smc-1039-20-Trig Function

Calculus, EXT1 C3 2022 SPEC1 10

Let `f(x)=\sec (4 x)`.

Sketch the graph of `f` for `x \in\left[-\frac{\pi}{4}, \frac{\pi}{4}\right]` on the set of axes below. Label any asymptotes with their equations and label any turning points and the endpoints with their coordinates.   (3 marks)
The graph of `y=f(x)` for `x \in\left[-\frac{\pi}{24}, \frac{\pi}{48}\right]` is rotated about the `x`-axis to form a solid of revolution.
Find the volume of this solid. Give your answer in the form `\frac{(a-\sqrt{b}) \pi}{c}`, where `a`, `b`, `c in R`. (3 marks)

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b. `\frac{(3-\sqrt{3}) \pi}{6}`

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♦ Mean mark (a) 49%.

b.	`V`	`=\pi \int_{-\frac{\pi}{24}}^{\frac{\pi}{48}} \sec ^2(4 x)\ dx`
		`=\frac{\pi}{4}[\tan (4 x)]_{-\frac{\pi}{24}}^{\frac{\pi}{48}}`
		`=\frac{\pi}{4} \tan \left(\frac{\pi}{12}\right)-\frac{\pi}{4} \tan \left(-\frac{\pi}{6}\right)`
		`=\frac{\pi}{4} \tan \left(\frac{\pi}{3}-\frac{\pi}{4}\right)-\frac{\pi}{4} \xx -\frac{1}{\sqrt{3}}`
		`=\frac{\pi}{4} \left(\frac{sqrt3-1}{1+sqrt3} xx \frac{1-sqrt3}{1-sqrt3}\right)+\frac{\pi}{4sqrt3}`
		`=\frac{\pi}{4} \left(\frac{sqrt3-3-1+sqrt3}{-2}\right)+\frac{\pi}{4sqrt3}`
		`=\frac{\pi}{4}(2-\sqrt{3}) +\frac{\pi}{4sqrt3}`
		`=\frac{\pi(2 \sqrt{3}-3+1)}{4 \sqrt{3}}`
		`=\frac{(6-2 \sqrt{3}) \pi}{12}`
		`=\frac{(3-\sqrt{3}) \pi}{6}`

♦ Mean mark (b) 55%.

Calculus, EXT1 C3 2022 HSC 13b

A solid of revolution is to be found by rotating the region bounded by the `x`-axis and the curve `y=(k+1) \sin (k x)`, where `k>0`, between `x=0` and `x=\frac{\pi}{2 k}` about the `x`-axis.

Find the value of `k` for which the volume is `pi^2`. (3 marks)

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`k=1`

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`y=(k+1) \sin (k x)`

`V`	`=pi int_0^((pi)/(2k)) (k+1)^2sin^(2)(kx)\ dx`
	`=pi(k+1)^2 int_0^((pi)/(2k)) 1/2[1-cos(2kx)]\ dx`
	`=(pi/2)(k+1)^2[x-(sin(2kx))/(2k)]_0^((pi)/(2k)) `
	`=(pi/2)(k+1)^2[(pi/(2k)- sin(pi)/(2k))-(0-sin0/(2k))]`
	`=(pi/2)(k+1)^2(pi/(2k))`
	`=pi^2/(4k)(k+1)^2`

`text{Given}\ \ V=pi^2:`

`pi^2/(4k)(k+1)^2`	`=pi^2`
`(k+1)^2`	`=4k`
`k^2+2k+1`	`=4k`
`k^2-2k+1`	`=0`
`(k-1)^2`	`=0`

`:.k=1`

Calculus, EXT1 C3 2021 SPEC1 4

The shaded region in the diagram below is bounded by the graph of `y = sin(x)` and the `x`-axis between the first two non-negative `x`-intercepts of the curve, that is interval `[0, pi]`. The shaded region is rotated about the `x`-axis to form a solid of revolution.

Find the volume, `V_s` of the solid formed. (3 marks)

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`(pi^2)/2\ text(u)³`

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	`V_s`	`= pi int_0^pi sin^2(x)\ dx`
		`= pi int_0^pi 1/2(1 – cos(2x))\ dx`
		`= pi/2 int_0^pi 1 – cos(2x)\ dx`
		`= pi/2 [x – 1/2 sin(2x)]_0^pi`
		`= pi/2[pi – 1/2 sin(2pi) – (0 – 1/2 sin 0)]`
		`= (pi^2)/2\ text(u)³`

Calculus, EXT1 C3 2020 HSC 13b

The region `R` is bounded by the `y`-axis, the graph of `y = cos(2x)` and the graph of `y = sin x`, as shown in the diagram.

Find the volume of the solid of revolution formed when the region `R` is rotated about the `x`-axis. (4 marks)

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`(3sqrt3 pi)/16\ text(u)³`

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`text(Find intersection:)`

Mean mark 53%.

`sin x = cos 2x`

`sin x = 1 – 2sin^2 x`

`2sin^2 x + sinx – 1`	`= 0`
`(2 sinx – 1)(sinx + 1)`	`= 0`

`sin x`	`= 1/2`	`text(or)`	`sin x`	`= −1`
`x`	`= pi/6`		`x`	`= (3pi)/2`

`V`	`= pi int_0^(pi/6) (cos 2x)^2\ dx – pi int_0^(pi/6)(sin x)^2\ dx`
	`= pi int_0^(pi/6) cos^2 2x – sin^2 x\ dx`
	`= pi int_0^(pi/6) 1/2 (1 + cos 4x) – 1/2 (1 – cos 2x)\ dx`
	`= pi/2 int_0^(pi/6) cos 4x + cos 2x\ dx`
	`= pi/2 [1/4 sin 4x + 1/2 sin 2x]_0^(pi/6)`
	`= pi/8 [sin\ (2pi)/3 + 2sin\ pi/3]`
	`= pi/8 (sqrt3/2 + 2 xx sqrt3/2)`
	`= (3sqrt3 pi)/16\ text(u)³`

Calculus, EXT1* C3 2004 HSC 4c

In the diagram, the shaded region is bounded by the curve `y = 2 sec x`, the coordinate axes and the line `x = pi/3`. The shaded region is rotated about the `x`-axis.

Calculate the exact volume of the solid of revolution formed. (3 marks)

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`4sqrt3\ pi\ \ text(u³)`

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`y`	`= 2 sec x`
`:. y^2`	`= 4 sec^2 x`

`V`	`= pi int_0^(pi/3) y^2\ dx`
	`= pi int_0^(pi/3) 4 sec^2 x\ dx`
	`= 4pi int_0^(pi/3) sec^2 x\ dx`
	`= 4pi[tan x]_0^(pi/3)`
	`= 4pi(tan\ pi/3 − tan 0)`
	`= 4pi(sqrt3 − 0)`
	`= 4sqrt3\ pi\ \ text(u³)`

Calculus, EXT1 C3 2005 HSC 5a

Find the exact value of the volume of the solid of revolution formed when the region bounded by the curve `y = sin 2x`, the `x`-axis and the line `x = pi/8` is rotated about the `x`-axis. (3 marks)

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`pi/16 (pi – 2)\ \ \ text(u³)`

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`y`	`= sin 2x`
`y^2`	`= sin^2 2x`

`text(Using:)\ \ sin^2 x= 1/2 (1 – cos 2x)`

COMMENT: Michael Wells (1st in state Ext2) would derive this formula in his working from the `sin^2x+cos^2=1` identity in ~5 seconds every time he used it in an exam.

`:. V`	`=pi int_0^(pi/8) y^2 \ dx`
	`= pi int_0^(pi/8) sin^2 2x \ dx`
	`= pi/2 int_0^(pi/8) 1 – cos\ 4x\ dx`
	`= pi/2 [x – 1/4 sin\ 4x]_0^(pi/8)`
	`= pi/2 [(pi/8 – 1/4 sin\ pi/2) – 0]`
	`= pi/2 (pi/8 – 1/4)`
	`= pi/2 ((pi – 2)/8)`
	`= pi/16 (pi – 2)\ \ \ text(u³)`

Calculus, EXT1 C3 2014 HSC 12b

The region bounded by `y = cos 4x` and the `x`-axis, between `x = 0` and `x = pi/8`, is rotated about the `x`-axis to form a solid.

Find the volume of the solid. (3 marks)

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`(pi^2)/16\ \ text(u³)`

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COMMENT: The identities `cos 2theta=` `cos^2 theta-sin^2 theta =` ` 2cos^2 theta-1=1-2sin^2 theta` are tested every year – know them.

`V`	`= pi int_0^(pi/8) y^2\ dx`
	`= pi int_0^(pi/8) cos^2 4x\ dx`
	`= pi int_0^(pi/8) 1/2 (cos 8x + 1)\ dx`
	`= pi/2 [1/8 sin 8x + x]_0^(pi/8)`
	`= pi/2 [(1/8 sin pi + pi/8)] – 0]`
	`= (pi^2)/16\ \ text(u³)`

Calculus, EXT1 C3 2013 HSC 12b

The region bounded by the graph `y = 3 sin\ x/2` and the `x`-axis between `x = 0` and `x = (3pi)/2` is rotated about the `x`-axis to form a solid.

Find the exact volume of the solid. (3 marks)

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`(9pi)/2 ((3pi)/2 + 1) text(u³)`

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`y`	`= 3 sin\ x/2`
`y^2`	`= 9 sin^2\ x/2`

`text(Using:)\ \ sin^2x= 1/2 (1 – cos 2x)`

`:. V`	`= pi int_0^((3pi)/2) 9 sin^2\ x/2\ dx`
	`= (9pi)/2 int_0^((3pi)/2) (1\ – cosx)\ dx`
	`= (9pi)/2 [x\ – sinx]_0^((3pi)/2)`
	`= (9pi)/2 [((3pi)/2\ – sin\ (3pi)/2)\ – 0]`
	`= (9pi)/2 ((3pi)/2 + 1)\ text(u³)`

Calculus, EXT1* C3 2009 HSC 6a

The diagram shows the region bounded by the curve `y = sec x`, the lines `x = pi/3` and `x = -pi/3`, and the `x`-axis.

The region is rotated about the `x`-axis. Find the volume of the solid of revolution formed. (3 marks)

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`2 sqrt 3 pi\ text(u³)`

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`V`	`= pi int_(-pi/3)^(pi/3) y^2\ dx`
	`= pi int_(-pi/3)^(pi/3) sec^2x\ dx`
	`= pi [tanx]_(-pi/3)^(pi/3)`
	`= pi[tan(pi/3) – tan(-pi/3)]`
	`= pi [sqrt3\ – (-sqrt3)]`
	`= 2 sqrt3 pi\ text(u³)`