Fin h’(t) when h(t) = 0.8sin[(pi/2)(t-1)] +1.8

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!

Discover how to calculate the derivative h'(t) of the function h(t) = 0.8sin[(pi/2)(t-1)] + 1.8. This solution includes the step-by-step process of differentiation using the chain rule and trigonometric derivatives, suitable for students in Grades 11-12.

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