To find $h'(t)$ for the function

$h(t) = 0.8 \sin\left(\frac{\pi}{2}(t-1)\right) + 1.8,$

we will differentiate it with respect to $t$.

Step 1: Differentiate $h(t)$

Using the chain rule and the derivative of sine:

$h'(t) = 0.8 \cdot \cos\left(\frac{\pi}{2}(t-1)\right) \cdot \frac{d}{dt}\left(\frac{\pi}{2}(t-1)\right).$

Step 2: Calculate the derivative of the inner function

The derivative of $\frac{\pi}{2}(t-1)$ is $\frac{\pi}{2}$.

Step 3: Substitute back into the derivative

Thus, we have:

$h'(t) = 0.8 \cdot \cos\left(\frac{\pi}{2}(t-1)\right) \cdot \frac{\pi}{2}.$

Step 4: Simplify the expression

So, we can simplify:

$h'(t) = 0.4\pi \cdot \cos\left(\frac{\pi}{2}(t-1)\right).$

Final Result

Thus, the derivative $h'(t)$ is:

$h'(t) = 0.4\pi \cos\left(\frac{\pi}{2}(t-1)\right).$

If you have any questions or need further details, feel free to ask! Here are some related questions you might consider:

1. What is the value of $h'(t)$ at specific values of $t$?
2. How does the graph of $h(t)$ look, and what are its key features?
3. Can you explain the significance of the parameters in $h(t)$?
4. How would you find the second derivative $h''(t)$?
5. What applications can this type of function have in real-world scenarios?

Tip: Always double-check your differentiation rules, especially with trigonometric functions!