Q. 33 - Calculus

Question

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

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

Step-by-Step Solution

Answer

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

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 

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

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

So we have 

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

Step 3. Calculation.

Use these above values in (1) we get, 

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

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

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

Step 4. Conclusion.

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