The location of an object moving along the number line at time t seconds is given by

d(t) =

100

5 + 4 sin(t)

where t is assumed to be non-negative. (h) On the time interval [4,15], what is maximum distance from the origin to the object?

To find the maximum distance from the origin to the object on the time interval \([4, 15]\), we need to maximize the function:

\[
d(t) = \frac{100}{5 + 4 \sin(t)}
\]

### Steps:

1. **Find the derivative \( d'(t) \):**
   We'll use the quotient rule to differentiate \( d(t) \). The quotient rule is:

   \[
   \frac{d}{dt} \left( \frac{f(t)}{g(t)} \right) = \frac{f'(t)g(t) - f(t)g'(t)}{g(t)^2}
   \]

   Here, \( f(t) = 100 \) (a constant) and \( g(t) = 5 + 4 \sin(t) \). So:

   - \( f'(t) = 0 \)
   - \( g'(t) = 4 \cos(t) \)

   Applying the quotient rule:

   \[
   d'(t) = \frac{0 \cdot (5 + 4 \sin(t)) - 100 \cdot 4 \cos(t)}{(5 + 4 \sin(t))^2}
   \]

   Simplifying:

   \[
   d'(t) = \frac{-400 \cos(t)}{(5 + 4 \sin(t))^2}
   \]

2. **Set \( d'(t) = 0 \) to find critical points:**

   The derivative is zero when the numerator is zero:

   \[
   -400 \cos(t) = 0
   \]

   This simplifies to:

   \[
   \cos(t) = 0
   \]

   The values of \( t \) where \( \cos(t) = 0 \) are:

   \[
   t = \frac{\pi}{2}, \frac{3\pi}{2}, \frac{5\pi}{2}, \dots
   \]

   In the interval \([4, 15]\), the relevant values of \( t \) are approximately:

   \[
   t = \frac{3\pi}{2} \approx 4.71 \quad \text{and} \quad t = \frac{7\pi}{2} \approx 10.99
   \]

3. **Check the endpoints and critical points:**

   We now evaluate \( d(t) \) at the critical points and the endpoints \( t = 4 \) and \( t = 15 \):

   - At \( t = 4 \):

     \[
     d(4) = \frac{100}{5 + 4 \sin(4)} \approx \frac{100}{5 + 4( -0.7568)} = \frac{100}{5 - 3.0272} = \frac{100}{1.9728} \approx 50.68
     \]

   - At \( t = 15 \):

     \[
     d(15) = \frac{100}{5 + 4 \sin(15)} \approx \frac{100}{5 + 4(0.6503)} = \frac{100}{5 + 2.6012} = \frac{100}{7.6012} \approx 13.16
     \]

   - At \( t = \frac{3\pi}{2} \approx 4.71 \):

     \[
     d\left(\frac{3\pi}{2}\right) = \frac{100}{5 + 4 \sin\left(\frac{3\pi}{2}\right)} = \frac{100}{5 + 4(-1)} = \frac{100}{1} = 100
     \]

   - At \( t = \frac{7\pi}{2} \approx 10.99 \):

     \[
     d\left(\frac{7\pi}{2}\right) = \frac{100}{5 + 4 \sin\left(\frac{7\pi}{2}\right)} = \frac{100}{5 + 4(1)} = \frac{100}{9} \approx 11.11
     \]

4. **Conclusion:**

   The maximum value of \( d(t) \) in the interval \([4, 15]\) is at \( t = \frac{3\pi}{2} \approx 4.71 \), where the distance is:

   \[
   d\left(\frac{3\pi}{2}\right) = 100
   \]

Thus, the maximum distance from the origin to the object is **100 units**.

Would you like further details on any step? Here are 5 related questions to explore:

1. How would the result change if the interval were extended beyond [4, 15]?
2. Can you explain why \( \cos(t) = 0 \) gives critical points?
3. What is the significance of the derivative in finding maximum values?
4. How does the function \( \sin(t) \) influence the behavior of \( d(t) \)?
5. What role do endpoints play in optimization problems?

**Tip:** Always check both the critical points and the endpoints when looking for absolute maximum or minimum values on a closed interval.

The location of an object moving along the number line at time t seconds is given by d(t) = 100 / (5 + 4 sin(t)), where t is assumed to be non-negative. On the time interval [4,15], what is the maximum distance from the origin to the object?

Learn how to find the maximum distance of an object moving along the number line using calculus. We apply the quotient rule and trigonometric functions to solve d(t) = 100 / (5 + 4 sin(t)) over the interval [4,15]. Ideal for college-level calculus students.

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