Rates in Implicitly Defined Curves

When a Moving Point Must Stay on a Curve

A point moving along a curve has coordinates that change with time. If the curve is given by an equation relating \(x\) and \(y\), that equation remains true throughout the motion. Differentiating it with respect to time gives a relationship between the rates \(\frac{dx}{dt}\) and \(\frac{dy}{dt}\).

This is useful when a curve is known implicitly, or when neither coordinate is best treated as an independent variable. The key is to remember that both coordinates may depend on time. In particular, differentiating a term involving \(y\) requires the chain rule.

The Rate Equation for an Implicit Curve

Suppose \(F\) is differentiable and a moving point satisfies \(F(x(t),y(t))=0\). The multivariable chain rule gives

Here, \(F_x\) and \(F_y\) are the partial derivatives of \(F\) with respect to its first and second variables. This equation is the instantaneous rate constraint. If \(F_y\ne 0\) at the point, it can be rearranged to find the \(y\)-rate:

The division is valid only when \(F_y\ne 0\). When \(F_y=0\), the rate equation may still be useful, but this rearrangement is not allowed. The equation might impose a condition on \(\frac{dx}{dt}\) without determining \(\frac{dy}{dt}\).

Applying the Rule to a Parabola

Consider the parabola \(y^2=4x\). Both coordinates can vary with time, so differentiate both sides with respect to \(t\). The right side differentiates to \(4\frac{dx}{dt}\). On the left, the chain rule gives \(2y\frac{dy}{dt}\), not just \(2y\).

At a point where \(y\ne 0\), division by \(2y\) gives

This relation makes the rate depend on the point as well as the horizontal rate. The sign of \(y\) matters: the upper and lower branches of the parabola can have different vertical rates for the same horizontal rate.

What Happens at the Vertex?

The vertex of \(y^2=4x\) is \((0,0)\). At this point \(y=0\), so the equation \(2y\frac{dy}{dt}=4\frac{dx}{dt}\) becomes

Consequently, any differentiable motion on this parabola that passes through the vertex must have \(\frac{dx}{dt}=0\) at that instant. But the equation says nothing about \(\frac{dy}{dt}\) there. Dividing by \(y\) to use \(\frac{dy}{dt}=\frac{2}{y}\frac{dx}{dt}\) would mean dividing by zero.

This is a singularity of the rate relation when solving for \(\frac{dy}{dt}\), not evidence that the parabola itself has a break or sharp corner. At the vertex the parabola has a vertical tangent, so a small change in \(y\) corresponds to a change in \(x\) that is especially small. The horizontal rate alone cannot tell us the vertical rate at that instant.

A Reliable Differentiation and Solving Process

Common Mistake: Dividing at a Zero Coefficient

A tempting but invalid step is to differentiate \(y^2=4x\), write \(\frac{dy}{dt}=\frac{2}{y}\frac{dx}{dt}\), and then use that formula at the vertex. At \(y=0\), the expression \(\frac{2}{y}\) is undefined. It is not repaired by substituting \(\frac{dx}{dt}=0\), since that would produce an undefined quotient rather than a rate.

The correction is to return to the undivided equation \(2y\frac{dy}{dt}=4\frac{dx}{dt}\). At the vertex it yields \(\frac{dx}{dt}=0\), but does not determine \(\frac{dy}{dt}\). Additional information about the motion would be needed to find that rate.

Why the Rate Relation Matters

The curve equation connects the two coordinate changes, while the derivative describes how that connection works at an instant. Away from the vertex, knowing one coordinate rate and the point determines the other. At the vertex, the coefficient needed to solve for the vertical rate vanishes, revealing a limitation of the available information rather than a calculation to force through.

For any implicitly defined curve, the same check applies. From \(F_x\frac{dx}{dt}+F_y\frac{dy}{dt}=0\), solve for \(\frac{dy}{dt}\) only when \(F_y\ne 0\). At a point where \(F_y=0\), keep the original rate equation and interpret what it actually implies.