Interpretation of the Derivative — Question 9

Question 9 - Solution

We are given: $$s(t) = \frac{3t^2 + 2}{t + 1}, \quad t > 0$$

(a) Instantaneous Velocity at $t = 1$:

Velocity is the derivative $s'(t)$. Using the quotient rule:

$$s'(t) = \frac{(t+1)(6t) - (3t^2 + 2)(1)}{(t+1)^2}$$

$$= \frac{6t^2 + 6t - 3t^2 - 2}{(t+1)^2} = \frac{3t^2 + 6t - 2}{(t+1)^2}$$

Evaluate at $t = 1$: $$s'(1) = \frac{3 + 6 - 2}{4} = \frac{7}{4}$$

$$\boxed{s'(1) = \frac{7}{4}}$$

(b) Interpretation of the Sign of Velocity:

Since $s'(1) = \frac{7}{4} > 0$, the particle is moving in the positive direction at $t = 1$.

$$\boxed{\text{The particle is moving to the right (increasing position) at } t = 1}$$

(c) Critical Points and Direction Change:

Critical points occur where $s'(t) = 0$ or where $s'(t)$ is undefined.

The derivative is undefined at $t = -1$, which is outside the domain $t > 0$ and therefore ignored.

Set the numerator equal to zero: $$3t^2 + 6t - 2 = 0$$

$$t = \frac{-6 \pm \sqrt{36 + 24}}{6} = \frac{-6 \pm \sqrt{60}}{6} = \frac{-3 \pm \sqrt{15}}{3}$$

This gives two solutions: $$t = \frac{-3 - \sqrt{15}}{3} \quad \text{and} \quad t = \frac{-3 + \sqrt{15}}{3}$$

Since time must satisfy $t > 0$, the negative solution is rejected. The only valid critical time is: $$\boxed{t = \frac{-3 + \sqrt{15}}{3}}$$

To determine whether the particle changes direction, test the sign of $s'(t)$ on either side of this value (within $t > 0$). The derivative changes from negative to positive, indicating a direction change.

$$\boxed{\text{The particle changes direction at } t = \frac{-3 + \sqrt{15}}{3}}$$