Two Conditions, and the Curve Has to Contain the Point

You'll learnto read an implicit slope as a quotient — numerator zero for a horizontal tangent, denominator zero for a vertical one — and to finish the job by substituting the condition back into the original equation, which is the step that turns a line of candidates into a short list of points.

Everything in Unit 5 so far assumed a function: one output per input, ready to differentiate. An implicit relation gives that up and keeps almost all the machinery. Two things change. The derivative arrives as a fraction in two variables, so a slope belongs to a point rather than to an x. And a candidate now has to pass two tests instead of one — the condition that makes the fraction behave, and membership of the curve itself. Skipping the second is the error this topic exists to prevent.

8 STEPS · 6 QUICK CHECKS · TOP ZERO IS HORIZONTAL · BOTTOM ZERO IS VERTICAL · THE POINT MUST BE ON THE CURVE · v1

x² + y² = 25 ⇒ dy/dx = −x/y · top zero ⇒ horizontal · bottom zero ⇒ vertical

Before you start

What you're looking at

The circle x² + y² = 25, whose implicit derivative is dy/dx = −x/y. It is a fraction in two variables, so it needs a whole point before it can produce a number — and it has two ways to be interesting, because a fraction has a top and a bottom.

The question

Where does this curve have a horizontal tangent, and where a vertical one? Each answer takes two steps, and the second one is the one that gets skipped.

Watch for

Step 7, where the point (0, 3) satisfies the condition perfectly and the line drawn through it slices the circle in two places instead of touching it once.

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