The given vector-valued function is $r(t) = \langle t, t^2 + 1 \rangle$.

We are asked to find $r(t + 2) - r(t)$.

Step 1: Find $r(t + 2)$

Substitute $t + 2$ into the vector function:

$r(t + 2) = \langle t + 2, (t + 2)^2 + 1 \rangle$

First, calculate each component:

• First component: $t + 2$
• Second component: $(t + 2)^2 + 1 = t^2 + 4t + 4 + 1 = t^2 + 4t + 5$

Thus, we have: $r(t + 2) = \langle t + 2, t^2 + 4t + 5 \rangle$

Step 2: Find $r(t + 2) - r(t)$

Now subtract $r(t)$ from $r(t + 2)$:

$r(t + 2) - r(t) = \langle t + 2, t^2 + 4t + 5 \rangle - \langle t, t^2 + 1 \rangle$

Subtract component-wise:

• First component: $(t + 2) - t = 2$
• Second component: $(t^2 + 4t + 5) - (t^2 + 1) = 4t + 4$

Thus, we get: $r(t + 2) - r(t) = \langle 2, 4t + 4 \rangle$

Final Answer:

$r(t + 2) - r(t) = \langle 2, 4t + 4 \rangle$

Would you like more details or clarification on any part?

Here are five related questions:

1. How would you find $r'(t)$, the derivative of the vector-valued function?
2. What is the magnitude of $r(t + 2) - r(t)$?
3. Can you generalize this process for $r(t + h) - r(t)$?
4. How does the second component of $r(t)$ influence the behavior of the vector?
5. What is the physical interpretation of $r(t)$ and its change over time?

Tip: In vector calculus, operations on vector-valued functions follow the same rules as scalar functions, but each component is handled separately.