How Much Coffee for One More Tea?

An indifference curve joins every mix of two goods you'd like equally. Its slope says how much of one you'd trade for a bit more of the other. The curve isn't solved for y, but implicit differentiation finds its slope anyway.

How it works

Implicit differentiation

In one sentence

If $F(x, y) = c$ defines $y$ implicitly as a differentiable function of $x$ , differentiating both sides with the chain rule gives $F_x + F_y\frac{dy}{dx} = 0$ , so $\frac{dy}{dx} = -F_x/F_y$ .

General formula

{\htmlData{sym=0}{x}}^{\htmlData{sym=2}{{a}}}{\htmlData{sym=1}{y}}^{1-{\htmlData{sym=2}{a}}} = {\htmlData{sym=3}{c}} \quad\Rightarrow\quad \frac{\htmlData{sym=2}{{a}}}{\htmlData{sym=0}{{x}}} + \frac{1-{\htmlData{sym=2}{a}}}{\htmlData{sym=1}{{y}}}\,{\htmlData{sym=4}{\frac{dy}{dx}}} = 0 \quad\Rightarrow\quad {\htmlData{sym=4}{\frac{dy}{dx}}} = -\frac{\htmlData{sym=2}{{a}}}{1-{\htmlData{sym=2}{a}}}\cdot\frac{\htmlData{sym=1}{{y}}}{\htmlData{sym=0}{{x}}}

Moving along the curve keeps $F$ constant, so whatever change $x$ causes, $y$ must cancel. Solving that balance gives the slope.

What the symbols mean

$x$
Coffee
Cups of coffee in the bundle.
$y$
Tea
Cups of tea in the bundle.
$a$
Preference weight
How much you value coffee relative to tea.
$c$
Satisfaction level
Every bundle on the curve gives this much satisfaction.
$\frac{dy}{dx}$
Trade-off rate
Tea you'd give up for one more coffee, staying equally happy.

Take logs: $a\ln x + (1 - a)\ln y = \ln c$ . Treat $y$ as a function of $x$ and differentiate both sides.
The chain rule turns $\ln y$ into $\frac{1}{y}\frac{dy}{dx}$ ; then solve for $\frac{dy}{dx}$ .
The slope $-\frac{a}{1-a}\frac{y}{x}$ is the marginal rate of substitution: tea given up per extra coffee, keeping you equally happy.
It changes along the curve: when you have little coffee, you'd give up a lot for more.

Learn the math

Implicit DifferentiationCalc I · Lesson 23, Derivatives