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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)

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

  3. 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)

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  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, 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-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, 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`