Related Rates Along an Implicit Curve

Motion Constrained to an Implicit Curve

A point moving in the plane may be required to stay on a curve that is given by an equation rather than an explicit formula. For example, the relation $x^2+y^2=25$ describes a circle without specifying $y$ as a single function of $x$ over the entire curve. If the point moves along such a curve, its coordinates $x$ and $y$ change together.

Related rates along an implicit curve use the constraint to connect those changing coordinates. If one coordinate rate is known, differentiating the constraint with respect to time can determine the other—provided the relevant coefficient is nonzero. The resulting rate pair must also agree with the curve’s tangent direction at the point of motion.

Central equation: If a differentiable motion $(x(t),y(t))$ satisfies $F(x(t),y(t))=0$, and $F$ is continuously differentiable near the point being considered, then $$ F_x(x(t),y(t))\frac{dx}{dt} + F_y(x(t),y(t))\frac{dy}{dt} =0. $$

Setting Up the Rate Equation

Let a curve be defined by $F(x,y)=0$, where $F$ has continuous first partial derivatives near the point of interest. Suppose a point moves along the curve, with differentiable coordinates $x=x(t)$ and $y=y(t)$. The constraint then holds throughout the motion: $F(x(t),y(t))=0$.

Differentiate this identity with respect to time. The multivariable chain rule says that the derivative of $F$ along the path is the sum of each partial derivative multiplied by the rate of its corresponding coordinate. Since the constraint is constantly zero, its derivative is zero:

\frac{d}{dt}F(x(t),y(t)) = F_x(x(t),y(t))\frac{dx}{dt} + F_y(x(t),y(t))\frac{dy}{dt} =0.

Here $F_x$ and $F_y$ are evaluated at the current point on the curve. In a calculation at a particular time $t_0$, first identify $(x(t_0),y(t_0))$, evaluate these partial derivatives there, and then substitute the known coordinate rate.

When this equation applies: The path coordinates must be differentiable at the time under consideration, and the composition $F(x(t),y(t))$ must be differentiable there. Continuous first partial derivatives of $F$ in a neighborhood are a standard sufficient condition. The chain-rule equation is a necessary condition for motion satisfying the constraint; it is not by itself a complete description of all possible motion.

If $F_y\ne 0$ at the point, the equation can be solved for the rate $dy/dt$. If $F_x\ne 0$, it can instead be solved for $dx/dt$. A zero coefficient does not permit division, and may mean that this first-derivative equation cannot determine that coordinate rate.

Why the Rate Equation Is a Tangency Condition

At a regular point of the curve, the vector of partial derivatives is nonzero: at least one of $F_x$ and $F_y$ is nonzero. When $F$ is continuously differentiable near that point, the implicit function theorem gives a local graph description. The rate equation then says precisely that the velocity of a constrained motion points along the tangent to that graph.

Theorem: Suppose $F$ is continuously differentiable near a point $(x_0,y_0)$ on $F(x,y)=0$, and at least one of $F_x(x_0,y_0)$ and $F_y(x_0,y_0)$ is nonzero. Then the tangent vectors to the curve at $(x_0,y_0)$ are exactly the vectors $(a,b)$ satisfying $$ F_x(x_0,y_0)a+F_y(x_0,y_0)b=0. $$ In particular, the velocity $(dx/dt,dy/dt)$ of any differentiable motion along the curve satisfies this equation.

To justify the statement, first suppose $F_y(x_0,y_0)\ne 0$. The implicit function theorem applies because $F$ is continuously differentiable and the indicated partial derivative is nonzero. It gives a local representation $y=g(x)$ of the curve. Differentiating $F(x,g(x))=0$ gives $F_x+F_y g'(x)=0$, so at the point $g'(x_0)=-F_x/F_y$. The tangent vectors to the graph have the form $(a,g'(x_0)a)$, and every such vector satisfies the displayed equation.

Conversely, if $(a,b)$ satisfies that equation, then $b=-F_xa/F_y=g'(x_0)a$. Thus it has the form of a tangent vector to the graph. If $F_y(x_0,y_0)=0$, regularity requires $F_x(x_0,y_0)\ne 0$. The implicit function theorem instead gives a local graph $x=h(y)$; the same argument shows that its tangent vectors are exactly the vectors satisfying the equation. This explains both the tangency check and why continuous differentiability and regularity matter for this interpretation.

Worked Examples

Worked Example: A Point Moving on a Circle

A point moves on the circle $x^2+y^2=25$. At the instant when $(x,y)=(3,4)$, its $x$-coordinate is increasing at $2$ centimeters per second. Find $dy/dt$, and check tangency.

Set $F(x,y)=x^2+y^2-25$. The point is on the curve because $3^2+4^2-25=9+16-25=0$. The partial derivatives are $F_x=2x$ and $F_y=2y$. Differentiating the constraint along the motion gives

2x\frac{dx}{dt}+2y\frac{dy}{dt}=0.

At $(3,4)$, this becomes $6(dx/dt)+8(dy/dt)=0$. Substitute $dx/dt=2$:

6(2)+8\frac{dy}{dt}=0, \qquad 12+8\frac{dy}{dt}=0, \qquad \frac{dy}{dt}=-\frac{12}{8}=-\frac{3}{2}.

Thus $y$ is decreasing at $1.5$ centimeters per second. Check the rate equation: $6(2)+8(-3/2)=12-12=0$. The velocity vector $(2,-3/2)$ therefore satisfies the tangent condition at $(3,4)$.

Worked Example: A Product Constraint

A point moves on the curve $xy=12$. When $(x,y)=(3,4)$, $x$ is decreasing at $1$ meter per second. Find $dy/dt$ and verify that the velocity is tangent to the curve.

Write the constraint as $F(x,y)=xy-12=0$. The point satisfies it because $3\cdot4-12=0$. Its partial derivatives are $F_x=y$ and $F_y=x$. Differentiating along the motion gives

y\frac{dx}{dt}+x\frac{dy}{dt}=0.

At the stated point, $y=4$, $x=3$, and $dx/dt=-1$. Hence

4(-1)+3\frac{dy}{dt}=0, \qquad -4+3\frac{dy}{dt}=0, \qquad \frac{dy}{dt}=\frac{4}{3}.

The $y$-coordinate is increasing at $4/3$ meters per second. For the tangency check, the partial derivatives at $(3,4)$ are $F_x=4$ and $F_y=3$, so

F_x\frac{dx}{dt}+F_y\frac{dy}{dt} = 4(-1)+3\left(\frac{4}{3}\right) =-4+4=0.

The computed velocity $(-1,4/3)$ satisfies the curve’s tangent condition.

Worked Example: A Curve with Mixed Terms

A point follows $x^2+xy+y^2=7$. At $(1,2)$, its $x$-coordinate is increasing at $5$ units per second. Find and check the rate $dy/dt$.

Define $F(x,y)=x^2+xy+y^2-7$. The specified point lies on the curve since $1^2+(1)(2)+2^2-7=1+2+4-7=0$. Its partial derivatives are $F_x=2x+y$ and $F_y=x+2y$. Therefore, differentiation along the path yields

(2x+y)\frac{dx}{dt}+(x+2y)\frac{dy}{dt}=0.

At $(1,2)$, $2x+y=2+2=4$ and $x+2y=1+4=5$. Substitution of $dx/dt=5$ gives

4(5)+5\frac{dy}{dt}=0, \qquad 20+5\frac{dy}{dt}=0, \qquad \frac{dy}{dt}=-4.

Check the result in the rate equation: $4(5)+5(-4)=20-20=0$. Thus the velocity vector $(5,-4)$ is tangent to the curve at the specified point.

When the Equation Does Not Determine a Rate

The rate equation is linear in the two coordinate rates, but it does not always determine whichever rate is requested. Consider the circle $x^2+y^2=25$ at $(0,5)$. There, $F_x=0$ and $F_y=10$, so the rate equation is

0\frac{dx}{dt}+10\frac{dy}{dt}=0.

This equation requires $dy/dt=0$, but it places no restriction on $dx/dt$. For example, attempting to solve for $dx/dt$ by writing $dx/dt=-(F_y/F_x)(dy/dt)$ would divide by $F_x=0$, so that calculation is invalid. If a differentiable constrained motion at this point is known to have $dy/dt=0$, this first-derivative equation alone cannot determine its $x$-rate.

If instead $dy/dt=1$ at this point is claimed, the rate equation gives $10(1)=10\ne0$. That claimed rate is incompatible with differentiable motion that stays on the circle. Checking the original equation before dividing can therefore reveal either an undetermined rate or inconsistent data.

Important distinction: The chain-rule equation is a necessary condition for a differentiable path to remain on the implicit curve. At a regular point with $F$ continuously differentiable, it also describes the tangent directions. If both partial derivatives vanish, the point is not regular, and the equation may provide no useful tangent information; do not infer a regular tangent description from it alone.

A Reliable Calculation Sequence

For a related-rates problem on an implicit curve, keep the constraint and the instantaneous rates distinct until the equation has been differentiated. A useful sequence is:

Record the constraint.
Write the curve as $F(x,y)=0$, and check that the specified point satisfies it.

Differentiate with respect to time.
Apply the chain rule to obtain $F_x(dx/dt)+F_y(dy/dt)=0$.

Evaluate at the instant.
Substitute the point’s coordinates into the partial derivatives, then insert the known rate.

Check whether solving is valid.
Divide only by a coefficient that is nonzero; otherwise interpret the equation without dividing.

Verify the result.
Substitute both rates into the rate equation and confirm that its left side is zero.

Why the Tangency Check Matters

The curve constraint restricts the point to a one-dimensional path in the plane near a regular point. Its two coordinate rates are therefore not independent: together they must give a velocity parallel to the curve’s tangent. The rate equation expresses that restriction without requiring the curve to be solved explicitly for $y$ in terms of $x$.

A zero result in the verification confirms that the calculated rates satisfy the instantaneous constraint. It does not, on its own, determine the full motion or show that the point follows a particular physical law. The calculation answers the related-rates question using the stated constraint and known rate; additional information may be needed to describe the motion over time.

Check Your Understanding

Use the chain-rule rate equation and, where appropriate, check the tangent condition.

For $x^2+y^2=10$, find $dy/dt$ at $(1,3)$ if $dx/dt=2$.
For $xy=15$, find $dy/dt$ at $(5,3)$ if $dx/dt=3$.
For $x^2+2y^2=18$, what does the rate equation imply about $dx/dt$ at $(3, 3/\sqrt2)$ if $dy/dt=2$?
At $(0,5)$ on $x^2+y^2=25$, can the first-derivative constraint determine $dx/dt$ from $dy/dt=0$? Explain.
What regularity and nonzero-derivative conditions justify interpreting the rate equation as describing exactly the tangent directions of the curve?