Calculus-Sec-13-3-Solution

Published 다른 공개된 워크시트들...

13.3 Partial Derivatives by SGLee - HSKim, VLang, DYKim

http://youtu.be/rSYLp1mSMXY

1. Find partial derivatives with respect to and for the function .

2. Find partial derivatives with respect to and for the function .

3. Find all second order partial derivatives of the function .

4. Find all second order partial derivatives of the function .

5. Take an example of the solution for the .

http://matrix.skku.ac.kr/cal-lab/cal-12-2-6.html

6-8. Laplace’s equation A classical equation of mathematics is Laplace’s equation,

which arises in both theory and applications. It governs ideal fluid flow, electrostatic potentials,

and the steady-state distribution of heat in a conducting medium.

In two dimensions, Laplace’s equation is

Show that the following functions are harmonic; that is, they satisfy Laplace’s equation.

7. for any real number .

9. The volume of a right circular cone of radius and height is .

Show that if the height remains constant while the radius changes the volume satisfies .

10. Show that the function deﬁned by is not continuous at but its first order partial derivatives exist at .

참고 ( http://calculus.subwiki.org/wiki/Existence_of_partial_derivatives_not_implies_differentiable)

By the definition of partial derivatives, we have

$f_x(x,y) = \frac{-4*x^2*y}{(2*x^2 + 3*y^2)^2}+\frac{y}{(2*x^2 + 3*y^2)}$ , $f_x(0,0)=\lim_{h->0}{\frac{f(0+h,0)-f(0,0)}{h}}=0$

$f_y(x,y) = \frac{-6*x*y^2}{(2*x^2 + 3*y^2)^2}+\frac{x}{(2*x^2 + 3*y^2)}$ , $f_y(0,0)=\lim_{h->0}{\frac{f(0,0+h)-f(0,0)}{h}}=0$

However, f(x,y) is not continuos at (0,0), that is,

$\lim_{(x,y)->(0,0)}{f(x,y)}\neq f(0,0)=0$

Let x->0, and y->0 along the line y=mx, so the limit of f(x,y) does not exist.

For m=0, f(x,y)->0 and for m=1, f(x,y) -> 1/5

$f(x,mx) = \frac{mx^2}{2x^2+3m^2x^2}$

Show that the function deﬁned by is continuous at but its first order partial derivatives do not exit at .

Since the definition of partial derivatives, we have

$f_x(x,y) = \frac{2x}{|x| + |y|} - \frac{(x^2 + 2y^2)x}{(|x| + |y|)^2|x|}$ , $f_x(0,0)=\lim_{h->0}{\frac{f(0+h,0)-f(0,0)}{h}}$ (undefined)

$f_y(x,y) = \frac{4y}{|x| + |y|} - \frac{(x^2 + 2y^2)y}{(|x| + |y|)^2|y|}$ , $f_y(0,0)=\lim_{h->0}{\frac{f(0,0+h)-f(0,0)}{h}}$

However, f(x,y) is continuos at (0,0), that is,

$\lim_{(x,y)->(0,0)}{f(x,y)} = f(0,0) = 0$

12. Let us consider the function . Show that at .

$f_x(x,y) = \frac{2*x}{|x| + |y|}- \frac{(x^2 + 2*y^2)*x}{((|x| + |y|)^2*|x|}$ , $f_{yx}(x,y) = \lim_{h->0}{\frac{f(h+0,0)-f(0,0)}{h}}$

$f_y(x,y) = \frac{4*y}{|x| + |y|}- \frac{(x^2 + 2*y^2)*y}{((|x| + |y|)^2*|y|}$ , $f_{xy}(x,y) = \lim_{h->0}{\frac{f(0,0+h)-f(0,0)}{h}}$ (undefined)

So, at .

3. Let us consider the function . Show that at .

$f_x(x,y) = \frac{-2(x^2 - 2y^2)x^2y}{(x^2 + y^2)^2} + \frac{2x^2y}{x^2 + y^2} + \frac{(x^2 -2y^2)y}{x^2 + y^2}$ , $f_{yx}(x,y) = \lim_{h->0}{\frac{f(h+0,0)-f(0,0)}{h}}$

$f_y(x,y) = \frac{-2(x^2 - 2y^2)xy^2}{(x^2 + y^2)^2} + \frac{-4xy^2}{x^2 + y^2} + \frac{(x^2 -2y^2)x}{x^2 + y^2}$ , $f_{xy}(x,y) = \lim_{h->0}{\frac{f(0,0+h)-f(0,0)}{h}}$ (undefined)

14. Let . Show that .

Back to Part II