Chain rule for partials: if V = f(eta) then (dV/dy) = (df/d eta)/x^{1/2} and (dV/dx) = - (eta/2x)...

Question:

Chain rule for partials: if {eq}V = f(\eta) {/eq} then {eq}\frac {\partial V}{\partial y} = \frac {df/d \eta}{x^{1/2}} {/eq} and {eq}\frac{\partial V}{\partial x}= - \frac \eta{2x} \frac{\mathrm{d} f}{\mathrm{d} \eta} {/eq}.

Given: {eq}\frac{C_a}{C_S} = 1 - erf (\beta) {/eq}, where {eq}C_S {/eq} = constant, {eq}\beta \equiv \frac y{(4D_{AB}x/U_S)^{1/2}} {/eq}; and {eq}erf(\beta) \equiv \frac 2{\pi^{1/2}} \int^\beta \exp(-t^2) dt {/eq}.

Show that at y = 0 : {eq}\frac{dC_A}{dy} = -\frac{C_S}{(\pi D_{AB} x/U_S)^{1/2}} {/eq}

The Fundamental Theorem of Calculus; Derivatives:

We'll use the identity below by means of the Fundamental Theorem of Calculus and the Chain Rule:

For any continuous function {eq}f, {/eq} and any differentiable function {eq}g, {/eq} we have:

{eq}\begin{align*} \frac{d\left(\int_{0}^{\beta(y)}f(t)\,dt\right)}{dy}****=f(\beta(y))\cdot\frac{d(\beta(y))}{dy}. \end{align*} {/eq}

Answer and Explanation:

We want to find {eq}\frac{dC_A}{dy} {/eq} at {eq}y=0. {/eq}

We have {eq}\frac{C_A}{C_S} = 1 - erf (\beta) {/eq}, with

{eq}C_S=\text{a constant} {/eq}, or equivalently,:

{eq}C_A=C_S\left(1 - erf (\beta)\right). {/eq}

{eq}\begin{align*} \frac{dC_A}{dy}&=\frac{d\left( C_S\left(1 - erf (\beta)\right) \right)}{dy}\\ &=C_S\left(\frac{d\left( 1 - erf (\beta)\right)}{dy} \right)\\ &=-C_S\left(\frac{d\left( \frac 2{\pi^{1/2}} \int^\beta \exp(-t^2) dt\right)}{dy} \right)\\ &=\frac {-2C_S}{\pi^{1/2}}\left(\frac{d\left( \int^\beta \exp(-t^2) dt\right)}{dy} \right),\text{ using the fundamental theorem of calculus}\\ &=\frac {-2C_S}{\pi^{1/2}}\cdot \exp(-\beta^2) \left(\frac{d\left( \beta \right)}{dy} \right),\text{ using that }\beta=\frac{ y}{(4D_{AB}x/U_S)^{1/2}}\\ &=\frac {-2C_S}{\pi^{1/2}}\cdot \exp\left (-\left(\frac{ y}{(4D_{AB}x/U_S)^{1/2}}\right)^2\right ) \left(\frac{d\left( \frac{ y}{(4D_{AB}x/U_S)^{1/2}} \right)}{dy} \right),\text{ taking }D_{AB},\;U_S\text{ as constants}\\ &=\frac {-2C_S}{\pi^{1/2}}\cdot \exp\left (-\left(\frac{ y}{(4D_{AB}x/U_S)^{1/2}}\right)^2\right ) \left( \frac{ 1}{(4D_{AB}x/U_S)^{1/2}} \right). \end{align*} {/eq}

Thus, evaluating at {eq}y=0 {/eq} we get:

{eq}\begin{align*} \left.\frac{dC_A}{dy}\right |_{y=0}&=\left( \frac {-2C_S}{\pi^{1/2}}\cdot \exp\left (-\left(\frac{ y}{(4D_{AB}x/U_S)^{1/2}}\right)^2\right ) \left( \frac{ 1}{(4D_{AB}x/U_S)^{1/2}} \right)\right |_{y=0}\\ &=\left( \frac {-2C_S}{\pi^{1/2}}\cdot \exp\left (-0\right)^2\right ) \left( \frac{ 1}{(4D_{AB}x/U_S)^{1/2}} \right)\\ &= \frac {-2C_S}{\pi^{1/2}(4D_{AB}x/U_S)^{1/2}}=\frac {-2C_S}{2\pi^{1/2}(D_{AB}x/U_S)^{1/2}} \\ &=- \frac {C_S}{\pi^{1/2}(D_{AB}x/U_S)^{1/2}} \\ &=- \frac {C_S}{(\pi\,D_{AB}x/U_S)^{1/2}} , \end{align*} {/eq}

thus

{eq}\boxed{\left .\frac{dC_A}{dy}\right |_{y=0}=- \frac {C_S}{(\pi\,D_{AB}x/U_S)^{1/2}}} {/eq}

as desired.


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The Fundamental Theorem of Calculus

from Math 104: Calculus

Chapter 12 / Lesson 10
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