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Calculus, 2ADV C1 EQ-Bank 32

The graph of  \(y=x^4\)  is dilated horizontally by a factor of \(k\) where \(k>0\). The normal to this dilated graph at  \(x=k\)  intersects the \(y\)-axis at \((0,5)\).

Find the value of \(k\).   (4 marks)

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\(k=4\)

Show Worked Solution

\(\text{Transformed graph has equation} \ y=\dfrac{x^4}{k^4}\)

\(\dfrac{d y}{d x}=\dfrac{4 x^3}{k^4}\)

\(\text{At }\ x=k:\)

\(m_{\text{tang}} = \dfrac{4 k^3}{k^4}=\dfrac{4}{k}\)

\(m_{\text{norm}} =-\dfrac{k}{4}\ (m_1m_2=-1)\)
 

\(\text{Equation of normal:}\)

\(y-1=-\dfrac{k}{4}(x-k)\ \ \Rightarrow\ \ y=-\dfrac{k}{4}x+1+\dfrac{k^2}{4}\)

\(\text{When }\ x=0, \ y=1+\dfrac{k^2}{4}\)

\(\text{\(y\)-intercept at }\left(0,1+\dfrac{k^2}{4}\right)\)

\(1+\dfrac{k^2}{4}=5\ \ \Rightarrow\ \ k^2=16\)

\(\therefore k=4\ \ (k>0)\).

Calculus, 2ADV C1 EQ-Bank 18

Let  \(f(x)=6 \sqrt{x+1}+5\).

Find the gradient of the tangent to  \(y=f(x)\) at  \(x=8\).   (2 marks)

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\(\text{Gradient of tangent }=1\)

Show Worked Solution
\(f(x)\) \(=6 \sqrt{x+1}+5\)
\(f^{\prime}(x)\) \(=6 \times \dfrac{1}{2} \times(x+1)^{-\tfrac{1}{2}}=\dfrac{3}{\sqrt{x+1}}\)

\(\text{At} \ \ x=8:\)

\(f^{\prime}(x)=\dfrac{3}{\sqrt{8+1}}=1\)

\(\therefore \ \text{Gradient of tangent }=1\)

Calculus, 2ADV C1 EQ-Bank 13

  1.  Use differentiation by first principles to find \(y^{′}\), given  \(y = 2x^2 + 5x\).   (2 marks)

  2.  Find the equation of the tangent to the curve when  \(x = 1\).   (1 mark)

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a.    `y^{′} = 4x+5`

b.    `y = 9x-2`

Show Worked Solution
a.    `f(x)` `= 2x^2 + 5x`
  `f^{′}(x)` `= lim_(h->0) (f(x + h)-f(x))/h`
    `= lim_(h->0) ((2(x + h)^2 +5(x + h))-(2x^2+5x))/h`
    `= lim_(h->0)(2x^2 + 4xh + 2h^2+5x+5h-2x^2-5x)/h`
    `= lim_(h->0)(4xh + 2h^2+5h)/h`
    `= lim_(h->0)(h(4x+5 +2h))/h`

 
`:.\ y^{′} = 4x+5`
 

b.   `text(When)\ \ x = 1, y = 7`

`y^{′} = 4+5 = 9`

`y-7` `= 9(x-1)`
`y` `= 9x-2`

Calculus, 2ADV C1 EQ-Bank 7 MC

At which point on the curve  \(y=2x^{2}-11x+3\)  can a tangent be drawn such that it is inclined at 45° when it crosses the positive \(x\)-axis?

  1. \((-3,54)\)
  2. \((-2,33)\)
  3. \((2,-11)\)
  4. \((3,-12)\)
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\(D\)

Show Worked Solution
\(y\) \(=2x^{2}-11x+3\)  
\(y^{′}\) \(=4x-11\)  

 
\(\text{If a tangent crosses (positive) x-axis at 45}^{\circ},\)

\(m_{\text{tang}}=1  \ \ (\tan 45^{\circ}=1) \)

\(\text{Find}\ x\ \text{when}\ \ y^{′}=1: \)

\(4x-11\) \(=1\)  
\(4x\) \(=12\)  
\(x\) \(=3\)  

 
\(\text{Tangent at point}\ (3,-12) \)

\(\Rightarrow D\)

Calculus, 2ADV C1 2023 HSC 14

Find the equation of the tangent to the curve  `y=(2x+1)^3`  at the point `(0,1)`.   (3 marks)

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`y=6x+1`

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`y` `=(2x+1)^3`  
`dy/dx` `=3xx2(2x+1)^2=6(2x+1)^2`  

  
`text{At}\ x=0\ \ =>\ \ dy/dx=6xx1^2=6`

`text{Find equation of line}\ m=6, text{through}\ (0,1):`

`y-y_1` `=m(x-x_1)`  
`y-1` `=6(x-0)`  
`y` `=6x+1`  

Calculus, 2ADV C1 2022 HSC 5 MC

Let  `h(x)=(f(x))/(g(x))`, where

`{:[f(1)=2, qquad f^{′}(1)=4],[g(1)=8, qquad g^{′}(1)=12]:}`

What is the gradient of the tangent to the graph of  `y=h(x)`  at  `x=1` ?

  1. `-8`
  2. `\ \ \ 8`
  3. `- 1/8`
  4. `\ \ \ 1/8`
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`D`

Show Worked Solution

`text{Using the quotient rule:}`

`h^{′}(1)` `=(g(1)\ f^{′}(1)-f(1)\ g^{′}(1))/[g(1)]^2`  
  `=(8xx4-2xx12)/(8^2)`  
  `=(32-24)/64`  
  `=1/8`  

 
`=>D`

Calculus, 2ADV C1 2019 HSC 14d

The equation of the tangent to the curve  `y = x^3 + ax^2 + bx + 4`  at the point where  `x = 2`  is  `y = x-4`.

Find the values of `a` and `b`.   (3 marks)

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`b = -3,\ \ a = -2`

Show Worked Solution
`y ` `= x^3 + ax^2 + bx + 4`
`(dy)/(dx)` `= 3x^2 + 2ax + b`

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

♦ Mean mark 46%.

`12 + 4a + b` `= 1`
`4a + b` `= -11\ …\ (1)`

 
`text(The point)\ (2, -2)\ text(lies on)\ y:`

`8 + 4a + 2b + 4` `=-2`
`4a + 2b` `= -14\ …\ (2)`

  
`text(Subtract)\ \ (2)-(1):`

`b = -3`

`text(Substitute into)\ (1):`

`4a-3` `= -11`
`4a` `= -8`
`a` `= -2`

Calculus, 2ADV C1 EQ-Bank 16

  1.  Find the equations of the tangents to the curve  `y = x^2-3x`  at the points where the curve cuts the `x`-axis.   (2 marks)

  2.  Where do the tangents intersect?   (2 marks)

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a.    `y = -3x, \ y = 3x-9`

b.    `(3/2, -9/4)`

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a.    `y= x^2-3x= x(x-3)`

`text(Cuts)\ xtext(-axis at)\ \ x = 0\ \ text(or)\ \ x = 3`

`(dy)/(dx) = 2x-3`

`text(At)\ \ x = 0 \ => \ (dy)/(dx) = -3`

`T_1\ text(has)\ \ m = -3,\ text{through (0, 0):}`

`y-0` `= -3(x-0)`
`y` `= -3x`

 
`text(At)\ \ x = 3 \ => \ (dy)/(dx) = 3`

`T_2\ text(has)\ \ m = 3,\ text{through (3, 0):}`

`y-0` `= 3(x-3)`
`y` `= 3x-9`

 

b.   `text(Intersection occurs when:)`

`3x-9` `= -3x`
`6x` `= 9`
`x` `= 3/2`

  

`y = -3 xx 3/2 = -9/2`

`:.\ text(Intersection at)\ \ (3/2, -9/2)`

Calculus, 2ADV C1 2017 HSC 12a

Find the equation of the tangent to the curve  `y = x^2 + 4x - 7`  at the point  `(1, -2)`.  (2 marks)

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`y = 6x – 8`

Show Worked Solution
`y` `= x^2 + 4x – 7`
`(dy)/(dx)` `= 2x + 4`

 
`text(When)\ x = 1,\ \ (dy)/(dx) = 6`

`text(Equation of tangent through)\ (1, -2)`

`y + 2` `= 6 (x – 1)`
`y` `= 6x – 8`

Calculus, 2ADV C1 2009 HSC 6c

The diagram illustrates the design for part of a roller-coaster track. The section `RO` is a straight line with slope 1.2, and the section `PQ` is a straight line with slope  – 1.8. The section `OP` is a parabola  `y = ax^2 + bx`. The horizontal distance from the `y`-axis to `P` is 30 m.
 

2009 6c

In order that the ride is smooth, the straight line sections must be tangent to the parabola at `O` and at `P`.  

  1. Find the values of `a` and `b` so that the ride is smooth.   (3 marks)

  2. Find the distance `d`, from the vertex of the parabola to the horizontal line through `P`, as shown on the diagram.   (2 marks)

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  1. `text(For a smooth ride,)\ a = -0.05\ text(and)\ b = 1.2`
  2. `16.2\ text(m)`
Show Worked Solution
i.    `y` `= ax^2 + bx`
  `dy/dx` `= 2ax + b`

 
`text{At}\ x=0\ \ \=>\ \ dy/dx=b`

`text(We need)\ m\ text(at)\ O = 1.2`

`:.\ b = 1.2`
 

`text(At)\ P,\ x = 30:`

`dy/dx= 2 xx a xx 30 + 1.2= 60a + 1.2`

`text(We need)\ m\ text(at)\ P = -1.8`

`60a + 1.2` `= -1.8`
`60a` `= -3`
`a` `= -3/60= -0.05`

 
`:.\ text(For a smooth ride,)\ a = -0.05\ \ text(and)\ \ b = 1.2`

 

ii.    `y` `= -0.05x^2 + 1.2x`
  `dy/dx` `= -0.1x + 1.2`

 
`text(Find)\ x\ text(when)\ dy/dx = 0`

`-0.1x + 1.2` `= 0`
`x` `= 1.2/0.1= 12`

 
`text(MAX when)\ x = 12`

`text(When)\ x = 12:`

`y= -0.05 xx 12^2 + 1.2 xx 12= -7.2 + 14.4= 7.2`

`text(When)\ x = 30:`

`y=-0.05 xx 30^2 + 1.2 xx 30= -45 + 36= -9`

`:.\ d= 7.2 + |-9|= 16.2\ text(m)`

Calculus, 2ADV C1 2010 HSC 7b

The parabola shown in the diagram is the graph  `y = x^2`. The points `A (–1,1)` and `B (2, 4)` are on the parabola.
 

 

  1.  Find the equation of the tangent to the parabola at `A`.   (2 marks)

  2.  Let `M` be the midpoint of `AB`.

     

    There is a point `C` on the parabola such that the tangent at `C` is parallel to `AB`.

     

    Show that the line `MC` is vertical.   (2 marks)  

  3. The tangent at `A` meets the line `MC` at `T`.

     

    Show that the line `BT` is a tangent to the parabola.   (2 marks)

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  1. `2x + y + 1 = 0`
  2. `text(Proof)  text{(See Worked Solutions)}`
  3. `text(Proof)  text{(See Worked Solutions)}`
Show Worked Solution
i.

`y=x^2\ \ =>\ \ dy/dx=2x`

`text(At)\ \ A(-1,1)\ \ =>\ \ dy/dx = -2`

`text(T)text(angent has)\ \ m=text(–2),\ text(through)\ text{(–1,1):}`

`y-y_1` `= m(x-x_1)`
`y-1` `= -2 (x + 1)`
`y-1` `= -2x-2`
`2x + y + 1` `= 0`

 
`:.\ text(T)text(angent at)\ A\ text(is)\ \ 2x + y + 1 = 0`

 

♦ Mean mark 37%.
IMPORTANT: Key strategy for solution: the gradient of `AB` needs to be equated to the gradient function (i.e. `dy/dx`).

ii.   `Atext{(–1,1)}\ \ \ B(2,4)`

`M= ((-1+2)/2 , (1+4)/2)= (1/2, 5/2)`

`m_(AB)= (y_2-y_1)/(x_2-x_1)= (4-1)/(2 + 1)=1`

`text(When)\ \ dy/dx=1:`

`2x=1\ \ =>\ \ x=1/2\ \ =>\ \ C \ (1/2, 1/4)`
 
`=>M\ text(and)\ C\ text(both have)\ x text(-value)=1/2`

`:. MC\ text(is vertical  … as required)`
 

iii.  `T\ text(is point on tangent when) \ x=1/2`

♦♦ Mean mark 29%.

`text(T)text(angent)\ \ \ 2x + y + 1 = 0`

`text(At)\ x = 1/2`

`2 xx (1/2) + y + 1=0\ \ =>\ \ y=–2`

`:.\ T (1/2, –2)`

 
`text (Given)\ B (2, 4):`

`m_(BT)= (4+2)/(2-1/2)=4`

  
`text(At)\ \ B(2,4),\ text(find gradient of tangent:)`

`dy/dx = 2x=2 xx2=4`

`m_text(tangent) = 4=m_(BT)`

`:.BT\ text(is a tangent)`

Calculus, 2ADV C1 2011 HSC 2c

Find the equation of the tangent to the curve  `y = (2x + 1)^4`   at the point where  `x = -1`.   (3 marks)

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`8x + y + 7 = 0`

Show Worked Solution

`y = (2x + 1)^4`

`text{Using the Chain Rule:}`

`dy/dx= 4 xx (2x + 1)^3 xx d/dx (2x + 1)= 8 (2x + 1)^3`

`text(At)\ \ x = -1:`

MARKER’S COMMENT: The best setting out should show the derivative function, the gradient, the point and then finally, calculations for the equation of the tangent.

`dy/dx= 8 (2(-1) + 1)^38 (-1)^3= -8`

`text{Tangent has}\ m = -8\ text(through)\ (-1,1):`

`y-y_1` `= m (x-x_1)`
`y-1` `= -8 (x + 1)`
`y-1` `= -8x -8`
`8x + y + 7` `= 0`

 
`:.\ text(Equation of tangent is)\ 8x + y + 7 = 0`

Calculus, 2ADV C1 2012 HSC 11c

Find the equation of the tangent to the curve  `y = x^2`  at the point where  `x = 3`.   (2 marks)

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`6x-y-9 = 0 `

Show Worked Solution

`y=x^2\ \ =>\ \ dy/dx=2x`

`text{Tangent equation has}\ m = 6, \text{through}\ (3,9):`

`text(When) \  x = 3, y = 9\ \ =>\ \ dy/dx = 6`

`y-y_1`  `= m (x-x_1)`
`y-9`  `= 6(x-3)`
`y-9`  `= 6x-18`
`6x-y-9` `=0`