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Calculus, MET2 2020 VCAA 1

Let  `f:R rarr R, \ f(x)=a(x+2)^(2)(x-2)^(2)`, where `a in R`. Part of the graph of `f` is shown below. 
 

  1. Show that  `a = 1/4`.    (1 mark)
  2. Express  `f(x)=(1)/(4)(x+2)^(2)(x-2)^(2)`  in the form  `f(x)=(1)/(4)x^(4)+bx^(2)+c`  where `b` and `c` are integers.   (1 mark)

Part of the graph of the derivative function  `f^{′}` is shown below.
 
     
 

  1.  i. Write the rule for `f^{′}` in terms of `x`.   (1 mark)
  2. ii. Find the minimum value of the graph of `f^{′}` on the interval  `x in (0, 2)`.   (2 marks)  

Let  `h:R rarr R, \ h(x)=-(1)/(4)(x+2)^(2)(x-2)^(2)+2`. Parts of the graph of `f` and `h` are shown below.
 
 
       
 

  1. Write a sequence of two transformations that map the graph of `f` onto the graph of `h`.   (1 mark)
     
       
     
  2.   i. State the values of `x` for which the graphs of `f`and `h` intersect.   (1 mark)
  3.  ii. Write down a definite integral that will give the total area of the shaded regions in the graph above.   (1 mark)
  4. iii. Find the total area of the shaded regions in the graph above. Give your answer correct to two decimal places.   ( 1 mark)
  5. Let `D` be the vertical distance between the graphs of `f`and`h`.
  6. Find all values of `x` for which `D` is at most 2 units. Give your answers correct to two decimal places.   (2 marks)

Show Answers Only
  1. `text(See Worked Solutions)`
  2. `f(x)=(1)/(4)x^(4)-2x^(2)+4`
  3.  i. `f^{′}(x)=x(x-2)(x+2)=x^(3)-4x`
  4. ii. `-(16sqrt3)/(9)`
  5. `text{Reflect in the x-axis then translate 2 units up, or translate 2 units down then reflect in the x-axis.}`
  6.   i. `x in{-sqrt6,-sqrt2,sqrt2,sqrt6}`
  7.  ii. `2int_(sqrt2)^(sqrt6)(h(x)-f(x))\ dx” or “int_(-sqrt6)^(-sqrt2)(h(x)-f(x))\ dx+int_(sqrt2)^(sqrt6)(h(x)-f(x))\ dx`
  8. iii. `text(2.72 u²)`
  9. `-2.61 <= x <= -1.08,1.08 <= x <= 2.61`
Show Worked Solution

a.   `text(S)text(ince)\ f(x)\ text{passes through (0, 4):}`

`4` `=a(2)^2(-2)^2`  
`4` `=16a`  
`:.a` `=1/4\ \ text(… as required)`  

 

b.  `f(x)` `=(1)/(4)(x+2)^(2)(x-2)^(2)`  
  `=(1)/(4)x^4-2x^2+4`  

 
c.i.  `f^{′}(x)=x(x-2)(x+2)=x^(3)-4x`
 

c.ii.  `text{Solve}\ \ f^{″}(x)=0\ \ \text{for}\ x:`

`3x^2-4=0`

`=> \ x=(2sqrt3)/(3)`

`f^{′}((2sqrt3)/(3))=-(16sqrt3)/(9)`
 

d.  `text{Reflect in the x-axis then translate 2 units up, or}`

`text{translate 2 units down then reflect in the x-axis.}`
 

e.i.  `text{By CAS, solve}\ \ f(x)=h(x)\ \ text{for}\ x:}`

`x in{-sqrt6,-sqrt2,sqrt2,sqrt6}`
 

e.ii.  `2int_(sqrt2)^(sqrt6)(h(x)-f(x))\ dx” or”`

`int_(-sqrt6)^(-sqrt2)(h(x)-f(x))\ dx+int_(sqrt2)^(sqrt6)(h(x)-f(x))\ dx`
 

e.iii.  `text(Area = 2.72 u²)`

♦♦ Mean mark part (f) 26%.

 

f.   `text{By CAS, solve}\ -2 <= h(x)-f(x) <= 2\ \ \text{for}\ x:`

`-2.61 <= x <= -1.08, \ 1.08 <= x <= 2.61,” or”`

`-sqrt(4+2sqrt2) <= x <= -sqrt(4-2sqrt2), \ sqrt(4-2sqrt2) <= x <= sqrt(4+2sqrt2)`

Calculus, MET2 2019 VCAA 2

An amusement park is planning to build a zip-line above a hill on its property.

The hill is modelled by  `y = (3x(x - 30)^2)/2000,  x in [0, 30]`, where  `x`  is the horizontal distance, in metres, from an origin and  `y`  is the height, in metres, above this origin, as shown in the graph below.
  


 

  1. Find  `(dy)/(dx)`.  (1 mark)
  2. State the set of values for which the gradient of the hill is strictly decreasing.  (1 mark)

The cable for the zip-line is connected to a pole at the origin at a height of 10 m and is straight for  `0 <= x <= a`, where  `10 <= a <= 20`. The straight section joins the curved section at  `A(a, b)`. The cable is then exactly 3 m vertically above the hill from  `a <= x <= 30`, as shown in the graph below.
 


 

  1. State the rule, in terms of  `x`, for the height of the cable above the horizontal axis for  `x in [a, 30]`.  (1 mark)
  2. Find the values of  `x`  for which the gradient of the cable is equal to the average gradient of the hill for  `x in [10, 30]`.  (3 marks)

The gradients of the straight and curved sections of the cable approach the same value at  `x = a`, so there is a continuous and smooth join at `A`.

    1. State the gradient of the cable at  `A`, in terms of  `a`.  (1 mark)
    2. Find the coordinates of  `A`, with each value correct to two decimal places.  (3 marks)
    3. Find the value of the gradient at  `A`, correct to one decimal place.  (1 mark)
Show Answers Only
  1. `(9(x – 10)(x – 30))/2000`
  2. `x in [0, 20]`
  3. `h(x) = 3 + (3x(x – 30)^2)/2000`
  4. `x = (60 +- 10 sqrt 3)/3`
    1. `(9(a – 10)(a – 30))/2000`
    2. `A(11.12, 8.95)`
    3. `-0.1`
Show Worked Solution

a.   `d/(dx)((3x(x – 30)^2)/2000) = (9(x – 10)(x – 30))/2000\ \ text{(by CAS)}`
 

b.   `text(Sketch)\ \ dy/dx\ \ text{(by CAS)}:`

`(dy)/(dx)\ \ text(strictly decreasing for)\ \ x in [0, 20]`
 

c.   `text(Cable height is a vertical shift of +3 of)\ \ y:`

`h(x) = 3 + (3x(x – 30)^2)/2000\ \ text(for)\ \ x in [a, 30]`
 

d.   `text(Let)\ \ f(x) = (9(x – 10)(x – 30))/2000`

`text(Average gradient) = (f(30) – f(10))/(30 – 10) = -3/10\ \ text{(by CAS)}`

`text(Solve for)\ \ x\ \ text{(by CAS):}`

`(9(x – 10)(x – 30))/2000 = -3/10`

`x = (60 +- 10 sqrt 3)/3`

 

e.i.   `text(S)text(ince)\ \ h(x)\ \ text(is a vertical shift of)\ \ f(x),`

`=> h prime(a) = f prime(a)`

`h prime(a) = (9(a – 10)(a – 30))/2000`

 

e.ii.   `h(a) = (3a(a – 30)^2)/2000 + 3`

`:. A(a, (3a(a – 30)^2)/2000 + 3),\ B(0, 10)`
 

`m_(AB) = ((3a(a – 30)^2)/2000 + 3 – 10)/a = (3a(a – 30)^2 – 14000)/(2000 a)`

 
`text(Equating gradients:)`

`text(Solve): \ f prime(a) = m_(AB)\ \ text{(by CAS)}`

`(9(a – 10)(a – 30))/2000 = (3a(a – 30)^2 – 14000)/(2000 a)`

`a ~~ 11.12, quad b ~~ 8.95`

`:. A(11.12, 8.95)`

 

e.iii.    `M_A` `= h prime(11.12)`
    `~~ -0.1\ text{(by CAS)}`

Calculus, MET2 2017 VCAA 1

Let  `f : R → R,\  f (x) = x^3 - 5x`. Part of the graph of  `f` is shown below.

 

  1. Find the coordinates of the turning points.  (2 marks)
  2. `A(−1, f (−1))`  and  `B(1, f (1))`  are two points on the graph of  `f`.

     

    1. Find the equation of the straight line through `A` and `B`.  (2 marks)
    2. Find the distance `AB`.  (1 mark)

Let  `g : R → R, \ g(x) = x^3 - kx, \ k ∈ R^+`.

  1. Let  `C(–1, g(−1))` and `D(1, g(1))` be two points on the graph of `g`.

     

    1. Find the distance `CD` in terms of `k`.  (2 marks)
    2. Find the values of `k` such that the distance `CD` is equal to  `k + 1`.  (1 mark)
  2. The diagram below shows part of the graphs of `g` and  `y = x`. These graphs intersect at the points with the coordinates `(0, 0)` and `(a, a)`.
  3.  
  4.  
    1. Find the value of `a` in terms of `k`.  (1 mark)
    2. Find the area of the shaded region in terms of `k`.  (2 marks)
Show Answers Only
  1. `(- sqrt15/3,(10sqrt15)/9)\ text(and)\ (sqrt15/3,-(10sqrt15)/9)`
    1. `y =-4x`
    2. `2sqrt17`
    1. `2sqrt(k^2 – 2k + 2)`
    2. `k = 1quadtext(or)quadk = 7/3`
    1. `sqrt(k + 1)`
    2. `((k + 1)^2)/4\ text(units)²`
Show Worked Solution
a.   
`text(Solve)\ \ f′(x)` `= 0\ \ text(for)\ x:`
`x` `= ± sqrt15/3`

 

`f(sqrt15/3) = -(10sqrt15)/9`

`f(−sqrt15/3) = (10sqrt15)/9`

 

`:.\ text(Turning points:)`

`(- sqrt15/3,(10sqrt15)/9)\ text(and)\ (sqrt15/3,-(10sqrt15)/9)`

 

b.i.   `A(-1,4),\ \ B(1,–4)`

`m_(AB) = (4 – (−4))/(−1 – (1)) = −4`

 

`text(Equation of)\ AB, m=-4, text(through)\ \ (-1,4):`

` y – 4` `= −4(x – (−1))`
`:. y` `= −4x`

 

MARKER’S COMMENT: Students skilled in the use of technology will be much more efficient and minimise errors here.
b.ii.    `d_(text(AB))` `=sqrt((x_2-x_1)^2+(f(x_2)-f(x_1))^2)`
    `= sqrt((1- (- 1))^2 + (f(1) – f(−1))^2)`
    `= 2sqrt17`

 

c.i.    `d_(text(CD))` `=sqrt((x_2-x_1)^2+(g(x_2)-g(x_1))^2)`
    `=sqrt((1-(-1))^2+(g(1)-g(-1))^2)`
    `= 2sqrt(k^2 – 2k + 2)`

 

c.ii.   `text(Solve:)quad2sqrt(k^2 – 2k + 2) = k + 1quadtext(for)quadk > 0`

`:. k = 1quadtext(or)quadk = 7/3`

 

d.i.   `text(Solve:)quada^3 – ka = a\ \ \ text(for)quad a,\ \ (a > 0)`

`:. a = sqrt(k + 1)`

 

d.ii.    `text(Area)` `= int_0^(sqrt(k + 1))(x – g(x))\ dx`
    `= ((k + 1)^2)/4\ text(units)²`

Calculus, MET1 2017 VCAA 3

Let  `f: [-3, 0] -> R,\ f(x) = (x + 2)^2 (x - 1).`

  1. Show that  `(x + 2)^2 (x - 1) = x^3 + 3x^2 - 4`.  (1 mark)
  2. Sketch the graph of  `f` on the axes below. Label the axis intercepts and any stationary points with their coordinates.  (3 marks)

Show Answers Only
  1. `text(Proof)\ \ text{(See Worked Solutions)}`
  2.  
Show Worked Solution

a.  `text(Expand LHS:)`

`(x^2 + 4x + 4) (x – 1)`

`=x^3 – x^2 + 4x^2 – 4x + 4x – 4`

`=x^3 + 3x^2 – 4\ \ text(… as required.)`

 

b.  

`x text(-intercepts:)\ \ (–2, 0), (1, 0)\ \ text(but)\ \ x ∈ [–3, 0]`

`:. x text(-int occurs at)\ (–2, 0)`

 

`text(Endpoints:)\ \ (–3, –4),\ \ (0, –4)`

`text(Stationary points occur when)\ \ f prime (x)=0`

`3x^2 + 6x` `= 0`
`3x(x + 2)` `= 0`
`:. x=0 or -2`  

Calculus, MET2 2009 VCAA 2

VCAA 2009 2a

A train is travelling at a constant speed of `w` km/h along a straight level track from `M` towards `Q.`

The train will travel along a section of track `MNPQ.`

Section `MN` passes along a bridge over a valley.

Section `NP` passes through a tunnel in a mountain.

Section `PQ` is 6.2 km long.

From `M` to `P`, the curve of the valley and the mountain, directly below and above the train track, is modelled by the graph of
 

`y = 1/200 (ax^3 + bx^2 + c)` where `a, b` and `c` are real numbers.
 

All measurements are in kilometres.

  1. The curve defined from `M` to `P` passes through `N (2, 0)`. The gradient of the curve at `N` is – 0.06 and the curve has a turning point at  `x = 4`.
  2.  i. From this information write down three simultaneous equations in `a`, `b` and `c`.  (3 marks)
  3. ii. Hence show that  `a = 1`, `b = – 6` and `c = 16`.  (2 marks)
  4. Find, giving exact values
  5.   i. the coordinates of `M and P`  (2 marks)
  6.  ii. the length of the tunnel  (1 mark)
  7. iii. the maximum depth of the valley below the train track.  (1 mark)

The driver sees a large rock on the track at a point `Q`, 6.2 km from `P`. The driver puts on the brakes at the instant that the front of the train comes out of the tunnel at `P`.

From its initial speed of `w` km/h, the train slows down from point `P` so that its speed `v` km/h is given by

`v = k log_e ({(d + 1)}/7)`,

where `d` km is the distance of the front of the train from `P` and `k` is a real constant.

  1. Find the value of `k` in terms of `w`.  (1 mark)
  2. If  `v = (120 log_e (2))/(log_e (7))`  when  `d = 2.5`, find the value of `w`.  (2 marks)
  3. Find the exact distance from the front of the train to the large rock when the train finally stops.  (2 marks)
Show Answers Only
    1. `1/200 (8a + 4b + c) = 0`
       
      `1/200 (12a + 4b) = (– 3)/50`
       
      `1/200 (48a + 8b) = 0`
    2. `text(Proof)\ \ text{(See Worked Solutions)}`
    1. `M (2 + 2 sqrt 3, 0),\ \ P (2 – 2 sqrt 3, 0)`
    2. `2 sqrt 3\ text(km)`
    3. `2/25\ text(km)`
  3. `(– w)/(log_e (7))`
  4. `120`
  5. `0.2\ text(km)`
Show Worked Solution

 a.i.   `N (2, 0),`

`1/200 (8a + 4b + c) = 0\ \ text{… (1)}`

 

`{:frac (dy) (dx)|:}_(x = 2) = (– 3)/50,`

`1/200 (12a + 4b) = (– 3)/50\ \ text{… (2)}`

 

`{:frac (dy) (dx)|:}_(x = 4) = 0,`

`1/200 (48a + 8b) = 0\ \ text{… (3)}`

 

  ii.   `text(Using the above equations,)`

♦ Mean mark part (a)(ii) 41%.
`12a + 4b` `=-12` `\ \ \ …\ (2′)`
`24 a + 4b` `=0` `\ \ \ …\ (3′)`

 

`text(Solve simultaneous equations:)`

`(3′) – (2′):`

`12a` `= 12`
`:. a` `= 1`

 

`text(Substitute)\ \ a = 1\ \ text(into)\ \ (3′):`

`24(1) + 4b` `= 0`
`:. b` `= – 6`

`text(Substitute)\ \ a = 1,\ \ b = – 6\ \ text(into)\ \ (1),`

`8(1) + 4 (– 6) + c` `= 0`
`:. c` `= 16`

 

b.i.   `text(For)\ \ x text(-intercepts),`

`text(Solve:)\ \ x^3 + -6x^2 + 16` `= 0\ \ text(for)\ x,`
 `x= 2 – 2 sqrt 3, 2, 2 + 2 sqrt 3`  

 

`:. M (2 + 2 sqrt 3, 0),\ \ P (2 – 2 sqrt 3, 0)`

 

  ii.   `N (2, 0)`

♦ Mean mark part (a) 41%.
`bar (NP)` `=\ text(Tunnel length)`
  `= 2 – (2 – 2 sqrt 3)`
  `= 2 sqrt 3\ text(km)`

 

   iii.  `text(Solve)\ \ frac (dy) (dx) = 0\ \ text(for)\ \ x in (2, 2 + 2 sqrt 3)`

♦ Mean mark part (b) 50%.

`x = 4`

`text(When)\ \ x=4,\ \ y = – 2/25\ text(km) = 80\ text{m (below track)}`

`:.\ text(Max depth below is)\ \ 80\ text(m.)`

 

c.  `text(Solution 1)`

♦ Mean mark part (c) 35%.

`text(Let)\ \ v(d) = k log_e ({(d + 1)}/7)`

`text(S)text(ince)\ \ v=w\ \ text(when)\ \ d=0\ \ text{(given),}`

`k log_e ({(0 + 1)}/7)` `=w`
`k` `=w/log_e (1/7)`
  `=(-w)/log_e7`

 

`text(Solution 2)`

`text(Solve:)\ \ v (0)` `= w\ \ text(for)\ \ k`
`:. k` `= (– w)/(log_e (7))`

 

d.  `v (2.5) = k log_e(1/2)`

♦ Mean mark part (d) 40%.
`text(Solve)\ \ v(2.5)` `= (120 log_e (2))/(log_e (7))\ \ text(for)\ \ k,`
`:. k` `= (– 120)/(log_e (7))`
`(– w)/(log_e (7))` `= (– 120)/(log_e (7))`
`:. w` `= 120`

 

e.   `text(Define)\ \ v (d) = (– 120)/(log_e (7)) log_e ((d + 1)/7)`

♦♦ Mean mark part (e) 32%.
`text(Solve:)\ \ v(d)` `= 0\ \ text(for)\ d,`
`:. d` `= 6\ text(km from)\ \ P`

 

`:.\ text(Distance between train and)\ \ Q`

`= 6.2 – 6`

`= 0.2\ text(km)`

Calculus, MET2 2011 VCAA 3

  1. Consider the function  `f: R -> R, f(x) = 4x^3 + 5x-9`.

     

    1. Find  `f^{′}(x)`  (1 mark)
    2. Explain why  `f^{′}(x) >= 5` for all `x`.  (1 mark)
  2. The cubic function `p` is defined by  `p: R -> R, p(x) = ax^3 + bx^2 + cx + k`, where `a`, `b`, `c` and `k` are real numbers.

     

    1. If `p` has `m` stationary points, what possible values can `m` have?  (1 mark)
    2. If `p` has an inverse function, what possible values can `m` have?  (1 mark)
  3. The cubic function `q` is defined by  `q:R -> R, q(x) = 3-2x^3`.

     

    1. Write down a expression for  `q^(−1)(x)`.  (2 marks)
    2. Determine the coordinates of the point(s) of intersection of the graphs of  `y = q(x)`  and  `y = q^(−1)(x)`.  (2 marks)
  4. The cubic function `g` is defined by  `g: R -> R, g(x) = x^3 + 2x^2 + cx + k`, where `c` and `k` are real numbers.

     

    1. If `g` has exactly one stationary point, find the value of `c`.  (3 marks)
    2. If this stationary point occurs at a point of intersection of  `y = g(x)`  and  `g^(−1)(x)`, find the value of `k`.  (3 marks)
Show Answers Only
    1. `f^{′}(x) = 12x^2 + 5`
    2. `text(See Worked Solutions)`
    1. `m = 0, 1, 2`
    2. `m = 0, 1`
    1. `q^(−1)(x) = root(3)((3-x)/2), x ∈ R`
    2. `(1, 1)`
    1. `4/3`
    2. `−10/27`
Show Worked Solution

a.i.   `f^{′}(x) = 12x^2 + 5`

 

a.ii.  `text(S)text(ince)\ \ x^2>=0\ \ text(for all)\ x,`

♦ Mean mark 47%.
` 12x^2` `>= 0`
`12x^2 + 5` `>=  5`
`f^{′}(x)` `>=  5\ \ text(for all)\ x`

 

b.i.   `p(x) = text(is a cubic)`

♦♦♦ Mean mark part (b)(i) 9%, and part (b)(ii) 20%.
MARKER’S COMMENT: Good exam strategy should point students to investigate earlier parts for direction. Here, part (a) clearly sheds light on a solution!

`:. m = 0, 1, 2`

`text{(Note: part a.ii shows that a cubic may have no SP’s.)}`

 

b.ii.   `text(For)\ p^(−1)(x)\ text(to exist)`

`:. m = 0, 1`

 

c.i.   `text(Let)\ y = q(x)`

`text(Inverse: swap)\ x ↔ y`

`x` `= 3-2y^3`
`y^3` `= (3-x)/2`

`:. q^(−1)(x) = root(3)((3-x)/2), \ x ∈ R`

 

c.ii.  `text(Any function and its inverse intersect on)`

   `text(the line)\ \ y=x.`

`text(Solve:)\ \ 3-2x^3` `= xqquadtext(for)\ x,`
`x` `= 1`

 

`:.\ text{Intersection at (1, 1)}`

 

♦ Mean mark part (d)(i) 44%.
d.i.    `g^{′}(x)` `= 0`
  `3x^2 + 4x + c` `= 0`
  `Delta` `= 0`
  `16-4(3c)` `= 0`
  `:. c` `= 4/3`

 

d.ii.   `text(Define)\ \ g(x) = x^3 + 2x^2 + 4/3x + k`

♦♦♦ Mean mark part (d)(ii) 14%.

  `text(Stationary point when)\ \ g^{′}(x)=0`

`g^{′}(x) = 3x^2+4x+4/3`

`text(Solve:)\ \ g^{′}(x)=0\ \ text(for)\ x,`

`x = −2/3`

`text(Intersection of)\ g(x)\ text(and)\ g^(−1)(x)\ text(occurs on)\ \ y = x`

`text(Point of intersection is)\  (−2/3, −2/3)`

`text(Find)\ k:`

`g(−2/3)` `= −2/3\ text(for)\ k`
`:. k` ` = −10/27`

Calculus, MET1 2014 VCAA 5

Consider the function  `f:[−1,3] -> R`,  `f(x) = 3x^2 - x^3`.

  1. Find the coordinates of the stationary points of the function.  (2 marks)
  2. On the axes  below, sketch the graph of  `f`.

     

    Label any end points with their coordinates.  (2 marks)

     

     
    met1-2014-vcaa-q5
     

  3. Find the area enclosed by the graph of the function and the horizontal line given by `y = 4`.  (3 marks)
Show Answers Only
  1. `(0, 0) and (2, 4)`
  2.  
    met1-2014-vcaa-q5-answer3
  3. `27/4\ text(u²)`
Show Worked Solution
a.    `text(SP’s occur when)\ \ f′(x)` `= 0`
`6x – 3x^2`  `= 0` 
 `3x(2 – x)` `=0`

`x = 0,\ \ text(or)\ \ 2`

 

`:.\ text{Coordinates are (0, 0) and (2, 4)}`

 

b.    met1-2014-vcaa-q5-answer3

 

♦ Part (c) mean mark 48%.
c.    met1-2014-vcaa-q5-answer4

`text(Solution 1)`

`text(Area)` `= int_(−1)^2 4 – (3x^2 – x^3)dx`
  `= int_(−1)^2 4 – 3x^2 + x^3dx`
  `= [4x – x^3 + 1/4x^4]_(−1)^2`
  `= (8 – 8 + 4) – (−4 – (−1) + 1/4)`
   
`:.\ text(Area)` `= 27/4 text(units²)`

 

`text(Solution 2)`

`text(Area)` `= 12 – int_(−1)^2(3x^2 – x^3)dx`
  `= 12 – [x^3 – 1/4 x^4]_(−1)^2`
  `= 12 – [(8 – 4) – (−1 – 1/4)]`
  `= 27/4\ text(units²)`

Calculus, MET1 2015 VCAA 4

Consider the function  `f:[−3,2] -> R, \ \ f(x) = 1/2(x^3 + 3x^2 - 4)`.

  1. Find the coordinates of the stationary points of the function.  (2 marks)

The rule for  `f` can also be expressed as  `f(x) = 1/2(x - 1)(x + 2)^2`.

  1. On the axes below, sketch the graph of  `f`, clearly indicating axis intercepts and turning points.

     

    Label the end points with their coordinates.  (2 marks)

      

    met1-2015-vcaa-q4
     

  2. Find the average value of  `f` over the interval  `0 <= x <=2`.  (2 marks)
Show Answers Only
  1. `(0, − 2), ( − 2,0)`
  2.  

     

    met1-2015-vcaa-q4-answer

  3. `1`
Show Worked Solution

a.   `text(Stationary points when)\ \ f´(x)=0,`

`1/2(3x^2 + 6x)` `= 0`
`3x(x + 2)` `= 0`

 

`:. x = 0, − 2`

 

`:.\ text(Coordinates of stationary points:)`

`(0, − 2), (− 2,0)`

 

b.    met1-2015-vcaa-q4-answer
♦ Part (c) mean mark 50%.
MARKER’S COMMENT: Most students recalled the average value definition but then did not integrate correctly.

 

c.    `text(Avg value)` `= 1/(2 – 0) int_0^2 f(x) dx`
    `= 1/2 int_0^2 1/2(x^3 + 3x^2 – 4)dx`
    `= 1/4[1/4x^4 + x^3 – 4x]_0^2`
    `= 1/4[(16/4 + 2^3 – 4(2))-0]`
    `= 1`

Calculus, MET2 2013 VCAA 3

Tasmania Jones is in Switzerland. He is working as a construction engineer and he is developing a thrilling train ride in the mountains. He chooses a region of a mountain landscape, the cross-section of which is shown in the diagram below.

VCAA 2013 2a

The cross-section of the mountain and the valley shown in the diagram (including a lake bed) is modelled by the function with rule
 

`f(x) = (3x^3)/64 - (7x^2)/32 + 1/2.`
 

Tasmania knows that  `A (0, 1/2)`  is the highest point on the mountain and that `C(2, 0)` and `B(4, 0)` are the points at the edge of the lake, situated in the valley. All distances are measured in kilometres.

  1. Find the coordinates of `G`, the deepest point in the lake.  (3 marks)

Tasmania’s train ride is made by constructing a straight railway line `AB` from the top of the mountain, `A`, to the edge of the lake, `B`. The section of the railway line from `A` to `D` passes through a tunnel in the mountain.

  1. Write down the equation of the line that passes through `A` and `B.`  (2 marks)
  2.  i. Show that the `x`-coordinate of `D`, the end point of the tunnel, is `2/3.`  (1 mark)
  3. ii. Find the length of the tunnel `AD.`  (2 marks)

In order to ensure that the section of the railway line from `D` to `B` remains stable, Tasmania constructs vertical columns from the lake bed to the railway line. The column `EF` is the longest of all possible columns. (Refer to the diagram above.)

  1.  i. Find the `x`-coordinate of `E.`  (2 marks)
  2. ii. Find the length of the column `EF` in metres, correct to the nearest metre.  (2 marks)

Tasmania’s train travels down the railway line from `A` to `B`. The speed, in km/h, of the train as it moves down the railway line is described by the function.

`V: [0, 4] -> R, \ V(x) = k sqrt x - mx^2,`

where `x` is the `x`-coordinate of a point on the front of the train as it moves down the railway line, and `k` and `m` are positive real constants.

The train begins its journey at  `A (0, 1/2)`. It increases its speed as it travels down the railway line.

The train then slows to a stop at  `B(4, 0)`, that is  `V(4) = 0.`

  1. Find `k` in terms of `m.`  (1 mark)
  2. Find the value of `x` for which the speed, `V`, is a maximum.  (2 marks)

Tasmania is able to change the value of `m` on any particular day. As `m` changes, the relationship between `k` and `m` remains the same.

  1. If, on one particular day, `m = 10`, find the maximum speed of the train, correct to one decimal place.  (2 marks)
  2. If, on another day, the maximum value of `V` is 120, find the value of `m.`  (2 marks)
Show Answers Only
  1. `G(28/9,−50/243)`
  2. `y = −1/8x + 1/2`
  3.  i. `text(See Worked Solutions)`
  4. ii. `sqrt65/12`
  5.  i. `(2(sqrt31 + 7))/9`
  6. ii. `336\ text(m)`
  7. `8\ text(m)`
  8. `2^(2/3)`
  9. `75.6\ text(km/h)`
  10. `10 xx 2^(2/3)`
Show Worked Solution

a.   `G\ text(occurs when)\ \ f′(x)=0.`

`text(Solve:)\ \ (3x^3)/64 – (7x^2)/32 + 1/2=0\ \ text(for)\ x`

`x= 28/9quadtext(for)quadx ∈ (2,4)`

`f(28/9) = −50/243`

`:. G(28/9,−50/243)`

 

b.    `m_(AB)` `= (1/2 – 0)/(0 – 4)`
    `= −1/8`

 

`text(Equation using point gradient formula,)`

`y – y_1` `= m(x – x_1)`
`y – 1/2` `= −1/8(x – 0)`
`:. y` `= −1/8x + 1/2`

 

c.i.   `text(Solution 1)`

`text(Intersection occurs when:)`

`3/64x^3 – 7/32x^2 + 1/2` `= −1/8x + 1/2`
`3x^3 – 14x^2 + 8x` `= 0`
`x(3x^2 – 14x + 8)` `= 0`
`x(x – 4)(3x – 2)` `= 0`

`x = 2/3,\ \ (0<x<4)`

`:. x text(-coordinate of)\ D = 2/3`

 

`text{Solution 2 (using technology)}`

`text(Solve:)\ \ ` `3/64x^3 – 7/32x^2 + 1/2` `= −1/8x + 1/2\ \ text(for)\ x`

`x=0, 2/3 or 4`

`:.x=2/3,\ \ (0<x<4)`

 

c.ii.   `D (2/3, 5/12)qquadA(0,1/2)`

 

  `text(By Pythagoras,)`

`AD` `= sqrt((2/3 – 0)^2 + (5/12 – 1/2)^2)`
  `= sqrt65/12`

 

d.i.  `text(Let)\ \ z =\ EF`

♦♦♦ Mean mark part (d)(i) 24%.
MARKER’S COMMENT: Many students unfamiliar with this type of question.
 `z` `= (−1/8x + 1/2) – ((3x^2)/64 – 7/32 x^2 + 1/2)`
  `=-1/8x-(3x^2)/64 + 7/32 x^2`

 

`text(Solve:)\ \ d/dx (-1/8x-(3x^2)/64 + 7/32 x^2) =0\ \ text(for)\ x`

`x= (2(sqrt31 + 7))/9,\ \ \ x > 0`

 

♦♦♦ Mean mark part (d)(ii) 22%.
d.ii.    `z((2sqrt31 + 14)/9)` `= 0.3360…\ text(km)`
    `= 336\ text{m  (nearest m)}`

 

e.    `V(4)` `= 0quadtext{(given)}`
  ` 0` `= k sqrt4 – m xx 4^2`
  `:.k` `= 8m`

 

f.  `V(x) = 8msqrtx – mx^2`

`text(Solve:)\ \ V′(x)` `= 0quadtext(for)quadx`

`x= 2^(2/3)`

 

g.  `text(When)\ \ m=10,\ \ k=8 xx 10=80`

♦ Mean mark 43%.

`:.V(x) = 80sqrtx – 10x^2`

`text(Solve:)\ \ V′(x)` `= 0quadtext(for)quadx`

`x= 2^(2/3)`

`V(2^(2/3))` `=75.59…`
  `= 75.6\ text(km/h)`

 

h.   `text(Maximum occurs at)\ x = 2^(2/3)`

♦ Mean mark 38%.

`V(x) = 8msqrtx – mx^2`

`text(Solve:)\ \ V(2^(2/3))` `= 120quadtext(for)quadm`
`:. m` `= 10 xx 2^(2/3)`