NCEA Level 2 Mathematics, Calculus Exam 2024

on this page: one (c) | one (d) | two (d) | three (d) | conclusion

QUESTION ONE (c)

A curve is given by `y = 3x^3 - 9x^2 - 27x + 4`.

Using calculus methods:
(i) Find the x-coordinate of the local minimum.
(ii) Explain how you know this is a minimum point.

Differentiate to find stationary points, let it equal zero, etc.

`(dy)/(dx) = 9x^2 - 18x - 27 = 0` @ stationary points.

`x^2 - 2x - 3 = 0`

`(x + 1)(x - 3) = 0`

`x = -1 or x = 3`

`x = 3` is a local minimum.

There are two ways we know this, but the question specifically requests calculus methods be used for (i) and (ii).

We know the `x^3` coefficient is positive, meaning the first unique `x` value, `x = -1`, is a local maximum, while the second, `x = 3`, is a local minimum. (There could only be a point of inflexion stationary point if the two `x` values were the same.) But this explanation is not adequate, since calculus methods have been requested.

Using calculus methods, differentiate again to get the second derivative (the rate of change of the gradient) and test the two `x` values.

`(d^2y)/(dx^2) = 18x - 18`

`18(-1) - 18 < 0 =>` local maximum

`18(3) - 18 > 0 =>` local minimum

Comment: Pretty straight forward, but read the question carefully so an adequate reason is given for part (ii).

QUESTION ONE (d)

A drink manufacturer would like to start a new line of cylindrical cups that are designed to keep hot liquids warm far as long as possible. To do this, they wish to minimise the surface area of the new cup, which includes a lid.

They would like the cup to hold a volume of 500 mL [corrected from ml], and be cylindrical in shape.

Calculate the dimensions of the cylinder that would satisfy the above conditions.

Volume of a cylinder: `V = pir^2h`

Surface area of a cylinder: `SA = 2pir^2 + 2pirh`

How exciting. I'm betting `2r = h` (diameter = height) gives minimum surface area.

Start with an expression for h, substitute that into the expression for surface area, then rewrite to make it easier to differentiate.

`V = pir^2h = 500`

`h = 500/(pir^2)`

`SA = 2pir^2 + 2pir500/(pir^2)`

`SA = 2pir^2 + 1000/r`

`SA = 2pir^2 + 1000r^(-1)`

`(d(SA))/(dr) = 4pir - 1000r^(-2) = 0` at minimum surface area

`4pir = 1000/r^2`

`r^3 = 1000/(4pi)`

`r = root(3)(1000/(4pi))`

`r = 10/root(3)(4pi)`

`r = 4.301` (4sf)

`h = 500/(pir^2) = 8.603` (4sf)

Yep, `2r = h`.

Don't forget to reread the question after getting a result; we are asked for the cup's dimensions.

The cup is (internally) 8.6 cm in diameter and 8.6 cm high.

Comment: This seems a pretty standard question (quite predictable, even).

QUESTION TWO (d)

A trapezium is drawn within a parabola given by `y = 8x - x^2` and the `x`-axis, such that the two parallel sides are positioned at `x` and `2x`, as shown in the graph below, where `0 <= x <= 4`.

Given that the area of a trapezium is given by `A = 1/2 (a + b) h`, where `a` and `b` are the lengths of the parallel sides, and `h` is the perpendicular distance between them, use calculus to find the maximum area of the trapezium.

I note that `x` can not be greater than 4 because the parabola returns to the `x`-axis at `x = 8` (so `2x <= 8`).

Keep the working nice and methodical. Taking shortcuts is the best way to make mistakes.

We want to find the maximum area of the trapezium, which means we'll need to differentiate an expression for the area, so that expression is what we have to find first.

`A = 1/2 (a + b) h`

`A = 1/2 (f(x) + f(2x)) (2x - x)`

`A = 1/2 (8x - x^2 + 8(2x) - (2x)^2) x`

`A = 1/2 (8x - x^2 + 16x - 4x^2) x`

`A = 1/2 (24x - 5x^2) x`

`A = 1/2 (24x^2 - 5x^3)`

`A = 12x^2 - 5/2x^3`

At this point we want to differentiate that, factorise, solve for x.

`(dA)/(dx) = 24x - 15/2x^2 = 0` at maximum or minimum area

`x(24 - 15/2x) = 0`

`x = 0` gives minimum area

`24 - 15/2x = 0` gives maximum area

`15/2x = 24`

`x = 48/15 = 16/5 = 3.2`

We have a result. Whenever we get a result, remember to reread the question; we are asked to find the maximum area, not just find the value of `x` which gives the maximum area. Substitute the `16/5` or 3.2 into the expression for the area.

maximum area `= 12(16/5)^2 - 5/2(16/5)^3`

`= 1024/25 = 40.96` (exact value)

QUESTION THREE (d)

If `y = k - x`, where `k` is a number, use calculus to show that the minimum value of `x^2 + 2y^2` is equal to `(2k^2)/3`.

The biggest problem here might be what we're going to call the expression `x^2 + 2y^2`. I definitely want to be able to refer to it. How about `E` for expression? We want to find the minimum value of `E`.

`E = x^2 + 2y^2`

This is a calculus exam, so to find a minimum we'll no doubt have to differentiate. But to do that I don't really want both `x` and `y`. I'll replace `y` using the first equation given.

`E = x^2 + 2(k - x)^2`

`E = x^2 + 2(k^2 - 2kx + x^2)`

`E = x^2 + 2k^2 - 4kx + 2x^2`

`E = 3x^2 - 4kx + 2k^2`

Time to differentiate.

`(dE)/(dx) = 6x - 4k = 0` at minimum*

`6x = 4k`

`x = (4k)/6`

`x = (2k)/3`

Don't mistake this for the `(2k^2)/3` which is the minimum value of `E`. They're not the same thing. This is simply the value of `x` where the minimum value of `E` occurs.

* We know this is going to be a minimum value of `E` because the coefficient of the `x^2` term is positive, so the parabola vertex is a minimum. It would be good to state this, but we could also confirm it by differentiating again to get the second derivative, the rate of change of the gradient.

If the question says something like "use calculus methods to show it is a minimum" then do work out the second differential.

`(d^2E)/(dx^2) = 6 > 0 =>` minimum

Right, that works. Let's substitute that value of `x` back into the expression. I'll use my simplified expression from earlier.

`E = 3x^2 - 4kx + 2k^2`

`E_"min" = 3((2k)/3)^2 - 4k(2k)/3 + 2k^2`

`E_"min" = 3(4k^2)/9 - (8k^2)/3 + 2k^2`

`E_"min" = (4k^2)/3 - (8k^2)/3 + (6k^2)/3`

`E_"min" = (4k^2)/3 - (8k^2)/3 + (6k^2)/3`

`E_"min" = (2k^2)/3`

And that's it. QED.

Comment: This one is also pretty straight forward.

Conclusion

The calculus questions that I looked at were not particularly difficult or unclear. On that basis it seems like an OK exam (apart from using the unit symbol "ml" instead of "mL").

The calculus exam was held on 5 November. The NCEA Level 2 calculus exam in 2016 and in 2017 were both held on 24 November. Why such a big loss of learning time?

I've always thought it extremely strange that the NCEA exam rules/guidelines on the front page of each exam do not mention the duration of the exam, or mention how long all three should take.