Solutions to Homework 6

13.7

#7

z_{x} = \frac{x}{\sqrt{x^{2} + y^{2}}} , z_{y} = \frac{y}{\sqrt{x^{2} + y^{2}}}

z_{xx} = \frac{\sqrt{x^{2} + y^{2}} − x (\frac{x}{\sqrt{x^{2}+y^{2}}} )}{x^{2} + y^{2}} = \frac{y^{2}}{(x^{2} + y^{2})^{3/2}}

z_{xy} = \frac{−xy}{(x^{2} + y^{2})^{3/2}}

z_{yx} = \frac{−xy}{(x^{2} + y^{2})^{3/2}}

z_{yy} = \frac{\sqrt{x^{2} + y^{2}} − y (\frac{y}{\sqrt{x^{2}+y^{2}}} )}{x^{2} + y^{2}} = \frac{x^{2}}{(x^{2} + y^{2})^{3/2}}

#13 z_{y} = g(x), so that z_{yy} = 0, since both f and g depend only on x, but not on y. #16 +, 0, –, 0, 0. #25 +, –, –, –, +.

13.9

#2 f_{x}(x, y) = 2 cos 2x, f_{y}(x, y) = − sin y, f_{xx}(x, y) = −4 sin 2x, f_{xy}(x, y) = 0 and f_{yy}(x, y) = − cos y. So,

T (x, y) = 1 + 2x and P (x, y) = 1 + 2x − \frac{1}{2}y^{2}.

#4 Compute the partials and evaluate: z_{x} = \frac{1}{y} cos(x/y), z_{y} = \frac{−x}{y^{2}} cos(x/y), z_{xx} =

\frac{−1}{y^{2}} sin(x/y), z_{xy} = \frac{x}{y^{3}} sin(x/y) − \frac{1}{y^{2}} cos(x/y) and z_{yy} = \frac{−x^{2}}{y^{4}} sin(x/y) + \frac{2x}{y^{3}} cos(x/y). So,

T (x, y) = sin 1 + (cos 1)(x − 1) + (− cos 1)(y − 1) and
P (x, y) = sin 1 + (cos 1)(x − 1) + (− cos 1)(y − 1) + (− sin 1)(x − 1)^{2}
+(sin 1 − cos 1)(x − 1)(y − 1) + (− sin 1 + 2 cos 1)(y − 1)^{2}.

f(1.1, 1.1) = sin 1 ≈ 0.84147098, T (1.1, 1.1) = sin 1 and P (1.1, 1.1) = sin 1 so that all approximations are exact.

#8 z_{x} = x/\sqrt{x^{2} + y^{2}}, z_{y} = y/\sqrt{x^{2} + y^{2}}, z_{xx} = y^{2}/(x^{2} + y^{2})^{3/2}, z_{xy} = −xy/(x^{2} + y^{2})^{3/2} and z_{yy} = x^{2}/(x^{2} + y^{2})^{3/2}. So,

T (x, y) = \sqrt{2} + (1/\sqrt{2})(x − 1) + (1/\sqrt{2})(y − 1) and
P (x, y) = \sqrt{2} + (1/\sqrt{2})(x − 1) + (1/\sqrt{2})(y − 1) + (1/4\sqrt{2})(x − 1)^{2}
+(−1/2\sqrt{2})(x − 1)(y − 1) + (1/4\sqrt{2})(y − 1)^{2}.

Accidentally again we get z(1.1, 1.1) = P (1.1, 1.1) = T (1.1, 1.1) = 1.1\sqrt{2} ≈ 1.5556349.