6.8 Answer

6.8

1.

(a) Let $f(x,y) = 2x^2 + 5xy - 3y^2 - 1 = 0$. Then find the total differential of $f(x,y)$.

$$df = f_{x}dx + f_{y}dy = (4x + 5y)dx + (5x - 6y)dy = 0$$

Thus，

$$\frac{dy}{dx} = -\frac{4x + 5y}{5x - 6y}.$$

$$\frac{d^2 y}{dx^2} = -\frac{(4 + 5\frac{dy}{dx})(5x - 6y) - (4x + 5y)(5 - 6\frac{dy}{dx})}{(5x - 6y)^2}$$

$$= -\frac{4(5x-6y)^2 - 10(4x+5y)(5x-6y) - 6(4x+5y)^2}{(5x - 6y)^3}$$

(b)

Let $f(x,y) = y - e^{x+y} = 0$. Then find the total differential of $f(x,y)$.

$$df = f_{x}dx + f_{y}dy = -e^{x+y}dx + (1 - e^{x+y})dy = 0$$

Then，

$$\frac{dy}{dx} = -\frac{e^{x+y}}{1 - e^{x+y}}.$$

Next we find $\frac{d^2 y}{dx^2}$. Here we use

$$\frac{d^2 y}{dx^2} = - \frac{f_{xx}f_{y}^2 - 2f_{xy}f_{x}f_{y} + f_{xx}f_{y}^2}{f_{y}^{3}}$$

$$f_{xx} = -e^{x+y}, f_{xy} = -e^{x+y}, f_{yy} = -e^{x+y}$$

Then

$$\frac{d^2 y}{dx^2} = -\frac{-e^{x+y}(1-e^{x+y})^2 + 2e^{x+y}(-e^{x+y})(1 - e^{x+y}) - e^{x+y}e^{2(x+y)}}{(1 - e^{x+y})^3}$$

$$= -\frac{-e^{x+y} + 2e^{2(x+y)} - e^{3(x+y)} - 2e^{2(x+y)} + 2e^{3(x+y)} - e^{3(x+y)}}{(1 - e^{x+y})^3}$$

$$= \frac{e^{x+y}}{(1 - e^{x+y})^{3}}$$

(c)

Let $f(x,y) = x^2 - y^2 - xy= 0$. Then find the total differntial of $f(x,y)$.

$$df = f_{x}dx + f_{y}dy = (2x-y)dx + (-2y-x)dy = 0$$

これよりThus,

$$\frac{dy}{dx} = \frac{2x - y}{2y+x}.$$

Next we find $\frac{d^2 y}{dx^2}$. We use

$$\frac{d^2 y}{dx^2} = - \frac{f_{xx}f_{y}^2 - 2f_{xy}f_{x}f_{y} + f_{xx}f_{y}^2}{f_{y}^{3}}$$

$$f_{xx} = 2, f_{xy} = -1, f_{yy} = -2$$

Then

$$\frac{d^2 y}{dx^2} = -\frac{2(-2y-x)^2 + 2(2x-y)(-2y-x) -2(2x-y)^2}{(-2y-x)^3}$$

$$= \frac{2(x+2y)^2 -2(2x-y)(x+2y) -2(2x-y)^2}{(x+2y)^3}$$

(d)

Let $f(x,y) = \log{\sqrt{x^2 + y^2}} - \tan^{-1}{\frac{y}{x}} = 0$, Then find the total differential of $f(x,y)$.

$$f_{x} = \frac{1}{2}(\frac{2x}{x^2 + y^2}) - \frac{-\frac{y}{x^2}}{1 + (\frac{y}{x})^2}$$

$$= \frac{x}{x^2 + y^2} + \frac{y}{x^2 + y^2} = \frac{x+y}{x^2 + y^2}$$

$$f_{y} = \frac{1}{2}(\frac{2y}{x^2 + y^2} - \frac{\frac{1}{x}}{1 + (\frac{y}{x})^2}$$

$$= \frac{y}{x^2 + y^2} - \frac{x}{x^2 + y^2} = \frac{y - x}{x^2 + y^2}$$

より

$$\frac{dy}{dx} = - \frac{f_{x}}{f_{y}} = - \frac{x+y}{y-x} = \frac{x+y}{x-y}$$

Next we find $\frac{d^2 y}{dx^2}$．

$$\frac{d^2 y}{dx^2} = \frac{(1 + \frac{dy}{dx})(x-y) - (x+y)(1 - \frac{dy}{dx})}{(x-y)^2}$$

$$= \frac{(x - y + x + y)(x-y) - (x-y)(x-y-x-y) }{(x - y)^3}$$

$$= \frac{2x^2 - 2xy + 2xy + 2y^2}{(x-y)^3}$$

$$= \frac{2(x^2 + y^2)}{(x-y)^3}$$

2.

(a) Let $f(x,y,z) = x^2 + y^2 + z^2 - 4 = 0, g(x,y) = x^2 + y^2 - 4x = 0$. Then find the total differential $df$.

$$df = f_{x}dx + f_{y}dy + f_{z}dz = 2xdx + 2ydy + 2zdz = 0$$ (1.5)

$$dg = g_{x}dx + g_{y}dy = (2x -4)dx + 2y dy = 0$$ (1.6)

式(1.6)より，

$$\frac{dy}{dx} = \frac{4-2x}{2y} = \frac{2-x}{y}$$

Next we have

$$4dx + 2zdz = 0$$

Then

$$\frac{dz}{dx} = -\frac{2}{z}.$$

(b) Let $f(x,y,z) = xyz -1 = 0, g(x,y,z) = xy + yz + zx - 1 = 0$. Then find the total differential.

$$df = f_{x}dx + f_{y}dy + f_{z}dz = yzdx + xzdy + xydz = 0$$ (1.7)

$$dg = g_{x}dx + g_{y}dy + g_{z}dz = (y + z)dx + (x + z) dy + (y + x)dz = 0$$ (1.8)

Now

$$((y+x)yz - xy(y+z))dx + ((y+x)xz - xy(x+z))dy = 0$$

Thus

$$\frac{dy}{dx} = -\frac{y^2 z - xy^2}{x^2 z - x^2 y} = \frac{y^2 (x-z)}{x^2 (z-y)}.$$

Also, using equation (1.7) and the equation (1.8), we eliminate $dy$. Then

$$((x+z)yz - xz(y+z))dx + ((x+z)xy - xz(y+x))dz = 0$$

$$z^2(y-x)dx + x^2(y-z)dz = 0$$

よって

$$\frac{dz}{dx} = \frac{z^2 (x-y)}{x^2 (y-z)}.$$

3.

From the equation $2x^2 + 5y^2 = 12$, we find $\frac{dy}{dx}$. Then

$$\frac{dy}{dx} = -\frac{f_{x}}{f_{y}} = -\frac{4x}{10y} = -\frac{2x}{5y}$$

Thus, the slope of the tangent line is

$$\frac{dy}{dx}\mid_{(1,\sqrt{2})} = -\frac{2}{5\sqrt{2}} = -\frac{\sqrt{2}}{5}$$

Thus the equation of the tangent line goes through a point $(1,\sqrt{2})$ is

$$y - \sqrt{2} = -\frac{\sqrt{2}}{5}(x - 1).$$

Also, the tangent line and the normal line are perpendicular，. Thus the slope of the normal line is $-\frac{5\sqrt{2}}{2}$．Thus the equation of the normal line goew through $(1,\sqrt{2})$ is

$$y - \sqrt{2} = -\frac{5\sqrt{2}}{2}(x-1).$$

Let $f(x,y,z) = \tan^{-1}{\frac{y}{x}} - z = 0$. Then

$$\nabla f(x,y,z) = (\frac{\frac{-y}{x}}{1 + (\frac{y}{x})^2}, \fra... ...x}}{1 + (\frac{y}{x})^2}, -1) = (\frac{-y}{x^2 + y^2}, \frac{x}{x^2 + y^2}, -1)$$

Thus,

$$\nabla f(1,1,\frac{\pi}{2}) = (-\frac{1}{2}, \frac{1}{2}, -1)$$

Note that $\nabla f$ is orthogonal to the surface $f(x,y,z) = 0$. Now take any point $(x,y,z)$ on the tangent plane $\Gamma$. Then the vectors $(x - 1, y - 1, z - \frac{\pi}{2})$ and $\nabla f(1,1,\frac{\pi}{2})$ are orthogonal．Thus, the equation of the tangent plane is

$$(x - 1, y - 1, z - \frac{\pi}{2}) \cdot (-\frac{1}{2}, \frac{1}{2}, -1) = 0$$

or，

$$-\frac{1}{2}(x-1) + \frac{1}{2}(y-1) - (z - \frac{\pi}{2}) = 0$$

Note that the direction of the normal line is the same as the direction of $\nabla f$. Thus for any point $(x,y,z)$ on the normal line, we have，

$$(x - 1, y - 1, z - \frac{\pi}{2}) = t(-\frac{1}{2}, \frac{1}{2}, -1)$$

or，

$$t = \frac{x-1}{-\frac{1}{2}} = \frac{y-1}{\frac{1}{2}} = \frac{z - \frac{\pi}{2}}{-1}$$