Q. 32 - Calculus

Question

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

$\frac{dx}{dt} \mathrm{when} x=\rho \sin \varphi \sin \theta , \rho =\sqrt{t},\varphi =\sqrt[3]{t}, \mathrm{and} \theta =\sqrt[4]{t}$

Step-by-Step Solution

Answer

The value of $\frac{dx}{dt}=\frac{1}{2t^{{\displaystyle \raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}}}\sin \left(t^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$3$}\right.}\right)\sin \left(t^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$4$}\right.}\right)+\frac{1}{3t^{{\displaystyle \raisebox{1ex}{$2$}\!\left/ \!\raisebox{-1ex}{$3$}\right.}}}\rho \cos \left(t^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$3$}\right.}\right)\sin \left(t^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$4$}\right.}\right)+\frac{1}{4t^{{\displaystyle \raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$4$}\right.}}}\rho \sin \left(t^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$3$}\right.}\right)\cos \left(t^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$4$}\right.}\right)$

Step 1. Given Information.

$$\begin{gathered} x=\rho \sin \varphi \sin \theta \\ \rho =\sqrt{t}=t^{\frac{1}{2}} \\ \varphi =\sqrt[3]{t}=t^{\frac{1}{3}} \\ \theta =\sqrt[4]{t}=t^{\frac{1}{4}} \end{gathered}$$

Step 2. Calculation.

By Theorem 12.34, we have

$\frac{dx}{dt}=\frac{\partial x}{\partial \rho }\frac{d\rho }{dt}+\frac{\partial x}{\partial \varphi }\frac{d\varphi }{dt}+\frac{\partial x}{\partial \theta }\frac{d\theta }{dt}-------\left(1\right)$

So first we find $\frac{\partial x}{\partial \rho },\frac{d\rho }{dt},\frac{\partial x}{\partial \varphi },\frac{d\varphi }{dt},\frac{\partial x}{\partial \theta },\frac{d\theta }{dt}$

So we have 

$$\begin{gathered} \frac{\partial x}{\partial \rho }=\sin \varphi \sin \theta \\ \frac{\partial x}{\partial \varphi }=\rho \cos \varphi \sin \theta \\ \frac{\partial x}{\partial \theta }=\rho \sin \varphi \cos \theta \\ \frac{d\rho }{dt}=\frac{1}{2}{t}^{-\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.} \\ \frac{d\varphi }{dt}=\frac{1}{3}{t}^{\raisebox{1ex}{$-2$}\!\left/ \!\raisebox{-1ex}{$3$}\right.} \\ \frac{d\theta }{dt}=\frac{1}{4}{t}^{\raisebox{1ex}{$-3$}\!\left/ \!\raisebox{-1ex}{$4$}\right.} \end{gathered}$$

Step 3. Calculation.

Use these above values in (1) we get,

$$\begin{gathered} \frac{dx}{dt}=\frac{\partial x}{\partial \rho }\frac{d\rho }{dt}+\frac{\partial x}{\partial \varphi }\frac{d\varphi }{dt}+\frac{\partial x}{\partial \theta }\frac{d\theta }{dt} \\ \frac{dx}{dt}=\sin \varphi \sin \theta \left(\frac{1}{2}{t}^{\raisebox{1ex}{$-1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}\right)+\rho \cos \varphi \sin \theta \left(\frac{1}{3}{t}^{\raisebox{1ex}{$-2$}\!\left/ \!\raisebox{-1ex}{$3$}\right.}\right)+\rho \sin \varphi \cos \theta \left(\frac{1}{4}{t}^{\raisebox{1ex}{$-3$}\!\left/ \!\raisebox{-1ex}{$4$}\right.}\right) \end{gathered}$$

So from here, putting the value of $\rho ,\varphi ,\theta$ in terms of $t$ we will get,

$$\begin{gathered} \frac{dx}{dt}=\sin \varphi \sin \theta \left(\frac{1}{2}{t}^{\raisebox{1ex}{$-1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}\right)+\rho \cos \varphi \sin \theta \left(\frac{1}{3}{t}^{\raisebox{1ex}{$-2$}\!\left/ \!\raisebox{-1ex}{$3$}\right.}\right)+\rho \sin \varphi \cos \theta \left(\frac{1}{4}{t}^{\raisebox{1ex}{$-3$}\!\left/ \!\raisebox{-1ex}{$4$}\right.}\right) \\ \frac{dx}{dt}=\frac{1}{2t^{{\displaystyle \raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}}}\sin \left(t^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$3$}\right.}\right)\sin \left(t^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$4$}\right.}\right)+\frac{1}{3t^{{\displaystyle \raisebox{1ex}{$2$}\!\left/ \!\raisebox{-1ex}{$3$}\right.}}}\rho \cos \left(t^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$3$}\right.}\right)\sin \left(t^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$4$}\right.}\right)+\frac{1}{4t^{{\displaystyle \raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$4$}\right.}}}\rho \sin \left(t^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$3$}\right.}\right)\cos \left(t^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$4$}\right.}\right) \end{gathered}$$

Step 4. Conclusion.

The value of $\frac{dx}{dt}=\frac{1}{2t^{{\displaystyle \raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}}}\sin \left(t^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$3$}\right.}\right)\sin \left(t^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$4$}\right.}\right)+\frac{1}{3t^{{\displaystyle \raisebox{1ex}{$2$}\!\left/ \!\raisebox{-1ex}{$3$}\right.}}}\rho \cos \left(t^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$3$}\right.}\right)\sin \left(t^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$4$}\right.}\right)+\frac{1}{4t^{{\displaystyle \raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$4$}\right.}}}\rho \sin \left(t^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$3$}\right.}\right)\cos \left(t^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$4$}\right.}\right)$