Practice Problems 12

Lecture 12: Implicit Functions

Problem numbers refer to the textbook (Chiang & Wainwright, 4th edition). These are not graded, but you should attempt them all before the quiz. Try each question yourself first, then expand the Solution to check.

Exercise 8.5

1. For each $F (x, y) = 0$ , find $d y / d x$ :

$y - 6 x + 7 = 0$

Solution

With $F_{x} = - 6$ and $F_{y} = 1$ , the implicit-function rule gives

$\frac{d y}{d x} = - \frac{F_{x}}{F_{y}} = - \frac{- 6}{1} = 6$

Equivalently, solving for $y = 6 x - 7$ and differentiating gives $d y / d x = 6$ .

$3 y + 12 x + 17 = 0$

Solution

With $F_{x} = 12$ and $F_{y} = 3$ ,

$\frac{d y}{d x} = - \frac{F_{x}}{F_{y}} = - \frac{12}{3} = - 4$

$x^{2} + 6 x - 13 - y = 0$

Solution

With $F_{x} = 2 x + 6$ and $F_{y} = - 1$ ,

$\frac{d y}{d x} = - \frac{F_{x}}{F_{y}} = - \frac{2 x + 6}{- 1} = 2 x + 6$

2. For each $F (x, y) = 0$ use the implicit-function rule to find $d y / d x$ :

$F (x, y) = 6 x^{3} - 3 y = 0$

Solution

$F_{x} = 18 x^{2} F_{y} = - 3$

$\frac{d y}{d x} = - \frac{F_{x}}{F_{y}} = - \frac{18 x^{2}}{- 3} = 6 x^{2}$

3. For each $F (x, y, z) = 0$ use the implicit-function rule to find $\partial y / \partial x$ and $\partial y / \partial z$ :

$F (x, y, z) = x^{2} y^{3} + z^{2} + x y z = 0$

Solution

$F_{x} = 2 x y^{3} + y z F_{y} = 3 x^{2} y^{2} + x z F_{z} = 2 z + x y$

$\frac{\partial y}{\partial x} = - \frac{F_{x}}{F_{y}} = - \frac{2 x y^{3} + y z}{3 x^{2} y^{2} + x z}$

$\frac{\partial y}{\partial z} = - \frac{F_{z}}{F_{y}} = - \frac{2 z + x y}{3 x^{2} y^{2} + x z}$