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Calculus, SPEC2 2022 VCAA 10 MC

Consider the curve given by  `5 x^2 y-3 x y+y^2=10`.

The equation of the tangent to this curve at the point `(1, m)`, where `m` is a real constant, will have a negative gradient when

  1. `m \in R \backslash[-1,0]`
  2. `m=-\sqrt{11}-1 \ text {only}`
  3. `m \in R \backslash(-1,0]`
  4. `m=\sqrt{11}-1 \ text[only]`
  5. `m=-\sqrt{11}-1 or m=\sqrt{11}-1`
Show Answers Only

`E`

Show Worked Solution

`5 x^2 y-3 x y+y^2=10`

`text{At}\ (1,m):\ \ 5m-3m+m^2=10`

`text{Solve (by CAS):}\ \ m=-1+-sqrt{11}`

`text{Check the value of the derivative function at both values of}\ m\ (x=1)`

`text{By CAS, both values (gradients) are negative.}`

`=>E`

Calculus, SPEC1 2022 VCAA 7

A curve has equation `x cos(x+y)=(pi)/(48)`.

Find the gradient of the curve at the point `((pi)/(24),(7pi)/(24))`. Give your answer in the form `(asqrtb-pi)/(pi)`, where `a,b in Z`.   (3 marks)

Show Answers Only

`(8sqrt3-pi)/pi`

Show Worked Solution

`text{Using implicit differentiation and the chain and product rules:}`

`d/(dx)(cos(x + y))`

`text{Let}\ \ u = x + y\ \ \=>\ \ y = cos(u)` 

`(du)/(dx) = 1 +dy/dx,\ \ (dy)/(du) = -sin(u)`

`cos (x+y)-x\ sin (x+y)(dy/dx+1)` `=0`  
`cos (x+y)-x\ sin(x+y)(dy/dx)-x\ sin(x+y)` `=0`  
`x\ sin(x+y)(dy/dx)` `=cos (x+y)-x\ sin(x+y)`  
`dy/dx` `=(cos (x+y))/(x\ sin(x+y)) – (x\ sin(x+y))/(x\ sin(x+y))`  
  `=(cos (x+y))/(x\ sin(x+y))-1`  

 
`text{At}\ ((pi)/(24),(7pi)/(24))\ \ \=>\ \ x+y = (pi)/(24)+(7pi)/(24) = (pi)/(3)`

`dy/dx` `=cos(pi/3)÷[(pi)/(24)sin(pi/3)]-1`  
  `=(1/2)/((pi)/(24) xx sqrt3/2)-1`  
  `=24/(sqrt3pi)-1`  
  `=(24-sqrt3pi)/(sqrt3pi) xx (sqrt3/sqrt3)`  
  `=(24sqrt3-3pi)/(3pi)`  
  `=(8sqrt3-pi)/pi`  

♦ Mean mark 50%.

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, SPEC2 2023 VCAA 9 MC

The position of a particle moving in the Cartesian plane, at time \(t\), is given by the parametric equations

\(x(t)=\dfrac{6 t}{t+1}\)  and  \(y(t)=\dfrac{-8}{t^2+4}\), where  \(t \geq 0\).

What is the slope of the tangent to the path of the particle when  \(t=2\) ?

  1. \(-\dfrac{1}{3}\)
  2. \(-\dfrac{1}{4}\)
  3. \(\dfrac{1}{3}\)
  4. \(\dfrac{3}{4}\)
  5. \(\dfrac{4}{3}\)
Show Answers Only

\(D\)

Show Worked Solution

\(\text{At}\ \ t=2\ \text{(by calc):}\)

\(\dfrac{dx}{dt}=\dfrac{2}{3}, \ \dfrac{dy}{dt}=\dfrac{1}{2} \)

\(\dfrac{dy}{dx} = \dfrac{dy}{dt} \times \dfrac{dt}{dx} = \dfrac{1}{2} \times \dfrac{3}{2} = \dfrac{3}{4} \)

\(\Rightarrow D\)

Calculus, SPEC1 2021 VCAA 5

Find the gradient of the curve with equation  `e^x e^(2y) = 2e^4`  at the point  `(2, 1)`.  (3 marks)

Show Answers Only

`-1/10`

Show Worked Solution

`e^x e^(2y) + e^(4y^2) = 2e^4`

`e^x · 2e^(2y) · (dy)/(dx) + e^x · e^(2y) + e^(4y^2) · 8y · (dy)/(dx)` `= 0`
`(dy)/(dx)(e^x ·2e^(2y) + e^(4y^2) · 8y)` `= -e^x · e^(2y)`

 
`(dy)/(dx) = (-e^x · e^(2y))/(e^x · 2e^(2y) + e^(4y^2) ·8y)`

 
`text(At)\ \ (2, 1):`

`(dy)/(dx)` `= (-e^2 · e^2)/(e^2 · 2e^2 + e^4 · 8)`
  `= (-e^4)/(e^4(2 + 8))`
  `= -1/10`

Calculus, SPEC1 2011 VCAA 10

Consider the relation  `y log_e (x) = e^(2y) + 3x - 4.`

Evaluate  `(dy)/(dx)`  at the point  `(1, 0).`  (4 marks)

Show Answers Only

`-3/2`

Show Worked Solution

`y log_e x = e^(2y) + 3x – 4`

`text(Using implicit differentiation:)`

`d/(dx)(y ln(x))` `= d/(dx)(e^(2y)) + d/(dx)(3x) – d/(dx)(4)`
`dy/dx*ln(x) + y(1/x)` `= 2e^(2y)*dy/dx + 3`

 
`text(At)\ \ (1,0):`

`dy/dx xx ln1 + 0` `= 2 e^0 * dy/dx+ 3`
`2*dy/dx` `= -3`
`:. dy/dx` `=- 3/2`

Calculus, SPEC1 2011 VCAA 5

For the curve with parametric equations

`x = 4 sin (t) - 1`

`y = 2 cos (t) + 3`

Find  `(dy)/(dx)`  at the point  `(1, sqrt 3 + 3).`  (3 marks)

Show Answers Only

`– sqrt 3/6`

Show Worked Solution

`(dx)/(dt) = 4cos(t)`

`(dy)/(dt) = −2sin(t)`

`(dy)/(dx)` `= ((dy)/(dt))/((dx)/(dt))`
  `= (-sin t)/(2cos t)`

 
`x = 4 sin (t) – 1\ \ text{(given)}\ \ =>\ \ sin(t)=(x+1)/4`

`y = 2 cos (t) + 3\ \ text{(given)}\ \ =>\ \ 2cos(t)=(y-3)`

`=> dy/dx= (-(x + 1)/4)/(y – 3)`
 

`text(At)\ \ (1, sqrt 3 + 3):`

`:. (dy)/(dx)` `= (-1/2)/(sqrt3+3 -3)`
  `= -1/(2sqrt3)`
  `= – sqrt3/6`

Calculus, SPEC1 2012 VCAA 6

Find the gradient of the tangent to the curve  `xy^2 + y + (log_e (x - 2))^2 = 14`  at the point  `(3, 2).`  (3 marks)

Show Answers Only

`(dy)/(dx) = -4/13`

Show Worked Solution

`xy^2 + y + (log_e (x – 2))^2 = 14`

`text(Using implicit differentiation:)`

`d/(dx) (xy^2) + d/(dx) (y) + d/(dx) ((ln (x – 2))^2) = d/(dx)(14)`

`d/(dx) (x) ⋅ y^2 + d/(dx) (y^2) ⋅ x + (dy)/(dx) + 2 xx 1/(x – 2) xx ln(x – 2) = 0`

`y^2 + 2xy* (dy)/(dx) + (dy)/(dx) + (2 ln (x – 2))/(x – 2) = 0`

`text{At (3,2):}`

`2^2 + 2(2)(3) m + m + (2 ln(1))/1` `= 0`
`4 + 12m + m` `= 0`
`13m` `= -4`
`:. m` `= -4/13`

Graphs, SPEC1 2013 VCAA 4

  1. State the maximal domain and the range of  `y = arccos(1-2x).`   (2 marks)

  2. Sketch the graph of  `y = arccos(1-2x)`  over its maximal domain. Label the endpoints with their coordinates.   (2 marks)

     

     
              VCAA 2013 spec 4b
     

  3. Find the gradient of the tangent to the graph of  `y = arccos (1 – 2x)`  at  `x = 1/4.`   (2 marks)

Show Answers Only
  1. `text(Maximal domain)\ \ [0, 1];\ \ \ text(Range)\ \ [0, pi]`
  2.  

  3. `4/sqrt3`
Show Worked Solution

a.  `text(Maximal domain:)`

`1-2x` `∈ [−1, 1]`
`-2x` `∈ [−2, 0]`
`x` `∈ [0, 1]`

 
`text(When)\ \ x = 0,\ \ y = cos^-1 1= 0;`

`text(When)\ \ x = 1,\ \ y = cos^-1 (-1)= pi`

`:.\ text(Range is)\ \ [0, pi]`

 

b.   

 

c.   `y = cos^-1 (1-2x),`

`(dy)/(dx)` `= (-1(-2))/sqrt(1-(1-2x)^2)`
  `= 2/sqrt(1-(1-2x)^2)`

 
`text(At)\ \ x = 1/4,`

`m_T` `= 2/sqrt(1-(1-1/2)^2)`
  `=2/sqrt(3/4)`
  `= 4/sqrt 3`
  `= (4 sqrt 3)/3`

Calculus, SPEC1 2013 VCAA 6

Find the value of `c`, where  `c in R`, such that the curve defined by

`y^2 + (3e^{(x - 1)})/(x - 2) = c`

has a gradient of 2 where  `x = 1.`   (4 marks)

Show Answers Only

`-3/4`

Show Worked Solution

`text(Using implicit differentiation:)`

`d/(dx)(y^2) + d/(dx)((3e^(x – 1))/(x – 2))` `= d/(dx)(c)`
`2y(dy)/(dx) + ((x – 2)3e^(x – 1) – 3e^(x – 1))/(x – 2)^2` `= 0`
`2y(dy)/(dx) +(3e^(x – 1) (x – 3))/(2y(x – 2)^2)` `=0`

 
`:. (dy)/(dx) = -(3e^(x – 1) (x – 3))/(2y(x – 2)^2)`

`text(When)\ \ x=1,\ \ dy/dx=2`

`2=(-6)/(-2y)\ \ =>\ \ y=3/2`

`text(Substituting into curve equation:)`

`(3/2)^2 + (3e^0)/(1 – 2)` `= c`
`9/4 – 3` `= c`
`c` `= -3/4`

Calculus, SPEC1-NHT 2017 VCAA 3

Find the gradient of the curve with equation  `x = sin (y/15)`  when  `x = 1/4`. Give your answer in the form  `a sqrt b`, where  `a, b \ in Z^+`.  (3 marks)

Show Answers Only

 `m = 4 sqrt 15`

Show Worked Solution
`x` `= sin (y/15)`
`sin^(-1)x` `= y/15`
`y` `= 15 sin^(-1)x`
`dy/dx` `= 15/(sqrt(1-x^2))`

 

`text(When)\ \ x=1/4:`

`dy/dx` `= 15/(sqrt(1-(1/4)^2))`
  `= 15/(sqrt(15/16))`
  `= 60/sqrt15`
  `=4sqrt15`

 
`:. m = 4 sqrt 15`

Calculus, SPEC1 2015 VCAA 9

Consider the curve represented by  `x^2 - xy + 3/2 y^2 = 9.`

  1. Find the gradient of the curve at any point  `(x, y).`  (2 marks)
  2. Find the equation of the tangent to the curve at the point  `(3, 0)`  and find the equation of the tangent to the curve at the point `(0, sqrt 6).`

     

    Write each equation in the form  `y = ax + b.`  (2 marks)

  3. Find the acute angle between the tangent to the curve at the point  `(3, 0)`  and the tangent to the curve at the point  `(0, sqrt 6).`

     

    Give your answer in the form  `k pi`, where `k` is a real constant  (2 marks)

Show Answers Only
  1. `(dy)/(dx) = (2x – y)/(x – 3y)`
  2. `y = 2(x – 3);\ \ \ y = 1/3 x + sqrt 6`
  3. `theta = pi/4`
Show Worked Solution
a.    `d/(dx)(x^2) – d/(dx)(xy) + 3/2* d/(dx) (y^2)` `= 0`
  `2x – x*(dy)/(dx) – y + 3/2(2y)*(dy)/(dx)` `= 0`
  `(dy)/(dx)(−x + 3y)` `= y – 2x`
  `:. (dy)/(dx)` `= (y – 2x)/(3y – x)`

 

b.   `m_{(3,0)} = (0 – 6)/(0 – 3) = 2`

`:.\ text{Equation of tangent at (3, 0):}`

`y = 2(x – 3)`

  `= 2x – 6`

 

`m_{(0,sqrt6)} = (sqrt6 – 0)/(3sqrt6 – 0) = 1/3`

`:.\ text{Equation of tangent at}\ (0,sqrt6):`

`y -sqrt6` `= 1/3(x – 0)`  
`y` `=1/3 x + sqrt6`  

 

c.   `m_1 = 2 = tan(theta_1), \ \ m_2 = 1/3 = tan(theta_2)`

`alpha` `= theta_1 – theta_2`
  `= tan^(−1)(2) – tan^(−1)(1/3)`
`tan(alpha)` `= tan(tan^(−1)(2) – tan^(−1)(1/3))`
  `= (tan(tan^(−1)(2)) – tan(tan^(−1)(1/3)))/(1 + tan(tan^(−1)(2)tan(tan^(−1)(1/3))))`
  `= (2 – 1/3)/(1 + 2/3)`
  `= 1`

 
`:. alpha = pi/4\ \ \ (alpha ∈ (0, pi/2))`

Calculus, SPEC1-NHT 2018 VCAA 9

  1.    i. Given that  `cot(2 theta) = a`, show that  `tan^2(theta) + 2a tan(theta)-1 = 0`.   (2 marks)

  2.  ii. Show that  `tan(theta) = -a +- sqrt(a^2 + 1)`.  (1 mark)

  3. iii. Hence, show that  `tan(pi/12) = 2-sqrt 3`, given that  `cot(2 theta) = sqrt 3`, where  `theta in (0, pi)`.   (1 mark)

  4.  Find the gradient of the tangent to the curve  `y = tan (theta)`  at  `theta = pi/12`.   (2 marks)

  5.  A solid of revolution is formed by rotating the region between the graph of  `y = tan(theta)`, the horizontal axis, and the lines  `theta = pi/12`  and  `theta = pi/3`  about the horizontal axis.
  6. Find the volume of the solid of revolution.   (3 marks)

Show Answers Only

    1. `text(Proof)\ \ text{(See Worked Solutions)}`
    2. `text(Proof)\ \ text{(See Worked Solutions)}`
    3. `text(Proof)\ \ text{(See Worked Solutions)}`
  2. `8-4 sqrt 3`
  3. `pi (2 sqrt 3-2-pi/4)`

Show Worked Solution

a.i.   `1/(tan 2 theta)` `= a`
  `1` `= a tan (2 theta)`
  `1` `= a ((2 tan (theta))/(1-tan^2(theta)))`
  `1/a (1-tan^2 (theta))` `= 2 tan (theta)`
  `1-tan^2 (theta)` `= 2a tan (theta)`
  `:. tan^2 (theta) + 2 a tan (theta)-1` `=0\ \ text(… as required)`

 

a.ii.   `[tan^2 (theta) + 2a tan (theta) + a^2]-a^2-1` `= 0`
  `(tan (theta) + a)^2` `= a^2 + 1`
  `tan (theta) + a` `= +- sqrt(a^2 + 1)`
  `:. tan (theta)` `= -a +- sqrt(a^2 + 1)`

 

a.iii.   `theta in (0, pi) \ => \ 2 theta in (0, 2 pi)`

`text(S)text(ince)\ \ cot(2theta)=sqrt3\ \ \ =>\ \ \ tan(2theta)=1/sqrt3`

  `:. 2 theta` `= pi/6, pi + pi/6`
  `theta` `= pi/12, (7 pi)/12`
  `tan (theta)` `=-sqrt 3 +- sqrt((sqrt 3)^2 + 1)`
    `=-sqrt 3 +- sqrt(3 + 1)`
    `=-sqrt 3 +- 2`

 
`:. tan (pi/12) = 2-sqrt 3,\ \ \ \ (tan (pi/12) > 0)`

 

b.   `y` `=tan(theta)`
  `y prime` `= sec^2 (theta)`
    `= 1 + tan^2 (theta)`

 

`y prime (pi/12)` `= 1 + tan^2 (pi/12)`
  `= 1 + (2-sqrt 3)^2`
  `= 1 + 4-4 sqrt 3 + 3`
  `= 8-4 sqrt 3`

 

c.   `V` `= pi int_(pi/12)^(pi/3) y^2\ d theta`
    `= pi int_(pi/12)^(pi/3) tan^2 (theta)\ d theta`
    `= pi int_(pi/12)^(pi/3) (1 + tan^2 (theta)-1)\ d theta`
    `= pi int_(pi/12)^(pi/3) (sec^2 (theta)-1)\ d theta`
    `= pi [tan (theta)-theta]_(pi/12)^(pi/3)`
    `= pi (tan (pi/3)-pi/3-(tan (pi/12)-pi/12))`
    `= pi (sqrt 3-(4 pi)/12-(2-sqrt 3) + pi/12)`
    `= pi (2 sqrt 3-2-pi/4)\ \ text(u³)`

Calculus, SPEC1 2018 VCAA 3

Find the gradient of the curve with equation  `2x^2 sin(y) + xy = pi^2/18`  at the point  `(pi/6, pi/6)`.

Give your answer in the form  `a/(pi sqrt b + c)`, where `a, b` and `c` are integers.  (4 marks)

Show Answers Only

`m = (-18)/(pi sqrt 3 + 6)`

Show Worked Solution

`d/(dx) (2x^2 sin (y)) + d/(dx) (xy) = d/(dx) (pi^2/18)`

`4x sin(y) + 2x^2 cos(y) (dy)/(dx) + y + x (dy)/(dx)` `=0`  
`(4 pi)/6 sin (pi/6) + (2 pi^2)/36 cos (pi/6) m + pi/6 + pi/6 m` `=0`  
`(2 pi)/3 xx 1/2 + pi^2/18 xx sqrt 3/2 m + pi/6 + pi/6 m` `=0`  
`pi/3 + pi/6 + ((pi^2 sqrt 3)/36 + pi/6) m` `=0`  
`pi/2 + pi/36 (pi sqrt 3 + 6) m` `=0`  
`(pi m)/36 (pi sqrt 3 + 6)` `= (-pi)/2`
`m (pi sqrt 3 + 6)` `= -18`
`:. m` `= (-18)/(pi sqrt 3 + 6)`