Implicit Constraints Along a Parametrized Path

Introduction

A constraint such as \(F(x,y,z)=0\) restricts which triples of coordinates are allowed. If a point moves while continuing to satisfy the constraint, its coordinates may change, but their rates cannot vary independently. Differentiating the constraint along the path reveals the relation among those rates.

The essential step is to treat each coordinate as a function of the same parameter. The multivariable chain rule then converts the changing equation into a single equation involving the partial derivatives of \(F\) and the coordinate rates. If one of those partial derivatives is nonzero, this equation can often be solved for the corresponding unknown rate.

The rate relation for a constrained path

Let \(F\) be a differentiable real-valued function of three variables, and let \(x(t)\), \(y(t)\), and \(z(t)\) be differentiable coordinate functions on an interval. Suppose the path remains on the constraint \(F(x,y,z)=0\): that is, the equation holds for every parameter value in the interval, not merely at one instant.

Definition: A differentiable path \(\mathbf{r}(t)=(x(t),y(t),z(t))\) satisfies the constraint \(F(x,y,z)=0\) on an interval if \(F(x(t),y(t),z(t))=0\) for every \(t\) in that interval. Its velocity is \(\mathbf{r}'(t)=(x'(t),y'(t),z'(t))\).

Here, \(F_x\), \(F_y\), and \(F_z\) denote the partial derivatives of \(F\) with respect to its first, second, and third variables. Applying the chain rule to the identity along the path gives the central rate equation.

Theorem: If \(F\) and the coordinate functions are differentiable, and \(F(x(t),y(t),z(t))=0\) throughout an interval, then at every parameter value in that interval,

$$ F_x(x(t),y(t),z(t))x'(t) + F_y(x(t),y(t),z(t))y'(t) + F_z(x(t),y(t),z(t))z'(t) =0. $$

Equivalently, the gradient of \(F\) dotted with the path’s velocity is zero:

\nabla F(x(t),y(t),z(t))\mathbin{\cdot}\mathbf{r}'(t)=0, \qquad \nabla F=(F_x,F_y,F_z).

When, for example, \(F_z\ne 0\) at the point under consideration, the equation can be rearranged to give

z'(t) = -\frac{F_x(x(t),y(t),z(t))x'(t) +F_y(x(t),y(t),z(t))y'(t)} {F_z(x(t),y(t),z(t))}.

The nonzero condition matters: this division is valid only when \(F_z\ne 0\). If \(F_z=0\), the differentiated constraint may still give a useful relation, but it does not determine \(z'\) by this formula.

Why the equation holds

Define a one-variable function \(h(t)=F(x(t),y(t),z(t))\). The path satisfies the constraint throughout the interval, so \(h(t)=0\) there. The derivative of a constant function is zero, and hence \(h'(t)=0\). The chain rule for a composition of a function of several variables with a parametrized path gives

$$ h'(t) = F_x(x(t),y(t),z(t))x'(t) + F_y(x(t),y(t),z(t))y'(t) + F_z(x(t),y(t),z(t))z'(t). $$

Since \(h'(t)=0\), this is exactly the rate equation. This also explains why the equation is a local restriction on the velocity: at each point of the path, the weighted sum of the coordinate rates must vanish.

In particular, the condition that the constraint holds along an interval is essential. Knowing only that \(F(x(t_0),y(t_0),z(t_0))=0\) at one value \(t_0\) does not establish that the composed function is constant near \(t_0\), so it does not by itself imply that its derivative is zero.

Solving for an unknown rate

To find a missing rate, evaluate the partial derivatives at the stated point, multiply each by its coordinate’s rate, and set their sum equal to zero. The constraint equation identifies the point on the surface; the differentiated equation relates the rates at that point. Keep those two checks distinct.

Worked Example: Find the rate of the third coordinate

A path satisfies \(x^2+2y^2+z^2=7\). At \(t=t_0\), its coordinates are \((1,1,2)\), with \(x'(t_0)=3\) and \(y'(t_0)=-1\). Find \(z'(t_0)\).

First check that the point satisfies the constraint:

$$ 1^2+2(1^2)+2^2=1+2+4=7. $$

Let \(F(x,y,z)=x^2+2y^2+z^2-7\). Its partial derivatives are \(F_x=2x\), \(F_y=4y\), and \(F_z=2z\). The rate equation is therefore

$$ 2xx'+4yy'+2zz'=0. $$

At the specified point and with the known rates, this becomes

2(1)(3)+4(1)(-1)+2(2)z'=0, $$ $$ 6-4+4z'=0, \qquad 4z'=-2, \qquad z'=-\frac{1}{2}.

Check by substituting all three rates into the differentiated constraint:

2(1)(3)+4(1)(-1)+2(2)\left(-\frac{1}{2}\right) =6-4-2=0.

The constraint is satisfied at the point, and the rates satisfy its differentiated equation.

Worked Example: Solve for the rate of the second coordinate

A path satisfies \(xy+yz+zx=11\). At \(t=t_0\), the coordinates are \((1,2,3)\), and the rates \(x'(t_0)=2\) and \(z'(t_0)=-1\) are known. Find \(y'(t_0)\).

Check the constraint at the given point:

$$ (1)(2)+(2)(3)+(3)(1)=2+6+3=11. $$

Define \(F(x,y,z)=xy+yz+zx-11\). Differentiating with respect to each coordinate gives \(F_x=y+z\), \(F_y=x+z\), and \(F_z=x+y\). Thus the rate equation is

$$ (y+z)x'+(x+z)y'+(x+y)z'=0. $$

Substitute the point and the known rates:

(2+3)(2)+(1+3)y'+(1+2)(-1)=0, $$ $$ 10+4y'-3=0, \qquad 4y'=-7, \qquad y'=-\frac{7}{4}.

Check the rate equation with this value:

(5)(2)+(4)\left(-\frac{7}{4}\right)+(3)(-1) =10-7-3=0.

The point satisfies the original constraint, and the computed rate makes the differentiated equation hold.

Choosing which rate to solve for

The differentiated equation is one linear equation in the coordinate rates. Which rate it can determine depends on the coefficient of that rate. For instance, \(z'\) can be isolated directly when \(F_z\ne 0\). If a different partial derivative is nonzero, it may be more appropriate to solve for that coordinate’s rate instead.

Worked Example: Solve for a different coordinate rate

A path satisfies \(x^2+y^2+z^2=25\). At \(t=t_0\), the coordinates are \((3,4,0)\), and \(y'(t_0)=1\), \(z'(t_0)=2\). Find \(x'(t_0)\).

First, verify the point:

$$ 3^2+4^2+0^2=9+16=25. $$

For \(F(x,y,z)=x^2+y^2+z^2-25\), the rate equation is

$$ 2xx'+2yy'+2zz'=0. $$

At the given point, this becomes

2(3)x'+2(4)(1)+2(0)(2)=0, $$ $$ 6x'+8+0=0, \qquad 6x'=-8, \qquad x'=-\frac{4}{3}.

Substitution checks the result:

2(3)\left(-\frac{4}{3}\right)+2(4)(1)+2(0)(2) =-8+8+0=0.

Here \(F_z=2z=0\) at the point, so the formula for solving \(z'\) by dividing by \(F_z\) would not apply. But \(F_x=2x=6\ne 0\), and the differentiated equation determines \(x'\) without difficulty.

What the rate equation can and cannot determine

At a given point, the equation \(F_xx'+F_yy'+F_zz'=0\) is a restriction on the possible rates, not a guarantee that every unspecified rate can be uniquely found. To solve for \(z'\), for example, the other rates must be known and \(F_z\) must be nonzero. If \(F_z=0\), the coefficient of \(z'\) vanishes; the equation may constrain the other rates without determining \(z'\).

The gradient form gives a useful interpretation. The gradient \(\nabla F\) is perpendicular to the surface \(F=0\) at a regular point, where the gradient is nonzero. The path’s velocity must have zero dot product with that gradient, so its instantaneous direction lies in the tangent plane to the constraint surface. This is precisely what the rate equation expresses, without requiring the coordinates to be written explicitly as functions of one another.

Key takeaway: Differentiate the identity along the path, not just the equation at one point. Then evaluate the partial derivatives and rates at the point of interest. Divide by a partial derivative only when it is nonzero.

A common mistake

A frequent error is to differentiate the constraint as though \(x\), \(y\), and \(z\) were fixed numbers, rather than functions of \(t\). For the first example, that incorrect approach might write

\frac{d}{dt}(x^2+2y^2+z^2) \ \text{incorrectly as}\ 2x+4y+2z.

At \((1,1,2)\), this gives \(2+4+4=10\), which is not zero. The error is that the derivative of \(x(t)^2\), for example, is \(2x(t)x'(t)\), not \(2x(t)\). The correct differentiation includes the rate of every changing coordinate:

\frac{d}{dt}\bigl(x(t)^2+2y(t)^2+z(t)^2\bigr) = 2xx'+4yy'+2zz'.

Using the rates from that example gives \(6-4-2=0\), as required. The coordinate rates are essential parts of the chain rule.

Check Your Understanding

Use the differentiated constraint in each question. State any nonzero condition needed when solving for a rate.

A path satisfies \(x^2+y^2+z^2=14\) and passes through \((1,2,3)\). Write the rate equation at that point.
If \(F_z=5\), \(F_x=2\), \(F_y=-1\), \(x'=3\), and \(y'=4\), what value of \(z'\) satisfies the constraint rate equation?
Why is it not enough to know that \(F(x(t_0),y(t_0),z(t_0))=0\) at just one parameter value?
If \(F_z=0\) at a point, does that prove that no differentiable constrained path passes through the point? Explain.
For a differentiable constrained path, what does the equation \(\nabla F\mathbin{\cdot}\mathbf{r}'=0\) say about the gradient and velocity?