The problem asks to find $\frac{\partial w}{\partial t}$ using the chain rule, where $w = \sqrt{x^2 + y^2 + z^2}$, $x = \cos(st)$, $y = \sin(st)$, and $z = s^2 t$. The final answer should be expressed in terms of $s$ and $t$.

AnalysisPartial DerivativesChain RuleMultivariable Calculus
2025/5/10

1. Problem Description

The problem asks to find wt\frac{\partial w}{\partial t} using the chain rule, where w=x2+y2+z2w = \sqrt{x^2 + y^2 + z^2}, x=cos(st)x = \cos(st), y=sin(st)y = \sin(st), and z=s2tz = s^2 t. The final answer should be expressed in terms of ss and tt.

2. Solution Steps

First, write down the chain rule for wt\frac{\partial w}{\partial t}:
wt=wxxt+wyyt+wzzt\frac{\partial w}{\partial t} = \frac{\partial w}{\partial x}\frac{\partial x}{\partial t} + \frac{\partial w}{\partial y}\frac{\partial y}{\partial t} + \frac{\partial w}{\partial z}\frac{\partial z}{\partial t}
Next, compute the partial derivatives:
wx=12x2+y2+z22x=xx2+y2+z2\frac{\partial w}{\partial x} = \frac{1}{2\sqrt{x^2 + y^2 + z^2}} \cdot 2x = \frac{x}{\sqrt{x^2 + y^2 + z^2}}
wy=12x2+y2+z22y=yx2+y2+z2\frac{\partial w}{\partial y} = \frac{1}{2\sqrt{x^2 + y^2 + z^2}} \cdot 2y = \frac{y}{\sqrt{x^2 + y^2 + z^2}}
wz=12x2+y2+z22z=zx2+y2+z2\frac{\partial w}{\partial z} = \frac{1}{2\sqrt{x^2 + y^2 + z^2}} \cdot 2z = \frac{z}{\sqrt{x^2 + y^2 + z^2}}
xt=ssin(st)\frac{\partial x}{\partial t} = -s\sin(st)
yt=scos(st)\frac{\partial y}{\partial t} = s\cos(st)
zt=s2\frac{\partial z}{\partial t} = s^2
Now, substitute these into the chain rule formula:
wt=xx2+y2+z2(ssin(st))+yx2+y2+z2(scos(st))+zx2+y2+z2(s2)\frac{\partial w}{\partial t} = \frac{x}{\sqrt{x^2 + y^2 + z^2}}(-s\sin(st)) + \frac{y}{\sqrt{x^2 + y^2 + z^2}}(s\cos(st)) + \frac{z}{\sqrt{x^2 + y^2 + z^2}}(s^2)
Next, substitute x=cos(st)x = \cos(st), y=sin(st)y = \sin(st), and z=s2tz = s^2 t:
wt=cos(st)cos2(st)+sin2(st)+(s2t)2(ssin(st))+sin(st)cos2(st)+sin2(st)+(s2t)2(scos(st))+s2tcos2(st)+sin2(st)+(s2t)2(s2)\frac{\partial w}{\partial t} = \frac{\cos(st)}{\sqrt{\cos^2(st) + \sin^2(st) + (s^2 t)^2}}(-s\sin(st)) + \frac{\sin(st)}{\sqrt{\cos^2(st) + \sin^2(st) + (s^2 t)^2}}(s\cos(st)) + \frac{s^2 t}{\sqrt{\cos^2(st) + \sin^2(st) + (s^2 t)^2}}(s^2)
Since cos2(st)+sin2(st)=1\cos^2(st) + \sin^2(st) = 1, we have
wt=scos(st)sin(st)+ssin(st)cos(st)+s4t1+s4t2\frac{\partial w}{\partial t} = \frac{-s\cos(st)\sin(st) + s\sin(st)\cos(st) + s^4 t}{\sqrt{1 + s^4 t^2}}
wt=s4t1+s4t2\frac{\partial w}{\partial t} = \frac{s^4 t}{\sqrt{1 + s^4 t^2}}

3. Final Answer

wt=s4t1+s4t2\frac{\partial w}{\partial t} = \frac{s^4 t}{\sqrt{1 + s^4 t^2}}

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