Implicitly Defined Motion And Acceleration

Motion Along an Implicit Constraint

A moving point in the plane has coordinates \(x(t)\) and \(y(t)\), both depending on time. Sometimes its path is described not by giving \(y\) explicitly as a function of \(x\), but by a constraint such as \(F(x,y)=0\). As the point moves while remaining on that path, its coordinates, velocities, and accelerations cannot vary independently.

The chain rule makes those dependencies precise. Differentiating the constraint once gives a relation between the velocity components. Differentiating a second time gives a relation involving both velocities and accelerations. The second relation is especially useful when one acceleration component is known and the other must be found.

The First Derivative: A Relation Between Velocities

Write \(x'(t)\) and \(y'(t)\) for the coordinate velocities. The partial derivatives \(F_x\) and \(F_y\) describe how \(F\) changes with its first and second inputs. Applying the chain rule to the constraint gives

This equation is an instantaneous relation: at a particular time, the partial derivatives are evaluated at the point’s current position. If \(F_y\ne 0\) there, it can be rearranged to determine the vertical velocity from the horizontal velocity:

The division is valid only when \(F_y\ne 0\). The undivided equation remains the appropriate velocity relation even when \(F_y=0\).

The Second Derivative: A Relation Between Accelerations

Assume \(F\) has continuous second partial derivatives near the point under consideration, and \(x(t)\) and \(y(t)\) are twice differentiable. Differentiate the first-derivative relation with respect to time. The derivatives \(F_x\) and \(F_y\) also vary with time because their inputs \(x(t)\) and \(y(t)\) vary.

The acceleration terms \(x''\) and \(y''\) appear multiplied by the first partial derivatives. The other terms account for how the constraint’s slope and curvature interact with the point’s velocities. In particular, the mixed term \(2F_{xy}x'y'\) must be included when \(F_{xy}\) is nonzero.

Why the Second-Derivative Formula Holds

The derivation follows by differentiating the first-derivative identity, applying the chain rule to each partial derivative. Starting from

differentiate the first product. The chain rule gives \(\frac{d}{dt}F_x=F_{xx}x'+F_{xy}y'\), so the product rule yields

For the second product, \(\frac{d}{dt}F_y=F_{yx}x'+F_{yy}y'\), and therefore

Adding these expressions gives zero because the derivative of the identically zero constraint relation is zero. Since \(F_{yx}=F_{xy}\), the two mixed terms add to \(2F_{xy}x'y'\), proving the formula. The factor of \(2\) comes from differentiating both products; it is not an extra assumption.

Worked Example: Motion on a Circle

Worked Example: A Product Constraint

Worked Example: A Nonlinear Constraint

A Common Pitfall: Losing the Mixed Term

When differentiating twice, it is easy to differentiate the first-derivative equation as if \(F_x\) and \(F_y\) were constants. They are generally not constant along the motion. For the product constraint, this mistake would omit the term \(2x'y'\).

In the product example, omitting that term would give the incorrect equation \(yx''+xy''=0\), and hence \(4(-1)+3y''=0\), or \(y''=4/3\). But substituting this into the correct acceleration relation gives

Thus the proposed value fails the constraint’s second-derivative test. The mixed term is required because both partial derivatives change along the path.

Using the Relations at a Specified Instant

A practical calculation keeps the general equations intact until the known position and rates are substituted. This helps preserve every term and makes it easier to check signs.

Each equation describes an instantaneous requirement imposed by the path. If \(F_y\ne0\), the acceleration relation can be solved for \(y''\) after the velocities are known. If \(F_y=0\), do not divide by it: the equation still constrains the motion, but it may not determine \(y''\) by itself.

Why These Equations Matter

The first relation tells how the velocity vector must be oriented relative to the constraint. The second tells how the velocity and acceleration components must combine to keep the point on the same curve. This is useful whenever a path is given implicitly and the question concerns motion at a particular instant, rather than a full explicit formula for the trajectory.

Units provide a useful consistency check. If \(x\) and \(y\) are lengths and \(t\) is time, then \(x'\) and \(y'\) have units of length per time, while \(x''\) and \(y''\) have units of length per time squared. The terms in the second-derivative equation need not each have the same units if \(F\) has been chosen with different units or scaling, but they must all have the units of the second time derivative of \(F(x(t),y(t))\). In a concrete problem, keep the units attached to the resulting velocity and acceleration.