twoflower
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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
<br /> f(x,y) = \sqrt[3]{x^3+y^3}<br />
have total differential in [0,0]?
First I computed partial derivatives:
<br /> \frac{\partial f}{\partial x} = \frac{x^3}{\sqrt[3]{(x^3 + y^3)^2}}<br />
<br /> \frac{\partial f}{\partial y} = \frac{y^3}{\sqrt[3]{(x^3 + y^3)^2}}<br />
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:
<br /> \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<br />
<br /> \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<br />
So in the case that total differential in the point [0,0] exists, it must be of form:
<br /> L(h) = \frac{\partial f}{\partial x}(0,0) h_1 + \frac{\partial f}{\partial y}(0,0) h_2 = h_1 + h_2<br />
for any
<br /> h = (h_1, h_2) \in \mathbb{R}^2<br />
and must satisfy the limit
<br /> \lim_{||h|| \rightarrow 0} \frac{f(0,0) + h) - f(0,0) - L(h)}{||h||} = 0<br />
I can write it this way:
<br /> \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}}<br />
When I put
<br /> h_2 = kh_1<br />
I can write
<br /> \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<br />
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
<br /> f(x,y) = \sqrt[3]{x^3+y^3}<br />
have total differential in [0,0]?
First I computed partial derivatives:
<br /> \frac{\partial f}{\partial x} = \frac{x^3}{\sqrt[3]{(x^3 + y^3)^2}}<br />
<br /> \frac{\partial f}{\partial y} = \frac{y^3}{\sqrt[3]{(x^3 + y^3)^2}}<br />
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:
<br /> \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<br />
<br /> \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<br />
So in the case that total differential in the point [0,0] exists, it must be of form:
<br /> L(h) = \frac{\partial f}{\partial x}(0,0) h_1 + \frac{\partial f}{\partial y}(0,0) h_2 = h_1 + h_2<br />
for any
<br /> h = (h_1, h_2) \in \mathbb{R}^2<br />
and must satisfy the limit
<br /> \lim_{||h|| \rightarrow 0} \frac{f(0,0) + h) - f(0,0) - L(h)}{||h||} = 0<br />
I can write it this way:
<br /> \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}}<br />
When I put
<br /> h_2 = kh_1<br />
I can write
<br /> \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<br />
And thus I say that f doesn't have total differential in [0,0].
Is this correct approach?
Thank you for checking this out.