SENSEI AI
About App

The problem states that if $w = f(r-s, s-t, t-r)$, then we need to show that $\frac{\partial w}{\partial r} + \frac{\partial w}{\partial s} + \frac{\partial w}{\partial t} = 0$.

AnalysisPartial DerivativesChain RuleMultivariable Calculus
2025/5/10

1. Problem Description

The problem states that if w=f(r−s,s−t,t−r)w = f(r-s, s-t, t-r), then we need to show that ∂w∂r+∂w∂s+∂w∂t=0\frac{\partial w}{\partial r} + \frac{\partial w}{\partial s} + \frac{\partial w}{\partial t} = 0.

2. Solution Steps

Let u=r−su = r-s, v=s−tv = s-t, and p=t−rp = t-r. Then w=f(u,v,p)w = f(u, v, p).
We will use the chain rule to find the partial derivatives of ww with respect to rr, ss, and tt.
∂w∂r=∂f∂u∂u∂r+∂f∂v∂v∂r+∂f∂p∂p∂r\frac{\partial w}{\partial r} = \frac{\partial f}{\partial u} \frac{\partial u}{\partial r} + \frac{\partial f}{\partial v} \frac{\partial v}{\partial r} + \frac{\partial f}{\partial p} \frac{\partial p}{\partial r}
∂u∂r=∂(r−s)∂r=1\frac{\partial u}{\partial r} = \frac{\partial (r-s)}{\partial r} = 1
∂v∂r=∂(s−t)∂r=0\frac{\partial v}{\partial r} = \frac{\partial (s-t)}{\partial r} = 0
∂p∂r=∂(t−r)∂r=−1\frac{\partial p}{\partial r} = \frac{\partial (t-r)}{\partial r} = -1
So, ∂w∂r=∂f∂u(1)+∂f∂v(0)+∂f∂p(−1)=∂f∂u−∂f∂p\frac{\partial w}{\partial r} = \frac{\partial f}{\partial u}(1) + \frac{\partial f}{\partial v}(0) + \frac{\partial f}{\partial p}(-1) = \frac{\partial f}{\partial u} - \frac{\partial f}{\partial p}.
∂w∂s=∂f∂u∂u∂s+∂f∂v∂v∂s+∂f∂p∂p∂s\frac{\partial w}{\partial s} = \frac{\partial f}{\partial u} \frac{\partial u}{\partial s} + \frac{\partial f}{\partial v} \frac{\partial v}{\partial s} + \frac{\partial f}{\partial p} \frac{\partial p}{\partial s}
∂u∂s=∂(r−s)∂s=−1\frac{\partial u}{\partial s} = \frac{\partial (r-s)}{\partial s} = -1
∂v∂s=∂(s−t)∂s=1\frac{\partial v}{\partial s} = \frac{\partial (s-t)}{\partial s} = 1
∂p∂s=∂(t−r)∂s=0\frac{\partial p}{\partial s} = \frac{\partial (t-r)}{\partial s} = 0
So, ∂w∂s=∂f∂u(−1)+∂f∂v(1)+∂f∂p(0)=−∂f∂u+∂f∂v\frac{\partial w}{\partial s} = \frac{\partial f}{\partial u}(-1) + \frac{\partial f}{\partial v}(1) + \frac{\partial f}{\partial p}(0) = -\frac{\partial f}{\partial u} + \frac{\partial f}{\partial v}.
∂w∂t=∂f∂u∂u∂t+∂f∂v∂v∂t+∂f∂p∂p∂t\frac{\partial w}{\partial t} = \frac{\partial f}{\partial u} \frac{\partial u}{\partial t} + \frac{\partial f}{\partial v} \frac{\partial v}{\partial t} + \frac{\partial f}{\partial p} \frac{\partial p}{\partial t}
∂u∂t=∂(r−s)∂t=0\frac{\partial u}{\partial t} = \frac{\partial (r-s)}{\partial t} = 0
∂v∂t=∂(s−t)∂t=−1\frac{\partial v}{\partial t} = \frac{\partial (s-t)}{\partial t} = -1
∂p∂t=∂(t−r)∂t=1\frac{\partial p}{\partial t} = \frac{\partial (t-r)}{\partial t} = 1
So, ∂w∂t=∂f∂u(0)+∂f∂v(−1)+∂f∂p(1)=−∂f∂v+∂f∂p\frac{\partial w}{\partial t} = \frac{\partial f}{\partial u}(0) + \frac{\partial f}{\partial v}(-1) + \frac{\partial f}{\partial p}(1) = -\frac{\partial f}{\partial v} + \frac{\partial f}{\partial p}.
Now, we add the partial derivatives:
∂w∂r+∂w∂s+∂w∂t=(∂f∂u−∂f∂p)+(−∂f∂u+∂f∂v)+(−∂f∂v+∂f∂p)=∂f∂u−∂f∂p−∂f∂u+∂f∂v−∂f∂v+∂f∂p=0\frac{\partial w}{\partial r} + \frac{\partial w}{\partial s} + \frac{\partial w}{\partial t} = (\frac{\partial f}{\partial u} - \frac{\partial f}{\partial p}) + (-\frac{\partial f}{\partial u} + \frac{\partial f}{\partial v}) + (-\frac{\partial f}{\partial v} + \frac{\partial f}{\partial p}) = \frac{\partial f}{\partial u} - \frac{\partial f}{\partial p} - \frac{\partial f}{\partial u} + \frac{\partial f}{\partial v} - \frac{\partial f}{\partial v} + \frac{\partial f}{\partial p} = 0.

3. Final Answer

∂w∂r+∂w∂s+∂w∂t=0\frac{\partial w}{\partial r} + \frac{\partial w}{\partial s} + \frac{\partial w}{\partial t} = 0

Related problems in "Analysis"

We are asked to evaluate the infinite sum $\sum_{k=2}^{\infty} (\frac{1}{k} - \frac{1}{k-1})$.

Infinite SeriesTelescoping SumLimits
2025/6/7

The problem consists of two parts. First, we are asked to evaluate the integral $\int_0^{\pi/2} x^2 ...

IntegrationIntegration by PartsDefinite IntegralsTrigonometric Functions
2025/6/7

The problem asks us to find the derivatives of six different functions.

CalculusDifferentiationProduct RuleQuotient RuleChain RuleTrigonometric Functions
2025/6/7

The problem states that $f(x) = \ln(x+1)$. We are asked to find some information about the function....

CalculusDerivativesChain RuleLogarithmic Function
2025/6/7

The problem asks us to evaluate two limits. The first limit is $\lim_{x\to 0} \frac{\sqrt{x+1} + \sq...

LimitsCalculusL'Hopital's RuleTrigonometry
2025/6/7

We need to find the limit of the expression $\sqrt{3x^2+7x+1}-\sqrt{3}x$ as $x$ approaches infinity.

LimitsCalculusIndeterminate FormsRationalization
2025/6/7

We are asked to find the limit of the expression $\sqrt{3x^2 + 7x + 1} - \sqrt{3}x$ as $x$ approache...

LimitsCalculusRationalizationAsymptotic Analysis
2025/6/7

The problem asks to evaluate the definite integral: $J = \int_0^{\frac{\pi}{2}} \cos(x) \sin^4(x) \,...

Definite IntegralIntegrationSubstitution
2025/6/7

We need to evaluate the definite integral $J = \int_{0}^{\frac{\pi}{2}} \cos x \sin^4 x \, dx$.

Definite IntegralIntegration by SubstitutionTrigonometric Functions
2025/6/7

We need to evaluate the definite integral: $I = \int (\frac{1}{x} + \frac{4}{x^2} - \frac{5}{\sin^2 ...

Definite IntegralsIntegrationTrigonometric Functions
2025/6/7