$$ r ( v ) = \frac { v \sin v } { 1 + v } , a = - \frac { \pi } { 2 } $$

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Evaluate $$ r\left(-\frac{\pi}{2}\right) $$ or find $$ r'(v) $$ at $$ a = -\frac{\pi}{2} $$. It looks like you have a function $$ r(v) = \frac{v \sin v}{1 + v} $$ and a specific value $$ a = -\frac{\pi}{2} $$. If you need to evaluate the function at this point or perform some analysis, please specify what you would like to do with this function and the value of $$ a $$. Here are a few possibilities: 1. **Evaluate $$ r(a) $$**: Calculate $$ r\left(-\frac{\pi}{2}\right) $$. 2. **Find the derivative**: Compute $$ r'(v) $$ and evaluate it at $$ a $$. 3. **Analyze the behavior**: Discuss the limits or behavior of $$ r(v) $$ as $$ v $$ approaches certain values. Let me know how you would like to proceed!

\( r ( v ) = \frac { v \sin v } { 1 + v } , a = - \frac { \pi } { 2 } \)

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\( r ( v ) = \frac { v \sin v } { 1 + v } , a = - \frac { \pi } { 2 } \)

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