Mathematics
functions of
several variables
functions of
several variables
Partial derivatives
Differential
Linear approximation
Error calculation
Extrema of a function
© The scientific sentence. 2010
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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 fv=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 fv=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)
≈ [fv=15(25) - fv=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
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