IB DP Math AA Topic: AHL 5.14 : Implicit differentiation: IB Style Questions HL Paper 2

Question

The population, P, of a particular species of marsupial on a small remote island can be modelled by the logistic differential equation

\(\frac{dP}{dt}= kP\left ( 1-\frac{P}{N} \right )\)

where t is the time measured in years and k , N are positive constants.
The constant N represents the maximum population of this species of marsupial that the island can sustain indefinitely.
(a) In the context of the population model, interpret the meaning of \(\frac{dP}{dt}\).
(b) Show that \(\frac{d^{2}P}{dt^{2}}= k^{2}P\left ( 1-\frac{P}{N} \right )\left ( 1-\frac{2P}{N} \right )\)
(c) Hence show that the population of marsupials will increase at its maximum rate when \(P = \frac{N}{2}\). Justify your answer.
(d) Hence determine the maximum value of \(\frac{dP}{dt}\) in terms of k and N.
Let P0 be the initial population of marsupials.
(e) By solving the logistic differential equation, show that its solution can be expressed in the form

\(kt = In\frac{P}{P_{0}}\left ( \frac{N-P_{0}}{N-P} \right )\).

After 10 years, the population of marsupials is 3P0 . It is known that N = 4P0 .
(f) Find the value of k for this population model.

▶️Answer/Explanation

Ans:

(a) rate of growth (change) of the (marsupial) population (with respect to time)
Note: Do not accept growth (change) in the (marsupials) population per year.

OR
a correct and clearly labelled sign diagram (table) showing \(P = \frac{N}{2}\) corresponding to a local maximum point for \(\frac{dP}{dt}\)

Question

Find the gradient of the tangent to the curve \({x^3}{y^2} = \cos (\pi y)\) at the point (−1, 1) .

▶️Answer/Explanation

Markscheme

METHOD 1

\(3{x^2}{y^2} + 2{x^3}y\frac{{{\text{d}}y}}{{{\text{d}}x}} = – \pi \sin (\pi y)\frac{{{\text{d}}y}}{{{\text{d}}x}}\)     A1A1A1

At \(( – 1,{\text{ }}1),{\text{ }}3 – 2\frac{{{\text{d}}y}}{{{\text{d}}x}} = 0\)     M1A1

\(\frac{{{\text{d}}y}}{{{\text{d}}x}} = \frac{3}{2}\)     A1

[6 marks]

METHOD 2

\(3{x^2}{y^2} + 2{x^3}y\frac{{{\text{d}}y}}{{{\text{d}}x}} = – \pi \sin (\pi y)\frac{{{\text{d}}y}}{{{\text{d}}x}}\)     A1A1A1

\(\frac{{{\text{d}}y}}{{{\text{d}}x}} = \frac{{3{x^2}{y^2}}}{{ – \pi \sin (\pi y) – 2{x^3}y}}\)     A1

At \(( – 1,{\text{ }}1),{\text{ }}\frac{{{\text{d}}y}}{{{\text{d}}x}} = \frac{{3{{( – 1)}^2}{{(1)}^2}}}{{ – \pi \sin (\pi ) – 2{{( – 1)}^3}(1)}} = \frac{3}{2}\)     M1A1

[6 marks]

Examiners report

A large number of candidates obtained full marks on this question. Some candidates missed \(\pi \) and/or \(\frac{{{\text{d}}y}}{{{\text{d}}x}}\) when differentiating the trigonometric function. Some candidates attempted to rearrange before differentiating, and some made algebraic errors in rearranging.

Question

Find the gradient of the curve \({{\text{e}}^{xy}} + \ln \left( {{y^2}} \right) + {{\text{e}}^y} = 1 + {\text{e}}\) at the point (0, 1) .

▶️Answer/Explanation

Markscheme

\({{\text{e}}^{xy}} + \ln \left( {{y^2}} \right) + {{\text{e}}^y} = 1 + {\text{e}}\)

\({{\text{e}}^{xy}}\left( {y + x\frac{{{\text{d}}y}}{{{\text{d}}x}}} \right) + \frac{2}{y}\frac{{{\text{d}}y}}{{{\text{d}}x}} + {{\text{e}}^y}\frac{{{\text{d}}y}}{{{\text{d}}x}} = 0\) , at (0, 1)      A1A1A1A1A1

\(1\left( {1 + 0} \right) + 2\frac{{{\text{d}}y}}{{{\text{d}}x}} + {\text{e}}\frac{{{\text{d}}y}}{{{\text{d}}x}} = 0\)

\(1 + 2\frac{{{\text{d}}y}}{{{\text{d}}x}} + {\text{e}}\frac{{{\text{d}}y}}{{{\text{d}}x}} = 0\)

\(\frac{{{\text{d}}y}}{{{\text{d}}x}} = – \frac{1}{{2 + {\text{e}}}}\)   (\( = -0.212\))     M1A1     N2

[7 marks]

Examiners report

Implicit differentiation is usually found to be difficult, but on this occasion there were many correct solutions. There were also a number of errors in the differentiation of \({e^{xy}}\) , and although these often led to the correct final answer, marks could not be awarded.

Question

The point P, with coordinates \((p,{\text{ }}q)\) , lies on the graph of \({x^{\frac{1}{2}}} + {y^{\frac{1}{2}}} = {a^{\frac{1}{2}}}\) , \(a > 0\) .

The tangent to the curve at P cuts the axes at (0, m) and (n, 0) . Show that m + n = a .

▶️Answer/Explanation

Markscheme

\({x^{\frac{1}{2}}} + {y^{\frac{1}{2}}} = {a^{\frac{1}{2}}}\)

\(\frac{1}{2}{x^{ – \frac{1}{2}}} + \frac{1}{2}{y^{ – \frac{1}{2}}}\frac{{{\text{d}}y}}{{{\text{d}}x}} = 0\)     M1

\(\frac{{{\text{d}}y}}{{{\text{d}}x}} = – \frac{{\frac{1}{{2\sqrt x }}}}{{\frac{1}{{2\sqrt y }}}} = – \sqrt {\frac{y}{x}} \)     A1

Note: Accept \(\frac{{{\text{d}}y}}{{{\text{d}}x}} = 1 – \frac{{{a^{\frac{1}{2}}}}}{{{x^{\frac{1}{2}}}}}\) from making y the subject of the equation, and all correct subsequent working

 

therefore the gradient at the point P is given by

\(\frac{{{\text{d}}y}}{{{\text{d}}x}} = – \sqrt {\frac{q}{p}} \)     A1

equation of tangent is \(y – q = – \sqrt {\frac{q}{p}} (x – p)\)     M1

\((y = – \sqrt {\frac{q}{p}} x + q + \sqrt q \sqrt p )\)

x-intercept: y = 0, \(n = \frac{{q\sqrt p }}{{\sqrt q }} + p = \sqrt q \sqrt p  + p\)     A1

y-intercept: x = 0, \(m = \sqrt q \sqrt p  + q\)     A1

\(n + m = \sqrt q \sqrt p  + p + \sqrt q \sqrt p  + q\)     M1

\( = 2\sqrt q \sqrt p  + p + q\)

\( = {\left( {\sqrt p  + \sqrt q } \right)^2}\)     A1

\( = a\)     AG

[8 marks]

Examiners report

Many candidates were able to perform the implicit differentiation. Few gained any further marks.

Question

A particle moves along a straight line so that after t seconds its displacement s , in metres, satisfies the equation \({s^2} + s – 2t = 0\) . Find, in terms of s , expressions for its velocity and its acceleration.

▶️Answer/Explanation

Markscheme

\(2s\frac{{{\text{d}}s}}{{{\text{d}}t}} + \frac{{{\text{d}}s}}{{{\text{d}}t}} – 2 = 0\)     M1A1

\(v = \frac{{{\text{d}}s}}{{{\text{d}}t}} = \frac{2}{{2s + 1}}\)     A1

EITHER

\(a = \frac{{{\text{d}}v}}{{{\text{d}}t}} = \frac{{{\text{d}}v}}{{{\text{d}}s}}\frac{{{\text{d}}s}}{{{\text{d}}t}}\)     (M1)

\(\frac{{{\text{d}}v}}{{{\text{d}}s}} = \frac{{ – 4}}{{{{(2s + 1)}^2}}}\)     (A1)

\(a = \frac{{ – 4}}{{{{(2s + 1)}^2}}}\frac{{{\text{d}}s}}{{{\text{d}}t}}\)

OR

\(2{\left( {\frac{{{\text{d}}s}}{{{\text{d}}t}}} \right)^2} + 2s\frac{{{{\text{d}}^2}s}}{{{\text{d}}{t^2}}} + \frac{{{{\text{d}}^2}s}}{{{\text{d}}{t^2}}} = 0\)     (M1)

\(\frac{{{{\text{d}}^2}s}}{{\underbrace {{\text{d}}{t^2}}_a}} = \frac{{ – 2{{\left( {\frac{{{\text{d}}s}}{{{\text{d}}t}}} \right)}^2}}}{{2s + 1}}\)     (A1)

THEN

\(a = \frac{{ – 8}}{{{{(2s + 1)}^3}}}\)     A1

[6 marks]

Examiners report

Despite the fact that many candidates were able to calculate the speed of the particle, many of them failed to calculate the acceleration. Implicit differentiation turned out to be challenging in this exercise showing in many cases a lack of understanding of independent/dependent variables. Very often candidates did not use the chain rule or implicit differentiation when attempting to find the acceleration. It was not uncommon to see candidates trying to differentiate implicitly with respect to t rather than s, but getting the variables muddled.

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