# W = xy2 + x2z + yz2, x = t2, y = 9t, z = 9 (a) Find dw/dt using the appropriate Chain Rule?

Jul 24, 2018

Please see the explanation below.

#### Explanation:

First start by calculating the following derivatives

$x = {t}^{2}$, $\implies$, $\frac{\mathrm{dx}}{\mathrm{dt}} = 2 t$

$y = 9 t$, $\implies$, $\frac{\mathrm{dy}}{\mathrm{dt}} = 9$

$z = 9$, $\implies$, $\frac{\mathrm{dz}}{\mathrm{dt}} = 0$

Then,

$w = x {y}^{2} + {x}^{2} z + y {z}^{2}$

$\frac{\mathrm{dw}}{\mathrm{dt}} = \frac{d}{\mathrm{dt}} \left(x {y}^{2} + {x}^{2} z + y {z}^{2}\right)$

$= \frac{d}{\mathrm{dt}} \left(x {y}^{2}\right) + \frac{d}{\mathrm{dt}} \left({x}^{2} z\right) + \frac{d}{\mathrm{dt}} \left(y {z}^{2}\right)$

$= x \frac{d}{\mathrm{dt}} \left({y}^{2}\right) + {y}^{2} \frac{\mathrm{dx}}{\mathrm{dt}} + z \frac{d}{\mathrm{dt}} \left({x}^{2}\right) + {x}^{2} \frac{\mathrm{dz}}{\mathrm{dt}} + y \frac{d}{\mathrm{dt}} \left({z}^{2}\right) + {z}^{2} \frac{\mathrm{dy}}{\mathrm{dt}}$

$= 2 y x \frac{\mathrm{dy}}{\mathrm{dt}} + {y}^{2} \frac{\mathrm{dx}}{\mathrm{dt}} + 2 x z \frac{\mathrm{dx}}{\mathrm{dt}} + {x}^{2} \frac{\mathrm{dz}}{\mathrm{dt}} + 2 y z \frac{\mathrm{dz}}{\mathrm{dt}} + {z}^{2} \frac{\mathrm{dy}}{\mathrm{dt}}$

$= 18 y x + 2 {y}^{2} t + 4 x z t + + 9 {z}^{2}$

$= 18 x y + \frac{2}{9} {y}^{3} + \frac{4}{9} x y z + 9 {z}^{2}$

$= 162 {t}^{3} + 162 {t}^{3} + 36 {t}^{3} + 729$

$= 360 {t}^{3} + 729$

Hope that this will help!!

Jul 28, 2018

Same answer, but using differentials (which can compact the notation, and sometimes also the algebra):

$w = x {y}^{2} + {x}^{2} z + y {z}^{2}$

$\mathrm{dw} = \mathrm{dx} \setminus {y}^{2} + 2 x y \setminus \mathrm{dy} + 2 x \setminus \mathrm{dx} \setminus z + {x}^{2} \setminus \mathrm{dz} + \mathrm{dy} \setminus {z}^{2} + 2 y z \setminus \mathrm{dz}$

$= \mathrm{dx} \left({y}^{2} + 2 x z\right) + \mathrm{dy} \left(2 x y + {z}^{2}\right) + \mathrm{dz} \left({x}^{2} + 2 y z\right)$

$\left\{\begin{matrix}x = {t}^{2} q \quad \mathrm{dx} = 2 t \setminus \mathrm{dt} \\ y = 9 t q \quad \mathrm{dy} = 9 \setminus \mathrm{dt} \\ z = 9 q \quad \mathrm{dz} = 0\end{matrix}\right.$

$\therefore \mathrm{dw} = 2 t \setminus \mathrm{dt} {\underbrace{\left(81 {t}^{2} + 18 {t}^{2}\right)}}_{= 99 {t}^{2}} + 9 \setminus \mathrm{dt} \left(18 {t}^{3} + 81\right) + \cancel{\mathrm{dz} \left({x}^{2} + 2 y z\right)}$

$\implies \frac{\mathrm{dw}}{\mathrm{dt}} = 2 t \left(99 {t}^{2}\right) + 9 \left(18 {t}^{3} + 81\right)$

 = (9 ( 2t ( 11t^2) + 18t^3+ 81 )

$= 9 \left(40 {t}^{3} + 81\right)$