To determine the rate of change of $h(t)$ when $t = 5$, we need to find the derivative of $h(t) = 3t(t^2 - 1)$ with respect to $t$, and then evaluate it at $t = 5$.

Step 1: Differentiate $h(t)$

We will apply the product rule of differentiation since $h(t)$ is a product of two functions: $3t$ and $(t^2 - 1)$.

$h(t) = 3t(t^2 - 1)$

Let:

• $u(t) = 3t$
• $v(t) = t^2 - 1$

Using the product rule $\frac{d}{dt} [u(t)v(t)] = u'(t)v(t) + u(t)v'(t)$:

1. Differentiate $u(t) = 3t$: $u'(t) = 3$

2. Differentiate $v(t) = t^2 - 1$: $v'(t) = 2t$

Now, apply the product rule: $h'(t) = u'(t)v(t) + u(t)v'(t)$ $h'(t) = 3(t^2 - 1) + 3t(2t)$ $h'(t) = 3(t^2 - 1) + 6t^2$ $h'(t) = 3t^2 - 3 + 6t^2$ $h'(t) = 9t^2 - 3$

Step 2: Evaluate $h'(t)$ at $t = 5$

Now, substitute $t = 5$ into $h'(t)$:

$h'(5) = 9(5^2) - 3$ $h'(5) = 9(25) - 3$ $h'(5) = 225 - 3$ $h'(5) = 222$

Thus, the rate of change of $h(t)$ when $t = 5$ is 222.

Would you like more details or clarifications on any step?

Here are 5 related questions you might find interesting:

1. What is the physical meaning of the derivative in this context?
2. How does the product rule work when more than two functions are involved?
3. Can you find the second derivative $h''(t)$ for further analysis?
4. How would the result change if $h(t)$ was a function of more complex terms?
5. What other rules of differentiation are useful in real-world problems like this one?

Tip: Always check if you can simplify a function before applying the product or chain rules; sometimes expanding helps!