Q. 31 - Calculus

Question

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

$\frac{dx}{dt} when x = \rho \sin \phi \cos \theta , \rho = t^2, \phi = t^3, and \theta = t^4.$

Step-by-Step Solution

Answer

$$\begin{gathered} \mathrm{The} \mathrm{single} \mathrm{variable} \mathrm{function} \mathrm{is} , \\ \boxed{\frac{dx}{dt}= (\sin t^3 \cos t^4)\left(2t\right) +(\hspace{0.17em}{t}^2 \cos t^3 \cos t^4)(3t^2\hspace{0.17em})+(t^2 \sin t^3 (-\sin t^4\left)\right)(4 t^3).} \end{gathered}$$

Step 1. Given

$x = \rho \sin \phi \cos \theta , \rho = t^2, \phi = t^3, and \theta = t^4.$

Step 2. Simplification

$$\begin{gathered} \mathrm{Consider} \mathrm{the} \mathrm{following} \mathrm{function} , \\ x = \rho \sin \phi \cos \theta , \rho = t^2, \phi = t^3, and \theta = t^4. \\ \mathrm{Objective} \mathrm{is} \mathrm{to} \mathrm{find} \frac{dx}{dt}. \\ \mathrm{By} \mathrm{using} \mathrm{chain} \mathrm{rule} , \\ \frac{dx}{dt}= \frac{\partial x}{\partial \rho }.\frac{d\rho }{dt}+\frac{\partial x}{\partial \phi }.\frac{d\phi }{dt}+\frac{\partial x}{\partial \theta }.\frac{d\theta }{dt}. \\ \frac{\partial x}{\partial \rho }=\sin \phi \cos \theta \\ \frac{\partial x}{\partial \phi }=\rho \cos \phi \cos \theta \\ \frac{\partial x}{\partial \theta }=\rho \sin \phi (-\sin \theta ) \\ \mathrm{On} \mathrm{proceeding} \mathrm{the} \mathrm{next} \mathrm{step} , \\ \frac{d\rho }{dt}=2t, \frac{d\phi }{dt}=3t^2, \frac{d\theta }{dt}=4 t^3. \\ \mathrm{On} \mathrm{proceeding} \mathrm{the} \mathrm{next} \mathrm{step} , \\ \frac{dx}{dt}= \frac{\partial x}{\partial \rho }.\frac{d\rho }{dt}+\frac{\partial x}{\partial \phi }.\frac{d\phi }{dt}+\frac{\partial x}{\partial \theta }.\frac{d\theta }{dt}. \\ \frac{dx}{dt} =(\sin \phi \cos \theta )\left(2t\right) +(\hspace{0.17em}\rho \cos \phi \cos \theta )(3t^2\hspace{0.17em})+(\rho \sin \phi (-\sin \theta \left)\right)(4 t^3) . \\ \frac{dx}{dt}= (\sin t^3 \cos t^4)\left(2t\right) +(\hspace{0.17em}{t}^2 \cos t^3 \cos t^4)(3t^2\hspace{0.17em})+(t^2 \sin t^3 (-\sin t^4\left)\right)(4 t^3). \\ \mathrm{The} \mathrm{single} \mathrm{variable} \mathrm{function} \mathrm{is} , \\ \boxed{\frac{dx}{dt}= (\sin t^3 \cos t^4)\left(2t\right) +(\hspace{0.17em}{t}^2 \cos t^3 \cos t^4)(3t^2\hspace{0.17em})+(t^2 \sin t^3 (-\sin t^4\left)\right)(4 t^3).} \end{gathered}$$