Find the region before selecting radii · source reference

Printed question

Curve C: $y=3-10/(x^2-2x+4)$.

(i) Without a calculator, determine its exact stationary point. [2]

(ii) Sketch C, stating clearly any asymptotes. [2]

R is bounded by C, $3y+16x=15$ and the $y$-axis.

(iii) Find its exact area. [3]

(iv) Find the volume when R is rotated $2\pi$ radians about the $y$-axis. [3]

[2021/EJC/Prelim/II/4]

Printed working

(i)

$$dy/dx=\frac{10(2x-2)}{(x^2-2x+4)^2}.$$

At a stationary point $dy/dx=0$, so $x=1$, $y=3-10/3=-1/3$: $(1,-1/3)$. Alternatively $y=3-10/[(x-1)^2+3]$ is symmetric about $x=1$, hence minimum $(1,-1/3)$.

(ii)

Horizontal asymptote $y=3$, $y$-intercept $1/2$, origin O and minimum $(1,-1/3)$ are labelled in the source graph.

(iii)

$$\begin{aligned}\text{Area}&=\int_0^1[(5-16x/3)-(3-10/(3+(x-1)^2))]dx\\&=\int_0^1[2-16x/3+10/(3+(x-1)^2)]dx\\&=[2x-8x^2/3+(10/\sqrt3)\tan^{-1}((x-1)/\sqrt3)]_0^1\\&=-2/3+(10/\sqrt3)(0-(-\pi/6))\\&=\frac{5\pi}{3\sqrt3}-\frac23=\frac13(5\pi/\sqrt3-2)\quad\text{unit}^2.\end{aligned}$$

(iv)

$$y=3-\frac{10}{(x-1)^2+3}\Rightarrow(x-1)^2+3=\frac{10}{3-y}\Rightarrow x-1=\pm\sqrt{10/(3-y)-3}.$$

The region is in $x<1$, so $x=1-\sqrt{10/(3-y)-3}$.

$$V=\pi\int_{-1/3}^5\left(\frac{15-3y}{16}\right)^2dy-\pi\int_{-1/3}^{1/2}(1-\sqrt{10/(3-y)-3})^2dy\approx5.0693=5.07\quad\text{unit}^3\ (3\text{ s.f.}).$$

Alternatively, using cone volume $\frac13\pi r^2h$:

$$V=\frac13\pi(1)^2[5-(-1/3)]-\pi\int_{-1/3}^{1/2}(1-\sqrt{10/(3-y)-3})^2dy=\frac{16}9\pi-0.16416\pi\approx5.0693=5.07\quad\text{unit}^3.$$

Teacher notes

Area R $=\int_0^1(y_1-y_2)dx=\int_0^1(f(x)-g(x))dx$, with upper $f$ and lower $g$ on $[0,1]$.

The blue region rotated about $y$-axis is labelled in the sidebar as “volume of cylinder of radius 1 unit and height $(5-(-1/3))$ units”. As printed, this says cylinder; the working correctly uses a cone. The native subtraction panels retain $x=(15-3y)/16$, $x=1-\sqrt{10/(3-y)-3}$, $5,1/2,-1/3,1$ and rotation arrows.

Printed answer key

(i) $(1,-1/3)$; (ii) labelled native graph below; (iii) $\frac13(5\pi/\sqrt3-2)$; (iv) $5.07$ units$^3$.

Source diagrams in native form

Mathematical review

The sidebar calls the blue triangular region a cylinder, but its rotation is a cone. The working uses the correct cone formula. This printed note is retained and explicitly corrected here.