G5 - Implicit and parametric differentiation

Differentiating an implicit relation

How can we find a tangent when a curve has not been rearranged into $y=f(x)$? For $x^2+y^2=25$, points above and below the horizontal axis have different $y$-values at the same $x$. We can still follow a local branch and differentiate its equation.

An implicit relation links $x$ and $y$ without necessarily isolating $y$. Locally, where a branch gives a differentiable $y(x)$, every occurrence of $y$ changes with $x$. For example,

These implicit derivative results follow from the recalled chain and product rules; they are not supplied as separate formulae. For $\ln y$, the derivative is $y'/y$, with $y>0$, also a recall consequence. For $x/y$, where $y\ne0$, the supplied quotient rule gives $(y-xy')/y^2$.

Differentiate every term, collect the terms containing $y'$, then divide only when its coefficient is nonzero.

Change only the power of the final $y$-term from a square to a cube; choose the constant so that the same point remains on the curve. The chain derivative now includes $3y^2$.

A simpler relation follows the same rule: $x^2+y^3=9$ gives $2x+3y^2y'=0$, hence $y'=-2x/(3y^2)$ where $y\ne0$. No rearrangement into cube roots is needed.

Change the final power once more. Verify the supplied point before interpreting the derivative.

Valid points, tangents and zero denominators

The original equation decides which points exist; the local geometry decides whether a finite gradient is available.

For a valid point $(a,b)$ with finite gradient $m$, the tangent is $y-b=m(x-a)$. If $m\ne0$, the normal has gradient $-1/m$. A horizontal tangent has a vertical normal; a vertical tangent has a horizontal normal. These line and perpendicular-gradient rules are not supplied in the formula booklet: learn them.

For the circle $x^2+y^2=25$, differentiation gives

This first-derivative formula is derived using recall rules; it is not supplied. It cannot be evaluated with $y=0$.

Change only the point on the same circle to $(5,0)$. The finite-gradient formula now divides by zero, so use the circle's geometry or regard $x$ as a function of $y$.

More generally, differentiation may give $A(x,y)+B(x,y)y'=0$. Where $B\ne0$, division gives a finite candidate slope. If $B=0$ and $A\ne0$, no finite slope can satisfy that equation; at a regular smooth curve point this indicates a vertical tangent. If both coefficients vanish, this equation alone says nothing about the slope. For example, $y^2-x^2=0$ gives zero coefficients at the origin, yet consists of the two lines $y=x$ and $y=-x$. There is no unique tangent there. This reasoning is not a supplied formula.

Move to one more point on the same circle. This time the tangent is horizontal, so the normal must be written without a gradient.

Using a parameter to find the gradient

A parameter determines both coordinates, so the curve gradient compares the two coordinate rates at the same parameter value.

For $x=t+1$, $y=t^2-2t$, increasing $t$ changes both $x$ and $y$. The chain rule gives $dy/dt=(dy/dx)(dx/dt)$. Dividing is valid only when $dx/dt\ne0$.