Family of surfaces

At a Glance

• Frequency: 7 sub-parts across 7 of 13 years (2013, 2018, 2019, 2020, 2021, 2022, 2023)
• Priority tier: T2
• Marks (count): 10 (7)
• Average solve time: ~7 min
• Difficulty mix: easy 4, medium 3
• Section: B | Dominant type: derivation

Why This Chapter Matters

This atom is an almost-guaranteed 10-mark question in Paper 2, Q5(a), appearing in 7 of the last 13 years. The task is always the same: given a family of surfaces $z(x,y)$ depending on one or two arbitrary functions, derive the PDE by eliminating those functions through differentiation. The method is mechanical once you recognise the archetype. Mastering the two strategies below — ratio elimination for one function, linear-combination elimination for two separated functions — covers every variant UPSC has set.

Minimum Theory

Why differentiation eliminates arbitrary functions. A surface family parameterised by an arbitrary function $f$ (or two functions $f,g$) contains infinitely many surfaces. The PDE of the family is a relation among $x, y, z, z_x, z_y, z_{xx}, \ldots$ that holds for every surface in the family. Differentiating the surface relation $F(x,y,z)=0$ or $z=H(x,y;f)$ with respect to $x$ and $y$ introduces $f_u, f_v$ (or similar derivatives); combining the resulting equations to cancel these derivatives gives the PDE. Two arbitrary functions require two differentiations, so the PDE is second order.

Elimination of one arbitrary function. Suppose $f(u(x,y,z), v(x,y,z))=0$ defines the family implicitly. Differentiate $f(u,v)=0$ w.r.t. $x$: $f_u u_x + f_v v_x = 0$. Differentiate w.r.t. $y$: $f_u u_y + f_v v_y = 0$. Dividing eliminates the unknown ratio $f_u / f_v$, yielding the Lagrange-type first-order PDE $u_x v_y - u_y v_x = 0$ (after cross-multiplication), which expands to $Pp + Qq = R$ in the standard notation $p=z_x$, $q=z_y$.

Elimination of two separated arbitrary functions. For $z = f(x) + g(y)$ or $z = yf(x) + xg(y)$ and similar, compute $p = z_x$, $q = z_y$, $r = z_{xx}$, $s = z_{xy}$, $t = z_{yy}$ as needed. Form linear combinations of $xp$, $yq$, $z$, and the mixed partial $s = f'(x) + g'(y)$ to cancel $f, g, f', g'$ simultaneously. The canonical result for $z = yf(x)+xg(y)$ is $xyz_{xy} = xz_x + yz_y - z$.

Geometric archetype — cone. A cone with vertex $(a,b,c)$ is defined by $f\!\bigl(\tfrac{x-a}{z-c}, \tfrac{y-b}{z-c}\bigr)=0$. Eliminating $f$ by the ratio method gives the standard cone PDE: $(x-a)p + (y-b)q = z-c$. This is the only geometric PDE type UPSC has set for this atom.

Question Archetypes

eliminate-arbitrary-function (6 question(s); 2013, 2019, 2020, 2021, 2022, 2023)

Form a PDE by eliminating arbitrary functions from a surface relation

Recognition Cues

The question gives a surface $z=H(x,y;f,g)$ or an implicit relation $f(u,v)=0$, where $f$ and $g$ are unspecified functions. The instruction is always “form a partial differential equation” or “obtain the PDE.” One arbitrary function → first-order PDE (ratio elimination); two arbitrary functions → second-order PDE (linear combination or double differentiation).

Solution Template

For $f(u,v)=0$ (one arbitrary function):

1. Compute $u_x, v_x, u_y, v_y$ (with $z$ treated as a function of $x,y$ via the chain rule).
2. Differentiate w.r.t. $x$: $f_u u_x + f_v v_x = 0$.
3. Differentiate w.r.t. $y$: $f_u u_y + f_v v_y = 0$.
4. Equate ratios $f_u/f_v$ from both equations and cross-multiply to eliminate $f$.

For $z = yf(x)+xg(y)$ and similar (two separated functions):

1. Compute $p = z_x$, $q = z_y$, $s = z_{xy}$.
2. Form $xp + yq$ and identify that it equals $xyz_{xy} + z$.
3. Rearrange to eliminate $f, g, f', g'$ simultaneously.

Worked Example 1

2013 Paper 2, 2013-P2-Q5a (10 marks)

Form a PDE by eliminating the arbitrary functions $f$ and $g$ from $z = yf(x) + xg(y)$.

Partials: $p = z_x = yf'(x)+g(y), \quad q = z_y = f(x)+xg'(y), \quad s = z_{xy} = f'(x)+g'(y).$

Form $xp$ and $yq$, then use $z=yf+xg$: $xp = xyf'(x)+xg(y), \quad yq = yf(x)+xyg'(y).$ $xp+yq = xy\bigl(f'(x)+g'(y)\bigr)+\bigl(yf(x)+xg(y)\bigr) = xy\,s + z.$

Rearrange: $\boxed{xy\,z_{xy} = xz_x + yz_y - z.}$

Note (2020 extension): The principal part has $A=0$, $C=0$, $2B=xy$, so $B^2 - AC = x^2y^2/4 > 0$ for $x>0, y>0$ — the PDE is hyperbolic in that region.

Worked Example 2

2021 Paper 2, 2021-P2-Q5a (10 marks)

Obtain the PDE by eliminating $f$ from $f(x+y+z,\, x^2+y^2+z^2)=0$.

Let $u = x+y+z$, $v = x^2+y^2+z^2$, so $f(u,v)=0$ with $p=z_x$, $q=z_y$.

Differentiate w.r.t. $x$: $f_u(1+p) + f_v\cdot 2(x+zp) = 0$.

Differentiate w.r.t. $y$: $f_u(1+q) + f_v\cdot 2(y+zq) = 0$.

Equate $f_u/f_v$: $\frac{f_u}{f_v} = \frac{-2(x+zp)}{1+p} = \frac{-2(y+zq)}{1+q}.$ Cross-multiply and simplify (the $zpq$ terms cancel): $x + xq + zp = y + yp + zq,$ $\boxed{(y-z)\,p + (z-x)\,q = x-y.}$

This is a Lagrange-type first-order PDE with a symmetric, cyclic structure.

Worked Example 3

2022 Paper 2, 2022-P2-Q5a (10 marks)

The equation of any cone with vertex at $(a,b,c)$ is $f\!\left(\tfrac{x-a}{z-c}, \tfrac{y-b}{z-c}\right)=0$. Find its differential equation.

Let $u = \tfrac{x-a}{z-c}$, $v = \tfrac{y-b}{z-c}$. Differentiate $f(u,v)=0$:

w.r.t. $x$: $u_x = \tfrac{1-up}{z-c}$, $v_x = \tfrac{-vp}{z-c}$, giving $f_u(1-up) = f_v\,vp$.

w.r.t. $y$: $u_y = \tfrac{-uq}{z-c}$, $v_y = \tfrac{1-vq}{z-c}$, giving $f_v(1-vq) = f_u\,uq$.

Equate $f_u/f_v$: $\frac{vp}{1-up} = \frac{1-vq}{uq} \;\Rightarrow\; uvpq = 1 - vq - up + uvpq \;\Rightarrow\; up + vq = 1.$ Substitute $u=(x-a)/(z-c)$, $v=(y-b)/(z-c)$: $\boxed{(x-a)\,p + (y-b)\,q = z-c.}$

Worked Example 4

2023 Paper 2, 2023-P2-Q5a (10 marks)

Eliminate $f$ and $g$ from $z = f(x^2-y) + g(x^2+y)$.

Let $\alpha = x^2-y$, $\beta = x^2+y$.

First partials: $z_x = 2x\bigl(f'(\alpha)+g'(\beta)\bigr), \qquad z_y = -f'(\alpha)+g'(\beta).$

Second partials: $z_{xx} = \frac{z_x}{x} + 4x^2\bigl(f''(\alpha)+g''(\beta)\bigr), \qquad z_{yy} = f''(\alpha)+g''(\beta).$

Eliminate: From the $z_{xx}$ equation, $4x^2(f''+g'')=z_{xx}-z_x/x$. Using $f''+g''=z_{yy}$: $z_{xx} - \frac{z_x}{x} = 4x^2\,z_{yy}.$ $\boxed{x\,z_{xx} - z_x - 4x^3\,z_{yy} = 0.}$

Common Traps

• Chain-rule signs. When $u$ or $v$ involves $z$ (e.g., $u=x+y+z$), differentiating w.r.t. $x$ gives $u_x = 1+z_x = 1+p$, not just $1$. Dropping the $z_x$ contribution is the most common error in the ratio-elimination method.
• Count the order. One arbitrary function eliminates to a first-order PDE; two arbitrary functions eliminate to a second-order PDE. If you reach a first-order result from two functions, you have made an error.
• The $2B$ factor in classification. The PDE $xyz_{xy} = xz_x + yz_y - z$ has second-order part $xy\,z_{xy}$ only. Writing this as $A z_{xx}+2B z_{xy}+C z_{yy}$ gives $A=0$, $2B=xy$, $C=0$. The discriminant is $B^2-AC = (xy/2)^2 > 0$ for $x>0,y>0$ — hyperbolic. The factor of 2 in $2B$ is definitional; $B = xy/2$, not $xy$.
• Cone PDE shortcut. $(x-a)p+(y-b)q=z-c$ can be stated directly from the standard result; full derivation is only needed if the question explicitly asks for it.

pde-from-geometry (1 question(s); 2018)

Form a PDE describing a geometric family (e.g. tangent planes to a conicoid)

Recognition Cues

The question gives a geometric family (tangent planes to a conicoid, spheres through a fixed circle, etc.) and asks for the PDE. A plane not perpendicular to the $xy$-plane can be written $z = px + qy + r$; imposing the tangency or geometric condition on $(p,q,r)$ and eliminating $r = z-px-qy$ gives a PDE in $p$ and $q$.

Solution Template

1. Identify the geometric condition (e.g., tangency of a plane to a conicoid: $A^2u^2+B^2v^2+C^2w^2=k^2$).
2. Write the plane as $z=px+qy+r$, so $r=z-px-qy$.
3. Substitute $r$ into the geometric condition.
4. The result is the PDE in $p,q,x,y,z$ — often Clairaut form $z=px+qy+f(p,q)$.

Worked Example

2018 Paper 2, 2018-P2-Q5a (10 marks)

Find the PDE of all tangent planes to the ellipsoid $x^2+4y^2+4z^2=4$, not perpendicular to the $xy$-plane.

Divide through: $\tfrac{x^2}{4}+y^2+z^2=1$, so $A^2=4$, $B^2=C^2=1$.

Tangency condition for a plane $ux+vy+wz=k$ touching the standard-form ellipsoid: $A^2u^2+B^2v^2+C^2w^2=k^2$.

Write the plane as $z=px+qy+r$ (non-perpendicular means $w\ne0$, so this form is valid). Then $u=p$, $v=q$, $w=-1$, $k=-r$, and $r=z-px-qy$: $4p^2 + q^2 + 1 = r^2 = (z-px-qy)^2.$ $\boxed{(z-px-qy)^2 = 4p^2+q^2+1, \quad p=z_x,\; q=z_y.}$ This is Clairaut form with complete integral $z=px+qy+\sqrt{4p^2+q^2+1}$.

Common Traps

• Read the ellipsoid correctly. $x^2+4y^2+4z^2=4$ divided by 4 gives $A^2=4$, $B^2=1$, $C^2=1$ — not the naive $A^2=1$, $B^2=4$, $C^2=4$. The tangency coefficients are the semi-axes squared.
• “Not perpendicular to the $xy$-plane” is precisely the assumption that the plane can be solved for $z$ (non-vertical normal). Vertical tangent planes (at the equator $z=0$) are excluded.

Marks-Aware Writing

For a 10-mark elimination question: Show the partial derivatives explicitly (even if brief), display the key intermediate step (the equated ratio or the combined expression), and state the final PDE in a box. A correct PDE with no work shown earns 0–2 marks; show the computation.

Include a spot-check. A one-line verification (test a specific $f$ or $g$ against the PDE) guards against sign errors and earns the accuracy mark if the PDE is otherwise correct in structure.

Practice Set

• 2024-P2-Q7a (15 m) — — eliminate two arbitrary functions; classify the resulting PDE
• 2016-P2-Q5a (10 m) — — standard single-function elimination
• 2016-P2-Q5e (10 m) — — geometric family (cylinder/sphere type)