To find the first value of $t$ where the function $y = 3 \cos(5t^2 + 2)$ reaches its maximum, we can follow these steps:

Step 1: Understand the structure of the function

The given function is:

$y = 3 \cos(5t^2 + 2)$

The cosine function, $\cos(x)$, reaches its maximum value at 1. Therefore, we want to find when the argument of the cosine function, $5t^2 + 2$, produces a maximum for the overall function $y$.

Step 2: Set the cosine part to its maximum

The cosine function $\cos(\theta)$ is maximized when $\theta = 2k\pi$ for integer $k$, because $\cos(0) = 1$.

Thus, we want:

$5t^2 + 2 = 2k\pi$

Step 3: Solve for $t$ when $k = 0$

The first time the cosine function reaches its maximum is when $k = 0$. This gives us the equation:

$5t^2 + 2 = 0$

Solving for $t$:

$5t^2 = -2$

This equation has no real solution because $t^2$ cannot be negative for real numbers.

Step 4: Consider next $k = 1$

Let's try $k = 1$, so:

$5t^2 + 2 = 2\pi$

Solving for $t$:

$5t^2 = 2\pi - 2$ $t^2 = \frac{2\pi - 2}{5}$ $t = \sqrt{\frac{2\pi - 2}{5}}$

Step 5: Calculate the value

Now, we calculate the value numerically:

$t = \sqrt{\frac{2(3.14159) - 2}{5}} = \sqrt{\frac{6.28318 - 2}{5}} = \sqrt{\frac{4.28318}{5}} = \sqrt{0.85664}$

$t \approx 0.9256$

Conclusion:

The first value of $t$ where the function reaches its maximum is approximately $t \approx 0.926$.

---

Would you like more details or have any questions on this?

5 Related Questions:

1. How do we find the minimum value of the same function?
2. Can you explain how the derivative of the function would help locate maxima?
3. What happens when $k = 2$ in the same equation?
4. How does the amplitude (3 in this case) affect the maximum value of $y$?
5. What are the general properties of the cosine function that are useful for finding extrema?

Tip:

To locate maxima and minima, checking the first and second derivatives of the function is a reliable method.