{a}|^2 - a_{T}^2} \). First, find \(|\mathbf{a}(t)| = \sqrt{(e^{-t})^2 + 0^2 + (e^{t})^2} \) and then evaluate it.

Step 7: Evaluate \( a_{N} \) at \( t=t_{1} \)

Using the results from previous steps, evaluate \( a_{N} \) at \( t=0 \).

Key Concepts

Velocity Vector Calculation

Understanding how to calculate the velocity vector is fundamental in analyzing the motion of an object. By knowing the position vector mathematical expressions, we can differentiate these to find the velocity vector. For example, with a position vector given as \( \mathbf{r}(t) = \langle e^{-t}, 2t, e^{t} \rangle \), we differentiate each component.

• The derivative of \( x = e^{-t} \) is \( \frac{dx}{dt} = -e^{-t} \).
• The derivative of \( y = 2t \) is \( \frac{dy}{dt} = 2 \).
• The derivative of \( z = e^{t} \) is \( \frac{dz}{dt} = e^{t} \).

Thus, the velocity vector comes out to be \( \mathbf{v}(t) = \langle -e^{-t}, 2, e^{t} \rangle \). This vector tells you the direction and speed with which the point is moving at any time \( t \).

Acceleration Vector Calculation

The acceleration vector indicates how a velocity vector changes with time. Just like a velocity vector is derived from position components, the acceleration vector is determined by differentiating the velocity vector. In our case:

• Differentiate \( \mathbf{v}(t) = \langle -e^{-t}, 2, e^{t} \rangle \).
• The derivative of \( -e^{-t} \) becomes \( e^{-t} \).
• The constant \( 2 \) differentiates to \( 0 \).
• The derivative of \( e^{t} \) remains \( e^{t} \).

Thus, we find the acceleration vector \( \mathbf{a}(t) = \langle e^{-t}, 0, e^{t} \rangle \). This gives insight into how the object's velocity changes as time progresses.

Magnitude of Velocity

To find how fast a particle is moving at any time \( t \), we need to calculate the magnitude of the velocity vector. Given a velocity vector, \( \mathbf{v}(t) = \langle -e^{-t}, 2, e^{t} \rangle \), the magnitude is computed by:
\[|\mathbf{v}(t)| = \sqrt{(-e^{-t})^2 + 2^2 + (e^{t})^2}\]This results in:
\[|\mathbf{v}(t)| = \sqrt{e^{-2t} + 4 + e^{2t}}\]Understanding this concept helps in visualizing the speed of a particle, independent of its direction of travel.

Differentiation Techniques

Differentiation is crucial in finding various vectors and their components in physics. By applying differential calculus, we can retrieve information about velocity, acceleration, and tangential acceleration:

• Basic Rule Application: Each component of the position or velocity vector is differentiated separately.
• Chain Rule: Particularly helpful when dealing with composite functions like our magnitude of velocity formula.

For instance, the chain rule is used to differentiate \( \sqrt{e^{-2t} + 4 + e^{2t}} \), which is essential in finding the tangential component of acceleration.

Component Evaluation

Finally, understanding the different components of acceleration—tangential and normal—is essential in physics, as they provide deeper insights into object motion:

• Tangential Component \(a_T\): Measures how fast the speed of the object is changing. It is obtained as the derivative of the magnitude of velocity and evaluated at specific \( t \).
• Normal Component \(a_N\): Represents changes in direction. Calculated using the formula \( a_N = \sqrt{|\mathbf{a}|^2 - a_T^2} \).

These components allow us to understand and evaluate both the speed changes and direction alterations of an object's path.