M408M Learning Module Pages

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Chapter 10: Parametric Equations and Polar Coordinates

Chapter 12: Vectors and the Geometry of Space


Chapter 13: Vector Functions


Chapter 14: Partial Derivatives


Learning module LM 14.1: Functions of 2 or 3 variables:

Learning module LM 14.3: Partial derivatives:

Learning module LM 14.4: Tangent planes and linear approximations:

      Tangent planes
      Linearization
      Quadratic approximations and concavity

Learning module LM 14.5: Differentiability and the chain rule:

Learning module LM 14.6: Gradients and directional derivatives:

Learning module LM 14.7: Local maxima and minima:

Learning module LM 14.8: Absolute maxima and Lagrange multipliers:

Chapter 15: Multiple Integrals



Linearization

Linearization

Partial derivatives allow us to approximate functions just like ordinary derivatives do, only with a contribution from each variable. In one dimensional calculus we tracked the tangent line to get a linearization of a function. With functions of several variables we track the tangent plane. Since the equation of the tangent plane at $(a,\, b,\, f(a, \,b))$ is $$z\ = \ f(a, b) + (x-a)f_x(a, b) + (y-b)f_y(a, b)\,,$$

The Linearization of a function $f(x,y)$ at $(a,\, b)$ is $$L(x, y) \ = \ f(a, b) + (x-a) f_x(a, b) + (y-b) f_y(a, b)\,.$$

This is very similar to the familiar formula $L(x)= f(a) + f'(a) (x-a)$ functions of one variable, only with an extra term for the second variable. The corresponding formulas for functions of more than two variables are similar, with one term for each variable.

Now let's use the Linearization $L(x,\, y)$ of $f$ to estimate the change $\Delta f$ in $f$ near $(a,\, b)$: $$\Delta f \ = \ f(x, y) - f(a, b)\quad \approx \quad L(x, y) - f(a, b) \quad = \quad f_x(a,b) \Delta x + f_y(a,b) \Delta y\,.$$ This is sometimes written using differentials as $$ df \ = f_x(a, b) dx + f_y(a, b) dy\,.$$ The function $L(x,\,y)$ is also called the Linear Approximation to $f$ at $(a,\,b)$.

  Example: Find the Linearization, $L(x,\, y)$, of $$z\ = \ f(x,\, y) \ = \ y\sqrt{x}$$ at the point $(9,\, -2)$.
Solution: the Linearization of $f$ at $(a,\,b)$ is $$L(x,\,y)= f(a,\,b) + f_x \Bigl|_{(a,\,b)} (x-a) + f_y \Bigl|_{(a,\,b)} (y-b).$$ But when $f(x,\, y) \ = \ y\sqrt{x}$, $$ \frac{\partial f}{\partial x} \ = \ \frac{y}{2\sqrt{x}}\,, \quad \frac{\partial f}{\partial y} \ = \ \sqrt{x}\,.$$
At $(9,\,-2)$, therefore, $f(9,\,-2) = -6$, while $$f_x\Bigl|_{(9,\,-2)} \ = \ -\frac{1}{3}\,, \qquad f_y\Bigl|_{(9,\, -2)} \ = \ 3\,.$$ Thus $$L(x,\,y) \ = \ -6 -\frac{1}{3}(x-9) +3 (y +2)\,,$$ which after rearrangement becomes $$L(x,\,y) \ = \ 3 -\frac{1}{3} x + 3y \, .$$

Linearization, differentials and higher-order approximations are explained in the following video: