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Calculus, SPEC1 2024 VCAA 5

The curve with equation  \(y=\sqrt{k-\dfrac{1}{x^2}}\),  for  \(1 \leq x \leq \dfrac{k}{2}\)  where  \(k>2\), is rotated about the \(x\)-axis to form a solid of revolution that has volume  \(\dfrac{7 \pi}{2}\) units\(^3\).

Show that \(k\) satisfies the equation  \(k^3-2 k^2-9 k+4=0\).   (3 marks)

Show Answers Only

  \(V\) \(=\pi \displaystyle{\int}_1^{\frac{k}{2}}\left(k-\dfrac{1}{x^2}\right) dx\)
  \(\dfrac{7 \pi}{2}\) \(=\pi\left[kx+\dfrac{1}{x}\right]_1^{\frac{k}{2}}\)
  \(7\) \(=2\left[\left(\dfrac{k^2}{2}+\dfrac{2}{k}\right)-(k+1)\right]\)
  \(7\) \(=k^2+\dfrac{4}{k}-2 k-2\)
  \(7k\) \(=k^3+4-2 k^2-2 k\)
  \(0\) \(=k^3-2k^2-9k+4\)

Show Worked Solution

\(y=\sqrt{k-\dfrac{1}{x^2}}\)

  \(V\) \(=\pi \displaystyle{\int}_1^{\frac{k}{2}}\left(k-\dfrac{1}{x^2}\right) dx\)
  \(\dfrac{7 \pi}{2}\) \(=\pi\left[kx+\dfrac{1}{x}\right]_1^{\frac{k}{2}}\)
  \(7\) \(=2\left[\left(\dfrac{k^2}{2}+\dfrac{2}{k}\right)-(k+1)\right]\)
  \(7\) \(=k^2+\dfrac{4}{k}-2 k-2\)
  \(7k\) \(=k^3+4-2 k^2-2 k\)
  \(0\) \(=k^3-2k^2-9k+4\)

Calculus, SPEC2 2022 VCAA 1

Consider the family of functions \(f\) with rule  \(f(x)=\dfrac{x^2}{x-k}\), where \(k \in R \backslash\{0\}\).

  1. Write down the equations of the two asymptotes of the graph of \(f\) when \(k=1\).   (2 marks)

  2. Sketch the graph of  \(y=f(x)\)  for  \(k=1\)  on the set of axes below. Clearly label any turning points with their coordinates and label any asymptotes with their equations.   (3 marks)
     

  1.  i. Find, in terms of \(k\), the equations of the asymptotes of the graph of  \(f(x)=\dfrac{x^2}{x-k}\).   (1 mark)

  2. ii. Find the distance between the two turning points of the graph of  \(f(x)=\dfrac{x^2}{x-k}\) in terms of \(k\).   (2 marks)

  3. Now consider the functions \(h\) and \(g\), where  \(h(x)=x+3\)  and  \(g(x)=\abs{\dfrac{x^2}{x-1}}\).
  4. The region bounded by the curves of \(h\) and \(g\) is rotated about the \(x\)-axis.
    1. Write down the definite integral that can be used to find the volume of the resulting solid.   (2 marks)

    2. Hence, find the volume of this solid. Give your answer correct to two decimal places.   (1 mark)

Show Answers Only

a.  \(\text {Asymptotes: } x=1,\  y=x+1\)

b.   
       

c.i.   \(\text {Asymptotes: } x=k,\  y=x+k\)

c.ii.  \(\text {Distance }=2 \sqrt{5}|k|\)

d.i.  \(\displaystyle V=\pi \int_{\frac{-\sqrt{7}-1}{2}}^{\frac{\sqrt{7}-1}{2}}(x+3)^2-\left(\frac{x^2}{x-1}\right)^2 dx\)

d.ii.  \(V=51.42\ \text{u}^3 \)

Show Worked Solution

a.    \(\text {When } k=1 :\)

\(f(x)=\dfrac{x^2}{x-1}=\dfrac{(x+1)(x-1)+1}{(x-1)}=x+1+\dfrac{1}{x-1}\)

\(\text {Asymptotes: } x=1,\  y=x+1\)
 

b.    
       

 

c.i. \(f(x)=\dfrac{x^2}{x-k}=\dfrac{(x+k)(x-k)+k^2}{x-k}=x+k+\dfrac{k^2}{x-k}\)

\(\text {Using part a.}\)

\(\text {Asymptotes: } x=k,\  y=x+k\)
 

c.ii.  \(f^{\prime}(x)=1-\left(\dfrac{k}{x-k}\right)^2\)

\(\text {TP’s when } f^{\prime}(x)=0 \text { (by CAS):}\)

\(\Rightarrow(2 k, 4 k),(0,0)\)

\(\text {Distance }\displaystyle=\sqrt{(2 k-0)^2+(4 k-0)^2}=\sqrt{20 k^2}=2 \sqrt{5}|k|\)
 

d.i  \(\text {Solve for intersection of graphs (by CAS):}\)

\(\displaystyle x+3=\left|\frac{x^2}{x-1}\right|\)

\(\displaystyle \Rightarrow x=\frac{3}{2}, x=\frac{-1 \pm \sqrt{7}}{2}\)

\(\displaystyle V=\pi \int_{\frac{-\sqrt{7}-1}{2}}^{\frac{\sqrt{7}-1}{2}}(x+3)^2-\left(\frac{x^2}{x-1}\right)^2 dx\)
 

d.ii. \(V=51.42\ \text{u}^3 \text{ (by CAS) }\)

♦♦ Mean mark (d)(ii) 37%.

Calculus, SPEC1-NHT 2019 VCAA 6

Part of the graph of  `y = (2)/(sqrt(x^2-4x+3))`, where  `x > 3`, is shown below.
 


 

Find the volume of the solid of revolution formed when the graph of  `y = (2)/(sqrt(x^2-4x+3))`  from  `x = 4`  to  `x = 6`  is rotated about the `x`-axis. Give your answer in the form  `a log_e(b)`  where `a` and `b` are real numbers.   (5 marks)

Show Answers Only

`pi log_e ((9)/(5))`

Show Worked Solution

`V = pi int_4 ^6 (4)/(x^2 – 4x + 3)\ dx`
 

`text(Using partial fractions:)`

`(4)/(x^2 – 4x + 3) = (a)/((x-3)) + (b)/((x-1))`

`a(x -1) + b(x – 3)= 4`
 

`text(When)\ \ x = 1, \ -2b = 4  => \ b = -2`

`text(When)\ \ x = 3, \ 2a = 4  => \ a = 2`

`:. \ V` `= pi int_4 ^6 (2)/(x-3) – (2)/(x-1)\ dx`
  `= 2 pi [log_e(x-3) -log_e(x-1)]_4 ^6`
  `= 2 pi [log_e 3 – log_e 5 – (log_e 1 – log_e3)]`
  `= 2 pi (2log_e 3 – log_e 5)`
  `= 2pi log_e((9)/(5))`

Calculus, SPEC2 2017 VCAA 1

Let  `f:D ->R, \ f(x) = x/(1 + x^3)`, where `D` is the maximal domain of  `f`.

  1.   i. Find the equations of any asymptotes of the graph of  `f`.   (1 mark)

  2.  ii. Find  `f′(x)`  and state the coordinates of any stationary points of the graph of  `f`, correct to two decimal places.  (2 marks)

  3. iii. Find the coordinates of any points of inflection of the graph of  `f`, correct to two decimal places.  (2 marks)

  4. Sketch the graph of  `f(x) = x/(1 + x^3)`  from  `x=–3`  and  `x = 3`  on the axes provided below, marking all stationary points, points of inflection and intercepts with axes, labelling them with their coordinates. Show any asymptotes and label them with their equations.  (3 marks)

     

     

  5. The region `S`, bounded by the graph of  `f`, the `x`-axis and the line  `x = 3`, is rotated about the `x`-axis to form a solid of revolution. The line  `x = a`, where  `0 < a < 3`, divides the region `S` into two regions such that, when the two regions are rotated about the `x`-axis, they generate solids of equal volume.
  6. i.  Write down an equation involving definite integrals that can be used to determine `a`.  (2 marks)

  7. ii. Hence, find the value of `a`, correct to two decimal places.  (1 mark)

Show Answers Only

  1. i.  `text(vertical asymptote:)\ \ x = −1“text(horizontal asymptote:)\ \ y = 0`
  2. ii. `text(S.P.)\ \ (0.79, 0.53)`
  3. iii. `text(P.O.I.)\ (1.26, 0.42)`
  4.  
  5. i.  `pi int_a^3 (x^2)/((1 + x^3)^2) dx`
  6. ii. `~~ 0.98`

Show Worked Solution

a.i.   `text(Graphing the function on CAS:`

♦♦ Mean mark 36%.

`text(Vertical asymptote:)\ x = −1`

`text(Horizontal asymptote:)\ \ y = 0`
 

a.ii.   `u = x, \ u′ = 1,\ \ v = 1 + x^3, \ v′ = 3x^2\ \ \ text{(manual or by CAS)}`

`f′(x)` `= (1(1 + x^3)-x(3x^2))/((1 + x^3)^2)`
  `= (1-2x^3)/((1 + x^3)^2)`

 
`text(S.P. when)\ \ f′(x)=0:\ `

`=>  (0.79, 0.53)\ \ \ text{(by CAS)}`

 

a.iii.  `text(When)\ \ f″(x)=0,\ \ \ text{(by CAS)}`

`=> x = 0, \ x = -1, \ x = 2`

`x != -1`

`text(Check concavity changes:)`

`f″(−1/2) = −1632/343`

`f″(1) = −3/4`

`f″(3) = 675/(10\ 976)`
 

`text(P.O.I. at)\ \ x = sqrt2 ~~ 1.26\ \ text{(concavity changes)}`

`=> f(sqrt2) ~~ 0.42`

`:. text(P.O.I.)\ (1.26, 0.42)`

 

b.   

 

c.i.    `V_1` `= pi int_0^a y^2\ dx`
  `V_2` `= pi int_a^3 y^2\ dx`

 
`:.\ text(Equation to solve for)\ a:`

`int_0^a x^2/(1 + x^3)^2\ dx = int_a^3 x^2/(1 + x^3)^2\ dx`

 

c.ii.  `a=0.98\ \ \ text{(by CAS)}`

Calculus, SPEC1-NHT 2017 VCAA 5

Part of the graph of  `y = (sqrt(x + 1))/(root4(1 - x^2))`  is shown below.
 


 

Find the volume generated if the region bounded by the graph of  `y = (sqrt(x + 1))/(root4(1 - x^2))`, the lines  `x = -1/2`  and  `x = 1/2`, and the `x`-axis is rotated about the `x`-axis.  (4 marks)

Show Answers Only

`pi^2/3`

Show Worked Solution
`V` `= pi int_(-1/2)^(1/2) y^2\ dx`
  `= pi int_(-1/2)^(1/2) (x + 1)/sqrt(1 – x^2)\ dx`
  `= pi int_(-1/2)^(1/2) x/sqrt(1 – x^2)\ dx + int_(-1/2)^(1/2) 1/sqrt(1 – x^2)\ dx`

 
`text(Let)\ \ u = 1 – x^2`

`(du)/(dx) = -2x\ \ =>\ \ -1/2\ du = x\ dx`

`text(When)\ \ x=1/2\ \ =>\ \ u=3/4`

`text(When)\ \ x=- 1/2\ \ =>\ \ u=3/4`

`text(Same limit)\ =>\ text(1st integral = 0)`
 

`:. V` `= 0+pi int_(-1/2)^(1/2) 1/sqrt(1 – x^2)\ dx`
  `= pi [sin^(-1) (x)]_(-1/2)^(1/2)`
  `= pi (pi/6 – ((-pi)/6))`
  `= pi (pi/3)`
  `= pi^2/3`

Calculus, SPEC1 2017 VCAA 10

  1.  Show that  `d/dx(x arccos(x/a)) = arccos(x/a)−x/(sqrt(a^2-x^2))`, where  `a > 0`.   (1 mark)

  2.  State the maximum domain and the range of  `f(x) = sqrt(arccos(x/2))`.   (2 marks)

  3.  Find the volume of the solid of revolution generated when the region bounded by the graph of  `y = f(x)`, and the lines  `x = −2`  and  `y = 0`, is rotated about the `x`-axis.   (4 marks)

Show Answers Only
  1. `text(See Worked Solutions)`
  2. `x ∈ [−2, 2], \ y ∈ [0, sqrtpi]`
  3. `2pi^2`
Show Worked Solution
a.    `u` `= x,` `v` `= cos^(−1)(x/a)`
  `uprime` `= 1,`   `vprime` `= (−1)/sqrt(a^2-x^2)`
`d/(dx)(xcos^(−1)(x/a))` `= uprimev + vprimeu`
  `= cos^(−1)(x/a) + (x(−1))/sqrt(a^2-x^2)`
  `= arccos(x/a)-x/sqrt(a^2-x^2)`

 

b.   `arccos(x/2)>=0`

`text(Maximal domain:)\  x ∈ [−2, 2]`

`f(x) = (arccos(x/2))^(1/2)`

`text(Range:)\ \ y ∈ [0, sqrtpi]`

 

c.   `V` `= pi int_(−2)^2 y^2\ dx`
  `= pi int_(−2)^2 cos^(-1)(x/2)\ dx`
  `= pi int_(−2)^2 cos^(-1) (x/2)-x/sqrt(4-x^2) + x/sqrt(4-x^2)\ dx`
  `= pi [x cos^(-1)(x/2)]_(−2)^2 + pi int_(−2)^2 x/sqrt(4-x^2)\ dx`

 
`text(Let)\ \ u = 4-x^2`

♦ Mean mark part (c) 35%.

`(du)/(dx) = -2x\ \ =>\ \ -1/2 (du) = x\ dx`
 

`text(When)\ \ x=2\ \ => \ u=0`

`text(When)\ \ x=-2\ \ =>\ \ u=0`

`:. V` `= pi [2cos^(−1)(1)-(-2)cos^(−1)(−1)]-pi/2 int_0^0 1/sqrtu du`
  `= pi(2 xx 0 + 2 xx pi)`
  `= 2pi^2`