Find the velocity, acceleration, and speed of a particle with the given position function: $\mathbf{r}(t) = t^2 \mathbf{i} + 3t \mathbf{j} + 5 \ln(t) \mathbf{k}.$

Neetesh Kumar | December 11, 2024

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This is the solution to Math 1C
Assignment: 13.4 Question Number 4
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Step-by-step solution:

Step 1: Velocity, $\mathbf{v}(t)$

The velocity is the derivative of the position function: $\mathbf{v}(t) = \frac{d}{dt} \mathbf{r}(t).$

Differentiate each component of $\mathbf{r}(t)$:

1. The derivative of $t^2$ is $2t$.
2. The derivative of $3t$ is $3$.
3. The derivative of $5 \ln(t)$ is $\frac{5}{t}$.

Thus: $\mathbf{v}(t) = \langle 2t, 3, \frac{5}{t} \rangle.$

Step 2: Acceleration, $\mathbf{a}(t)$

The acceleration is the derivative of the velocity function: $\mathbf{a}(t) = \frac{d}{dt} \mathbf{v}(t).$

Differentiate each component of $\mathbf{v}(t)$:

1. The derivative of $2t$ is $2$.
2. The derivative of $3$ is $0$.
3. The derivative of $\frac{5}{t}$ is $-\frac{5}{t^2}$.

Thus: $\mathbf{a}(t) = \langle 2, 0, -\frac{5}{t^2} \rangle.$

Step 3: Speed, $|\mathbf{v}(t)|$

The speed is the magnitude of the velocity vector: $|\mathbf{v}(t)| = \sqrt{(2t)^2 + (3)^2 + \left(\frac{5}{t}\right)^2}.$

Simplify each term:

1. $(2t)^2 = 4t^2$,
2. $(3)^2 = 9$,
3. $\left(\frac{5}{t}\right)^2 = \frac{25}{t^2}$.

Thus: $|\mathbf{v}(t)| = \sqrt{4t^2 + 9 + \frac{25}{t^2}}.$

Final Answer:

1. Velocity: $\boxed{\mathbf{v}(t) = \langle 2t, 3, \frac{5}{t} \rangle.}$

   
   

2. Acceleration: $\boxed{\mathbf{a}(t) = \langle 2, 0, -\frac{5}{t^2} \rangle.}$

   
   

3. Speed: $\boxed{|\mathbf{v}(t)| = \sqrt{4t^2 + 9 + \frac{25}{t^2}}.}$

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