\( 1 \leftarrow \) Use the Chain Rule to find \( \frac{d Q}{d t} \), where \( Q=\sqrt{x^{2}+y^{2}+z^{2}}, x=\sin t, y=\cos t \), and \( z=\cos t \).
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To find \( \frac{dQ}{dt} \) using the Chain Rule, we first express \( Q \) in terms of \( t \): \[ Q = \sqrt{x^2 + y^2 + z^2} = \sqrt{(\sin t)^2 + (\cos t)^2 + (\cos t)^2} \] This simplifies to: \[ Q = \sqrt{\sin^2 t + \cos^2 t + \cos^2 t} = \sqrt{1 + \cos^2 t} \] Next, we apply the Chain Rule: \[ \frac{dQ}{dt} = \frac{1}{2\sqrt{1 + \cos^2 t}} \cdot \frac{d}{dt}(1 + \cos^2 t) \] Now we differentiate \( 1 + \cos^2 t \): \[ \frac{d}{dt}(1 + \cos^2 t) = 2\cos t(-\sin t) = -2\cos t \sin t \] Substituting back, we get: \[ \frac{dQ}{dt} = \frac{-2\cos t \sin t}{2\sqrt{1 + \cos^2 t}} = \frac{-\cos t \sin t}{\sqrt{1 + \cos^2 t}} \] Thus, the final result is: \[ \frac{dQ}{dt} = \frac{-\cos t \sin t}{\sqrt{1 + \cos^2 t}} \]