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Calculus, SPEC2 2023 VCAA 1

Viewed from above, a scenic walking track from point \(O\) to point \(D\) is shown below. Its shape is given by

\(f(x)= \begin{cases}-x(x+a)^2, & 0 \leq x \leq 1 \\ e^{x-1}-x+b, & 1<x \leq 2 .\end{cases}\)

The minimum turning point of section \(O A B C\) occurs at point \(A\). Point \(B\) is a point of inflection and the curves meet at point \(C(1,0)\). Distances are measured in kilometres.
 

  1. Show that  \(a=-1\)  and  \(b=0\).  (1 mark)

  2. Verify that the two curves meet smoothly at point \(C\).  (2 marks)

  3.  i. Find the coordinates of point \(A\).  (1 mark)

  4. ii. Find the coordinates of point \(B\).  (1 mark)

The return track from point \(D\) to point \(O\) follows an elliptical path given by

\(x=2 \cos (t)+2, y=(e-2) \sin (t)\), where \(t \in\left[\dfrac{\pi}{2}, \pi\right]\).

  1. Find the Cartesian equation of the elliptical path.  (2 marks)

  2. Sketch the elliptical path from \(D\) to \(O\) on the diagram above.  (1 mark)

  3.  i. Write down a definite integral in terms of \(t\) that gives the length of the elliptical path from \(D\) to \(O\).  (1 mark)

  4. ii. Find the length of the elliptical path from \(D\) to \(O\).
  5.     Give your answer in kilometres correct to three decimal places.   (1 mark)

Show Answers Only

a.    \(\text{At}\ C(1,0): \)

\(\underset{x \to 1^{-}}{\lim} f_1(x) = \underset{x \to 1^{-}}{\lim}-x(x+a)^2 = -1(1+a)^2 = 0 \)

    \(a=-1\)

\(\underset{x \to 1^{+}}{\lim} f_2(x)\ \underset{x \to 1^{+}}{\lim} (e^{x-1}-x+b) = e^0-1+b=b = 0 \)
 

b.    \(\text{Find derivative functions (by calc)}:\)

\(\dfrac{d}{dx} \big{(}-x(x-1)^2\big{)} = -(x-1)(3x-1)\ \ \Rightarrow \ \ \text{At}\ \ x=1, f_1^{′}(x)=0 \)

\(\dfrac{d}{dx} (e^{x-1}-x+b) = e^{x-1}-1\ \ \Rightarrow \ \ \text{At}\ \ x=1, f_2^{′}(x)=0 \)

\(\Rightarrow\ \text{Since gradients are equal, curves meet smoothly.}\)
 

c.i.  \(A\Big{(} \frac{1}{3}, -\frac{4}{27} \Big{)} \)
 

c.ii.  \(B\Big{(} \frac{2}{3}, -\frac{2}{27} \Big{)} \)
 

d.   \(\Big{(}\dfrac{x-2}{2}\Big{)}^2 + \Big{(}\dfrac{y}{e-2}\Big{)}^2=1\)
 

e.    
          
 

f.i.    \(\displaystyle{\int_{\frac{\pi}{2}}^{\pi}} \sqrt{4\,\sin^2(t)+(e-2)^2\cos^2(t)}\,dt\)
 

f.ii.  \(\text{Length}\ = 2.255\)

Show Worked Solution

a.    \(\text{At}\ C(1,0): \)

\(\underset{x \to 1^{-}}{\lim} f_1(x) = \underset{x \to 1^{-}}{\lim}-x(x+a)^2 = -1(1+a)^2 = 0 \)

    \(a=-1\)

\(\underset{x \to 1^{+}}{\lim} f_2(x)\ \underset{x \to 1^{+}}{\lim} (e^{x-1}-x+b) = e^0-1+b=b = 0 \)
 

b.    \(\text{Find derivative functions (by calc)}:\)

\(\dfrac{d}{dx} \big{(}-x(x-1)^2\big{)} = -(x-1)(3x-1)\ \ \Rightarrow \ \ \text{At}\ \ x=1, f_1^{′}(x)=0 \)

\(\dfrac{d}{dx} (e^{x-1}-x+b) = e^{x-1}-1\ \ \Rightarrow \ \ \text{At}\ \ x=1, f_2^{′}(x)=0 \)

\(\Rightarrow\ \text{Since gradients are equal, curves meet smoothly.}\)
 

c.i.  \(\dfrac{d}{dx} \big{(}-x(x-1)^2\big{)} = -(x-1)(3x-1)\ \Rightarrow \ \text{SP when}\ x=\dfrac{1}{3} \)

\(A\Big{(} \frac{1}{3}, -\frac{4}{27} \Big{)} \)
 

c.ii.  \(\text{POI at}\ B: \ f_1^{″}(x)=-6x+4=0\ \ \Rightarrow x=\dfrac{2}{3} \)

\(B\Big{(} \frac{2}{3}, -\frac{2}{27} \Big{)} \)
 

d.   \(\dfrac{x-2}{2}=\cos(t), \ \ \dfrac{y}{e-2}=\sin(t) \)

\(\Big{(}\dfrac{x-2}{2}\Big{)}^2 + \Big{(}\dfrac{y}{e-2}\Big{)}^2=1\)
 

e.    
          
 

f.i.    \(\dfrac{dx}{dt} = -2\sin(t),\ \ \dfrac{dy}{dt} = (e-2)\cos(t) \)

\(\displaystyle{\int_{\frac{\pi}{2}}^{\pi}} \sqrt{4\,\sin^2(t)+(e-2)^2\cos^2(t)}\,dt\)
 

f.ii.   \(\text{Evaluate the integral in part f.i.}\)

\(\Rightarrow \text{Length}\ = 2.255\)

Calculus, SPEC1 2020 VCAA 9

Consider the curve defined parametrically by

`x = arcsin (t)`

`y = log_e(1 + t) + 1/4 log_e (1 - t)`

where `t in [0, 1)`.

  1. `((dy)/(dt))^2` can be written in the form  `1/(a(1 + t)^2) + 1/(b(1 - t^2)) + 1/(c(1 - t)^2)` where  `a, b`  and  `c`  are real numbers.

     

     

    Show that  `a = 1, b = – 2`  and  `c = 16`.  (2 marks)

  2. Find the arc length, `s`, of the curve from  `t = 0`  to  `t = 1/2`. Give your answer in the form  `s = log_e(m) + n log_e (p)`, where  `m, n, p in Q`.  (3 marks)
Show Answers Only
  1. `text(Proof)\ text{(See Worked Solutions)}`
  2. `log_e(3/2) – 1/4 log_e(1/2)`
Show Worked Solution
a.   `y` `= log_e(1 + t) + 1/4 log_e (1 – t)`
  `(dy)/(dt)` `= 1/(1 + t) – 1/(4(1 – t))`
  `((dy)/(dt))^2` `= 1/(1 + t)^2 – 2 ⋅ 1/(1 + t) ⋅ 1/(4(1 – t)) + 1/(16(1 – t)^2)`
    `= 1/(1 + t)^2 – 1/(2(1 – t^2)) + 1/(16(1 – t)^2)`

 
`:. a = 1, b = – 2, c = 16`

♦♦ Mean mark part (b) 31%.

 

b.   `s` `= int_0^(1/2) sqrt(((dx)/(dt))^2 + ((dy)/(dt))^2)\ dt`
    `= int_0^(1/2) sqrt(1/(1 – t^2) + 1/(1 + t^2) – 1/(2(1 – t^2)) + 1/(16(1 – t)^2))\ dt`
    `= int_0^(1/2) sqrt(1/(1 + t^2) + 1/(2(1 – t^2)) + 1/(16(1 – t)^2))\ dt`
    `= int_0^(1/2) sqrt((1/(1 + t) + 1/(4(1 – t)))^2)\ dt`
    `= int_0^(1/2) 1/(1 + t) + 1/(4(1 – t))\ dt`
    `= [log_e(1 + t) – 1/4 log_e(1 – t)]_0^(1/2)`
    `= log_e(3/2) – 1/4 log_e(1/2)`

Calculus, SPEC1-NHT 2019 VCAA 8

Find the length of the arc of the curve defined by  `y = (x^4)/(4) + (1)/(8x^2) + 3`  from  `x = 1`  to  `x = 2`. Give your answer in the form  `(a)/(b)`, where `a` and `b` are positive integers.   (4 marks)

Show Answers Only

`(123)/(32)`

Show Worked Solution

`(dy)/(dx) = x^3 + ((-2)/(8) x^-3) = x^3 – (1)/(4x^3)`

`1 + ((dy)/(dx))^2` `= 1 + (x^6 – (1)/(2) + (1)/(16x^6))`
  `= (16x^6 + 16x^12 – 8x^6 + 1)/(16x^6)`
  `= (16x^12 + 8x^6 + 1)/(16x^6)`
  `= ((4x^6 + 1)^2)/(16x^6)`

 

`text(Arc Length)` `= int_1 ^2 sqrt(((4x^6 + 1)^2)/(16x^6))\ dx`
  `= int_1 ^2 (4x^6 + 1)/(4x^3)\ dx`
  `= int_1 ^2 x^3 + (1)/(4x^3)\ dx`
  `= [(x^4)/(4) – (1)/(8x^2)]_1 ^2`
  `= [4 – (1)/(32) – ((1)/(4) – (1)/(8))]`
  `= (123)/(32)`

Calculus, SPEC2 2019 VCAA 7 MC

The length of the curve defined by the parametric equations  `x = 3sin(t)`  and  `y = 4cos(t)`  for  `0 <= t <= pi`  is given by

  1. `int_0^pi sqrt(9cos^2(t) - 16sin^2(t))\ dt`
  2. `int_0^pi sqrt(9 + 7sin^2(t))\ dt`
  3. `int_0^pi sqrt(1 + 16sin^2(t))\ dt`
  4. `int_0^pi (3cos(t) - 4sin(t))\ dt`
  5. `int_0^pi sqrt(3cos^2(t) + 4sin^2(t))\ dt`
Show Answers Only

`B`

Show Worked Solution

`x = 3sin(t) \ => \ (dx)/(dt) = 3cos(t)`

`y = 4cos(t) \ => \ (dy)/(dt) = −4sin(t)`

`text(Length)` `= int_0^pi sqrt((3cos(t))^2 + (−4sin(t))^2)\ dt`
  `= int_0^pi sqrt(9cos^2(t) + 16sin^2(t))\ dt`
  `= int_0^pi sqrt(9cos^2(t) + 9sin^2(t) + 7sin^2(t))\ dt`
  `= int_0^pi sqrt(9 + 7sin^2(t))\ dt`

 
`=>B`

Calculus, SPEC1 2016 VCAA 7

Find the arc length of the curve  `y = 1/3 (x^2 + 2)^(3/2)`  from  `x = 0`  to  `x = 2`.  (4 marks)

Show Answers Only

`14/3`

Show Worked Solution
`y` `= 1/3 (x^2 + 2)^(3/2)`
`(dy)/(dx)` `= 1/3 xx 3/2 xx 2x (x^2 + 2)^(3/2 – 1)`
  `= x(x^2 + 2)^(1/2)`
`((dy)/(dx))^2` `= x^2 (x^2 + 2)`
  `= x^4 + 2x^2`

 

`l` `= int_0^2 sqrt(1 + ((dy)/(dx))^2)\ dx`
  `= int_0^2 sqrt(x^4 + 2x^2 + 1)\ dx`
  `= int_0^2 sqrt((x^2 + 1)^2)\ dx`
  `= int_0^2 x^2 + 1\ dx`
  `= [x^3/3 + x]_0^2`
  `= 8/3 + 2`
  `= 14/3`

Calculus, SPEC2 2017 VCAA 3

A brooch is designed using inverse circular functions to make the shape shown in the diagram below.
 


 

The edges of the brooch in the first quadrant are described by the piecewise function
 

`f(x){(3text(arcsin)(x/2)text(,), 0 <= x <= sqrt2),(3text(arccos)(x/2)text(,), sqrt2 < x <= 2):}`
 

  1. Write down the coordinates of the corner point of the brooch in the first quadrant.  (1 mark)
  2. Specify the piecewise function that describes the edges in the third quadrant.  (1 mark)
  3. Given that each unit in the diagram represents one centimetre, find the area of the brooch.

     

    Give your answer in square centimetres, correct to one decimal place.  (3 marks)

  4. Find the acute angle between the edges of the brooch at the origin. Give your answer in degrees, correct to one decimal place.  (3 marks)
  5. The perimeter of the brooch has a border of gold.
    Show that the length of the gold border needed is given by a definite integral of the form  `int_0^2 (sqrt(a + b/(4 - x^2)))dx`, where  `a, b ∈ R`. Find the values of `a` and `b`.  (2 marks)
Show Answers Only
  1. `(sqrt2,(3pi)/4)`
  2. `text(See Worked Solutions)`
  3. `9.9\ text(cm²)`
  4. `67.4°`
  5. `a = 16, b = 144`
Show Worked Solution

a.   `text(Corner point occurs when)\ \ x=sqrt2.`

`y=3 sin^(-1) (sqrt2/2) = (3pi)/4`

`:.\ text(Coordinates are:)\ \ (sqrt2, (3pi)/4)`
 

b.    `text(Reflect in the)\ xtext(-axis and then the)\ ytext(-axis:)`

♦ Mean mark 49%.

`g(x){(-3text(arccos)(- x/2)text(,), -2 <= x < -sqrt2),(-3text(arcsin)(- x/2)text(,), -sqrt2 <= x <= 0):}`

 

c.    `A` `= 4 xx (int_0^sqrt2 3sin^(−1)(x/2)dx + int_sqrt2^2 3cos^(−1)(x/2)dx)`
    `~~ 9.9\ text(cm²)`

 

d.   `text(Find the gradient of graph at)\ \ x=0:`

`f′(0) = 3/2`

`alpha = tan^(−1)(3/2) = 56.31…°`

`beta = pi/2 – tan^(−1)(1.5) = 33.69°`


 
`:.\ text(Acute angle between the edges)`

`=2 xx 33.69`

`=67.4°`
  

e.  `f′(x)\ {(3/sqrt(4-x^2)text(,), 0<= x <= sqrt2),((-3)/sqrt(4-x^2)text(,), sqrt2 < x <= 2):}`
 

`:.\ text(Length of border)`

♦♦ Mean mark 27%.

`= 4 int_0^sqrt2 sqrt(1 + (3/sqrt(4 – x^2))^2)\ dx + 4 int_sqrt2^2 sqrt(1 + ((-3)/sqrt(4 – x^2))^2)\ dx`

`= 4 int_0^2 sqrt(1 + (3/sqrt(4 – x^2))^2)\ dx`

`= int_0^2 sqrt(16 + 144/(4 – x^2)\ dx`

  
`:. a=16, \ b=144`

Calculus, SPEC1-NHT 2017 VCAA 11

Find the length of the curve specified parametrically by  `x = a theta - a sin (theta), \ y = a - a cos (theta)`  from  `theta = (2 pi)/3`  to  `theta = 2 pi`, where  `a in R^+`. Give your answer in terms of `a`.  (4 marks)

Show Answers Only

`6a`

Show Worked Solution
`(dx)/(d theta)` `= a – a cos (theta)`
`(dy)/(d theta)` `= a sin (theta)`
   
`((dx)/(d theta))^2` `= (a(1 – cos (theta)))^2`
  `= a^2(1 – 2 cos (theta) + cos^2 theta)`
`((dy)/(d theta))^2` `= a^2 sin^2 (theta)`

 

`l` `= int_((2 pi)/3)^(2pi) sqrt(a^2 sin^2(theta) + a^2 – 2a^2 cos (theta) + a^2 cos^2(theta))\ d theta`
  `= int_((2 pi)/3)^(2pi) sqrt(a^2(sin^2(theta) + cos^2(theta) + 1 – 2 cos(theta)))\ d theta`
  `= int_((2 pi)/3)^(2pi) a sqrt(1 + 1 – 2 cos (theta))\ d theta, \ \ a > 0`
  `= a int_((2 pi)/3)^(2pi) underbrace{sqrt(2 – 2cos (theta))}_{cos (theta) = 2 cos^2(theta/2) – 1}\ d theta`
  `= a int_((2 pi)/3)^(2pi) sqrt(2 – 4 cos^2 (theta/2) + 2)\ d theta`
  `= a int_((2 pi)/3)^(2pi) sqrt(4 – 4 cos^2(theta/2))\ d theta`
  `= a int_((2 pi)/3)^(2pi) sqrt(4 sin^2 (theta/2))\ d theta`
  `= a int_((2 pi)/3)^(2pi) 2 sin (theta/2)\ d theta`
  `(theta in [(2 pi)/3, 2 pi]\ \ =>\ \  theta/2 in [pi/3, pi]\ \ =>\ \ sin(theta/2) > 0)`
  `= a [-4 cos (theta/2)]_((2 pi)/3)^(2pi)`
  `= a(-4 cos (pi) – (-4)cos (pi/3))`
  `= a(-4 xx (-1) + 4 xx 1/2)`
  `= 6a`

Calculus, SPEC1-NHT 2018 VCAA 7

  1. Find  `d/(dx) ((1 - x^2)^(1/2))`.  (2 marks)
  2. Hence, find the length of the curve specified by  `y = sqrt (1 - x^2)`  from  `x = 1/2`  to  `x = sqrt 3/2`.

     

    Give your answer in the form  `k pi, k in R`.  (2 marks)

Show Answers Only
  1. `(-x)/sqrt(1 – x^2)`
  2. `pi/6`
Show Worked Solution

a.  `d/(dx) ((1 – x^2)^(1/2))`

`= (1/2 (-2x) (1 – x^2)^(-1/2))`

`= -x(1 – x^2)^(-1/2)`

`= (-x)/sqrt (1 – x^2)`

 

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

Calculus, SPEC2 2018 VCAA 7 MC

A curve is described parametrically by  `x = sin(2t), y = 2 cos (t)`  for  `0 <= t <= 2pi`.

The length of the curve is closest to

A.    9.2

B.    9.5

C.  12.2

D.  12.5

E.  38.3 

Show Answers Only

`C`

Show Worked Solution

`x prime (t) = 2 cos (2t)`

`y prime (t) = -2 sin(t)`

`ℓ` `= int_0^(2 pi) sqrt{(2 cos(2t))^2 + (-2 sin (t))^2 dt}`
  `~~ 12.2`

 
`=>  C`