NCEA Level 3 Mathematics 2021
Calculus: Integration exam
Excellence questions only.
QUESTION ONE
(d) A water tank developed a leak.
6 hours after the tank started to leak, the volume of water in the tank was 400 litres.
10 hours after the tank started to leak, the volume of water in the tank was 256 litres.
The rate at which the water leaks out of the tank at any instant is proportional to the square root of the volume of the water in the tank at that instant.
How much water was in the tank at the instant it started to leak?
You must use calculus and show the results of any integration needed to solve the problem.
First, understanding the question. We need to make an equation for the rate of change of the water volume, then integrate it to get a formula for the volume.
We need some variables to work with, such as `V`, the volume of water in the tank, and `t`, the time in hours from when the tank starts leaking. I like to sort out the given information nice and early so it's easier to see what I have to work with.
At `t=6, V=400` and at `t=10, V=256`
Those two data points are a hint we might have to use simultaneous equations rather than just find a single constant of integration. Anyway, we'll get back to them later.
Now to set up to equation which will be integrated.
`(dV)/dt prop sqrt(V)`
`(dV)/dt = -ksqrt(V)` for some constant `k`, which will be negative.
Rearrange to get `V` on the left with the `dV`, and the other stuff on the right.
`1/sqrt(V) dV = k dt`
Change the V term to have an exponent instead of a surd (the square root) so it is easier to integrate.
`V^(-1/2) dV = k dt`
`int V^(-1/2) dV = int k dt`
`2V^(1/2) = kt + c`
If I leave that `2` on the left I'm going to get numbers that will have to be tweaked later, so I'll replace my `k` and `c` with values half their size, `K` and `C`. (What symbols are used are actually irrelevant; it could be `k_1` and `c_1`, or `x` and `y`. They're still just constants which I need to find.)
`sqrtV = K t + C`
This looks OK to work with – I don't want to square both sides because that will make the equation much more complicated. It's much easier to use this equation for simultaneous equations.
Substitute in our values from above to get a pair of equations which get simplified before solving.
`{: [sqrt400, =, K, times, 6, +, C], [sqrt256, =, K, times, 10, +, C] }`
`{: [6K, +, C, =, 20], [10K, +, C, =, 16] }`
And because this is an integration exam not algebra, I'm just going to solve that off-screen using my calculator, which has that feature.
`{: [K, =, -1], [C, =, 26] }`
These values go back into the equation.
`sqrtV = -t + 26`
At time `t = 0`, `sqrtV = 26`
`V = 26^2 = 676` litres.
And that's it, the number to get an excellence grade for this question. (The unit is not required.)
Comment: This could be confusing, so be aware of clues as to the required approach, such as the number of given points on the curve. Also, because `k` is negative, rearranging to get an expression for `V` in terms of `k^2` etc is not desirable for simultaneous equations.
QUESTION TWO
(e) The diagram below shows the graph of a curve `y = f(x)`,
which satisfies the differential equation `dy/dx = 2 / (ye^(0.5x))`.
Points P and Q are the points on the graph of the curve that have x-coordinates of 3.
What is the vertical distance between points P and Q?
You must use calculus and show the results of any integration needed to solve the problem.
Again, we'll split the `dy/dx`, rearrange the terms, and integrate.
`dy/dx = 2 / (ye^(0.5x))`
`y dy = 2 e^(-0.5x) dx`
`int y dy = int 2 e^(-0.5x) dx`
`1/2 y^2 = -2 times 2 e^(-0.5x) + c_0`
Multiply both sides by 2, changing the constant to a different constant (but still just a constant) which we need to evaluate.
`y^2 = (-8) / e^(0.5x) + c`
`y = +- sqrt((-8) / e^(0.5x) + c)`
From the graph, at `x=0`, `y = +-1`. It seems a little strange that that information wasn't stated in the problem. Substitute in to get `c`.
`(-8) / e^(0.5(0)) + c = 1^2`
`c = 1 + 8 / e^0`
Any number to the power of 0 is 1.
`c = 9`
Substitute in to get `c`.
`y = +- sqrt((-8) / e^(0.5x) + 9)`
We have a result. Now reread the question to check what is actually required; the vertical separation of the two points at `x = 3`. The graph is symetrical about the `x`-axis so the distance above the `x`-axis is the same as the distance below it. The `Delta` is a capital Greek letter delta, the equivalent of our D. `Deltay` means the change in `y`. In other words, the difference between the two `y` values.
`Deltay = 2 sqrt((-8) / e^(0.5(3)) + 9)`
`Deltay = 2 sqrt((-8) / e^(1.5) + 9)`
`Deltay = 2 times 2.68606...`
`Deltay = 5.372...` (4sf)
And that's the value asked for.
Comment: This is a straight forward excellence question, and shouldn't be too hard for most students to do well with. 4 significant figures seems to be the most common precision used in NCEA maths, but in another exam I've seen three different amounts of significant figures used for answers within the same question, with no clues in the assessment schedule to why the different precisions were used.
QUESTION THREE
(e) The graph below shows the functions `y = (ke^x)^2` and `y = k`, where `k` is a constant greater than 1.
Show that the shaded area is `k/2 (k - 1 + ln{:1/k:} )`.
You must use calculus and show the results of any integration needed to solve the problem.
Clearly show each step of your working.
This is a definite integral problem, meaning we're calculating the area between two `x` values (generically, `a` and `b`), and also between two functions which give the upper and lower boundaries of the area (the `y` values).
In this case, the lesser `x` value, `a`, will need an expression worked out for the `x` coordinate of the intersection of the two functions; the higher `x` value, `b`, is simply the `y`-axis.
The general way to work out an area between two functions is to integrate the combined function formed from the top function minus the bottom function. Alternatively, the two functions can be integrated individually to get the total area beneath each of them separately, then take the area under the upper function and subtract the area under the lower function.
Shaded area `= int_a^b ("upper function" - "lower function") dx`
OR Shaded area `= int_a^b "upper function" dx - int_a^b "lower function" dx`
I'm going to use the first option.
The upper function is `y = (ke^x)^2` and the lower function is `y = k`. The `a` limit is the `x` value of where the two functions intersect (which we need to work out), and the `b` limit is the `y`-axis, hence an `x` value of zero.
We can work out where the intersection is by equating the two functions, because the `y` values of the two functions will have the same value at the point of intersection. Then rearrange to get an expression for the `x` value of the point of intersection.
`(ke^x)^2 = k`
`ke^x = sqrtk`
This should be the positive square root only (not the negative as well) because `e^x` is positive (and `k` > 1), which means the next line of working is possible.
`e^x = sqrtk / k = 1 / sqrtk`
`x = ln {:1/sqrtk:}`
Note that many calculators want brackets around log arguments (the input to the log function). I generally use brackets around a log argument if it's not just a simple number, so there's no ambiguity of what is being fed into the log. Desmos seems to agree and has limits to what it'll accept without brackets. It's fine with `ln {:1/k:}` but wants brackets for `ln (1/sqrtk)`.
Now back to the integral with everything substituted in, then a little rearrangement to make it easier to integrate.
`= int_ln {:1/sqrtk:}^0 ((ke^x)^2 - k) dx`
`= int_ln {:1/sqrtk:}^0 (k^2(e^x)^2 - k) dx`
`= int_ln {:1/sqrtk:}^0 (k^2e^(2x) - k) dx`
There's a common factor of the constant `k` in those terms, and taking another look at what we're aiming for, we'll need to take it out at some point, to make the `k/2` out the front of the expression. The `k` can be taken out now or after the integration.
`= k int_ln {:1/sqrtk:}^0 (ke^(2x) - 1) dx`
And now the integration. The `a` and `b` values become the limits of the interval the integration is taken over. The `1/2` comes from the inverse chain rule applied to `e^(2x)`.
`= k [k/2e^(2x) - x]_ln {:1/sqrtk:}^0`
Substitute in the limits (the `x` values of the integral).
`= k [(k/2e^(2(0)) - (0)) - (k/2e^(2ln {:1/sqrtk:}) - ln {:1/sqrtk:})]`
One of the log rules tells us `nlog(x) = log(x^n)`, and we really want a common factor of `1/2`, so force it on the last term.
`= k [k/2e^0 - k/2e^(ln ((1/sqrtk)^2)) + 2/2ln {:1/sqrtk:}]`
There's now a common factor of `1/2` which can be taken out, making the common factor complete (the `k/2` we need at the start of the expression we're aiming at).
`= k/2 [ke^0 - ke^(ln ((1/sqrtk)^2)) + 2ln {:1/sqrtk:}]`
Any number to the power of zero is 1, so the factor of `e^0` vanishes. And we can do the same thing to the factor of 2 on the last term as we did earlier with the exponent of the middle term.
`= k/2 [k - ke^(ln {:1/k:}) + ln ((1/sqrtk)^2)]`
`e^x` and `lnx` are inverse functions of each other and cancel out. `sqrtx` and `x^2` are inverse functions of each other and cancel out.
`= k/2 [k - k times 1/k + ln {:1/k:}]`
`= k/2 [k - 1 + ln {:1/k:}]`
And that's it.
Comment: This is a pretty good question for assessing a student's ability to calculate a definite integral while using various algebra essentials like logarithms.
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