Given the following position function: (position is in ft , time is in sec ) \[ \mathrm{s}(\mathrm{t})=\frac{\sqrt{\mathrm{t}}}{\sqrt{\mathrm{t}+1}} \] Find the velocity at \( \mathrm{t}=3 \). \( \mathrm{v}(3)=[?] \frac{\mathrm{ft}}{\mathrm{sec}} \) Round your answer to the nearest thousandth.

To find the velocity at \( t = 3 \), we first need to calculate the derivative of the position function \( s(t) \). The velocity function \( v(t) \) is given by the derivative \( \frac{ds}{dt} \). The position function is: \[ s(t) = \frac{\sqrt{t}}{\sqrt{t+1}} \] To differentiate it, we can use the quotient rule: If \( u(t) = \sqrt{t} \) and \( v(t) = \sqrt{t+1} \), then: \[ s(t) = \frac{u(t)}{v(t)} \] The quotient rule states that: \[ \frac{d}{dt}\left(\frac{u}{v}\right) = \frac{u'v - uv'}{v^2} \] Now, find \( u' \) and \( v' \): 1. \( u(t) = \sqrt{t} \) → \( u' = \frac{1}{2\sqrt{t}} \) 2. \( v(t) = \sqrt{t+1} \) → \( v' = \frac{1}{2\sqrt{t+1}} \) Now apply the quotient rule: \[ v(t) = \frac{\frac{1}{2\sqrt{t}} \sqrt{t+1} - \sqrt{t} \frac{1}{2\sqrt{t+1}}}{t+1} \] Simplifying gives us: \[ v(t) = \frac{\sqrt{t+1}}{2\sqrt{t}(t+1)} - \frac{\sqrt{t}}{2\sqrt{t+1}(t+1)} \] Bringing them under a common denominator: \[ v(t) = \frac{\sqrt{t+1}^2 - \sqrt{t}^2}{2\sqrt{t}\sqrt{t+1}(t+1)} \] Now substituting back in: \[ v(t) = \frac{(t+1) - t}{2\sqrt{t}\sqrt{t+1}(t+1)} = \frac{1}{2\sqrt{t}\sqrt{t+1}(t+1)} \] Now we can substitute \( t = 3 \): \[ v(3) = \frac{1}{2\sqrt{3}\sqrt{4}(4)} = \frac{1}{2\sqrt{3}\cdot 2 \cdot 4} = \frac{1}{16\sqrt{3}} \] Now to approximate \( \frac{1}{16\sqrt{3}} \): Calculating \( \sqrt{3} \) approximately as \( 1.732 \): \[ v(3) \approx \frac{1}{16 \cdot 1.732} \approx \frac{1}{27.712} \approx 0.0361 \] Rounding to the nearest thousandth gives us: \[ v(3) \approx 0.036 \text{ ft/sec} \] So, the velocity at \( t = 3 \) is \( \frac{0.036}{\text{ft/sec}} \).