Question

Find the velocity \(\mathbf{v}\), acceleration \(\mathbf{a}\), and speed \(s\) at the indicated time \(t=t_{1}\). $$ \mathbf{r}(t)=\ln t \mathbf{i}+\ln t^{2} \mathbf{j}+\ln t^{3} \mathbf{k} ; t_{1}=2 $$

Short answer

Velocity: \( \frac{1}{2} \mathbf{i} + 1 \mathbf{j} + \frac{3}{2} \mathbf{k} \); Acceleration: \(-\frac{1}{4} \mathbf{i} - \frac{1}{2} \mathbf{j} - \frac{3}{4} \mathbf{k} \); Speed: \(\sqrt{7/2}\).

Step-by-step solution

Find the Velocity Function

The velocity function is the derivative of the position function \( \mathbf{r}(t) \). To find the velocity \( \mathbf{v}(t) \), take the derivative of each component separately.\[\mathbf{v}(t) = \frac{d}{dt}(\ln t) \mathbf{i} + \frac{d}{dt}(\ln t^2) \mathbf{j} + \frac{d}{dt}(\ln t^3) \mathbf{k} \]Using the derivative \( \frac{d}{dt}(\ln t^n) = \frac{n}{t} \), we get:\[\mathbf{v}(t) = \frac{1}{t} \mathbf{i} + \frac{2}{t} \mathbf{j} + \frac{3}{t} \mathbf{k} \]This simplifies to \[\mathbf{v}(t) = \left(\frac{1}{t} \right) \mathbf{i} + \left(\frac{2}{t} \right) \mathbf{j} + \left(\frac{3}{t} \right) \mathbf{k} \].

Evaluate Velocity at Given Time

Substitute \( t_1 = 2 \) into the velocity function to find \( \mathbf{v}(2) \).\[\mathbf{v}(2) = \left(\frac{1}{2} \right) \mathbf{i} + \left(\frac{2}{2} \right) \mathbf{j} + \left(\frac{3}{2} \right) \mathbf{k} \]Thus, \( \mathbf{v}(2) = \frac{1}{2} \mathbf{i} + 1 \mathbf{j} + \frac{3}{2} \mathbf{k} \).

Find the Acceleration Function

The acceleration function is the derivative of the velocity function \( \mathbf{v}(t) \).\[\mathbf{a}(t) = \frac{d}{dt}\left( \frac{1}{t} \right) \mathbf{i} + \frac{d}{dt}\left( \frac{2}{t} \right) \mathbf{j} + \frac{d}{dt}\left( \frac{3}{t} \right) \mathbf{k} \]These derivatives result in:\[\mathbf{a}(t) = -\frac{1}{t^2} \mathbf{i} - \frac{2}{t^2} \mathbf{j} - \frac{3}{t^2} \mathbf{k} \].

Evaluate Acceleration at Given Time

Substitute \( t_1 = 2 \) into the acceleration function to find \( \mathbf{a}(2) \).\[\mathbf{a}(2) = -\frac{1}{2^2} \mathbf{i} - \frac{2}{2^2} \mathbf{j} - \frac{3}{2^2} \mathbf{k} \]This simplifies to \( \mathbf{a}(2) = -\frac{1}{4} \mathbf{i} - \frac{1}{2} \mathbf{j} - \frac{3}{4} \mathbf{k} \).

Calculate the Speed

The speed is the magnitude of the velocity vector \( \mathbf{v}(2) \). Calculate the magnitude as follows:\[ s = \left| \mathbf{v}(2) \right| = \sqrt{\left( \frac{1}{2} \right)^2 + (1)^2 + \left( \frac{3}{2} \right)^2 } \]Simplify the expression:\[ s = \sqrt{\frac{1}{4} + 1 + \frac{9}{4} } = \sqrt{\frac{1 + 4 + 9}{4}} = \sqrt{\frac{14}{4}} = \sqrt{\frac{7}{2}} \].

Key concepts

Derivatives of Vector Functions

In vector calculus, finding the derivative of a vector function is key to uncovering the velocity and acceleration of an object. A vector function, like \( \mathbf{r}(t) = \ln t \mathbf{i} + \ln t^2 \mathbf{j} + \ln t^3 \mathbf{k} \), describes the path of a moving object in space. Each component of this function is a function of time \( t \).

To find the velocity, you differentiate each component with respect to time. This means applying the rules of differentiation separately for \( \ln t \), \( \ln t^2 \), and \( \ln t^3 \). Using the formula \( \frac{d}{dt}(\ln t^n) = \frac{n}{t} \), we calculate the velocity as:

• \( \mathbf{v}(t) = \frac{1}{t} \mathbf{i} + \frac{2}{t} \mathbf{j} + \frac{3}{t} \mathbf{k} \)

Each derivative tells you how fast the object is moving along the x, y, and z axes. It boils down to the basic concept of derivatives, turning a position function into a velocity function, one component at a time.

Magnitude of Velocity Vector

The magnitude of the velocity vector represents the speed of an object. Once you've found the velocity vector, determining the speed is your next step. In simpler terms, speed is how fast something is moving regardless of direction.

To get the magnitude of the velocity vector, apply the Pythagorean theorem in three-dimensional space. For our velocity vector \( \mathbf{v}(2) = \frac{1}{2} \mathbf{i} + 1 \mathbf{j} + \frac{3}{2} \mathbf{k} \), calculate the magnitude as follows:

• \( s = \sqrt{ \left( \frac{1}{2} \right)^2 + (1)^2 + \left( \frac{3}{2} \right)^2 } \)
• \( s = \sqrt{ \frac{1}{4} + 1 + \frac{9}{4} } = \sqrt{\frac{7}{2}} \)

This magnitude gives you the speed of the object at time \( t_1=2 \). It’s a vital step when studying motion in physics as it translates motion into a more understandable scalar quantity.

Acceleration Vector

The acceleration vector indicates how the velocity of an object changes with time. Just as velocity is the derivative of position, acceleration is the derivative of velocity.

Starting with the velocity function \( \mathbf{v}(t) = \frac{1}{t} \mathbf{i} + \frac{2}{t} \mathbf{j} + \frac{3}{t} \mathbf{k} \), differentiate once more to find the acceleration:

• \( \mathbf{a}(t) = \frac{d}{dt}\left( \frac{1}{t} \right) \mathbf{i} + \frac{d}{dt}\left( \frac{2}{t} \right) \mathbf{j} + \frac{d}{dt}\left( \frac{3}{t} \right) \mathbf{k} \)
• These simplify to \( \mathbf{a}(t) = -\frac{1}{t^2} \mathbf{i} - \frac{2}{t^2} \mathbf{j} - \frac{3}{t^2} \mathbf{k} \)

Substituting \( t_1 = 2 \) gives you \( \mathbf{a}(2) = -\frac{1}{4} \mathbf{i} - \frac{1}{2} \mathbf{j} - \frac{3}{4} \mathbf{k} \).

Each component of this vector shows how the velocity along each axis is changing, pointing to whether an object is speeding up, slowing down, or changing direction.

Speed Calculation in Calculus

Speed, in the realm of calculus, focuses on analyzing scalar quantities when vectors break down into components. Unlike velocity, speed is directionless and only measures how quickly distance is covered.

After finding the velocity at a specific time \( t \), calculate the speed as the magnitude of the vector components at that time. In exercises, this involves taking the square root of the sum of squares of the velocity components. For example:

• Calculate \( s = \sqrt{ \left( \frac{1}{2} \right)^2 + (1)^2 + \left( \frac{3}{2} \right)^2 } \)
• Which simplifies to \( s = \sqrt{ \frac{7}{2} } \) at \( t = 2 \).

This approach transforms vector analysis into simpler, more intuitive calculations, allowing students to connect their mathematical work with real-world physical phenomena.