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Calculus, SPEC1 2025 VCAA 1

Consider the curve with equation  \(x e^{-2 y}+y^2 e^x=8 e^4\).

Find the equation of the tangent to the curve at the point \((4,-2)\).   (4 marks)

Show Answers Only

\(y=\dfrac{5}{12} x-\dfrac{11}{3}\)

Show Worked Solution

\(x e^{-2 y}+y^2 e^x=8 e^4\)

\(x\left(-2 e^{-2 y}\right) \dfrac{d y}{d x}+e^{-2 y}+2 y e^x \dfrac{d y}{d x}+y^2 e^x=0\)

\(\dfrac{d y}{d x}\left(-2 x e^{-2 y}+2 y e^x\right)=-e^{-2 y}-y^2 e^x\)

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

\(\text{Find  \(\ \dfrac{d y}{d x}\ \)  at  \(\ (4,-2)\):}\)

\(\dfrac{d y}{d x}=\dfrac{-e^4-4 e^4}{-8 e^4-4 e^4}=\dfrac{-5 e^4}{-12 e^4}=\dfrac{5}{12}\)
 

\(\text{Equation of tangent, \(m=\dfrac{5}{12} \) through \((4,-2)\):}\)

\(y+2\) \(=\dfrac{5}{12}(x-4)\)
\(y+2\) \(=\dfrac{5}{12} x-\dfrac{5}{3}\)
\(y\) \(=\dfrac{5}{12} x-\dfrac{11}{3}\)

Calculus, SPEC1 2024 VCAA 8

Consider the relation  \(x^2 y^2+x y=2\), where  \(x, y \in R\).

  1. Using implicit differentiation, show that  \(\dfrac{d y}{d x}=-\dfrac{y}{x}\)  given that  \(2 x y \neq-1\).   (2 marks)

  2. Find all points on the graph of  \(x^2 y^2+x y=2\)  where the slope of the tangent is equal to \(-1\).   (2 marks)

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a.   \(\text{See worked solutions}\)

b.   \((1, 1), (-1,-1)\)

Show Worked Solution

a.    \(x^2 y^2+x y=2\)

\(2 x y^2+2 x^2 y \cdot \dfrac{d y}{d x}+y+x \cdot \dfrac{d y}{d x}=0\)

\(\dfrac{d y}{d x} \cdot 2 x^2 y+\dfrac{d y}{d x} \cdot x\) \(=-2 x y^2-y\)
\(\dfrac{dy}{dx}\left(2 x^2 y+x\right)\) \(=-2 x y^2-y\)
\(\dfrac{d y}{d x}\) \(=\dfrac{-2 x y^2-y}{2 x^2 y+x}\)
  \(=\dfrac{-y(2xy+1)}{x(2xy+1)}\)
  \(=\dfrac{-y}{x}(2xy+1 \neq 0)\)

 
b.    \(\dfrac{dy}{dx}=-1 \ \ \text{when}\ \ y=x:\)

\(\text{Substitute} \ \  y=x \ \ \text{into} \ \  x^2 y^2+x y=2\)

\(x^4+x^2\) \(=2\) 
\(x^4+x^2-2\) \(=0\) 
\(\left(x^2+2\right)\left(x^2-1\right)\) \(=0\) 
\(x\) \(=\pm 1\) 

 
\(\therefore \text{ Points on graph where \(m=-1\) are} \ (1, 1), (-1,-1)\)

♦ Mean mark (b) 47%.

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)

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`(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, SPEC1 2023 VCAA 4

Consider the relation \(x\, \arcsin \left(y^2\right)=\pi\).

Use implicit differentiation to find \(\dfrac{d y}{d x}\) at the point \(\left(6, \dfrac{1}{\sqrt{2}}\right)\).

Give your answer in the form \(-\dfrac{\pi \sqrt{a}}{b}\), where  \(a, b \in Z^{+}\).   (3 marks)

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\(\dfrac{dy}{dx}=-\dfrac{\pi \sqrt6}{144} \)

Show Worked Solution
\(\dfrac{d}{dx} (x\, \sin^{-1} y^2) \) \(=0\)  
\(\sin^{-1} y^2+ \dfrac{2xy}{\sqrt{1-y^4}} \cdot \dfrac{dy}{dx} \) \(=0\)  

 
\(\dfrac{dy}{dx} =-\dfrac{\sin^{-1}y^2 \times \sqrt{1-y^4}}{2xy} \)

\(\text{At}\ \ \Big{(} 6, \dfrac{1}{\sqrt{2}} \Big{)}, \)

\(\dfrac{dy}{dx}\) \(= -\dfrac{\sin^{-1}(\frac{1}{2}) \times \sqrt{\frac{3}{4}}}{2 \times 6 \times \frac{1}{\sqrt 2}}\)  
  \(=-\dfrac{\pi}{6} \times \dfrac{\sqrt3}{2} \times \dfrac{\sqrt2}{12} \)  
  \(=-\dfrac{\pi \sqrt6}{144} \)  

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)

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`-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)

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`-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 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)

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`(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`

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)

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`-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 2016 VCAA 3

Find the equation of the line perpendicular to the graph of  `cos (y) + y sin(x) = x^2`  at  `(0, -pi/2)`.  (4 marks)

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`y_N = (-2)/pi x – pi/2`

Show Worked Solution

`d/(dx) (cos(y)) + d/(dx) (y sin(x)) = d/(dx) (x^2)`

`-sin(y) *(dy)/(dx) + sin(x) *(dy)/(dx) + y cos (x) = 2x`

`dy/dx(sin(x)-sin(y))` `=2x-ycos(x)`  
`dy/dx` `=(2x-ycos(x))/(sin(x)-sin(y))`  

 
`dy/dx |_(x=0, y=pi/2) =(0-(pi/2)(1))/(0-1)=pi/2`

`m_T = pi/2\ \ =>\ \ m_⊥ = (-2)/pi`
 

`:.\ text(Equation of ⊥ line:)`

`y – ((-pi)/2)` `=(-2)/pi (x-0)`  
`y` `=((-2)/pi) x – pi/2`  

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 2014 VCAA 4

Find the gradient of the line perpendicular to the tangent to the curve defined by  `y = -3e^(3x) e^y`  at the point  `(1, -3)`.  (3 marks)

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`4/9`

Show Worked Solution
`(dy)/(dx)` `= -3 d/(dx) (e^(3x) e^y)`
`(dy)/(dx)` `= -9e^(3x) e^y + -3 e^(3x) e^y (dy)/(dx)`
` -9e^(3x) e^y` `=(dy)/(dx) (1 + 3e^(3x) e^y)`
`(dy)/(dx)` `= (-9e^(3x) e^y)/(1 + 3e^(3x) e^(-3))`

 

`text{At  (1, –3):}`  
`:. m_text(norm)` `= (1 + 3e^(3 xx 1) e^(-3))/(9e^(3 xx 1) e^(-3))`
  `= (1 + 3e^3 e^(-3))/(9e^3 e^(-3))`
  `= 4/9`

Calculus, SPEC1 2017 VCAA 1

Find the equation of the tangent to the curve given by  `3xy^2 - 2y = x`  at the point (1, –1).  (3 marks)

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`y = 1/2x – 3/2`

Show Worked Solution
`d/dx(3xy^2) + d/dx(2y)` `= d/dx(x)`
`d/(dx)(3x) · y^2 + d/(dx)(y^2)(3x) + 2(dy)/(dx)` `= 1`
`3y^2 + 6xy*(dy)/(dx) + 2(dy)/(dx)` `= 1`
`text(At)\ \ (1, -1):`  
`3(−1)^2 + 6(1)(−1)m + 2m` `= 1`
`3 – 6m + 2m` `= 1`
`−4m` `= −2`
`m` `= 1/2`
`y+1` `=1/2(x-1)`  
`:. y` `=1/2x -3/2`  

Calculus, SPEC2-NHT 2018 VCAA 7 MC

The gradient of the line that is perpendicular to the graph of the relation  `3y^2 - 5xy - x^2 = 1`  at the point  `(1, 2)`  is

A.  `-1/12`

B.      `12/7`

C.       `21`

D.   `-7/12`

E.   `-7/13`

Show Answers Only

`D`

Show Worked Solution
`6y *(dy)/(dx) – (5y + 5x* (dy)/(dx)) – 2x` `= 0`
`6y *(dy)/(dx) – 5x *(dy)/(dx)` `= 2x + 5y`
`6 xx 2 xx m_T – 5 xx 1 xx m_T` `= 2 xx 1 + 5 xx 2`
`(12 – 5)m_T` `= 12`
`7 m_T` `= 12`
`m_T` `= 12/7`

 `:.m_N = -7/12`

`=>  D`

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)`