NCEA Level 3 Mathematics 2021
Calculus: Differentiation exam
Excellence questions only.
QUESTION ONE
(d) A curve is defined parametrically by the equations `x = t^2 + 3t` and `y = t^2ln(2t - 3)`, for `t > 3/2`.
Find the gradient of the tangent to the curve at the point (10,0).
You must use calculus and show any derivatives that you need to find when solving this problem.
This looks like a product rule question.
`dx/dt = 2t + 3`
`dy/dt = 2tln(2t - 3) + t^2 * 1/(2t - 3) * 2`
`dy/dt = 2tln(2t - 3) + (2t^2)/(2t - 3)`
`dy/dx = (dy/dt) / (dx/dt) = (2tln(2t - 3))/(2t + 3) + (2t^2)/((2t - 3)(2t + 3))`
`dy/dx = (2tln(2t - 3))/(2t + 3) + (2t^2)/(4t^2 - 9)`
This is the gradient of the function, and hence the gradient of the tangent at any particular point on the curve. Next step is figure out what the value of `t` is at (10,0).
`x = t^2 + 3t`
`t^2 + 3t = 10`
`t^2 + 3t - 10 = 0`
`(t + 5)(t - 2) = 0`
`t = -5 or t = 2`
But `t > 3/2` so `t = 2` only.
Check: `y = t^2ln(2t - 3) = 2^2ln(2*2-3) = 4ln1 = 0`
`dy/dx = (2(2)ln(2(2) - 3))/(2(2) + 3) + (2(2)^2)/(4(2)^2 - 9)`
`dy/dx = (4ln(1))/7 + 8/7`
`dy/dx = 8/7`
QUESTION ONE
(e) A cone has a height of 3 m and a radius of 1.5 m.
A cylinder in inscribed in the cone, as shown in the diagram below.
The base of the cylinder has the same centre as the base of the cone.
Prove that the maximum volume of the cylinder is `pi` m3.
You must use calculus and show any derivatives that you need to find when solving this problem.
This looks like a fairly typical maximum volume question. If the cylinder is the same width as the cone, the height of the cylinder will be zero, and therefore have zero volume. If the cylinder is the same height as the cone, the width of the cylinder will be zero, and therefore have zero volume. Somewhere between those extremes the cylinder volume will be a maximum, at which point the volume will not be changing, so the differential of the volume will be zero.
In other words, after understanding the question, we need to put together a formula for the volume of the cylinder. Let's start with a generic formula. Make a diagram of a triangle (a vertical cross-section from the centre of the cone to its edge) to figure out what the height of the cylinder is in terms of its radius.
`V = pir^2*h`
`V = pir^2(3 - 2r)`
`V = 3pir^2 - 2pir^3`
And now we differentiate. Remember the volume isn't changing when the volume is a maximum.
`(dV)/dr = 6pir - 6pir^2 = 0` at maximum volume
`r - r^2 = 0`
`r(1 - r) = 0`
`r = 0` gives a minimum (zero) volume
`=> r = 1` gives maximum volume
Maximum volume is `3pir^2 - 2pir^3 = 3pi(1)^2 - 2pi(1)^3 = pi` m3
Comment: This is a straight forward question if a student is used to obtaining expressions for one variable in terms of another (h in terms of r in this case).
QUESTION TWO
(e) The graph below shows the curve `y = sqrt(2x - 4)`, and the tangent to the curve at point P. The tangent passes through the point (-2,1).
Find the coordinates of point P.
You must use calculus and show any derivatives that you need to find when solving this problem.
We'll need to differentiate the equation of the curve to get its gradient function. Let's start with what we have.
Tangent: `y - 1 = m(x - -2)`
`y = m(x + 2) + 1`
Curve: `y = sqrt(2x - 4) = (2x - 4)^(1/2)`
We may need to differentiate – it is a differentiation exam, after all. The equation of the curve will be easier to differentiate using an exponent instead of a surd (the √ symbol).
Curve: `y = (2x - 4)^(1/2)`
`dy/dx = 1/2(2x - 4)^(-1/2) * 2`
`dy/dx = 1 / sqrt(2x - 4)`
Tangent: `m = (Delta y) / (Delta x) = (sqrt(2x - 4) - 1) / (x + 2)`
We want the point at which these gradients are equal!
`(sqrt(2x - 4) - 1) / (x + 2) = 1 / sqrt(2x - 4)`
`sqrt(2x - 4)(sqrt(2x - 4) - 1) = x + 2`
`2x - 4 - sqrt(2x - 4) = x + 2`
`2x - 4 - x - 2 = sqrt(2x - 4)`
`x - 6 = sqrt(2x - 4)`
Warning: squaring both sides may introduce a phantom solution.
`(x - 6)^2 = (sqrt(2x - 4))^2`
`x^2 - 12x + 36 = 2x - 4`
`x^2 - 14x + 40 = 0`
`(x - 4)(x - 10) = 0`
`x = 4 or x = 10`
Test both possible solutions. The quickest and easiest way is to throw the two possible solutions into the equation before both sides were squared.
`x = 4, (4) - 6 = sqrt(2(4) - 4)`
`=> -2 = 2,` which is not valid
`=> x = 4` is not a solution
`x = 10, (10) - 6 = sqrt(2(10) - 4)`
`=> 4 = 4,` which is valid
`=> x = 10` is a solution
Alternatively, check if the gradients match. If not, the line cannot be a tangent to the curve.
`x = 4, m = (sqrt(2(4) - 4) - 1) / ((4) + 2) = 1/6, y_"tangent" = 1/6((4) + 2) + 1 = 2`
`dy/dx = 1 / sqrt(2(4) - 4) = 1/2, y_"curve" = sqrt(2(4) - 4) = 2`
At `x = 4` the gradients are different, so `x = 4` is not a solution.
`x = 10, m = (sqrt(2(10) - 4) - 1) / ((10) + 2) = 1/4, y_"tangent" = 1/4((10) + 2) + 1 = 4`
`dy/dx = 1 / sqrt(2(10) - 4) = 1/4 , y_"curve" = sqrt(2(10) - 4) = 4`
Location and gradients agree.
`P` is `(10,4)`
Comment: There are lots of ways this one could go wrong – and I found several. Equations can easily spiral out of control if choosing the wrong approach. For example, it looks like it should be a discriminant = 0 question. It's not. Students simply need to be prepared for this sort of question.
QUESTION THREE
(c) For what values of `x` is the function `y = x / (x^2 + 4)` increasing?
You must use calculus and show any derivatives that you need to find when solving this problem.
Hm, a part c which is an excellence question. That's a tad strange... Oh, yeah – it's NCEA.
To differentiate this we can use the quotient rule. (BTW, a quotient is the result of a division, in the same way a product is the result of a multiplication.)
`y = x / (x^2 + 4)`
`dy/dx = ( (x^2 + 4) - x(2x) ) / (x^2 + 4)^2 = 0` @ stationary points
We can use just the numerator... which can be simplified.
`x^2 + 4 - 2x^2 = 0`
`-x^2 + 4 = 0`
`x^2 = 4`
`x = +-sqrt(4)`
`x = +-2`
I'll use test points to find the actual gradients around those points.
`x = -4, dy/dx = (-(-4)^2 + 4) / ((-4)^2 + 4)^2 = -12/400 = -3/100`
`x = 0, dy/dx = ((0)^2 + 4) / ((0^2 + 4)^2 = 4/16 = 1/4`
`x = 4, dy/dx = (-(4)^2 + 4) / ((4)^2 + 4)^2 = -12/400 = -3/100`
`y` is increasing for `-2 < x < 2`
Comment: This doesn't seem hard enough to be an excellence question.
QUESTION THREE
(e) A lamp is suspended above the centre of a round table of radius `r`.
The height, `h`, of the lamp above the table is adjustable.
Point `P` is on the edge of the table.
At point `P` the illumination `I` is directly proportional to the cosine of angle `theta` in the above diagram, and inversely proportional to the square of the distance, `S`, to the lamp.
i.e. `I = (kcostheta) / S^2`, where `k` is a constant.
Prove that the edge of the table will have maximum illumination when `h = r / sqrt(2)`.
You do not need to prove that your solution gives the maximum value.
You must use calculus and show any derivatives that you need to find when solving this problem.
This looks like an interesting question.
The values `k` and `r` are both constants, which is helpful to keep in mind when considering what we're trying to prove – a relationship between `h` and `r` – and which variable we'll try to get `I` in terms of. The value `S` depends on `h` and `r`, and is not referred to in `h = r / sqrt(2)`. This all implies that `h` is the variable we should use for expressing `I` in terms of.
We need to get rid of the `costheta`...
`I = (kcostheta) / S^2`
`costheta = "adj" / "hyp" = h / S`
`=> I = (kh) / S^3`
... and the `S`. Pythagoras to the rescue! Remember `r` is a constant.
`S^2 = h^2 + r^2 => S = sqrt(h^2 + r^2) => S^3 = (h^2 + r^2)^(3/2)`
`=> I = (kh) / (h^2 + r^2)^(3/2)`
This is it – the illumination expressed in terms of one variable, `h`, all ready to differentiate.
We want to differentiate to give the rate of change of illumination with respect to `h`. When the rate of change of the illumination is zero, the illumination will be at its maximum, and that will be the value of `h` we want.
The quotient rule will give the whole result as a single quotient (or fraction), which will be handy for the following step. Don't forget to apply the chain rule when differentiating the nested expression in the denominator. That's where the extra factor of `2h` comes from.
`(dI)/(dh) = ( k(h^2 + r^2)^(3/2) - 3/2 * kh(h^2 + r^2)^(1/2) * 2h ) / ( (h^2 + r^2)^(3/2) )^2 = 0` at maximum illumination
We can dump the denominator because a quotient equals zero when its numerator equals zero. (So it doesn't matter what its denominator is, as long as it's not also zero; this one is not. If `h` was zero the illumination at `P` would be zero – definitely not a maximum. If `r` was zero we wouldn't have a table.) And we can simplify the second term.
`k(h^2 + r^2)^(3/2) - 3kh^2(h^2 + r^2)^(1/2) = 0`
Now divide by the common factors of `k` and `(h^2 + r^2)^(1/2)`, which is OK because they are not zero.
`(h^2 + r^2)^(2/2) - 3h^2 = 0`
`h^2 + r^2 - 3h^2 = 0`
`-2h^2 + r^2 = 0`
`2h^2 = r^2`
`h^2 = r^2 / 2`
We can take just the positive square root because `h` and `r` are both positive. (Without a good excuse like that the negative root should be included.
`h = sqrt(r^2 / 2)`
`h = r / sqrt(2)`
Comment: I can imagine this might seem a bit intimidating to many students but it really boils down to picking the right variable to express `I` in terms of, figuring out how to eliminate other variables, and doing the differentiation and subsequent algebra correctly. I like this question.
Overall comment
Many students could find themselves pressed for time, given the difficulty of some of the excellence questions and the time they could take, especially if an incorrect or inefficient approach is chosen.
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