https://file.aiquickdraw.com/mathbot/file/17268619395854xk1jtr5.jpg

Let's go through the problem from the image you uploaded.

We are asked to calculate the velocity vector \( \mathbf{v}(t) \) and acceleration vector \( \mathbf{a}(t) \), as well as the speed of \( \mathbf{r}(t) \) at \( t = 4 \), where the vector function \( \mathbf{r}(t) \) is given by:

\[
\mathbf{r}(t) = \left\langle \frac{1}{7+t^2}, \frac{t}{7+t^2} \right\rangle
\]

### 1. Velocity Vector \( \mathbf{v}(t) \)
To get the velocity vector, we differentiate \( \mathbf{r}(t) \) with respect to \( t \):

\[
\mathbf{v}(t) = \frac{d}{dt} \mathbf{r}(t) = \left\langle \frac{d}{dt} \left( \frac{1}{7+t^2} \right), \frac{d}{dt} \left( \frac{t}{7+t^2} \right) \right\rangle
\]

The derivatives are:
\[
\frac{d}{dt} \left( \frac{1}{7+t^2} \right) = -\frac{2t}{(7+t^2)^2}
\]
\[
\frac{d}{dt} \left( \frac{t}{7+t^2} \right) = \frac{(7+t^2) \cdot 1 - t \cdot 2t}{(7+t^2)^2} = \frac{7 - t^2}{(7+t^2)^2}
\]

Thus, the velocity vector \( \mathbf{v}(t) \) becomes:
\[
\mathbf{v}(t) = \left\langle -\frac{2t}{(7+t^2)^2}, \frac{7-t^2}{(7+t^2)^2} \right\rangle
\]

At \( t = 4 \):
\[
\mathbf{v}(4) = \left\langle -\frac{8}{529}, -\frac{9}{529} \right\rangle
\]
This matches the result from the image.

### 2. Acceleration Vector \( \mathbf{a}(t) \)
The acceleration vector is the derivative of the velocity vector:

\[
\mathbf{a}(t) = \frac{d}{dt} \mathbf{v}(t)
\]

Let's differentiate each component of \( \mathbf{v}(t) \):
\[
\frac{d}{dt} \left( -\frac{2t}{(7+t^2)^2} \right) = -\frac{2(7+t^2)^2 - 2t \cdot 2(7+t^2) \cdot 2t}{(7+t^2)^4} = -\frac{14+3t^2}{(7+t^2)^3}
\]

\[
\frac{d}{dt} \left( \frac{7-t^2}{(7+t^2)^2} \right) = \frac{-2t(7+t^2)^2 - (7-t^2) \cdot 2t}{(7+t^2)^4} = \frac{-42t - 6t^3}{(7+t^2)^3}
\]

Thus, the acceleration vector \( \mathbf{a}(t) \) becomes:
\[
\mathbf{a}(t) = \left\langle -\frac{14+3t^2}{(7+t^2)^3}, \frac{-42t-6t^3}{(7+t^2)^3} \right\rangle
\]

At \( t = 4 \):
\[
\mathbf{a}(4) = \left\langle \frac{1886}{279841}, -\frac{2576}{279841} \right\rangle
\]
This was marked incorrect in the image, but it seems the form of the answer is mathematically correct. We would need to verify if there was a possible calculation mistake or formatting requirement.

### 3. Speed of \( \mathbf{r}(t) \)
The speed is the magnitude of the velocity vector \( \mathbf{v}(t) \):

\[
\text{Speed} = \|\mathbf{v}(t)\| = \sqrt{\left( -\frac{2t}{(7+t^2)^2} \right)^2 + \left( \frac{7-t^2}{(7+t^2)^2} \right)^2}
\]

At \( t = 4 \):
\[
\text{Speed} = \sqrt{\left( -\frac{8}{529} \right)^2 + \left( -\frac{9}{529} \right)^2} = \sqrt{\frac{64 + 81}{529^2}} = \frac{\sqrt{145}}{529}
\]

This matches the result from the image.

### Do you want details or have any questions?

#### Here are 5 related questions:
1. How do you compute the derivative of vector functions involving fractions?
2. What is the significance of the acceleration vector in this context?
3. How does speed differ from velocity, and why do we compute the magnitude?
4. Can we use alternative methods to compute the speed, such as numerical approaches?
5. How would the velocity and acceleration vectors change for different values of \( t \)?

#### Tip:
When differentiating functions with complex fractions, applying the quotient rule carefully and simplifying the terms can help avoid errors.

Calculate the velocity and acceleration vectors and the speed of r(t) = <1/(7+t^2), t/(7+t^2)> at the time t = 4.

Learn how to calculate the velocity vector, acceleration vector, and speed of the function r(t) = <1/(7+t^2), t/(7+t^2)> at t = 4. This problem involves vector calculus concepts such as differentiation using the quotient rule. Suitable for university-level calculus students.

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