Calculus II, Spring 2016 (http://www.math.nthu.edu.tw/ wangwc/)

Solutions to selected problems in HW for Week 10

1. Homework assignment: Problem 2.

Short summary:

• All nearby gradient vectors point towards the critical point: local maximum.

• All nearby gradient vectors point away from the critical point: local minimum.

• Some nearby gradient vectors point towards the critical point, some nearby gra-

dient vectors point away from the critical point: saddle point.

2. Section 14.10: Problem 12.

\{ \begin{matrix}\frac{∂f}{∂x} + \frac{∂f}{∂z} \frac{∂z}{∂x} + \frac{∂f}{∂w} \frac{∂w}{∂x} = 0 \\ \frac{∂g}{∂x} + \frac{∂g}{∂z} \frac{∂z}{∂x} + \frac{∂g}{∂w} \frac{∂w}{∂x} = 0 \end{matrix} ⇒ \frac{∂z}{∂x} = −\frac{f_{x}g_{w} − f_{w}g_{x}}{f_{z}g_{w} − f_{w}g_{z}}

\{ \begin{matrix}\frac{∂f}{∂y} + \frac{∂f}{∂z} \frac{∂z}{∂y} + \frac{∂f}{∂w} \frac{∂w}{∂y} = 0 \\ \frac{∂g}{∂y} + \frac{∂g}{∂z} \frac{∂z}{∂y} + \frac{∂g}{∂w} \frac{∂w}{∂y} = 0 \end{matrix} ⇒ \frac{∂w}{∂y} = −\frac{f_{z}g_{y} − f_{y}g_{z}}{f_{z}g_{w} − f_{w}g_{z}}

3. Homework assignment: Problem 5.

Note that

(\frac{∂u}{∂x})_{y} = \frac{∂u}{∂x} + \frac{∂u}{∂z} \frac{∂z}{∂x} + \frac{∂u}{∂w} \frac{∂w}{∂x}.

We’ve known from Problem.12 that

\frac{∂z}{∂x} = −\frac{f_{x}g_{w} − f_{w}g_{x}}{f_{z}g_{w} − f_{w}g_{z}} .

For another unknown term, consider

\{ \begin{matrix}\frac{∂f}{∂x} + \frac{∂f}{∂z} \frac{∂z}{∂x} + \frac{∂f}{∂w} \frac{∂w}{∂x} = 0 \\ \frac{∂g}{∂x} + \frac{∂g}{∂z} \frac{∂z}{∂x} + \frac{∂g}{∂w} \frac{∂w}{∂x} = 0 \end{matrix} ⇒ \frac{∂w}{∂x} = −\frac{f_{x}g_{z} − f_{z}g_{x}}{f_{w}g_{z} − f_{z}g_{w}} .

Therefore,

(\frac{∂u}{∂x})_{y} = \frac{∂u}{∂x} + \frac{∂u}{∂z} (−\frac{f_{x}g_{w} − f_{w}g_{x}}{f_{z}g_{w} − f_{w}g_{z}} ) + \frac{∂u}{∂w} (−\frac{f_{x}g_{z} − f_{z}g_{x}}{f_{w}g_{z} − f_{z}g_{w}} ) .

4. Homework assignment: Problem 6.

\frac{d}{dx}f(x, y(x), z(x))|_{x=1/\sqrt{2}} = 2x + 2y (−\frac{x}{y}) + 2z (−1 + \frac{x}{y}) |_{(1/\sqrt{2},1/\sqrt{2},1−\sqrt{2})}

= 2z (−1 + \frac{x}{y}) |_{(1/\sqrt{2},1/\sqrt{2},1−\sqrt{2})} = 0.