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Calculus, MET2 2020 VCAA 2

An area of parkland has a river running through it, as shown below. The river is shown shaded.

The north bank of the river is modelled by the function  `f_(1):[0,200]rarr R, \ f_(1)(x)=20 cos((pi x)/(100))+40`.

The south bank of the river is modelled by the function  `f_(2):[0,200]rarr R, \ f_(2)(x)=20 cos((pi x)/(100))+30`.

The horizontal axis points east and the vertical axis points north.

All distances are measured in metres.
 

A swimmer always starts at point `P`, which has coordinates  (50, 30).

Assume that no movement of water in the river affects the motion or path of the swimmer, which is always a straight line.

  1. The swimmer swims north from point `P`.
  2. Find the distance, in metres, that the swimmer needs to swim to get to get to the north bank of the river.   (1 mark)

  3. The swimmer swims east from point `P`.
  4. Find the distance, in metres, that the swimmer needs to swim to get to the north bank of the river.   (2 marks)

  5. On another occasion, the swimmer swims the minimum distance from point `P` to the north bank of the river.
  6. Find this minimum distance. Give your answer in metres, correct to one decimal place.   (2 marks)

  7. Calculate the surface area of the section of the river shown on the graph in square metres.   (1 mark)

  8. A horizontal line is drawn through point `P`. The section of the river that is south of the line is declared a no "no swimming" zone.
  9. Find the area of the "no swimming" zone, correct to the nearest square metre.   (3 marks)

  10. Scientists observe that the north bank of the river is changing over time. It is moving further north from its current position. They model its predicted new location using the function with rule  `y=kf_(1)(x)`, where `k >= 1`.
  11. Find the values of  `k` for which the distance north across the river, for all parts of the river, is strictly less than 20 m.   (2 marks)

Show Answers Only
  1. `10\ text{m}`
  2. `16 2/3\ text{m}`
  3. `8.5\ text{m}`
  4. `2000\ text{m}^2`
  5. `837\ text{m²}`
  6. `k in [1, 7/6)`
Show Worked Solution

a.   `text{Since swimmer swims due north,}`

`text{Distance}\ = 40-30=10\ text{m}`

 

b.  `text{Solve} \ f_(1)(x)=30 \ text{or} \ x in[50,100]`

`=> x=200/3`

`:.\ text{Distance to swim (east) to reach north bank}`

`=200/3-50`

`=16 2/3\ text{m}`
 

c.   `text{Let swimmer arrive at north bank at the point}\ \ (x,f_(1)(x))`

`text{By Pythagoras,}`

♦ Mean mark part (c) 39%.

`d(x)=sqrt((x-50)^(2)+(f_(1)(x)-30)^(2))`

`text{Solve} \ d/dx(d(x))=0 \ text{for} \ x:`

`x=54.47…`

`:. d_min=8.5\ text{m (to 1 d.p.)}`

 

d.   `text{Shaded Area}`

`=int_(0)^(200)(f_(1)(x)-f_(2)(x))\ dx`

`=2000\ text{m}^2`

 

e.   `text{Find}\ \ f_(1)(x) = 30 \ text{for} x in [50,150]:`

♦♦ Mean mark part (e) 35%.

`=>x=200/3, 400/3`

`text{Find}\ \ f_(2)(x) = 30 \ text{for} \ x in [50,150]:`

`=>x=50, 150`

`:.\ text{Area}` `=int_(50)^(150)(30-f_(2)(x))\ dx-int_((200)/(3))^((400)/(3))(30-f_(1)(x))\ dx`  
  `=837\ text{m² (to nearest m²)}`  

♦♦♦Mean mark part (f) 15%.

 

f.   `text{Let}\ \ D(x)=\ text{vertical distance between banks}`

`D(x)` `=kf_(1)(x)-f_(2)(x)`  
  `=20k cos((pi x)/(100))+40k-(20 cos((pi x)/(100))+30)`   
  `=(20k-20)cos((pi x)/(100)) +40k-30`  

  
`text{Given}\ \ D(x)<20 \ text{for}\ x in[0,200]`

`text{Maximum} \ cos((pi x)/(100)) = 1\  text{when}\ \ x=0, 200`

`text{Solve} \ 20k-20+40k-30<20\ \ text{for}\ k:`

`=> k<7/6`

`:. k in [1,7/6)`

Calculus, MET2 2021 VCAA 5

Part of the graph of  `f: R to R , \ f(x) = sin (x/2) + cos(2x)`  is shown below.
 

  1. State the period of `f`.   (1 mark)

  2. State the minimum value of `f`, correct to three decimal places.   (1 mark)

  3. Find the smallest positive value of `h` for which  `f(h-x) = f(x)`.   (1 mark)
Consider the set of functions of the form  `g_a : R to R, \ g_a (x) = sin(x/a) + cos(ax)`, where `a` is a positive integer.
  1. State the value of `a` such that  `g_a (x) = f(x)`  for all `x`.   (1 mark)

  2. i.  Find an antiderivative of `g_a` in terms of `a`.   (1 mark)

  3. ii. Use a definite integral to show that the area bounded by `g_a` and the `x`-axis over the interval  `[0, 2a pi]`  is equal above and below the `x`-axis for all values of `a`.  (3 marks)

  4. Explain why the maximum value of `g_a` cannot be greater than 2 for all values of `a` and why the minimum value of `g_a` cannot be less than –2 for all values of `a`.   (1 mark)

  5. Find the greatest possible minimum value of `g_a`.   (1 mark)

Show Answers Only
  1. `4 pi`
  2. `-1.722`
  3. `2 pi`
  4. `text{See Worked Solutions}`
  5. i.  `text{See Worked Solutions}`
    ii. `text{See Worked Solutions}`
  6. `text{See Worked Solutions}`
  7. `-sqrt2`
Show Worked Solution

a.    `text{By inspection, graph begins to repeat after 4pi.}`

`text{Period}\ = 4 pi`
 

b.    `text{By CAS: Sketch}\ \ f(x) = sin (x/2) + cos(2x)`

`f_min = -1.722`

♦ Mean mark part (c) 21%.
 

c.     `text{If} \ \ f(x)\ \ text{is reflected in the} \ y text{-axis and translated} \ 2 pi \ text{to the right} => text{same graph}`

`f(x) = f(-x + h) = f(2 pi-x)`

`:. h = 2 pi`
 

d.    `f(x) = sin(x/2) + cos(2x)`

`g_a(x) = sin(x/a) + cos(2a)`

`g_a(x) = f(x) \ \ text{when} \ \ a = 2`
 

e.i.  `int g_a (x)\ dx`

♦ Mean mark part (e)(i) 50%.

`= -a cos (x/a) + {sin (ax)}/{a} , \ (c = 0)`
 

e.ii.   `int_0^{2a pi} g_a(x)\ dx`

♦ Mean mark part (e)(ii) 29%.

`= {sin (2a^2 pi)}/{a}`

`= 0 \ \ (a ∈ ZZ^+)`
 

`text{When integral = 0, areas above and below the} \ x text{-axis are equal.}`
 

f.    `g_a (x) = sin(x/a) + cos (ax)`

♦ Mean mark part (f) 13%.

`-1 <= sin(x/a) <= 1 \ \ text{and}\ \ -1 <= cos(ax) <= 1`

`:. -2 <= g_a (x) <= 2`
 

g.    `text{Sketch}\ \ g_a (x) \ \ text{by CAS}`

♦ Mean mark part (g) 2%.

`text{Minimum access at}\ \ a = 1`

`g_a(x)_min = – sqrt2`

Calculus, MET1-NHT 2018 VCAA 7

Let  `f : [ 0, (pi)/(2)] → R, \ f(x) = 4 cos(x)`  and  `g : [0, (pi)/(2)]  → R, \ g(x) = 3 sin(x)`.

  1. Sketch the graph of `f` and the graph of `g` on the axes provided below.   (2 marks)

     
     
    `qquad qquad `
     

  2. Let `c` be such that  `f(c) = g(c)`,  where  `c∈[0, (pi)/(2)]`

     
    Find the value of  `sin(c)`  and the value of  `cos(c)`.   (3 marks)

     

  3.  Let `A` be the region enclosed by the horizontal axis, the graph of `f` and the graph of `g`.
    1. Shade the region `A` on the axes provided in part a. and also label the position of `c` on the horizontal axis.   (1 mark)

    2. Calculate the area of the region `A`.   (3 marks)

Show Answers Only
  1.  

     

  2. `sin(c) = (4)/(5), \ cos(c) = (3)/(5)`

     

  3. i.

     
    ii. `2 \ u^2`
Show Worked Solution

a.   

 

b.   `text(At intersection:)`

`4cos(c)` `= 3sin(c)`
`tan(c)` `= (4)/(3)`

`sin(c) = (4)/(5)`

`cos(c) = (3)/(5)`

 

c. i.

 

   ii.       `A` `= int_0^c g(x)\ dx + int_c^((pi)/(2)) f(x)\ dx`
  `= int_0^c 3sin x \ dx + int_c^((pi)/(2)) 4cos \ x \ dx`
  `= 3[-cos x]_0^c + 4[sin x]_c^((pi)/(2))`
  `= 3(-cos(c) + cos \ 0) + 4(sin \ (pi)/(2)-sin(c))`
  `= 3(-(3)/(5) + 1) + 4(1-(4)/(5))`
  `= (6)/(5) + (4)/(5)`
  `= 2 \ \ text(u²)`

Calculus, MET1-NHT 2019 VCAA 7

The shaded region in the diagram below is bounded by the vertical axis, the graph of the function with rule  `f(x) = sin(pix)`  and the horizontal line segment that meets the graph at  `x = a`, where  `1 <= a <= 3/2`.
 


 

Let  `A(a)`  be the area of the shaded region.

  1. Show that  `A(a) = 1/pi-1/pi cos(a pi)-a sin (a pi)`.   (3 marks)

  2. Determine the range of values of `A(a)`.   (2 marks)

    1. Express in terms of  `A(a)`, for a specific value of `a`, the area bounded by the vertical axis, the graph of  `y = 2(sin(pi x) + sqrt 3/2)`  and the horizontal axis.   (2 marks)

    2. Hence, or otherwise, find the area described in part c.i.   (1 mark)

Show Answers Only
  1. `text(Proof)\ \ text{(See Worked Solutions)}`
  2. `[2/pi, (2 + 3 pi)/(2pi)]`
    1. `2 A(a)`
    2. `(9 + 4 sqrt 3 pi)/(3 pi)`
Show Worked Solution

a.   `text(Lower border of shaded area:)\ y = f(a) = sin(a pi)`

`text(Area)` `= int_0^a sin(pi x)-sin (a pi)\ dx`
  `= [-1/pi cos (pi x)-x sin(a pi)]_0^a`
  `= [-1/pi cos(pi a)-a sin (a pi)-(-1/pi-0)]`
  `= 1/pi-1/pi cos (a pi)-a sin(a pi)`

 

b.   `text(S) text(ince)\ 1 <= a <= 3/2,`

`A(1) = 1/pi-1/pi cos(pi)-1 sin (pi) = 2/pi`

`A(3/2) = 1/pi-1/pi cos ((3 pi)/2)-3/2 sin ((3pi)/2) = 1/pi + 3/2 = (2 + 3pi)/(2 pi)`
 

`:.\ text(Range:)\ \ [2/pi, (2 + 3pi)/(2 pi)]`

 

c.i.  `A(a) = int_0^a sin(pi x)-sin (a pi)\ dx`

`A_1` `=2int _0^a sin(pi x) + sqrt 3/2\ dx`   
  `=2int _0^a sin(pi x)-sin((4pi)/3)\ dx,\ \ \ (a=4/3)`  
  `=2int _0^(4/3) sin(pi x)-sin((4pi)/3)\ dx`  
  `=2 xx A(a)`  

 
`:.\ text(When)\ \ a = 4/3,\ \ text(Area) = 2 xx A(a)`

 

c.ii.  `text(When)\ \ a = 4/3`

`text(Area)` `= 2 xx (1/pi-1/pi cos ((4 pi)/3)-4/3 sin ((4 pi)/3))`
  `= 2 xx (1/pi + 1/(2 pi) + 4/3 xx sqrt 3/2)`
  `= 2(3/(2pi) + (2 sqrt 3)/3)`
  `= 3/pi + (4 sqrt 3)/3`
  `= (9 + 4 sqrt 3 pi)/(3 pi)`

Calculus, MET2 2019 VCAA 3

During a telephone call, a phone uses a dual-tone frequency electrical signal to communicate with the telephone exchange.

The strength, `f`, of a simple dual-tone frequency signal is given by the function  `f(t) = sin((pi t)/3) + sin ((pi t)/6)`, where  `t`  is a measure of time and  `t >= 0`.

Part of the graph of `y = f(t)`  is shown below

  1. State the period of the function.   (1 mark)

  2. Find the values of  `t`  where  `f(t) = 0`  for the interval  `t in [0, 6]`.   (1 mark)

  3. Find the maximum strength of the dual-tone frequency signal, correct to two decimal places.   (1 mark)

  4. Find the area between the graph of  `f`  and the horizontal axis for  `t in [0, 6]`.   (2 marks)

Let  `g`  be the function obtained by applying the transformation  `T`  to the function  `f`, where
 

`T([(x), (y)]) = [(a, 0), (0, b)] [(x), (y)] + [(c), (d)]`
 

and `a, b, c` and `d` are real numbers.

  1. Find the values of `a, b, c` and `d` given that  `int_2^0 g(t)\ dt + int_2^6 g(t)\ dt`  has the same area calculated in part d.   (2 marks)

  2. The rectangle bounded by the line  `y = k, \ k in R^+`, the horizontal axis, and the lines  `x = 0`  and  `x = 12`  has the same area as the area between the graph of  `f`  and the horizontal axis for one period of the dual-tone frequency signal.

     

    Find the value of  `k`.   (2 marks)

Show Answers Only
  1. `12`
  2. `0, 4, 6`
  3. `1.760`
  4. `15/pi\ text(u²)`
  5. `a = 1,\ b =-1,\ c =-6,\ d = 0`
  6. `5/(2pi)`
Show Worked Solution

a.   `text(Period) = 12`
  

b.   `t = 0, 4, 6`
  

c.   `f(t) = sin ((pi t)/3) + sin ((pi t)/6)`

`f(t)_max ~~ 1.76\ \ text{(by CAS)}`

 

d.    `text(Area)` `= int_0^4 sin((pi t)/3) + sin ((pi t)/6) dt-int_4^6 sin ((pi t)/3) + sin ((pi t)/6) dt`
    `= 15/pi\ text(u²)`

 

e.   `text(Same area) => f(t)\ text(is reflected in the)\ x text(-axis and)`

`text(translated 6 units to the left.)`

`x′=ax+c`

`y′=by+d`

`text(Reflection in)\ xtext(-axis) \ => \ b=-1, \ d=0`

`text(Translate 6 units to the left) \ => \ a=1, \ c=-6`

`:. a = 1,\ b = -1,\ c = -6,\ d = 0`
  

f.    `text(Area of rectangle)` `= 2 xx text(Area between)\ f(t) and x text(-axis)\ \ t in [0, 6]`
  `12k` `= 2 xx 15/pi`
  `:. k` `=5/(2pi)`

Calculus, MET1 2018 VCAA 9

Consider a part of the graph of  `y = xsin(x)`, as shown below.

  1.  i. Given that  `int(xsin(x))\ dx = sin(x)-xcos(x) + c`, evaluate  `int_(npi)^((n + 1)pi)(xsin(x))\ dx`  when `n` is a positive even integer or 0.
      
    Give your answer in simplest form.   (2 marks)

    ii. Given that  `int(xsin(x))\ dx = sin(x)-xcos(x) + c`, evaluate  `int_(npi)^((n + 1)pi)(xsin(x))\ dx`  when `n` is a positive odd integer.
    Give your answer in simplest form.   (1 mark)

  2.  
  3. Find the equation of the tangent to  `y = xsin(x)`  at the point  `(−(5pi)/2,(5pi)/2)`.   (2 marks)

  4. The translation `T` maps the graph of  `y = xsin(x)`  onto the graph of  `y = (3pi-x)sin(x)`, where
  5. `qquad T: R^2 -> R^2, T([(x),(y)]) = [(x),(y)] + [(a),(0)]`
  6. and `a` is a real constant.
  7. State the value of `a`.   (1 mark)

  8. Let  `f:[0,3pi] -> R, f(x) = (3pi-x)sin(x)`  and  `g:[0,3pi] -> R, g(x) = (x-3pi)sin(x)`.
    The line `l_1` is the tangent to the graph of `f` at the point `(pi/2,(5pi)/2)` and the line `l_2` is the tangent to the graph of `g` at `(pi/2,-(5pi)/2)`, as shown in the diagram below.
     
         
    Find the total area of the shaded regions shown in the diagram above.   (2 marks)

Show Answers Only
  1.  i. `(2n + 1)pi`
    ii. `-(2n + 1)pi`
  2. `:. y =-x`
  3. `-3pi`
  4. `9pi(pi-2)`
Show Worked Solution

a.i.  `text(Given)\ \ n\ \ text(is a positive even integer:)`

♦♦ Mean mark 31%.

`int_(npi)^((n + 1)pi)(xsin(x))dx`

`= [sin(x)-xcos(x)]_(npi)^((n + 1)pi)`

`= [sin((n + 1)pi)-(n + 1)pi · cos((n + 1)pi)]-[sin(npi)-npi · cos(npi)]`

`= [0-(n + 1)pi (−1)]-[0-npi]`

`= (n + 1)pi + npi`

`= (2n + 1)pi`

 

a.ii. `text(Given)\ \ n\ \ text(is a positive odd integer:)`

♦♦♦ Mean mark 19%.

`int_(npi)^((n + 1)pi)(xsin(x))dx`

`= [sin((n + 1)pi)-(n + 1)pi · cos((n + 1)pi)]-[sin(npi)-npi · cos(npi)]`

`= [0-(n + 1)pi(1)]-[0-npi(−1)]`

`= -(n + 1)pi-npi`

`= -(2n + 1)pi`

 

b.    `y` `= xsin(x)`
  `(dy)/(dx)` `= x · cos(x) + sin(x)`

 

♦ Mean mark part (b) 49%.

`(dy)/(dx)` `= -(5pi)/2 · cos(-(5pi)/2) + sin(-(5pi)/2)`
  `= -(5pi)/2 · (0) + (-1)`
  `= -1`

 
`text(T)text(angent has equation)\ \ y =-x + c\ \ text(and passes through)\ \ (-(5pi)/2, (5pi)/2):`

`(5pi)/2` `= +(5pi)/2 + c`
`c` `= 0`

 
`:. y = -x`

 

♦♦ Mean mark part (c) 34%.

c.    `y` `= (3pi-x)sin(x)`
    `=-(x-3pi)sin(x)`

 
`:. a = 3pi`
 

d.   `f(x) = (3pi-x)sin(x)`

♦♦♦ Mean mark 7%.

  `-> l_1\ text(is the tangent)\ \ y =-x\ \ (text{using part (b)})`

`-> g(x)\ text(is)\ \ y=xsin(x)\ \ text(translated 3π to the right.)`

`-> f(x)\ text(is)\ \ g(x)\ \ text(reflected in the)\ \ x text(-axis.)`

 
`text(Area between)\ f(x)\ text(and)\ xtext(-axis)`

`= int_0^pi xsin(x)\ dx + | int_pi^(2pi) xsin(x)\ dx | + int_(2pi)^(3pi) xsin(x)\ dx`

`= (2 xx 0 + 1)pi + |-1(2 xx 1 + 1)pi | + (2 xx 2 + 1)pi\ \ (text{using part (a)})`

`= pi + 3pi + 5pi`

`= 9pi`
 

`:.\ text(Shaded Area)`

`= 2 xx (1/2 xx 3pi xx 3pi)-2 xx 9pi`

`= 9pi^2-18pi`

`= 9pi(pi-2)`

Calculus, MET2 2018 VCAA 19 MC

The graphs  `f: R -> R,\ f(x) = cos ((pi x)/2)`  and  `g: R -> R,\ g(x) = sin (pi x)` are shown in the diagram below.
 


 

An integral expression that gives the total area of the shaded regions is

A.   `int_0^3 (sin(pi x) - cos ((pi x)/2)) dx`

B.   `2 int_(5/3)^3 (sin(pi x) - cos((pi x)/2))dx`

C.   `int_0^(1/3)(cos((pi x)/2) - sin (pi x)) dx - 2 int_(1/3)^1(cos((pi x)/2) - sin (pi x)) dx`

                `- int_(5/3)^3 (cos ((pi x)/2) - sin (pi x)) dx`

D.   `2 int_1^(5/3)(cos((pi x)/2) - sin (pi x)) dx - 2 int_(5/3)^3 (cos ((pi x)/2) - sin (pi x)) dx`

E.   `int_0^(1/3) (cos((pi x)/2) - sin (pi x)) dx + 2 int_(1/3)^1(sin (pi x) - cos ((pi x)/2))dx`

                `+ int_(5/3)^3 (cos((pi x)/2) - sin (pi x)) dx`

Show Answers Only

`C`

Show Worked Solution

`text(The graph contains 2 symmetrical shaded areas:)`

♦ Mean mark 41%.
COMMENT: In each shaded area, it is critical to note which graph is “higher” and which side of the x-axis the shading is.
 

`:.\ text(Area can be represented by:)`

`int_0^(1/3)(cos((pi x)/2) – sin (pi x)) dx – 2 int_(1/3)^1(cos((pi x)/2) – sin (pi x)) dx`

         `- int_(5/3)^3 (cos ((pi x)/2) – sin (pi x)) dx`

 
`=>   C`

Calculus, MET2 2018 VCAA 16 MC

Jamie approximates the area between the `x`-axis and the graph of  `y = 2 cos(2x) + 3`, over the interval  `[0, pi/2]`, using the three rectangles shown below.
 


 

Jamie’s approximation as a fraction of the exact area is

A.   `5/9`

B.   `7/9`

C.   `9/11`

D.   `11/18`

E.   `7/3`

Show Answers Only

`B`

Show Worked Solution

`text(Area of rectangles)`

♦ Mean mark 49%.

`= pi/6 xx f(pi/6) + pi/6 xx f(pi/3) + pi/6 xx f(pi/2)`

`= (7pi)/6\ text(u²)`

`text(Actual Area) = int_0^(pi/2) 2 cos (2x) + 3\ dx = (3 pi)/2`

 
`:.\ text(Fraction of exact area)`

`= (7 pi)/6 ÷ (3 pi)/2`

`= 7/9`

 
`=>   B`

Calculus, MET2 2017 VCAA 20 MC

The graphs of  `f:[0,pi/2] -> R,  f(x) = cos(x)`  and  `g:[0,pi/2] -> R,  g(x) = sqrt3 sin(x)`  are shown below.

The graph intersect at `B`.

 

The ratio of the area of the shaded region to the area of triangle `OAB` is

  1. `9:8`
  2. `sqrt3 - 1 : (sqrt3 pi)/8`
  3. `8sqrt3 - 3 : 3pi`
  4. `sqrt3 - 1 : (sqrt3 pi)/4`
  5. `1 : (sqrt3 pi)/8`
Show Answers Only

`B`

Show Worked Solution

`text(Find intersection point)\ B:`

`cos(x)` `=sqrt3 sin(x)`
`tan x` `=1/sqrt3`
`x` `=pi/6`

`:. B(pi/6,sqrt3/2)`

 

`A_text(shaded) : A_Δ`

`int_0^(pi/6) sqrt3 sin(x)\ dx + int_(pi/6)^(pi/2) cos(x)\ dx\ :\ 1/2 xx pi/2 xx sqrt3/2`

`sqrt3-1\ :\ (sqrt3 pi)/8`

`=> B`

Calculus, MET2 2008 VCAA 1 MC

VCAA 2008 1mc

The area under the curve  `y = sin (x)`  between  `x = 0`  and  `x  = pi/2`  is approximated by two rectangles as shown.

This approximation to the area is

A.   `1`

B.   `pi/2`

C.   `((sqrt 3 + 1) pi)/12`

D.   `0.5`

E.   `((sqrt 3 + 1) pi)/6`

Show Answers Only

`C`

Show Worked Solution

`text(Rectangle width) = pi/6`

`text(Area)` `~~ pi/6 [sin (pi/6) + sin (pi/3)]`
  `~~pi/6(sqrt3/2 + 1/2)`
  `~~ (pi (sqrt 3 + 1))/12`

`=>   C`

Calculus, MET1 SM-Bank 28

The function  `f`  has the rule  `f(x) = 1 + 2 cos x`.

  1. Show that the graph of  `y = f(x)`  cuts the `x`-axis at  `x = (2 pi)/3`.   (1 mark)

  2. Sketch the graph  `y = f(x)`  for  `x  in [-pi,pi]`  showing where the graph cuts each of the axes.   (2 marks)

  3. Find the area under the curve  `y = f(x)`  between  `x = -pi/2`  and  `x = (2 pi)/3`.   (3 marks)

Show Answers Only
  1. `text(Proof)\ \ text{(See Worked Solutions)}` 
  2.  
       
  3. `((7 pi)/6 + sqrt 3 + 1)\ text(u²)`
Show Worked Solution

a.   `f(x) = 1 + 2 cos x`

`f(x)\ text(cuts the)\ x text(-axis when)\ f(x) = 0`

`1 + 2 cos x` `= 0`
`2 cos x` `=-1`
 `cos x` `= -1/2`

 
`:.  x = (2 pi)/3\ …\ text(as required)`

 

b.   2UA HSC 2006 7b

 

c.  `text(Area)` `= int_(-pi/2)^((2 pi)/3) 1 + 2 cos x\ \ dx`
  `= [x + 2 sin x]_(-pi/2)^((2 pi)/3)`
  `= [((2 pi)/3 + 2 sin­ (2 pi)/3)-((-pi)/2 + 2 sin­ (-pi)/2)]`
  `= ((2 pi)/3 + 2 xx sqrt 3/2)-((-pi)/2 +2(- 1))`
  `= (2 pi)/3 + sqrt(3) + pi/2 + 2`
  `= ((7 pi)/6 + sqrt(3) + 2)\ text(u²)`

Calculus, MET2 2015 VCAA 4

An electronics company is designing a new logo, based initially on the graphs of the functions

`f(x) = 2 sin (x) and g(x) = 1/2 sin (2x),\ text(for)\ 0 <= x <= 2 pi`

These graphs are shown in the diagram below, in which the measurements in the `x` and `y` directions are in metres.

VCAA 2015 4a

The logo is to be painted onto a large sign, with the area enclosed by the graphs of the two functions (shaded in the diagram) to be painted red.

  1. The total area of the shaded regions, in square metres, can be calculated as  `a int_0^pi sin(x)\ dx`.
  2. What is the value of `a`?   (1 mark)

The electronics company considers changing the circular functions used in the design of the logo.

Its next attempt uses the graphs of the functions  `f(x) = 2 sin(x) and h(x) = 1/3 sin (3x),\ text(for)\ \ 0 <= x <= 2 pi`.

  1. On the axes below, the graph of  `y = f(x)`  has been drawn.
  2. On the same axes, draw the graph of  `y = h(x)`.   (2 marks)

     

          VCAA 2015 4b

  3. State a sequence of two transformations that maps the graph of  `y = f (x)`  to the graph of  `y = h(x)`.   (2 marks)

The electronics company now considers using the graphs of the functions  `k(x) = m sin(x) and q (x) = 1/n sin (nx)`, where `m` and `n` are positive integers with  `m >= 2` and `0<= x <= 2pi`.

    1. Find the area enclosed by the graphs of  `y = k(x)`  and  `y = q(x)`  in terms of `m` and `n` if `n` is even.
    2. Give your answer in the form  `am + b/n^2`, where `a` and `b` are integers.   (2 marks)

    3. Find the area enclosed by the graphs of  `y = k(x)`  and  `y = q(x)`  in terms of `m` and `n` if `n`is odd.
    4. Give your answer in the form  `am + b/n^2`, where `a` and `b` are integers.   (2 marks)

Show Answers Only
  1. `4`

  2.  

    vcaa-2015-4b-answer   

  3. `text(See Worked Solutions)`
    1. `4m + 0/n^2`
    2. `4m + (−4)/(n^2)`
Show Worked Solution
a.   `text(Area)` `= 2 xx int_0^pi (2 sin (x))\ dx`
    `= 4 xx int_0^pi sin (x)\ dx`

`:. a = 4`

♦ Mean mark part (a) 36%.

 

b.   vcaa-2015-4b-answer

 

c.   `text(Find sequence that takes)\ f(x) -> h(x)`

`text(A dilation by factor of)\ 1/6\ text(from the)\ xtext(-axis.)`

`text(A dilation by factor of)\ 1/3\ text(from the)\ ytext(-axis.)`

 

d.i.  `text(Area when)\ n\ text(is even)`

♦♦♦ Mean mark 20%.
MARKER’S COMMENT: Note that `cos(npi)=1` for `n` even.
`= 2 int_0^pi (msin(x) – 1/n sin(nx))\ dx`
`=2 [-mcos x + 1/n^2 cos (nx)]_0^pi`
`=2[m+1/n^2 cos (npi) – (-m+1/n^2)]`
`= 2(2m+(cos(npi)-1)/n^2)`
`= 4m + 0/(n^2)`

 

d.ii.  `text(Area when)\ n\ text(is even)`

♦♦♦ Mean mark 14%.
MARKER’S COMMENT: Note that `cos(npi)=-1` for `n` odd.
`= 2 int_0^pi (msin(x) – 1/n sin(nx))\ dx`
`=2 [-mcos x + 1/n^2 cos (nx)]_0^pi`
`=2[m+1/n^2 cos (npi) – (-m+1/n^2)]`
`= 2(2m+(cos(npi)-1)/n^2)`
`= 4m + (-4)/(n^2)`