The math optional, made finite. Daily Practice

Curve tracing (cartesian and polar)

At a Glance

Why This Chapter Matters

Curve tracing carries 20 marks — the largest single question in the calculus cluster and twice the weight of a typical Section A item. The examiner asks for a complete systematic analysis: domain, symmetry, intercepts, asymptotes, and a rough sketch. Both past questions involve implicit equations that must be solved for y2y^2, followed by a sign analysis to find the real domain. Mastering the seven-step checklist below and the sign-table method for domain analysis turns this high-stakes item into a routine procedure.

Minimum Theory

Seven-step curve-tracing checklist (Cartesian). For any curve F(x,y)=0F(x,y)=0: (1) Rewrite as y2=f(x)y^2=f(x) or y=f(x)y=f(x) if possible. (2) Domain: find where y20y^2\ge 0 via a sign table; boundary points are curve endpoints or asymptotes. (3) Symmetry: yyy\to-y invariance means xx-axis symmetry; xxx\to-x means yy-axis symmetry; both means origin symmetry. (4) Intercepts: set x=0x=0 and y=0y=0. (5) Asymptotes: vertical asymptotes where denominator 0\to 0; horizontal asymptotes from y2cy^2\to c as x±x\to\pm\infty; oblique asymptotes from y/xmy/x\to m. (6) Tangent at special points: at a boundary point (a,0)(a,0) where y20y^2\to 0, compute dy/dxdy/dx to determine if the curve ends sharply or with a vertical tangent. (7) Monotonicity: whether y|y| increases or decreases as xx increases.

Sign table method. To determine the sign of P(x)Q(x)\frac{P(x)}{Q(x)}: list all real zeros of PP and QQ in order, test one value in each interval, and track sign changes. The curve exists precisely where the expression is 0\ge 0.

Vertical tangent criterion. From y2=f(x)y^2=f(x), differentiate: 2ydy/dx=f(x)2y\,dy/dx=f'(x), so dy/dx=f(x)/(2y)dy/dx=f'(x)/(2y). At a boundary point where y0+y\to 0^+ and f(x0)0f'(x_0)\ne 0, dy/dxdy/dx\to\infty — the curve has a vertical tangent at (x0,0)(x_0,0).

Anatomy of curve tracing: domain, symmetry, asymptotes, and vertical tangents illustrated for y^2=1-a^2/x^2

Question Archetypes

ArchetypeYou are seeing this when…
curve-tracing”Trace the curve F(x,y)=0F(x,y)=0” — full analysis required for 20 marks

curve-tracing (2 question(s); 2022, 2023)

Recognition Cues

Solution Template

  1. Rewrite as y2=f(x)y^2=f(x). Factor numerator and denominator.
  2. Sign table. List all zeros of ff; determine sign of f(x)f(x) in each interval; the curve exists where f0f\ge 0.
  3. Symmetry. State axis/origin symmetry.
  4. Intercepts. Find where y=0y=0 (zeroes of numerator) and where x=0x=0.
  5. Asymptotes. Vertical: where denominator 0\to 0; horizontal: limxf(x)\lim_{|x|\to\infty}f(x).
  6. Tangent at boundary points. Compute dy/dx=f(x)/(2y)dy/dx=f'(x)/(2y) at the endpoints; state “vertical tangent” if dy/dxdy/dx\to\infty.
  7. Monotonicity. Whether yy increases/decreases along each branch.
  8. Sketch. Label all key features.

Worked Example

2022 Paper 1, 2022-P1-Q4b (20 marks)

Trace the curve y2x2=x2a2y^2 x^2=x^2-a^2 (where aa is a real constant).

Step 1 — Rewrite.

y2=x2a2x2=1a2x2.y^2=\frac{x^2-a^2}{x^2}=1-\frac{a^2}{x^2}.

Step 2 — Domain. y201a2/x20x2a2xay^2\ge 0\Leftrightarrow 1-a^2/x^2\ge 0\Leftrightarrow x^2\ge a^2\Leftrightarrow|x|\ge|a|. Curve exists for xax\le -|a| and xax\ge|a|. These are two disjoint branches.

Step 3 — Symmetry. y2y^2 depends only on x2x^2: invariant under both xxx\to-x and yyy\to-y. Symmetric about both axes (and origin).

Step 4 — Intercepts. At x=±ax=\pm a: y2=0y^2=0, so y=0y=0. Curve passes through (±a,0)(\pm a,0).

Step 5 — Asymptotes. As x|x|\to\infty: y210=1y^2\to 1-0=1, so y±1y\to\pm 1. Horizontal asymptotes y=±1y=\pm 1. No vertical asymptote (curve has no denominator factor tending to zero; the exclusion is a gap, not an asymptote).

Step 6 — Tangent at (±a,0)(\pm a,0).

dydx=12y2a2x3=a2yx3.\frac{dy}{dx}=\frac{1}{2y}\cdot\frac{2a^2}{x^3}=\frac{a^2}{yx^3}.

At x=a+x=a^+, y0+y\to 0^+: dy/dx+dy/dx\to+\infty. Vertical tangent at (±a,0)(\pm a,0).

Step 7 — Monotonicity. For x>ax>a: y2=1a2/x2y^2=1-a^2/x^2 increases as xx increases (since a2/x2a^2/x^2 decreases). So y|y| increases from 00 at x=ax=a toward 11 as xx\to\infty.

Step 8 — Sketch description. Two symmetric “trumpet” pieces, each with upper and lower arcs. Left piece: from (,±1)(-\infty, \pm 1) narrowing to the point (a,0)(-a,0) with a vertical tangent. Right piece: mirror image, from (a,0)(a,0) upward and rightward to the asymptote y=±1y=\pm 1.

  Domain: xa;  symm. both axes;  intercepts (±a,0);  horiz. asymptotes y=±1;  vertical tangents at (±a,0).  \boxed{\;\text{Domain: }|x|\ge|a|;\;\text{symm. both axes;}\;\text{intercepts }(\pm a,0);\;\text{horiz. asymptotes }y=\pm 1;\;\text{vertical tangents at }(\pm a,0).\;}


2023 Paper 1, 2023-P1-Q4b (20 marks)

Trace the curve y2(x21)=2x1y^2(x^2-1)=2x-1.

Step 1 — Rewrite.

y2=2x1x21=2x1(x1)(x+1).y^2=\frac{2x-1}{x^2-1}=\frac{2x-1}{(x-1)(x+1)}.

Step 2 — Sign table. Zeros: x=1/2x=1/2 (numerator), x=±1x=\pm 1 (denominator).

Region2x12x-1(x1)(x+1)(x-1)(x+1)y2y^2Real?
x<1x<-1-++-no
1<x<12-1<x<\tfrac{1}{2}--++yes
x=12x=\tfrac{1}{2}00-00yes (y=0y=0)
12<x<1\tfrac{1}{2}<x<1++--no
x>1x>1++++++yes

Domain: x(1,12](1,)x\in(-1,\tfrac{1}{2}]\cup(1,\infty).

Step 3 — Symmetry. Only y2y^2 appears: symmetric about the xx-axis. No symmetry in xx.

Step 4 — Intercepts. At x=1/2x=1/2: y=0y=0. At x=0x=0: y2=(1)/(1)=1y^2=(-1)/(-1)=1, so y=±1y=\pm 1.

Step 5 — Asymptotes.

Vertical: as x1+x\to-1^+: (x+1)0+(x+1)\to 0^+, (x1)2(x-1)\to-2, 2x132x-1\to-3; so y2=(3)/(0(2))=(3)/0+y^2=(-3)/(0^-\cdot(-2))=(-3)/0^-\to+\infty. Vertical asymptote x=1x=-1. As x1+x\to 1^+: (x1)0+(x-1)\to 0^+, 2x112x-1\to 1; y2+y^2\to+\infty. Vertical asymptote x=1x=1.

Horizontal (right branch): as x+x\to+\infty: y22x/x2=2/x0y^2\sim 2x/x^2=2/x\to 0. The xx-axis is a horizontal asymptote for the right branch.

Step 6 — Tangent at (1/2,0)(1/2,0).

dydx=12yddx ⁣(2x1x21).\frac{dy}{dx}=\frac{1}{2y}\cdot\frac{d}{dx}\!\left(\frac{2x-1}{x^2-1}\right).

Derivative of f(x)=(2x1)/(x21)f(x)=(2x-1)/(x^2-1): f(x)=2(x21)(2x1)(2x)(x21)2=2x2+2x+2(x21)2f'(x)=\frac{2(x^2-1)-(2x-1)(2x)}{(x^2-1)^2}=\frac{-2x^2+2x+2}{(x^2-1)^2}.

At x=1/2x=1/2: f(1/2)=(1/2+1+2)/((3/4)2)=(5/2)/(9/16)=40/9>0f'(1/2)=(-1/2+1+2)/((-3/4)^2)=(5/2)/(9/16)=40/9>0. As y0y\to 0, dy/dxdy/dx\to\infty. Vertical tangent at (1/2,0)(1/2,0).

Step 7 — Monotonicity. Left branch (1,1/2](-1,1/2]: y|y| decreases from ++\infty (at x=1+x=-1^+) to 00 at x=1/2x=1/2, giving a closed loop. Right branch (1,)(1,\infty): y|y| decreases from ++\infty (at x=1+x=1^+) toward 00 as xx\to\infty.

Step 8 — Sketch description. Left piece: a closed “beak” between x=1x=-1 and x=1/2x=1/2, with y±y\to\pm\infty as x1+x\to-1^+ and pinching to a point at (1/2,0)(1/2,0). Right piece: two arcs from ±\pm\infty at x=1+x=1^+, asymptotically approaching the xx-axis.

  Domain: (1,1/2](1,);  vert. asymp. x=±1;  intercepts (1/2,0),(0,±1);  right branch x-axis.  \boxed{\;\text{Domain: }(-1,1/2]\cup(1,\infty);\;\text{vert. asymp. }x=\pm 1;\;\text{intercepts }(1/2,0),(0,\pm 1);\;\text{right branch }\to x\text{-axis.}\;}

Common Traps


Marks-Aware Writing

Both questions are 20 marks. Full marks require all seven features to be clearly addressed. A marking scheme breakdown: domain/sign analysis (4 marks), symmetry (2 marks), intercepts (2 marks), asymptotes — vertical and horizontal (4 marks), tangents at boundary points (3 marks), monotonicity/shape (2 marks), sketch with all features labelled (3 marks). An answer that correctly finds the domain and asymptotes but skips tangents and monotonicity earns about 12 marks. Always present the sign table explicitly — it shows method and earns marks even if a subsequent conclusion has an error.

Practice Set

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