Calculus I:
functions of several variables
linear approximation
Approximation of a function of several variables

Approximation of a function
of several variables

The previous ideas can also be applied to partial derivatives.

Indeed, we have seen that a partial derivative is nothing other than the derivative of a function to a single variable.

Let's see this on an example.


Exercise 1

The score of the high jump of Cindy depends mainly on her speed v and the time t during which she maintains this speed.

Values of the function h = f (v, t) are gathered in the following table:

  t 4  12  15  20  25  30  35
v
10  2  3   2   4   3   2   3
15  3  4   5   5   6   6   6
20  6  7   7   8   9   9   9
30  7  12  16  17  17  19  20

Calculate ∂/∂t f(15, 20).

Solution:

∂/∂t f(15, 20) is the derivative of the function f_v=15(t) when t = 20.

The values of this function are shown in the third line of the table.

We will therefore apply the Proposition 1 to the function f_v=15(t) in t = 20 with dt = 5.



Recall: Proposition 1

f'(x) ≈ [f(x + δx) - f (x)]/δx

So ∂/∂t f(15, 20) ≈ f'_v=15(20) ≈ [f_v=15(25) - f_v=15(20)]/5 ≈ [(15,25) - f(15,20)]/5 = (6 - 1)/5
≈ 0.2

In the previous example we have reduced the problem to the study of a function in one variable. We went back to the case where only t "moved" and v remaining constant. However, such an approach does not solve all the problems. We will therefore generalize the formula :

f(x + δx) ≈ f'(x) δx + f(x)

Proposition 2 (Affine approximation of a function of two variables).

Let f(x, y) be a function of two variables. We have the following approximation:

f(x + δx, y + δy) ≈ f(x,y) +
(∂f(x,y)/∂x) δx + (∂f(x,y)/∂y) δy

Note that the more dx and dy become smaller, the more approximation becomes better.


Exercise 2

Let f (x, y) = 3x2 - xy - y2 .

Calculate without calculator an approximate value of f(1.01; 2.98).

Solution: We will use Proposition 2 with x = 1, dx = 0.01, y = 3 and dy = - 0.02.

∂f(x,y)/∂x = 6x - 6 → ∂f(1,3)/∂x = 3

∂f(x,y)/∂y = -x - 2y → ∂f(1,3)/∂y = - 7

We get then:

f (1.01; 2.98) ≈ f(1;3) + (∂f(1,3)/∂x)(0.01) +
(∂f(1,3)/∂y)(- 0.02)
≈ - 9 + 3 x (0.01) - 7 x (- 0.02)
≈ - 8.87

Note: The exact value of f (1.01; 2.98) is - 8.8299