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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)

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

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 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)

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  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 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`