Given the following position function: (position is in ft , time is in sec) \( \mathrm{s}(\mathrm{t})=\tan \left(\mathrm{t}^{2}\right) \) Find the velocity at \( \mathrm{t}=0.5 \) \( \mathrm{v}(0.5)=[?] \frac{\mathrm{ft}}{\mathrm{sec}} \) Round your answer to the nearest thousandth

To find the velocity, we need to take the derivative of the position function \( s(t) = \tan(t^2) \) with respect to time \( t \). The velocity function is given by: \[ v(t) = s'(t) = \frac{d}{dt}(\tan(t^2)) \] Using the chain rule, we differentiate: \[ v(t) = \sec^2(t^2) \cdot \frac{d}{dt}(t^2) = \sec^2(t^2) \cdot 2t \] Now, we can find \( v(0.5) \): \[ v(0.5) = \sec^2((0.5)^2) \cdot 2(0.5) = \sec^2(0.25) \cdot 1 = \frac{1}{\cos^2(0.25)} \cdot 1 \] Calculating \( \cos(0.25) \): \[ \cos(0.25) \approx 0.968912 \] Now, calculate \( \sec^2(0.25) \): \[ \sec^2(0.25) \approx \frac{1}{(0.968912)^2} \approx 1.064 \] Thus, \[ v(0.5) \approx 1.064 \cdot 1 \approx 1.064 \] Rounding this to the nearest thousandth, we get: \[ \mathbf{v(0.5) \approx 1.064 \text{ ft/sec}} \]