https://file.aiquickdraw.com/mathbot/file/1728686332179lygv653t.jpg

The problem asks to calculate the velocity and acceleration vectors at time \( t = 3 \) for the given vector function \( \mathbf{r}(t) = \left\langle \frac{1}{5+t^2}, \frac{1}{5+t^2} \right\rangle \).

### 1. **Velocity Vector**:
The velocity vector \( \mathbf{v}(t) \) is the derivative of the position vector \( \mathbf{r}(t) \) with respect to \( t \):
\[
\mathbf{v}(t) = \frac{d}{dt}\left( \mathbf{r}(t) \right).
\]
For each component of \( \mathbf{r}(t) \):
- First component: \( r_1(t) = \frac{1}{5+t^2} \)
- Second component: \( r_2(t) = \frac{1}{5+t^2} \).

Using the chain rule:
\[
\frac{d}{dt} \left( \frac{1}{5+t^2} \right) = -\frac{2t}{(5+t^2)^2}.
\]
Thus, the velocity vector becomes:
\[
\mathbf{v}(t) = \left\langle -\frac{2t}{(5+t^2)^2}, -\frac{2t}{(5+t^2)^2} \right\rangle.
\]
Substituting \( t = 3 \):
\[
\mathbf{v}(3) = \left\langle -\frac{6}{(5+9)^2}, -\frac{6}{(5+9)^2} \right\rangle = \left\langle -\frac{6}{14^2}, -\frac{6}{14^2} \right\rangle = \left\langle -\frac{6}{196}, -\frac{6}{196} \right\rangle = \left\langle -\frac{3}{98}, -\frac{3}{98} \right\rangle.
\]
This matches the provided value for \( \mathbf{v}(3) \), except the second component should be \( -\frac{3}{98} \), not \( -\frac{2}{98} \).

### 2. **Acceleration Vector**:
The acceleration vector \( \mathbf{a}(t) \) is the derivative of the velocity vector \( \mathbf{v}(t) \):
\[
\mathbf{a}(t) = \frac{d}{dt}\left( \mathbf{v}(t) \right).
\]
Differentiate \( \mathbf{v}(t) = \left\langle -\frac{2t}{(5+t^2)^2}, -\frac{2t}{(5+t^2)^2} \right\rangle \) using the product and chain rules. For each component:
\[
\frac{d}{dt} \left( -\frac{2t}{(5+t^2)^2} \right) = -\frac{2(5+t^2)^2 - 2t \cdot 2t(5+t^2)}{(5+t^2)^4}.
\]
Simplify:
\[
\mathbf{a}(t) = \left\langle \frac{-2(5+t^2)^2 + 8t^2(5+t^2)}{(5+t^2)^4}, \frac{-2(5+t^2)^2 + 8t^2(5+t^2)}{(5+t^2)^4} \right\rangle.
\]
Substituting \( t = 3 \):
\[
\mathbf{a}(3) = \left\langle \frac{-2(5+9)^2 + 8(9)(14)}{(14)^4}, \frac{-2(5+9)^2 + 8(9)(14)}{(14)^4} \right\rangle = \left\langle \frac{-2(196) + 8(126)}{14^4}, \frac{-2(196) + 8(126)}{14^4} \right\rangle.
\]
Simplifying further:
\[
\mathbf{a}(3) = \left\langle \frac{-392 + 1008}{38416}, \frac{-392 + 1008}{38416} \right\rangle = \left\langle \frac{616}{38416}, \frac{616}{38416} \right\rangle = \left\langle \frac{77}{4816}, \frac{77}{4816} \right\rangle.
\]

Thus, the acceleration vector is:
\[
\mathbf{a}(3) = \left\langle \frac{77}{4816}, \frac{77}{4816} \right\rangle.
\]

Let me know if you'd like further clarification or details!

### 5 Related Questions:
1. How do you compute the speed from the velocity vector?
2. How do partial derivatives apply in vector calculus for multivariable functions?
3. What is the geometric interpretation of velocity and acceleration vectors?
4. Can velocity and acceleration vectors ever be orthogonal?
5. How does chain rule apply when differentiating vector-valued functions?

### Tip:
When differentiating complex functions, always check for opportunities to simplify the expression before plugging in specific values.

Calculate the velocity and acceleration vectors and the speed of r(t) = ⟨1/(5+t^2), 1/(5+t^2)⟩ at the time t = 3. (Use symbolic notation and fractions where needed. Give your answer in vector form.)

This page provides a detailed solution to calculate the velocity and acceleration vectors for the function r(t) = ⟨1/(5+t^2), 1/(5+t^2)⟩ at t = 3. Learn how to apply derivatives and the chain rule to find velocity and acceleration in vector calculus.

Math Problem Statement

Solution

Ask a new question for Free

By Image

Math Problem Analysis

Mathematical Concepts

Formulas

Theorems

Suitable Grade Level

Related Recommendation