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Calculus, MET2 2023 VCE SM-Bank 2

Jac and Jill have built a ramp for their toy car. They will release the car at the top of the ramp and the car will jump off the end of the ramp.

The cross-section of the ramp is modelled by the function \(f\), where

\(f(x)= \begin{cases}\displaystyle \ 40 & 0 \leq x<5 \\ \dfrac{1}{800}\left(x^3-75 x^2+675 x+30\ 375\right) & 5 \leq x \leq 55\end{cases}\)

\(f(x)\) is both smooth and continuous at \(x=5\).

The graph of  \(y=f(x)\)  is shown below, where \(x\) is the horizontal distance from the start of the ramp and \(y\) is the height of the ramp. All lengths are in centimetres.

  1. Find \(f^{\prime}(x)\) for \(0<x<55\).  (2 marks)

  2.   i. Find the coordinates of the point of inflection of \(f\).  (1 mark)

  3.  ii. Find the interval of \(x\) for which the gradient function of the ramp is strictly increasing.  (1 mark)

  4. iii. Find the interval of \(x\) for which the gradient function of the ramp is strictly decreasing.  (1 mark)

Jac and Jill decide to use two trapezoidal supports, each of width \(10 cm\). The first support has its left edge placed at \(x=5\) and the second support has its left edge placed at \(x=15\). Their cross-sections are shown in the graph below.

  1. Determine the value of the ratio of the area of the trapezoidal cross-sections to the exact area contained between \(f(x)\) and the \(x\)-axis between \(x=5\) and \(x=25\). Give your answer as a percentage, correct to one decimal place.  (3 marks)

  2. Referring to the gradient of the curve, explain why a trapezium rule approximation would be greater than the actual cross-sectional area for any interval \(x \in[p, q]\), where \(p \geq 25\).  (1 mark)

  3. Jac and Jill roll the toy car down the ramp and the car jumps off the end of the ramp. The path of the car is modelled by the function \(P\), where

      1. \(P(x)=\begin{cases}f(x) & 0 \leq x \leq 55 \\ g(x) & 55<x \leq a\end{cases}\)
  4. \(P\) is continuous and differentiable at \(x=55, g(x)=-\frac{1}{16} x^2+b x+c\), and \(x=a\) is where the car lands on the ground after the jump, such that \(P(a)=0\).
    1. Find the values of \(b\) and \(c\).  (2 marks)

    2. Determine the horizontal distance from the end of the ramp to where the car lands. Give your answer in centimetres, correct to two decimal places.  (1 mark)


Show Answers Only

a.    \(f^{\prime}(x)= \begin{cases}\displaystyle \ 0 & 0 < x<5 \\ \dfrac{1}{800}\left(3x^2-150 x+675 \right) & 5 \leq x < 55\end{cases}\)

b.i   \((25, 20)\)

b.ii  \([25, 55]\)

b.iii \([5, 25]\)

c.    \(98.1\%\)

d.    \(\text{In the interval }x\in [25, q]\text{ the curve is concave up.}\)

\(\therefore\ \text{The sloped edges of the trapeziums would be above }f(x)\)

\(\text{making the trapezium rule approximation greater than the}\)

\(\text{actual area.}\)

e.i   \(b=8.75, c=-283.4375\)

e.ii  \(34.10\ \text{cm}\)

Show Worked Solution

a.    \(\text{Using CAS: Define f(x) then}\)

\(\text{OR}\quad f^{\prime}(x)= \begin{cases}\displaystyle \ 0 & 0 < x<5 \\ \dfrac{1}{800}\left(3x^2-150 x+675 \right) & 5 \leq x < 55\end{cases}\)

b.i   \(\text{Using CAS: Find }f^”(x)\ \text{then solve }=0\)

\(\therefore\ \text{Point of inflection at }(25, 20)\)
  

b.ii  \(\text{Gradient function strictly increasing for }x\in [25, 55]\)

b.iii \(\text{Gradient function strictly decreasing for }x\in [5, 25]\)

c.    \(\text{Using CAS:}\)

\(\text{Area}\) \(=\dfrac{h}{2}\Big(f(5)+2f(15)+f(25)\Big)\)
  \(=\dfrac{10}{2}\Big(40+2\times 33.75+20\Big)=637.5\)

  

\(\therefore\ \text{Estimate:Exact}\) \(=637.5:650\)
  \(=\dfrac{637.5}{650}\times 100\)
  \(=98.0769\%\approx 98.1\%\)

   
d.    \(\text{In the interval }x\in [25, q]\text{ the curve is concave up.}\)

\(\therefore\ \text{The sloped edges of the trapeziums would be above }f(x)\)

\(\text{making the trapezium rule approximation greater than the}\)

\(\text{actual area.}\)

e.i   \(f(55)=\dfrac{35}{4}\ \text{(Using CAS)}\)

\(\text{and }f(55)=g(55)\rightarrow g(55)=\dfrac{35}{4}\)

\(\therefore -\dfrac{1}{16}\times 55^2+55b+c\) \(=\dfrac{35}{4}\)
\(c\) \(=\dfrac{35}{4}+\dfrac{3025}{16}-55b\)
\(c\) \(=\dfrac{3165}{16}-55b\quad (1)\)

  
\(g'(x)=-\dfrac{x}{8}+b\)

\(\text{and }g'(55)=f'(55)\)

\(\rightarrow\ f'(55)=\dfrac{1}{800}(3\times 55^2-150\times 55+675)=\dfrac{15}{8}\)

\(\rightarrow\ -\dfrac{55}{8}+b=\dfrac{15}{8}\)

\(\therefore b=\dfrac{35}{4}\quad (2)\)

\(\text{Sub (2) into (1)}\)

\(c=\dfrac{3165}{16}-55\times\dfrac{35}{4}\)

\(c=-\dfrac{4535}{16}\)

\(\therefore\ b=\dfrac{35}{4}\ \text{or}\ 8.75 , c=-\dfrac{4535}{16}\ \text{or}\ -283.4375\)

e.ii  \(\text{Using CAS: Solve }g(x)=0|x>55\)

\(\text{Horizontal distance}=\sqrt{365}+70-55=34.104\dots\approx 34.10\ \text{cm (2 d.p.)}\)

Calculus, MET1 2022 VCAA 7

A tilemaker wants to make square tiles of size 20 cm × 20 cm.

The front surface of the tiles is to be painted with two different colours that meet the following conditions:

  • Condition 1 - Each colour covers half the front surface of a tile.
  • Condition 2 - The tiles can be lined up in a single horizontal row so that the colours form a continuous pattern.

An example is shown below.
 

There are two types of tiles: Type A and Type B.

For Type A, the colours on the tiles are divided using the rule `f(x)=4 \sin \left(\frac{\pi x}{10}\right)+a`, where `a \in R`.

The corners of each tile have the coordinates (0,0), (20,0), (20,20) and (0,20), as shown below.
 

  1.  i. Find the area of the front surface of each tile.   (1 mark)

    ii. Find the value of `a` so that a Type A tile meets Condition 1.   (1 mark)


Type B tiles, an example of which is shown below, are divided using the rule `g(x)=-\frac{1}{100} x^3+\frac{3}{10} x^2-2 x+10`.
 

  1. Show that a Type B tile meets Condition 1.   (3 marks)

  2. Determine the endpoints of `f(x)` and `g(x)` on each tile. Hence, use these values to confirm that Type A and Type B tiles can be placed in any order to produce a continuous pattern in order to meet Condition 2.   (2 marks)

Show Answers Only

a.i.    `400` cm²

a.ii.   `a = 10`

b.     `200` cm²

c.     See worked solution

Show Worked Solution
a.i   Area `= 20 xx 20`  
  `= 400` cm²  

  
a.ii  `a = 10`
 

b.   Area `=\int_0^{20} \frac{-x^3}{100}+\frac{3 x^2}{10}-2 x+10\ d x`  
  `=\left[\frac{-x^4}{400}+\frac{x^3}{10}-x^2+10 x\right]_0^{20}`  
  `=\left[-\frac{20^4}{400}+\frac{20^3}{10}-20^2+10 xx 20\right]-\left[0\right]`  
  `= – 400 +800 -400 +200`  
  `= 200` cm²  

 
`:.` The area of the coloured section of the Type B tile is 200 cm² which is half the 400 cm² area of the tile.
 

c.   
`f(0)` `=4 \sin \left(\frac{\pi(0)}{10}\right)+10 = 10`  
  `f(20)` `=4 \sin (2 \pi)+10 = 10`  
  `g(0)` `=\frac{-0}{100}+\frac{3(0)}{10}-2(0)+10=10`  
  `g(20)` `=-\frac{8000}{100}+\frac{200}{10}-2(20)+10 = 10`  

 
→`\ f(0) = f(20) = g(0) = g(20) =10`
 

`:.`    The endpoints for `f(x)` are `(0,10)` and `(20,10)` and for `g(x)` are also `(0,10)` and `(20,10)`.

So the tiles can be placed in any order to make the continuous pattern.


♦♦ Mean mark (c) 30%.
MARKER’S COMMENT: Students often only found `f(20)` and `g(20)`, however, `f(0)` and `g(0)` also needed to be found to verify the pattern match.

Calculus, MET1 2022 VCAA 2b

Evaluate `\int_0^1(f(x)(2 f(x)-3))dx`, where `\int_0^1[f(x)]^2 dx=\frac{1}{5}` and `\int_0^1 f(x) dx=\frac{1}{3}`.   (3 marks)

Show Answers Only

`-\frac{3}{5}`

Show Worked Solution
`\int_0^1(f(x)(2 f(x)-3)) d x` `=\int_0^1\left(2[f(x)]^2-3 f(x)\right) d x`  
  `=2 \int_0^1[f(x)]^2 d x-3 \int_0^1 f(x) d x`  
  `=2 \cdot \frac{1}{5}-3 \cdot \frac{1}{3}`  
  `=-\frac{3}{5}`  

♦ Mean mark 50%.

Calculus, MET2 2018 VCAA 8 MC

If  `int_1^12 g(x)\ dx = 5`  and  `int_12^5 g(x)\ dx = -6`,  then  `int_1^5 g(x)\ dx`  is equal to

A.   −11

B.     –1

C.      1

D.      3

E.     11

Show Answers Only

`B`

Show Worked Solution

♦ Mean mark 41%.

`int_1^12 g(x)\ dx` `= int_1^5 g(x)\ dx + int_5^12 g(x)\ dx`
  `= int_1^5 g(x)\ dx – int_12^5 g(x)\ dx`
`5` `= int_1^5 g(x)\ dx – (-6)`
`int_1^5 g(x)\ dx` `= -1`

 
`=>   B`

Calculus, MET1 2010 ADV 2e

Given that `int_0^6 ( x + k ) \ dx = 30`, and `k` is a constant, find the value of `k`.   (2 marks)

Show Answers Only

`k = 2`

Show Worked Solution
`int_0^6 ( x + k ) \ dx` `= 30`
`int_0^6 ( x + k ) \ dx` `= [ 1/2\ x^2 + kx ]_0^6`
`30` `= [(1/2xx 6^2 + 6 xx k ) – 0 ]`
`30` `= 18 + 6k`
`6k` `= 12`
`:.  k` `= 2`

Calculus, MET1 2009 ADV 2biii

Evaluate  `int_1^4 x^2 + sqrtx\ dx`.   (3 marks)

 

 

Show Answers Only

`77/3`

 

Show Worked Solution

`int_1^4 x^2 + sqrtx\ \ dx`

`= int_1^4 (x^2 + x^(1/2))\ dx`

`= [1/3 x^3 + 1/(3/2) x^(3/2)]_1^4`

`= [(x^3)/3 + 2/3x^(3/2)]_1^4`

`= [((4^3)/3 + 2/3 xx 4^(3/2))\ – (1/3 + 2/3)]`

`= [(64/3 + 16/3) – 3/3]`

`= [80/3 – 3/3]`

`= 77/3`

Calculus, MET2 2008 VCAA 4 MC

If  `int_1^3 f(x)\ dx = 5`, then  `int_1^3 (2f(x) - 3)\ dx`  is equal to

A.   `4`

B.   `5`

C.   `7`

D.   `10`

E.   `16`

Show Answers Only

`A`

Show Worked Solution

`int_1^3 (2 f(x) – 3)\ dx`

♦ Mean mark 49%.
MARKER’S COMMENT: Over one third of students did not integrate –3.

`= 2 int_1^3 f(x)\ dx + int_1^3 (– 3)\ dx`

`=2(5) +[-3x]_1^3`

`= 10 + (– 6)`

`= 4`

 
`=>   A`

Calculus, MET2 2010 VCAA 20 MC

Let  `f` be a differentiable function defined for all real `x`, where  `f (x) >= 0`  for all  `x in [0, a].`

If  `int_0^a f(x)\ dx = a`, then  `2 int_0^(5a) (f (x/5) + 3)\ dx`  is equal to

A.   `2a + 6`

B.   `10a + 6`

C.   `20a`

D.   `40a`

E.   `50a`

Show Answers Only

`D`

Show Worked Solution

`2 int_0^(5a) f(x/5) + 3\ dx`

♦♦♦ Mean mark 25%.

`= 2 int_0^(5a) f(x/5)\ dx + 2 int_0^(5a) (3)\ dx`

`= 2 xx 5 [F(x/5)]_0^(5a) + [3x]_0^(5a)`

`= 10 [F(a) – F (0)] + 30a`

`= 10a + 30a`

`= 40a`

 
`=>   D`

Calculus, MET1 2015 VCAA 3

Evaluate  `int_1^4 (1/sqrtx)\ dx`.  (2 marks)

Show Answers Only

`2`

Show Worked Solution
MARKER’S COMMENT: This basic integral caused problems with many students answering with a log function.
`int_1^4 x^(−1/2)\ dx` `= 2[x^(1/2)]_1^4`
  `= 2[4^(1/2) – 1^(1/2)]`
  `= 2(2 – 1)`
  `= 2`

Calculus, MET2 2015 VCAA 19 MC

If  `f(x) = int_0^x (sqrt(t^2 + 4))\ dt`, then  `f prime (– 2)`  is equal to

A.   `sqrt 2`

B.   `- sqrt 2`

C.   `2 sqrt 2`

D.   `-2 sqrt 2`

E.   `4 sqrt 2`

Show Answers Only

`C`

Show Worked Solution
`f(x)` `= int_0^x (sqrt(t^2 + 4))\ dt`
`f′(x)` `= sqrt(t^2 + 4)`
`:. f′(− 2)` `= sqrt((-2)^2+4)`
  `=sqrt8`
  `=2 sqrt2`

`=>   C`

Calculus, MET2 2015 VCAA 15 MC

If  `int_0^5 g(x)\ dx = 20`  and  `int_0^5 (2g(x) + ax)\ dx = 90`, then the value of `a` is

A.       `0`

B.       `4`

C.       `2`

D.   `− 3`

E.       `1`

Show Answers Only

`B`

Show Worked Solution
`int_0^5 (2g(x) + ax)\ dx` `=90`
`2int_0^5 g(x)\ dx + int_0^5 ax\ dx` `= 90`
`2(20) + a/2[x^2]_0^5` `=90`
`25/2a` `= 50`
`:. a` `= 4`

`=>   B`