- #1
twoflower
- 368
- 0
Hi,
I started computing excercises on total differential and I would like to know if I'm doing it correctly. Could you please check it? Here it is:
Does the function
[tex]
f(x,y) = \sqrt[3]{x^3+y^3}
[/tex]
have total differential in [0,0]?
First I computed partial derivatives:
[tex]
\frac{\partial f}{\partial x} = \frac{x^3}{\sqrt[3]{(x^3 + y^3)^2}}
[/tex]
[tex]
\frac{\partial f}{\partial y} = \frac{y^3}{\sqrt[3]{(x^3 + y^3)^2}}
[/tex]
I see that partial derivatives are continuous everywhere with the exception of the point [0,0].
For the point [0,0] I have to compute partial derivatives from definition using the limit:
[tex]
\frac{\partial f}{\partial x}(0,0) = \lim_{t \rightarrow 0} \frac{f(t,0) - f(0,0)}{t} = \lim_{t \rightarrow 0} \frac{t}{t} = 1
[/tex]
[tex]
\frac{\partial f}{\partial y}(0,0) = \lim_{t \rightarrow 0} \frac{f(0,t) - f(0,0)}{t} = \lim_{t \rightarrow 0} \frac{t}{t} = 1
[/tex]
So in the case that total differential in the point [0,0] exists, it must be of form:
[tex]
L(h) = \frac{\partial f}{\partial x}(0,0) h_1 + \frac{\partial f}{\partial y}(0,0) h_2 = h_1 + h_2
[/tex]
for any
[tex]
h = (h_1, h_2) \in \mathbb{R}^2
[/tex]
and must satisfy the limit
[tex]
\lim_{||h|| \rightarrow 0} \frac{f(0,0) + h) - f(0,0) - L(h)}{||h||} = 0
[/tex]
I can write it this way:
[tex]
\lim_{[h_1,h_2] \rightarrow [0,0]} \frac{\sqrt[3]{h_1^3 + h_2^3} - 0 - h_1 - h_2}{\sqrt{h_1^2 + h_2^2}}
[/tex]
When I put
[tex]
h_2 = kh_1
[/tex]
I can write
[tex]
\lim_{h_1 \rightarrow 0} \frac{ \sqrt[3]{h_1^3 + k^3h_1^3} - h_1kh_1}{\sqrt{h_1^2 + k^2h_1^2}} = \frac{\sqrt[3]{1+k^3} - 1 - k}{\sqrt{1 + k^2}} \neq 0
[/tex]
And thus I say that f doesn't have total differential in [0,0].
Is this correct approach?
Thank you for checking this out.
I started computing excercises on total differential and I would like to know if I'm doing it correctly. Could you please check it? Here it is:
Does the function
[tex]
f(x,y) = \sqrt[3]{x^3+y^3}
[/tex]
have total differential in [0,0]?
First I computed partial derivatives:
[tex]
\frac{\partial f}{\partial x} = \frac{x^3}{\sqrt[3]{(x^3 + y^3)^2}}
[/tex]
[tex]
\frac{\partial f}{\partial y} = \frac{y^3}{\sqrt[3]{(x^3 + y^3)^2}}
[/tex]
I see that partial derivatives are continuous everywhere with the exception of the point [0,0].
For the point [0,0] I have to compute partial derivatives from definition using the limit:
[tex]
\frac{\partial f}{\partial x}(0,0) = \lim_{t \rightarrow 0} \frac{f(t,0) - f(0,0)}{t} = \lim_{t \rightarrow 0} \frac{t}{t} = 1
[/tex]
[tex]
\frac{\partial f}{\partial y}(0,0) = \lim_{t \rightarrow 0} \frac{f(0,t) - f(0,0)}{t} = \lim_{t \rightarrow 0} \frac{t}{t} = 1
[/tex]
So in the case that total differential in the point [0,0] exists, it must be of form:
[tex]
L(h) = \frac{\partial f}{\partial x}(0,0) h_1 + \frac{\partial f}{\partial y}(0,0) h_2 = h_1 + h_2
[/tex]
for any
[tex]
h = (h_1, h_2) \in \mathbb{R}^2
[/tex]
and must satisfy the limit
[tex]
\lim_{||h|| \rightarrow 0} \frac{f(0,0) + h) - f(0,0) - L(h)}{||h||} = 0
[/tex]
I can write it this way:
[tex]
\lim_{[h_1,h_2] \rightarrow [0,0]} \frac{\sqrt[3]{h_1^3 + h_2^3} - 0 - h_1 - h_2}{\sqrt{h_1^2 + h_2^2}}
[/tex]
When I put
[tex]
h_2 = kh_1
[/tex]
I can write
[tex]
\lim_{h_1 \rightarrow 0} \frac{ \sqrt[3]{h_1^3 + k^3h_1^3} - h_1kh_1}{\sqrt{h_1^2 + k^2h_1^2}} = \frac{\sqrt[3]{1+k^3} - 1 - k}{\sqrt{1 + k^2}} \neq 0
[/tex]
And thus I say that f doesn't have total differential in [0,0].
Is this correct approach?
Thank you for checking this out.