Problem 17 - Calculus III

Question

Given that \(\mathbf{r}(t)=\left\langle e^{-5 t} \sin t, e^{-5 t} \cos t, 4 e^{-5 t}\right\rangle\) is the position vector of a moving particle, find the following quantities: The velocity of the particle

Step-by-Step Solution

Answer

The velocity vector is \( \mathbf{v}(t) = e^{-5t} \left\langle \cos t - 5 \sin t, -\sin t - 5 \cos t, -20 \right\rangle \).

Step 1: Understanding the Position Vector

The position vector \( \mathbf{r}(t) \) represents the coordinates of the particle in 3D space at any time \( t \). It is given by \( \mathbf{r}(t)=\left\langle e^{-5 t} \sin t, e^{-5 t} \cos t, 4 e^{-5 t}\right\rangle \). To find the velocity, we need to differentiate this vector with respect to \( t \).

Step 2: Differentiation of the Position Vector

The velocity vector \( \mathbf{v}(t) \) is the derivative of the position vector \( \mathbf{r}(t) \) with respect to time \( t \). Let's differentiate each component of \( \mathbf{r}(t) \): - For the first component: \( \frac{d}{dt}(e^{-5t} \sin t) \)- For the second component: \( \frac{d}{dt}(e^{-5t} \cos t) \)- For the third component: \( \frac{d}{dt}(4 e^{-5t}) \).

Step 3: Applying the Product Rule

Each component of \( \mathbf{r}(t) \) includes a product of functions. Use the product rule \( \frac{d}{dt}[u(t)v(t)] = u'(t)v(t) + u(t)v'(t) \) to differentiate them:- First component: \( u(t) = e^{-5t} \), \( v(t) = \sin t \).- Second component: \( u(t) = e^{-5t} \), \( v(t) = \cos t \).- Third component: Only involves one function, so the derivative is \( \frac{d}{dt}(4 e^{-5t}) = -20 e^{-5t} \).

Step 4: Calculating Derivatives

Compute derivatives for each component:- For \( e^{-5t} \sin t \): \[ \mathbf{v}_1(t) = (-5 e^{-5t}) \sin t + e^{-5t} \cos t \]- For \( e^{-5t} \cos t \): \[ \mathbf{v}_2(t) = (-5 e^{-5t}) \cos t - e^{-5t} \sin t \]- For \( 4 e^{-5t} \): \[ \mathbf{v}_3(t) = -20 e^{-5t} \].

Step 5: Forming the Velocity Vector

Combine the derivatives into the velocity vector:\[ \mathbf{v}(t) = \left\langle (-5 e^{-5t}) \sin t + e^{-5t} \cos t, (-5 e^{-5t}) \cos t - e^{-5t} \sin t, -20 e^{-5t} \right\rangle \].

Step 6: Simplifying the Velocity Vector

Factor out \( e^{-5t} \) from the velocity components:\[ \mathbf{v}(t) = e^{-5t} \left\langle \cos t - 5 \sin t, -\sin t - 5 \cos t, -20 \right\rangle \].

Key Concepts

Product Rule in Differentiation

When differentiating functions, especially when they are composed of products of two or more functions, the product rule is an essential tool. The product rule states that if you have two functions, say \( u(t) \) and \( v(t) \), their derivative with respect to time \( t \) is given by:

• \( \frac{d}{dt}[u(t)v(t)] = u'(t)v(t) + u(t)v'(t) \)

This means you take the derivative of the first function and keep the second function as is, then add it to the product of the first function as is and the derivative of the second function.
In the context of our position vector \( \mathbf{r}(t) \), this rule is applied to each component where multiples of functions are present. For example:

• In \( e^{-5t} \sin t \), the product rule helps us compute the derivative as \( (-5 e^{-5t}) \sin t + e^{-5t} \cos t \).

Using the product rule simplifies integration, making it possible to derive beautiful expressions for movements in physics and engineering.

Differentiation and Its Role

Differentiation refers to the process of finding the derivative, which reveals the rate at which something changes. In our exercise, differentiation is crucial to find the velocity from the position vector. The derivative, in this case, turns the information about where the particle is at any time \( t \) into information about how fast and in what direction the particle moves.
By differentiating each component of \( \mathbf{r}(t) \), we effectively find the velocity vector \( \mathbf{v}(t) \). Each component involves applying rules such as the product rule. For instance, differentiating the first component involves applying the product rule since we have \( e^{-5t} \sin t \), a product of two functions. The derivative tells us how the position changes, an essential aspect for analyzing motion.
Differentiation is therefore fundamental not only in physics but also in various calculations where rates of change and slopes are vital.

Understanding the Position Vector

The position vector is key to describing the location of a particle or object in space relative to an origin point. Position vectors are usually denoted by letters like \( \mathbf{r}(t) \) when they depend on time, indicating they describe position changes over time.
In our given function \( \mathbf{r}(t)=\left\langle e^{-5 t} \sin t, e^{-5 t} \cos t, 4 e^{-5 t}\right\rangle \), each component expresses the position in a specific direction (often \( x \), \( y \), and \( z \)) at any time \( t \).
This vector forms the basis for finding other dynamic properties like velocity and acceleration by applying differentiation techniques. In particular, when we differentiate this vector with respect to \( t \), we obtain the velocity vector which offers insights into the motion characteristics such as speed and direction.