Q. 34 - Calculus

Question

Use Theorem 12.34 to find the indicated derivatives in Exercises 31–36. Be sure to simplify your answers. 

$\frac{\partial w}{\partial \theta }\mathrm{when} \mathrm{w}=\left({\mathrm{x}}^2+\mathrm{z}\right){\mathrm{e}}^{\mathrm{y}}, \mathrm{x}=\mathrm{\rho sin\varphi cos\theta }, \mathrm{y}=\mathrm{\rho sin\varphi sin\theta }, \mathrm{z}=\mathrm{\rho cos\varphi }$

Step-by-Step Solution

Answer

The value of $\frac{\partial w}{\partial \theta }=\rho e^{\rho \sin \varphi \sin \theta }\sin \varphi \left\{-2\sin \varphi \cos \theta \sin \theta +{\rho }^2{\sin }^2\varphi {\cos }^3\theta +\rho \cos \varphi \cos \theta \right\}$

Step 1. Given Information.

$$\begin{gathered} \mathrm{w}=\left({\mathrm{x}}^2+\mathrm{z}\right){\mathrm{e}}^{\mathrm{y}} \\ \mathrm{x}=\mathrm{\rho sin\varphi cos\theta } \\ y=\mathrm{\rho sin\varphi sin\theta } \\ \mathrm{z}=\mathrm{\rho cos\varphi } \end{gathered}$$

Step 2. Calculation.

By Theorem 12.34, we have 

$\frac{\partial w}{\partial \theta }=\frac{\partial w}{\partial x}·\frac{\partial x}{\partial \theta }+\frac{\partial w}{\partial y}·\frac{\partial y}{\partial \theta }+\frac{\partial w}{\partial z}·\frac{\partial z}{\partial \theta }-------\left(1\right)$

So first we find $\frac{\partial w}{\partial x},\frac{\partial x}{\partial \theta },\frac{\partial w}{\partial y},\frac{\partial y}{\partial \theta },\frac{\partial w}{\partial z},\frac{\partial z}{\partial \theta }$

So we have 

$$\begin{gathered} \frac{\partial w}{\partial x}=e^{y}2x \\ \frac{\partial w}{\partial y}=\left(x^2+z\right)e^y \\ \frac{\partial w}{\partial z}=e^y·1=e^y \\ \frac{\partial x}{\partial \theta }=-\rho \sin \varphi \sin \theta \\ \frac{\partial y}{\partial \theta }=\rho \sin \varphi \sin \theta \\ \frac{\partial z}{\partial \theta }=0 \end{gathered}$$

Step 3. Calculation.

Use these above values in (1) we get, 

$$\begin{gathered} \frac{\partial w}{\partial \theta }=\frac{\partial w}{\partial x}·\frac{\partial x}{\partial \theta }+\frac{\partial w}{\partial y}·\frac{\partial y}{\partial \theta }+\frac{\partial w}{\partial z}·\frac{\partial z}{\partial \theta } \\ \frac{\partial w}{\partial \theta }=2x{e}^y\left\{\left(-\rho \sin \varphi \sin \theta \right)+e^y\left(x^2+z\right)\rho \sin \varphi \cos \theta +e^y\left(0\right)\right\} \\ \frac{\partial w}{\partial \theta }=\rho e^y\sin \varphi \left\{-2x\sin \theta +\left(x^2+z\right)\cos \theta \right\} \end{gathered}$$

So from here, putting the value of  $x \mathrm{and} y$ in terms of $\rho ,\varphi ,\theta$ we get

$$\begin{gathered} \frac{\partial w}{\partial \theta }=\rho e^y\sin \varphi \left\{-2x\sin \theta +\left(x^2+z\right)\cos \theta \right\} \\ \frac{\partial w}{\partial \theta }=\rho e^{\rho \sin \varphi \sin \theta }\sin \varphi \left\{-2\sin \varphi \cos \theta \sin \theta +{\rho }^2{\sin }^2\varphi {\cos }^3\theta +\rho \cos \varphi \cos \theta \right\} \end{gathered}$$

Step 4. Conclusion.

The value of $\frac{\partial w}{\partial \theta }=\rho e^{\rho \sin \varphi \sin \theta }\sin \varphi \left\{-2\sin \varphi \cos \theta \sin \theta +{\rho }^2{\sin }^2\varphi {\cos }^3\theta +\rho \cos \varphi \cos \theta \right\}$