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Basically, an increment is a slight change in a function. 

MathJax TeX Test Page Let w = $f(x,y)$ and let $\Delta{}x$ and $\Delta{}y$ be increments of x and y, respectively. So, the increment, $\Delta{}w$, is $$\Delta{}w = f(x + \Delta{}x, y + \Delta{}y) - f(x,y)$$

Alternate Definition

Basically, our goal is to rewrite the expression above, because that is how we will define differentiation. 

MathJax TeX Test Page Let's start by thinking about this in 2D. Instead of $w$, let's let $u$ equal a differentiable function of one variable, x, and let $u = f(x)$. Let $x_0$ be an arbitrary value of x. By our last definition, the increment of $u$ is $$\Delta{}u = f(x_0 + \Delta{}x) - f(x_0)$$ Now, let's divide through by $\Delta{}x$ and see what happens. $$\dfrac{\Delta{}u}{\Delta{}x} = \dfrac{f(x_0 + \Delta{}x) - f(x_0)}{\Delta{}x}$$ The right side of the equation is the definition of a derivative, so if we take the limit of both sides as $\Delta{}x$ approaches 0, we get $$\lim_{\Delta{}x \to 0} \dfrac{\Delta{}u}{\Delta{}x} = f'(x_0)$$ We can move it all to one side and get $$lim_{\Delta{}x \to 0} \left( \dfrac{\Delta{}u}{\Delta{}x} - f'(x_0)\right) = 0$$ So, we can define $\varepsilon$ as $\dfrac{\Delta{}u}{\Delta{}x} - f'(x_0)$, which approaches 0 as $\Delta{}x$ approaches 0. We get $$\Delta{}u = f'(x_0)\Delta{}x + \varepsilon\Delta{}x$$ If we extend that to two variables, we get $$\Delta{}w = f_x(x_0, y_0)\Delta{}x + f_y(x_0, y_0)\Delta{}y + \varepsilon_1\Delta{}x + \varepsilon_2\Delta{}y$$ Both $\varepsilon$'s approach 0 as $(\Delta{}x, \Delta{}y) \to (0,0)$

You can prove that, but it's too long to type out. Basically, you just apply the Mean Value Theorem on both x and y, and you get that.


MathJax TeX Test Page Find values of $\varepsilon_1$ and $\varepsilon_2$ that satisfy Definition (16.12). $$f(x,y) = (2x-y)^2$$ Recall that 16.12, or our definition for $$\Delta{}w = f_x(x_0,y_0)\Delta{}x + f_y(x_0, y_0)\Delta{}y + \varepsilon_1\Delta{}x + \varepsilon_2\Delta{}y = f(x + \Delta{}x, y + \Delta{}y) - f(x,y)$$ So, what we have to do is find $f(x + \Delta{}x, y + \Delta{}y) - f(x,y)$, and make sure it equals our real definition. $$f(x + \Delta{}x, y + \Delta{}y) - f(x,y) = ((2x + 2\Delta{}x) - (y + \Delta{}y))^2 - (2x - y)^2$$ $$= (2x + 2\Delta{}x)^2 + (y + \Delta{}y)^2 - 2(2x + 2\Delta{}x)(y + \Delta{}y) - 4x^2 + 4xy - y^2$$ $$ = 4x^2 + 8x\Delta{}x + 4\Delta{}x^2 + y^2 + 2y\Delta{}y + \Delta{}y^2 - 2(2xy + 2x\Delta{}y + 2y\Delta{}x + 2\Delta{}x\Delta{}y) - 4x^2 + 4xy - y^2$$ $$=8x\Delta{}x + 4\Delta{}x^2 + 2y\Delta{}y + \Delta{}y^2 - 4x\Delta{}y - 4y\Delta{}x - 4\Delta{}x\Delta{}y$$ $$f_x(x,y) = 8x - 4y, f_y(x,y) = 2y - 4x$$ Therefore, we must have terms like $$(8x - 4y)\Delta{}x, (2y - 4x)\Delta{}y$$ So, we get $$(8x-4y)\Delta{}x + (2y-4x)\Delta{}y + 4\Delta{}x^2 + \Delta{}y^2 - 4\Delta{}x\Delta{}y$$ $$(8x-4y)\Delta{}x + (2y-4x)\Delta{}y + (4\Delta{}x - 4\Delta{}y)\Delta{}x + (\Delta{}y)\Delta{}y$$ $$\varepsilon_1 = 4\Delta{}x - 4\Delta{}y$$ $$\varepsilon_2 = \Delta{}y$$ Note that how you partition the epsilons doesn't matter, as long as they approach zero as $\Delta{}x,y \to 0$, then it's fine.
David Witten

Multivariable Differentials

Tangent and Normal Components of Acceleration

Tangent and Normal Components of Acceleration