Use the Chain Rule to find \( \frac{d Q}{d t} \), where \( Q=\sqrt{2 x^{2}+2 y^{2}+2 z^{2}}, x=\sin t, y=\cos t \), and \( z=\sin t \). \( \frac{\partial Q}{\partial x}=\frac{2 x}{\sqrt{2 x^{2}+2 y^{2}+2 z^{2}}} \) (Type an expression using \( x, y \), and \( z \) as the variables.) \( \frac{d x}{d t}=\square \) (Type an expression using \( t \) as the variable.)

To find \( \frac{d Q}{d t} \) using the chain rule, let’s first express \( Q \) in terms of \( t \) via the relations for \( x \), \( y \), and \( z \). Given: \[ Q = \sqrt{2 x^2 + 2 y^2 + 2 z^2} \] and substituting \( x = \sin t \), \( y = \cos t \), and \( z = \sin t \): \[ Q = \sqrt{2 (\sin t)^2 + 2 (\cos t)^2 + 2 (\sin t)^2} = \sqrt{2 \sin^2 t + 2 \cos^2 t + 2 \sin^2 t} = \sqrt{4 \sin^2 t + 2 \cos^2 t} \] Next, we use the chain rule to find \( \frac{d Q}{d t} \): \[ \frac{d Q}{d t} = \frac{\partial Q}{\partial x} \cdot \frac{d x}{d t} + \frac{\partial Q}{\partial y} \cdot \frac{d y}{d t} + \frac{\partial Q}{\partial z} \cdot \frac{d z}{d t} \] Given \( \frac{\partial Q}{\partial x} = \frac{2x}{\sqrt{2x^2 + 2y^2 + 2z^2}} \), we can compute \( \frac{\partial Q}{\partial y} \) and \( \frac{\partial Q}{\partial z} \) similarly, finding they will be analogous due to symmetry in \( y \) and \( z \). Next, for the derivatives: \[ \frac{d x}{d t} = \cos t, \quad \frac{d y}{d t} = -\sin t, \quad \frac{d z}{d t} = \cos t \] Finally, you can compute the entire expression for \( \frac{d Q}{d t} \): \[ \frac{d Q}{d t} = \frac{2 \sin t}{\sqrt{2 \sin^2 t + 2 \cos^2 t + 2 \sin^2 t}} \cdot \cos t + \frac{2 \cos t}{\sqrt{2 \sin^2 t + 2 \cos^2 t + 2 \sin^2 t}} \cdot (-\sin t) + \frac{2 \sin t}{\sqrt{2 \sin^2 t + 2 \cos^2 t + 2 \sin^2 t}} \cdot \cos t \] Summing these terms gives you the complete derivative \( \frac{d Q}{d t} \). Now, filling in the boxes: \[ \frac{d x}{d t} = \cos t \]