Exam-Style Problem

9231 P11 - Jun 2013 - Q11 - 11 marks

6396

Answer only one of the following two alternatives.

EITHER
The curve \(C\) has equation \(y=2 \sec x\), for \(0 \leqslant x \leqslant \frac{1}{4} \pi\). Show that the arc length \(s\) of \(C\) is given by
\(s=\int_{0}^{\frac{1}{4} \pi}\left(2 \sec ^{2} x-1\right) \mathrm{d} x\)

Find the exact value of \(s\).

The surface area generated when \(C\) is rotated through \(2 \pi\) radians about the \(x\)-axis is denoted by \(S\). Show that
(i) \(S=4 \pi \int_{0}^{\frac{1}{4} \pi}\left(2 \sec ^{3} x-\sec x\right) \mathrm{d} x\),
(ii) \(\frac{\mathrm{d}}{\mathrm{d} x}(\sec x \tan x)=2 \sec ^{3} x-\sec x\).

Hence find the exact value of \(S\).

The points \(A, B, C\) and \(D\) have coordinates as follows:
\(A(2,1,-2), \quad B(4,1,-1), \quad C(3,-2,-1) \quad \text { and } \quad D(3,6,2) .\)

The plane \(\Pi_{1}\) passes through the points \(A, B\) and \(C\). Find a cartesian equation of \(\Pi_{1}\).

Find the area of triangle \(A B C\) and hence, or otherwise, find the volume of the tetrahedron \(A B C D\).

[The volume of a tetrahedron is \(\frac{1}{3} \times\) area of base × perpendicular height.]
The plane \(\Pi_{2}\) passes through the points \(A, B\) and \(D\). Find the acute angle between \(\Pi_{1}\) and \(\Pi_{2}\).

Solution

Answer: EITHER

For \(y=2\sec x\),

\(\dfrac{dy}{dx}=2\sec x\tan x\).

The arc length is

\(s=\int_0^{\pi/4}\sqrt{1+\left(2\sec x\tan x\right)^2}\,dx\).

Using \(\tan^2x=\sec^2x-1\),

\(1+4\sec^2x\tan^2x=1+4\sec^2x(\sec^2x-1)=4\sec^4x-4\sec^2x+1=(2\sec^2x-1)^2\).

So on \(0\le x\le \pi/4\),

\(\sqrt{1+(y')^2}=2\sec^2x-1\),

hence

\(s=\int_0^{\pi/4}(2\sec^2x-1)\,dx=[2\tan x-x]_0^{\pi/4}=2-\dfrac{\pi}{4}.\)

Also, for the surface area about the \(x\)-axis,

\(S=2\pi\int_0^{\pi/4}y\sqrt{1+(y')^2}\,dx\)

\(=2\pi\int_0^{\pi/4}2\sec x(2\sec^2x-1)\,dx=4\pi\int_0^{\pi/4}(2\sec^3x-\sec x)\,dx.\)

Now

\(\dfrac{d}{dx}(\sec x\tan x)=\sec x\tan^2x+\sec^3x\)

\(=\sec x(\sec^2x-1)+\sec^3x=2\sec^3x-\sec x.\)

Therefore

\(S=4\pi[\sec x\tan x]_0^{\pi/4}=4\pi\sqrt2.\)

EITHER

Differentiate first:

\(y=2\sec x\implies \dfrac{dy}{dx}=2\sec x\tan x.\)

Using the arc-length formula,

\(s=\int_0^{\pi/4}\sqrt{1+\left(\dfrac{dy}{dx}\right)^2}\,dx=\int_0^{\pi/4}\sqrt{1+4\sec^2x\tan^2x}\,dx.\)

Since \(\tan^2x=\sec^2x-1\),

\(1+4\sec^2x\tan^2x=1+4\sec^2x(\sec^2x-1)=4\sec^4x-4\sec^2x+1=(2\sec^2x-1)^2.\)

Hence

\(\sqrt{1+4\sec^2x\tan^2x}=2\sec^2x-1\),

\(s=\int_0^{\pi/4}(2\sec^2x-1)\,dx.\)

Integrating,

\(s=[2\tan x-x]_0^{\pi/4}=2-\dfrac{\pi}{4}.\)

For the surface area formed by rotating about the \(x\)-axis,

\(S=2\pi\int_0^{\pi/4}y\sqrt{1+\left(\dfrac{dy}{dx}\right)^2}\,dx.\)

Substitute \(y=2\sec x\) and \(\sqrt{1+(y')^2}=2\sec^2x-1\):

\(S=2\pi\int_0^{\pi/4}2\sec x(2\sec^2x-1)\,dx\)

\(=4\pi\int_0^{\pi/4}(2\sec^3x-\sec x)\,dx.\)

To evaluate this integral, differentiate \(\sec x\tan x\):

\(\dfrac{d}{dx}(\sec x\tan x)=(\sec x)'\tan x+\sec x(\tan x)'\)

\(=\sec x\tan x\cdot\tan x+\sec x\sec^2x\)

\(=\sec x\tan^2x+\sec^3x\)

\(=\sec x(\sec^2x-1)+\sec^3x\)

\(=2\sec^3x-\sec x.\)

\(S=4\pi[\sec x\tan x]_0^{\pi/4}.\)

Now \(\sec(\pi/4)=\sqrt2\) and \(\tan(\pi/4)=1\), while at \(x=0\), \(\sec0\tan0=0\). Therefore

\(S=4\pi(\sqrt2-0)=4\pi\sqrt2.\)