91Ó°ÊÓ

We start by applying the chain rule to get \(\frac{dw}{dt}\). The given function is \(w = \ln(x^2 + y^2 + z^2)\). We have \(x = \cos t\), \(y = \sin t\), and \(z = 4\sqrt{t}\). We first find the derivatives of \(w\) with respect to \(x\), \(y\), and \(z\), then multiply each by the derivative of the respective variable with respect to \(t\). The partial derivatives are:\[\frac{\partial w}{\partial x} = \frac{2x}{x^2 + y^2 + z^2}\]\[\frac{\partial w}{\partial y} = \frac{2y}{x^2 + y^2 + z^2}\]\[\frac{\partial w}{\partial z} = \frac{2z}{x^2 + y^2 + z^2}\]The derivatives of \(x\), \(y\), and \(z\) with respect to \(t\) are:\[\frac{dx}{dt} = -\sin t\]\[\frac{dy}{dt} = \cos t\]\[\frac{dz}{dt} = \frac{2}{\sqrt{t}}\]By the chain rule:\[\frac{dw}{dt} = \frac{\partial w}{\partial x}\frac{dx}{dt} + \frac{\partial w}{\partial y}\frac{dy}{dt} + \frac{\partial w}{\partial z}\frac{dz}{dt}\]Substitute the partial derivatives and the derivatives of \(x\), \(y\), and \(z\) into the chain rule equation.

Substituting the expressions from Step 1 into the chain rule equation, we have:\[\frac{dw}{dt} = \frac{2x}{x^2 + y^2 + z^2}(-\sin t) + \frac{2y}{x^2 + y^2 + z^2}(\cos t) + \frac{2z}{x^2 + y^2 + z^2}\left(\frac{2}{\sqrt{t}}\right)\]Simplifying:\[\frac{dw}{dt} = \frac{-2x \sin t + 2y \cos t + \frac{4z}{\sqrt{t}}}{x^2 + y^2 + z^2}\]

Now, substitute \(x = \cos t\), \(y = \sin t\), and \(z = 4\sqrt{t}\) directly into \(w\) to express it in terms of \(t\):\[w = \ln((\cos t)^2 + (\sin t)^2 + (4\sqrt{t})^2)\]Given \((\cos t)^2 + (\sin t)^2 = 1\), this simplifies to:\[w = \ln(1 + 16t)\]Differentiate with respect to \(t\):\[\frac{dw}{dt} = \frac{1}{1 + 16t} \cdot 16\]Therefore, the derivative is:\[\frac{dw}{dt} = \frac{16}{1 + 16t}\]

Substitute \(t = 3\) into the expression for \(\frac{dw}{dt}\) obtained using both methods.Using chain rule:\[\frac{dw}{dt} = \frac{-2(\cos 3)(\sin 3) + 2(\sin 3)(\cos 3) + \frac{4(4\sqrt{3})}{\sqrt{3}}}{(\cos 3)^2 + (\sin 3)^2 + (4\sqrt{3})^2}\]Simplifies to:\[\frac{dw}{dt} = \frac{0 + \frac{16 \sqrt{3}}{\sqrt{3}}}{1 + 48} = \frac{16}{49}\]Using direct differentiation:\[\frac{dw}{dt} = \frac{16}{1 + 16 \times 3} = \frac{16}{49}\]Both methods should yield the same value.