Question

If (𝜑)𝑥, 𝑦, 𝑧) = 𝑥𝑦 2 𝑧 and 𝐴 = 𝑥𝑧𝑖 + 𝑥𝑦 2 𝑗 + 𝑦𝑧 2𝑘, find 𝜕 3 𝜕2𝑥𝜕𝑧 𝜑𝐴 at point 2, −1,1 .

Answer 1

$\displaystyle \varphi = (xy)^2z \\ \frac{\partial \varphi}{\partial x} = 2xy^2z\\ \frac{\partial^2\varphi}{\partial x^2} = 2y^2 z\\ \frac{\partial^3\varphi}{\partial x^2 \partial z} = 2y^2 \\ \textsf{At}\,\, y = -1, \\ \frac{\partial^3\varphi}{\partial x^2 \partial z} = 2\\ A= xz\hat{\textbf{i}} + xy^2\hat{\textbf{j}} + yz^2\hat{\textbf{k}} \\ \textsf{At}\,\,x=2, y = -1, z=1 \\ \begin{aligned} \frac{\partial^3\varphi}{\partial x^2 \partial z} \cdot A &= 2\left((2)(1)\hat{\textbf{i}} + 2(-1)^2\hat{\textbf{j}} + (-1)(1)^2\hat{\textbf{k}}\right) \\ &= 2\left(2\hat{\textbf{i}} + 2\hat{\textbf{j}} - \hat{\textbf{k}}\right) = 2(2,2 , -1) \end{aligned}$