$\text { Find } \frac { \partial w } { \partial t } \text { using the appropriate Chain Rule for } w = x ^ { 2 } + y ^ { 2 } + z ^ { 2 } \text { where }$ $x = 8 t \sin s , y = 8 t \cos s$, and $z = 5 s t ^ { 2 }$.
A) $128 t + 100 s ^ { 2 } t ^ { 3 }$
B) $50 t + 256 s ^ { 2 } t ^ { 3 }$
C) $128 t ^ { 2 } + 50 s ^ { 3 } t ^ { 2 }$
D) $128 t + 256 s ^ { 3 } t ^ { 2 }$
E) $128 t ^ { 2 } + 50 s ^ { 2 } t ^ { 3 }$

Question

Quizplus · Accepted Answer

To find $\frac{\partial w}{\partial t}$, we apply the chain rule to each of $x$, $y$, and $z$ as functions of $t$ and $s$, and then differentiate $w$ with respect to $x$, $y$, and $z$. Given $w = x^2 + y^2 + z^2$, we have:- $x = 8t\sin(s)$, so $\frac{\partial x}{\partial t} = 8\sin(s)$ and $\frac{\partial w}{\partial x} = 2x$.- $y = 8t\cos(s)$, so $\frac{\partial y}{\partial t} = 8\cos(s)$ and $\frac{\partial w}{\partial y} = 2y$.- $z = 5st^2$, so $\frac{\partial z}{\partial t} = 10st$ and $\frac{\partial w}{\partial z} = 2z$.Thus, using the chain rule, we get:$$\frac{\partial w}{\partial t} = 2x\cdot\frac{\partial x}{\partial t} + 2y\cdot\frac{\partial y}{\partial t} + 2z\cdot\frac{\partial z}{\partial t}$$Substituting the expressions for $x$, $y$, $z$, and their derivatives with respect to $t$, we get:$$\frac{\partial w}{\partial t} = 2(8t\sin(s))\cdot(8\sin(s)) + 2(8t\cos(s))\cdot(8\cos(s)) + 2(5st^2)\cdot(10st)$$Simplifying, we find:$$\frac{\partial w}{\partial t} = 128t(\sin^2(s) + \cos^2(s)) + 100s^2t^3$$Since $\sin^2(s) + \cos^2(s) = 1$, the expression simplifies to:$$\frac{\partial w}{\partial t} = 128t + 100s^2t^3$$Therefore, the correct answer is A) $128t + 100s^2t^3$.

x=8tsin⁡s,y=8tcos⁡sx = 8 t \sin s , y = 8 t \cos sx=8tsins,y=8tcoss

$x = 8 t \sin s , y = 8 t \cos s$