SmarterEd

Aussie Maths & Science Teachers: Save your time with SmarterEd

Calculus, MET2 2023 VCE SM-Bank 2

Jac and Jill have built a ramp for their toy car. They will release the car at the top of the ramp and the car will jump off the end of the ramp.

The cross-section of the ramp is modelled by the function \(f\), where

\(f(x)= \begin{cases}\displaystyle \ 40 & 0 \leq x<5 \\ \dfrac{1}{800}\left(x^3-75 x^2+675 x+30\ 375\right) & 5 \leq x \leq 55\end{cases}\)

\(f(x)\) is both smooth and continuous at \(x=5\).

The graph of  \(y=f(x)\)  is shown below, where \(x\) is the horizontal distance from the start of the ramp and \(y\) is the height of the ramp. All lengths are in centimetres.

  1. Find \(f^{\prime}(x)\) for \(0<x<55\).  (2 marks)

  2.   i. Find the coordinates of the point of inflection of \(f\).  (1 mark)

  3.  ii. Find the interval of \(x\) for which the gradient function of the ramp is strictly increasing.  (1 mark)

  4. iii. Find the interval of \(x\) for which the gradient function of the ramp is strictly decreasing.  (1 mark)

Jac and Jill decide to use two trapezoidal supports, each of width \(10 cm\). The first support has its left edge placed at \(x=5\) and the second support has its left edge placed at \(x=15\). Their cross-sections are shown in the graph below.

  1. Determine the value of the ratio of the area of the trapezoidal cross-sections to the exact area contained between \(f(x)\) and the \(x\)-axis between \(x=5\) and \(x=25\). Give your answer as a percentage, correct to one decimal place.  (3 marks)

  2. Referring to the gradient of the curve, explain why a trapezium rule approximation would be greater than the actual cross-sectional area for any interval \(x \in[p, q]\), where \(p \geq 25\).  (1 mark)

  3. Jac and Jill roll the toy car down the ramp and the car jumps off the end of the ramp. The path of the car is modelled by the function \(P\), where

      1. \(P(x)=\begin{cases}f(x) & 0 \leq x \leq 55 \\ g(x) & 55<x \leq a\end{cases}\)
  4. \(P\) is continuous and differentiable at \(x=55, g(x)=-\frac{1}{16} x^2+b x+c\), and \(x=a\) is where the car lands on the ground after the jump, such that \(P(a)=0\).
    1. Find the values of \(b\) and \(c\).  (2 marks)

    2. Determine the horizontal distance from the end of the ramp to where the car lands. Give your answer in centimetres, correct to two decimal places.  (1 mark)


Show Answers Only

a.    \(f^{\prime}(x)= \begin{cases}\displaystyle \ 0 & 0 < x<5 \\ \dfrac{1}{800}\left(3x^2-150 x+675 \right) & 5 \leq x < 55\end{cases}\)

b.i   \((25, 20)\)

b.ii  \([25, 55]\)

b.iii \([5, 25]\)

c.    \(98.1\%\)

d.    \(\text{In the interval }x\in [25, q]\text{ the curve is concave up.}\)

\(\therefore\ \text{The sloped edges of the trapeziums would be above }f(x)\)

\(\text{making the trapezium rule approximation greater than the}\)

\(\text{actual area.}\)

e.i   \(b=8.75, c=-283.4375\)

e.ii  \(34.10\ \text{cm}\)

Show Worked Solution

a.    \(\text{Using CAS: Define f(x) then}\)

\(\text{OR}\quad f^{\prime}(x)= \begin{cases}\displaystyle \ 0 & 0 < x<5 \\ \dfrac{1}{800}\left(3x^2-150 x+675 \right) & 5 \leq x < 55\end{cases}\)

b.i   \(\text{Using CAS: Find }f^”(x)\ \text{then solve }=0\)

\(\therefore\ \text{Point of inflection at }(25, 20)\)
  

b.ii  \(\text{Gradient function strictly increasing for }x\in [25, 55]\)

b.iii \(\text{Gradient function strictly decreasing for }x\in [5, 25]\)

c.    \(\text{Using CAS:}\)

\(\text{Area}\) \(=\dfrac{h}{2}\Big(f(5)+2f(15)+f(25)\Big)\)
  \(=\dfrac{10}{2}\Big(40+2\times 33.75+20\Big)=637.5\)

  

\(\therefore\ \text{Estimate:Exact}\) \(=637.5:650\)
  \(=\dfrac{637.5}{650}\times 100\)
  \(=98.0769\%\approx 98.1\%\)

   
d.    \(\text{In the interval }x\in [25, q]\text{ the curve is concave up.}\)

\(\therefore\ \text{The sloped edges of the trapeziums would be above }f(x)\)

\(\text{making the trapezium rule approximation greater than the}\)

\(\text{actual area.}\)

e.i   \(f(55)=\dfrac{35}{4}\ \text{(Using CAS)}\)

\(\text{and }f(55)=g(55)\rightarrow g(55)=\dfrac{35}{4}\)

\(\therefore -\dfrac{1}{16}\times 55^2+55b+c\) \(=\dfrac{35}{4}\)
\(c\) \(=\dfrac{35}{4}+\dfrac{3025}{16}-55b\)
\(c\) \(=\dfrac{3165}{16}-55b\quad (1)\)

  
\(g'(x)=-\dfrac{x}{8}+b\)

\(\text{and }g'(55)=f'(55)\)

\(\rightarrow\ f'(55)=\dfrac{1}{800}(3\times 55^2-150\times 55+675)=\dfrac{15}{8}\)

\(\rightarrow\ -\dfrac{55}{8}+b=\dfrac{15}{8}\)

\(\therefore b=\dfrac{35}{4}\quad (2)\)

\(\text{Sub (2) into (1)}\)

\(c=\dfrac{3165}{16}-55\times\dfrac{35}{4}\)

\(c=-\dfrac{4535}{16}\)

\(\therefore\ b=\dfrac{35}{4}\ \text{or}\ 8.75 , c=-\dfrac{4535}{16}\ \text{or}\ -283.4375\)

e.ii  \(\text{Using CAS: Solve }g(x)=0|x>55\)

\(\text{Horizontal distance}=\sqrt{365}+70-55=34.104\dots\approx 34.10\ \text{cm (2 d.p.)}\)