Determine which is the graph of the curve with vector equation r(t)=⟨5t,−2sin(t)⟩ for −π2≤t≤π2.

To determine the graph of the curve given by the vector equation \( \mathbf{r}(t) = \langle 5t, -2\sin(t) \rangle \) for \( -\frac{\pi}{2} \leq t \leq \frac{\pi}{2} \), let's analyze the components and how they relate to the plot.

### Step-by-Step Breakdown:

1. **x-component (horizontal)**: 
   The x-component of the vector is \( x(t) = 5t \). This indicates a linear relationship between \( x \) and \( t \), meaning the x-values are scaled by 5. As \( t \) increases, \( x \) will increase proportionally by 5 times the value of \( t \).

   Since \( t \) is bounded by \( -\frac{\pi}{2} \leq t \leq \frac{\pi}{2} \), the corresponding x-values will range from:
   \[
   x\left( -\frac{\pi}{2} \right) = 5 \times -\frac{\pi}{2} = -\frac{5\pi}{2} \quad \text{to} \quad x\left( \frac{\pi}{2} \right) = 5 \times \frac{\pi}{2} = \frac{5\pi}{2}.
   \]
   So, the x-values span from \( -\frac{5\pi}{2} \) to \( \frac{5\pi}{2} \).

2. **y-component (vertical)**: 
   The y-component is \( y(t) = -2\sin(t) \). This describes a sinusoidal behavior with an amplitude of 2, but it is reflected across the x-axis because of the negative sign. As \( t \) varies from \( -\frac{\pi}{2} \) to \( \frac{\pi}{2} \), the sine function ranges from \( -1 \) to \( 1 \), and thus the y-values will range from:
   \[
   y\left( -\frac{\pi}{2} \right) = -2 \times \sin\left( -\frac{\pi}{2} \right) = -2 \times (-1) = 2,
   \]
   to
   \[
   y\left( \frac{\pi}{2} \right) = -2 \times \sin\left( \frac{\pi}{2} \right) = -2 \times 1 = -2.
   \]
   Therefore, the y-values will range between -2 and 2.

### Summary of the curve:
- The x-component moves linearly from \( -\frac{5\pi}{2} \) to \( \frac{5\pi}{2} \).
- The y-component follows a sinusoidal oscillation between -2 and 2, with a reflection (negative sign) applied.

The resulting graph will be sinusoidal, stretched horizontally by a factor of 5 and vertically by a factor of 2, with the curve oscillating between 2 and -2 along the y-axis. This curve will have a wavelike pattern due to the sine function, but its horizontal extent is much larger due to the linear \( 5t \) component.

Would you like to proceed with graphing this function, or do you need further clarifications on the vector components?

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Here are some questions to explore further:
1. How would the curve change if the x-component were \( 3t \) instead of \( 5t \)?
2. What effect does changing the amplitude of the sine function have on the graph?
3. How would the curve look if the sine function were not reflected (i.e., positive sine)?
4. What happens to the graph if the parameter range for \( t \) is extended?
5. How can you express this curve in parametric form for different intervals of \( t \)?

**Tip**: For vector functions, the x and y components are independent, and analyzing them separately can help visualize how the curve behaves overall.

Determine which is the graph of the curve with vector equation r(t)=⟨5t,−2sin(t)⟩ for −π/2≤t≤π/2.

Explore the graph of the vector equation r(t) = ⟨5t,−2sin(t)⟩ for −π/2 ≤ t ≤ π/2. Learn how the x-component moves linearly while the y-component oscillates sinusoidally, creating a stretched wavelike curve. Suitable for students in Grades 10-12.

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