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Worked solution

Separate a vertical tangent from an excluded endpoint

The curve CC has x=t2(t+6)x=t^2(t+6), y=t2+t−6y=t^2+t-6, t<0t<0.

(i) Find the equation(s) of the tangent to CC parallel to the y-axis. marks

(ii) Sketch CC, showing its axial intercepts. marks

[2018/NYJC/I/8(i)&(ii)]

CJC · Review problems · Question 6

Answer

Vertical tangent x=32x=32 at (32,6)(32,6). Actual axial intercepts (0,24)(0,24) and (27,0)(27,0); (0,−6)(0,-6) is excluded.

(i) Check the velocity components, not just a zero denominator

dx/dt=3t(t+4)dx/dt=3t(t+4) and dy/dt=2t+1dy/dt=2t+1. A vertical tangent requires dx/dt=0dx/dt=0 and dy/dt≠0dy/dt\ne0. The candidates are t=−4,0t=-4,0; 00 is excluded. At t=−4t=-4, the point is (32,6)(32,6) and dy/dt=−7≠0dy/dt=-7\ne0, so the tangent is x=32x=32.

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(ii) Retain only actual axial intercepts

x=0x=0 gives t=−6t=-6 or 00. Only t=−6t=-6 is allowed, giving (0,24)(0,24). The source lists (0,−6)(0,-6) from t=0t=0, but its graph correctly marks this as an open endpoint: it is approached as t→0−t\to0^- and is not an intercept of the restricted curve. From y=(t+3)(t−2)=0y=(t+3)(t-2)=0, only t=−3t=-3 is allowed, giving (27,0)(27,0).

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(ii) Trace the turning shape

As tt increases from −∞-\infty, the curve enters from the upper left, crosses (0,24)(0,24), reaches its rightmost point (32,6)(32,6), passes through (27,0)(27,0) and then approaches the open point (0,−6)(0,-6). Before that endpoint it has a horizontal tangent at t=−1/2t=-1/2, namely (11/8,−25/4)(11/8,-25/4), which explains the shallow dip below y=−6y=-6. The endpoint itself is excluded despite its limiting vertical tangent.

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Compare your sketch

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For t<0 the curve comes from the upper left through (0,24), turns at (32,6) with vertical tangent x=32, crosses (27,0), dips to (11/8,−25/4), and ends at an open circle (0,−6).
For t<0 the curve comes from the upper left through (0,24), turns at (32,6) with vertical tangent x=32, crosses (27,0), dips to (11/8,−25/4), and ends at an open circle (0,−6).
For t<0 the curve comes from the upper left through (0,24), turns at (32,6) with vertical tangent x=32, crosses (27,0), dips to (11/8,−25/4), and ends at an open circle (0,−6).

Common mistakes

A coordinate obtained from an excluded parameter is a limiting point, not an included intercept.

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