Implicit Differentiation and Related Rates

A circle cannot be represented by one real function $y=f(x)$ over its entire horizontal span: most horizontal positions have two vertical coordinates. It can nevertheless have well-defined tangent lines on its local branches. An equation relating the variables can retain information that solving explicitly would split into cases.

Differentiating a constraint

For $x^2+y^2=25$, treat $y$ as a differentiable function of $x$ on a branch. Differentiation gives $2x+2yy'=0$, hence

$$y'=-x/y\qquad(y\ne0).$$

At $(3,4)$, the slope is $-3/4$, so the tangent is $y-4=-(3/4)(x-3)$. At $(5,0)$, the formula cannot produce a finite slope for $y$ as a function of $x$. The circle still has a tangent, namely the vertical line $x=5$. A failure of this coordinate representation is not a disappearance of the geometric curve.

For a general differentiable equation $F(x,y)=0$, the chain rule gives $F_x+F_yy'=0$ when the partial derivatives are defined, leading formally to $y'=-F_x/F_y$ where $F_y\ne0$. A local implicit function theorem supplies sufficient conditions for such a branch to exist. In this course, polynomial and elementary examples will also be checked directly when possible; the displayed ratio should not be treated as an unconditional existence theorem.

A second derivative needs the chain rule again

From the circle’s $y'=-x/y$, the quotient rule gives

$$y''=\frac{-y+xy'}{y^2} =-\frac{x^2+y^2}{y^3}=-\frac{25}{y^3}.$$

The first equality differentiates $y$ as a function rather than freezing it as a constant. At $(3,4)$, this is $-25/64$. The upper branch is concave down there, consistent with its geometry.

Rates share a clock

In a related-rate problem, several quantities depend on the same time variable. Write the constraint between them, then differentiate with respect to time. For a spherical balloon with radius $r(t)$ and volume $V(t)=4\pi r(t)^3/3$,

$$\frac{dV}{dt}=4\pi r^2\frac{dr}{dt}.$$

If the radius is $2$ centimetres and volume is increasing at $12\pi$ cubic centimetres per second, then $r'=12\pi/(16\pi)=3/4$ centimetres per second. Substituting $r=2$ before differentiating would turn volume into a constant and destroy the time dependence we need.

A sliding ladder

A rigid ladder of length $5$ metres touches a horizontal floor and vertical wall. Let $x(t)$ be the foot’s distance from the wall and $y(t)>0$ the top’s height. While the geometric model applies, $x^2+y^2=25$ and

$$2xx'+2yy'=0,\qquad y'=-\frac{x}{y}x'.$$

At $x=3$, $y=4$, and $x'=0.2$ metres per second, the top moves at $y'=-0.15$ metres per second. The sign describes downward motion. Its downward speed is $0.15$ metres per second.

As $y\to0^+$ with fixed positive $x'$, the formula predicts unbounded downward rate. That signals a limitation of extending this idealized contact-and-constant-foot-speed model to the final configuration. Real mechanics, contact forces, and finite energy may invalidate the imposed rate before that limit. Mathematics can expose a model’s consequences without certifying its physical continuation.

Organizing a modeling solution

Define variables and units; identify the instant; draw or describe the geometry; write the constraint; differentiate while quantities are still functions; substitute the instant’s values; solve for the requested rate; interpret its sign and domain. A clear equation often does more work than a memorized “related rates” formula.

Exercises

1. Find the tangent slope to $xy=6$ at $(2,3)$ by implicit differentiation.
2. Find $y'$ for $x^2+xy+y^2=7$, wherever the denominator is nonzero.
3. A circle’s radius increases at $0.4$ centimetres per second. Find its area rate when $r=5$ centimetres.
4. Repeat the ladder calculation when $x=4$, $x'=0.2$.
5. Explain why a rate measured in cubic metres per second cannot be reported as a radius rate without an additional geometric factor.
6. For the upper semicircle $y=\sqrt{25-x^2}$, verify the implicit derivative by explicit differentiation.

Worked solutions

1. $y+xy'=0$ gives $y'=-y/x=-3/2$.
2. $2x+y+xy'+2yy'=0$, so $y'=-(2x+y)/(x+2y)$.
3. $A'=2\pi rr'=4\pi$ square centimetres per second.
4. $y=3$, hence $y'=-(4/3)(0.2)=-4/15$ metres per second.
5. Volume and length are different quantities. For a sphere the conversion factor is $1/(4\pi r^2)$, with inverse-area units.
6. The chain rule gives $-x/\sqrt{25-x^2}=-x/y$ for $-5<x<5$.