$$ 1 \leftarrow $$ 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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We start with
$$
Q = \sqrt{x^2 + y^2 + z^2},
$$
where
$$
x = \sin t,\quad y = \cos t,\quad z = \cos t.
$$

**Step 1.** Differentiate $$Q$$ with respect to $$t$$ using the chain rule. Write $$Q$$ as
$$
Q = \left( x^2 + y^2 + z^2 \right)^{\frac{1}{2}}.
$$
Differentiating gives
$$
\frac{dQ}{dt} = \frac{1}{2}\left( x^2+y^2+z^2 \right)^{-\frac{1}{2}} \cdot \frac{d}{dt}\left( x^2+y^2+z^2 \right).
$$

**Step 2.** Compute the derivative $$\frac{d}{dt}\left( x^2+y^2+z^2 \right)$$. Differentiate each term:
$$
\frac{d}{dt}(x^2) = 2x\,\frac{dx}{dt},\quad \frac{d}{dt}(y^2) = 2y\,\frac{dy}{dt},\quad \frac{d}{dt}(z^2) = 2z\,\frac{dz}{dt}.
$$
Thus,
$$
\frac{d}{dt}\left( x^2+y^2+z^2 \right) = 2x\,\frac{dx}{dt} + 2y\,\frac{dy}{dt} + 2z\,\frac{dz}{dt}.
$$

**Step 3.** Substitute this back into the expression for $$\frac{dQ}{dt}$$:
$$
\frac{dQ}{dt} = \frac{1}{2}\left( x^2+y^2+z^2 \right)^{-\frac{1}{2}} \left( 2x\,\frac{dx}{dt} + 2y\,\frac{dy}{dt} + 2z\,\frac{dz}{dt} \right).
$$
Cancel the factor of $$2$$:
$$
\frac{dQ}{dt} = \frac{x\,\frac{dx}{dt}+y\,\frac{dy}{dt}+z\,\frac{dz}{dt}}{\sqrt{x^2+y^2+z^2}}.
$$

**Step 4.** Find the derivatives of $$x$$, $$y$$, and $$z$$:
$$
\frac{dx}{dt} = \cos t,\quad \frac{dy}{dt} = -\sin t,\quad \frac{dz}{dt} = -\sin t.
$$

**Step 5.** Substitute $$x$$, $$y$$, $$z$$ and their derivatives:
$$
x\,\frac{dx}{dt} = (\sin t)(\cos t) = \sin t\,\cos t,
$$
$$
y\,\frac{dy}{dt} = (\cos t)(-\sin t) = -\cos t\,\sin t,
$$
$$
z\,\frac{dz}{dt} = (\cos t)(-\sin t) = -\cos t\,\sin t.
$$
Then the numerator becomes:
$$
\sin t\,\cos t - \cos t\,\sin t - \cos t\,\sin t = -\cos t\,\sin t.
$$

**Step 6.** Write the final derivative:
$$
\frac{dQ}{dt} = \frac{-\cos t\,\sin t}{\sqrt{x^2+y^2+z^2}}.
$$
Recall that
$$
x^2+y^2+z^2 = \sin^2 t + \cos^2 t + \cos^2 t = \sin^2 t + 2\cos^2 t.
$$
Thus, the final answer is:
$$
\frac{dQ}{dt} = \frac{-\cos t\,\sin t}{\sqrt{\sin^2 t + 2\cos^2 t}}.
$$
$$ \frac{dQ}{dt} = \frac{-\cos t\,\sin t}{\sqrt{\sin^2 t + 2\cos^2 t}}. $$

\( 1 \leftarrow \) 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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