Let's tackle part (b) step by step:

Step 1: Find the tangent vector at $t = 2$

The path of the ball is given by: $\vec{r}(t) = \langle t^2 - 2, \cos(t^2 - 4) \rangle$

The derivative of $\vec{r}(t)$ is: $\vec{r}'(t) = \langle \frac{d}{dt}(t^2 - 2), \frac{d}{dt}(\cos(t^2 - 4)) \rangle = \langle 2t, -\sin(t^2 - 4) \cdot \frac{d}{dt}(t^2 - 4) \rangle$

Simplify each component:

• The derivative of $t^2 - 2$ is $2t$,
• The derivative of $\cos(t^2 - 4)$ is $-\sin(t^2 - 4) \cdot 2t$.

So, $\vec{r}'(t) = \langle 2t, -2t \sin(t^2 - 4) \rangle$

At $t = 2$: $\vec{r}'(2) = \langle 2(2), -2(2)\sin(2^2 - 4) \rangle = \langle 4, -4\sin(0) \rangle = \langle 4, 0 \rangle$

Thus, the tangent vector at $t = 2$ is: $\vec{r}'(2) = \langle 4, 0 \rangle$

Step 2: Parametrize the tangent line

The ball starts veering off along the tangent line at $t = 2$. The tangent line can be written as: $\vec{L}(t) = \vec{r}(2) + \lambda \vec{r}'(2)$

From part (a), $\vec{r}(2) = \langle 2, 1 \rangle$, and $\vec{r}'(2) = \langle 4, 0 \rangle$. Substituting: $\vec{L}(t) = \langle 2, 1 \rangle + \lambda \langle 4, 0 \rangle = \langle 2 + 4\lambda, 1 \rangle$

This is the equation of the tangent line.

Step 3: Predict the ball's position after 4 seconds

The ball starts veering off the tangent line at $t = 2$. Four seconds after it was initially hit corresponds to $t = 4$, so the time elapsed on the tangent line is $\lambda = 4 - 2 = 2$.

Substitute $\lambda = 2$ into $\vec{L}(t)$: $\vec{L}(2) = \langle 2 + 4(2), 1 \rangle = \langle 2 + 8, 1 \rangle = \langle 10, 1 \rangle$

Final Answer:

The location of the ball 4 seconds after it was hit is: $\boxed{\langle 10, 1 \rangle}$

Do you want a step-by-step explanation for the differentiation process or details on tangent line parametrization?

Related Questions:

1. What is the role of the tangent vector in predicting the ball's new path?
2. How would the solution change if the ball veered off at a different $t$-value?
3. How do we calculate the tangent vector for a general parametric path?
4. Why does the ball's $y$-coordinate remain constant after $t = 2$?
5. Can you graph this tangent line and path to confirm the result?

Tip:

For parametric equations, always calculate the derivative carefully for tangent lines to avoid errors in direction!