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Calculus, MET2 2024 VCAA 16 MC

Suppose that a differentiable function  \(  f: R \rightarrow R \)  and its derivative  \(f^{\prime}: R \rightarrow R\)  satisfy  \(f(4)=25\)  and  \(f^{\prime}(4)=15\).

Determine the gradient of the tangent line to the graph of  \( {\displaystyle y=\sqrt{f(x)} } \)  at  \( x=4 \).

  1. \(\sqrt{15}\)
  2. \(\dfrac{1}{10}\)
  3. \(\dfrac{15}{2}\)
  4. \(\dfrac{3}{2}\)
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\(D\)

Show Worked Solution
\(y\) \(=\sqrt{f(x)}={f(x)}^{\frac{1}{2}}\)
\(y^{\prime}\) \(=\dfrac{1}{2}\left({f(x)}^{-\frac{1}{2}}\right)\times f^{\prime}(x)=\dfrac{f^{\prime}(x)}{2\sqrt{f(x)}}\)

 
\(\text{Given}\ \ f(4)=25,\ f^{\prime}(4)=15\)

\(\therefore\ y^{\prime}=\dfrac{15}{2\sqrt{25}}=\dfrac{3}{2}\)

\(\Rightarrow D\)

♦♦ Mean mark 36%.

Calculus, MET1 2017 VCAA 9

The graph of  `f: [0, 1] -> R,\ f(x) = sqrt x (1-x)`  is shown below.
 

  1. Calculate the area between the graph of `f` and the `x`-axis.   (2 marks)

  2. For `x` in the interval `(0, 1)`, show that the gradient of the tangent to the graph of `f` is  `(1-3x)/(2 sqrt x)`.   (1 mark)

The edges of the right-angled triangle `ABC` are the line segments `AC` and `BC`, which are tangent to the graph of `f`, and the line segment `AB`, which is part of the horizontal axis, as shown below.

Let `theta` be the angle that `AC` makes with the positive direction of the horizontal axis, where  `45^@ <= theta < 90^@`.

  1. Find the equation of the line through `B` and `C` in the form  `y = mx + c`, for  `theta = 45^@`.   (2 marks)

  2. Find the coordinates of `C` when  `theta = 45^@`.   (4 marks)

Show Answers Only
  1. `4/15\ text(units)^2`
  2. `text(Proof)\ \ text{(See Workes Solutions)}`
  3. `y = -x + 1`
  4. `C (11/27, 16/27)`
Show Worked Solution
a.   `text(Area)` `= int_0^1 (sqrt x-x sqrt x)\ dx`
    `= int_0^1 (x^(1/2)-x^(3/2))\ dx`
    `= [2/3 x^(3/2)-2/5 x^(5/2)]_0^1`
    `= (2/3-2/5)-(0-0)`
    `= 10/15-6/15`
    `= 4/15\ text(units)^2`

 

♦♦ Mean mark part (b) 35%.
MARKER’S COMMENT: Establishing the common denominator in the working was required!
b.   `f (x)` `= x^(1/2)-x^(3/2)`
  `f^{′}(x)` `= 1/2 x^(-1/2)-3/2 x^(1/2)`
    `= 1/(2 sqrt x)-(3 sqrt x)/2`
    `= (1-3x)/(2 sqrt x)\ \ text(.. as required.)`

 

c.  `m_(AC) = tan 45^@=1`

♦♦♦ Mean mark part (c) 20%.
MARKER’S COMMENT: Most successful answers introduced a pronumeral such as `a=sqrtx` to solve.

`=> m_(BC) = -1\ \ (m_text(BC) _|_ m_(AC))`

 

`text(At point of tangency of)\ BC,\  f^{prime}(x) = -1`

`(1-3x)/(2 sqrt x)` `=-1`
`1-3x` `=-2sqrtx`
`3x-2sqrt x-1` `=0`

 

`text(Let)\ \ a=sqrtx,`

`3a^2-2a-1` `=0`
`(3a+1)(a-1)` `=0`
`a=1 or -1/3`   
`:. sqrt x` `=1` `or`   `sqrt x=- 1/3\ \ text{(no solution)}`
`x` `=1`    

 
`f(1)=sqrt1(1-1)=0\ \ =>B(1,0)`
 

`:.\ text(Equation of)\ \ BC, \ m=-1, text{through (1,0) is:}`

`y-0` `=-1(x-1)`
`y` `=-x+1`

 

d.  `text(Find Equation)\ AC:`

♦♦♦ Mean mark part (d) 17%.

`m_(AC) =1`

`text(At point of tangency of)\ AC,\  f^{prime}(x) = 1`

`(1-3x)/(2 sqrt x)` `=1`
`1-3x` `=2sqrtx`
`3x+2sqrt x-1` `=0`
`(3 sqrtx-1)(sqrtx+1)` `=0`
   
`:. sqrt x` `=1/3` `or`   `sqrt x=-1\ \ text{(no solution)}`
`x` `=1/9`    

 
`f(1/9)=sqrt(1/9)(1-1/9)=1/3 xx 8/9 = 8/27\ \ =>P(1/9,8/27)`
 

`:.\ text(Equation of)\ AC, m=1, text(through)\ \ P\ \ text(is):`

`y-8/27` `= 1 (x-1/9)`
`y` `= x + 5/27`

 
`C\ text(is at intersection of)\ AC and CB:`

`-x + 1` `= x + 5/27`
`2x` `= 22/27`
`:. x` `= 11/27`
`y` `= -11/27 + 1 = 16/27`

 
`:. C (11/27, 16/27)`

Calculus, MET1 2016 VCAA 2

Let  `f: (-∞,1/2] -> R`, where  `f(x) = sqrt(1-2x)`.

  1. Find  `f^{prime}(x)`.   (1 mark)

  2. Find the angle `theta` from the positive direction of the `x`-axis to the tangent to the graph of  `f` at  `x =-1`, measured in the anticlockwise direction.   (2 marks)

Show Answers Only
  1. `(-1)/(sqrt(1-2x))`
  2. `(5pi)/6`
Show Worked Solution
a.    `f(x)` `= (1-2x)^(1/2)`
  `f^{prime}(x)` `= 1/2(1-2x)^(-1/2) (-2)qquadtext([Using Chain Rule])`
    `= (-1)/(sqrt(1-2x))`

 

♦ Mean mark 40%.
b.    `tan theta` `= f^{prime}(-1)`
    `= (-1)/(sqrt(1-2(-1)))`
    `= (-1)/(sqrt3)`

 

`text(S)text(ince)\ theta ∈ [0,pi],`

`=> theta = (5pi)/6`

Calculus, MET1 2006 VCAA 8

A normal to the graph of  `y = sqrt x`  has equation  `y = -4x + a`, where `a` is a real constant. Find the value of `a.`   (4 marks)

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`a = 18`

Show Worked Solution

`text(Normal) = text(line) _|_ text(to tangent)`

♦ Mean mark 40%.

`m_{text(norm)}=-4\ \ => \ m_tan=1/4`

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

`text(Find)\ x\ text(when:)`

`1/(2 sqrt x)` `=1/4`
`sqrt x` `=2`
`x` `=4`

 

`:.\ text(Point of tangency is)\ \ (4, 2)`
 

`text(Find equation of normal:)`

`y-y_1` `= m (x-x_1)`
`y-2` `= -4(x-4)`
`:. y` `= -4x + 18`
`:. a` `= 18`

Calculus, MET2 2012 VCAA 9 MC

The normal to the graph of  `y = sqrt (b - x^2)`  has a gradient of 3 when  `x = 1.`

The value of `b` is

A.   `-10/9`

B.   `10/9`

C.   `4`

D.   `10`

E.   `11`

Show Answers Only

`D`

Show Worked Solution
`y` `=sqrt (b – x^2)`
`dy/dx` `=(-2x)/(2 sqrt(b-x^2))`
   

`text(When)\ \ x=1,`

`dy/dx` `=(-1)/sqrt(b-1)`
   

`text(S)text(ince)\ \ m_(norm) xx m_(tan) = -1,`

`dy/dx=- 1/3`

 

`(-1)/sqrt(b-1)` `=- 1/3`
`sqrt(b-1)` `=3`
`b-1` `=9`
`b` `=10`

 
`=>   D`