To solve this problem, we need to use the chain rule for partial derivatives. First, we find the partial derivatives of \(x\), \(y\), and \(z\) with respect to \(r\) and \(t\). Then, we apply the chain rule to find \(\frac{\partial w}{\partial r}\) and \(\frac{\partial w}{\partial t}\). The chain rule states that:
\[
\frac{\partial w}{\partial r} = \frac{\partial w}{\partial x} \cdot \frac{\partial x}{\partial r} + \frac{\partial w}{\partial y} \cdot \frac{\partial y}{\partial r} + \frac{\partial w}{\partial z} \cdot \frac{\partial z}{\partial r}
\]
\[
\frac{\partial w}{\partial t} = \frac{\partial w}{\partial x} \cdot \frac{\partial x}{\partial t} + \frac{\partial w}{\partial y} \cdot \frac{\partial y}{\partial t} + \frac{\partial w}{\partial z} \cdot \frac{\partial z}{\partial t}
\]
We calculate each of these derivatives and substitute them into the chain rule formulas.
Given:
\[ x = 5r e^t, \quad y = 2t e^r, \quad z = e^{rt} \]
Calculate:
\[
\frac{\partial x}{\partial r} = \frac{\partial}{\partial r}(5r e^t) = 5e^t
\]
\[
\frac{\partial y}{\partial r} = \frac{\partial}{\partial r}(2t e^r) = 2t e^r
\]
\[
\frac{\partial z}{\partial r} = \frac{\partial}{\partial r}(e^{rt}) = t e^{rt}
\]
Calculate:
\[
\frac{\partial x}{\partial t} = \frac{\partial}{\partial t}(5r e^t) = 5r e^t
\]
\[
\frac{\partial y}{\partial t} = \frac{\partial}{\partial t}(2t e^r) = 2 e^r
\]
\[
\frac{\partial z}{\partial t} = \frac{\partial}{\partial t}(e^{rt}) = r e^{rt}
\]
The function \(w(x, y, z) = \sqrt{x^2 + y^2 + z^2}\).
Using the chain rule:
\[
\frac{\partial w}{\partial r} = \frac{\partial w}{\partial x} \cdot \frac{\partial x}{\partial r} + \frac{\partial w}{\partial y} \cdot \frac{\partial y}{\partial r} + \frac{\partial w}{\partial z} \cdot \frac{\partial z}{\partial r}
\]
First, find \(\frac{\partial w}{\partial x}\), \(\frac{\partial w}{\partial y}\), and \(\frac{\partial w}{\partial z}\):
\[
\frac{\partial w}{\partial x} = \frac{x}{\sqrt{x^2 + y^2 + z^2}}, \quad \frac{\partial w}{\partial y} = \frac{y}{\sqrt{x^2 + y^2 + z^2}}, \quad \frac{\partial w}{\partial z} = \frac{z}{\sqrt{x^2 + y^2 + z^2}}
\]
Substitute:
\[
\frac{\partial w}{\partial r} = \frac{x}{\sqrt{x^2 + y^2 + z^2}} \cdot 5e^t + \frac{y}{\sqrt{x^2 + y^2 + z^2}} \cdot 2t e^r + \frac{z}{\sqrt{x^2 + y^2 + z^2}} \cdot t e^{rt}
\]
Using the chain rule:
\[
\frac{\partial w}{\partial t} = \frac{\partial w}{\partial x} \cdot \frac{\partial x}{\partial t} + \frac{\partial w}{\partial y} \cdot \frac{\partial y}{\partial t} + \frac{\partial w}{\partial z} \cdot \frac{\partial z}{\partial t}
\]
Substitute:
\[
\frac{\partial w}{\partial t} = \frac{x}{\sqrt{x^2 + y^2 + z^2}} \cdot 5r e^t + \frac{y}{\sqrt{x^2 + y^2 + z^2}} \cdot 2 e^r + \frac{z}{\sqrt{x^2 + y^2 + z^2}} \cdot r e^{rt}
\]
\[
\frac{\partial w}{\partial r} = \frac{x \cdot 5e^t + y \cdot 2t e^r + z \cdot t e^{rt}}{\sqrt{x^2 + y^2 + z^2}}
\]
\[
\frac{\partial w}{\partial t} = \frac{x \cdot 5r e^t + y \cdot 2 e^r + z \cdot r e^{rt}}{\sqrt{x^2 + y^2 + z^2}}
\]
\[
\boxed{\frac{\partial w}{\partial r} = \frac{x \cdot 5e^t + y \cdot 2t e^r + z \cdot t e^{rt}}{\sqrt{x^2 + y^2 + z^2}}}
\]
\[
\boxed{\frac{\partial w}{\partial t} = \frac{x \cdot 5r e^t + y \cdot 2 e^r + z \cdot r e^{rt}}{\sqrt{x^2 + y^2 + z^2}}}
\]
Answered
