Question
A small bead is free to move along a smooth wire in the shape of the curve y = \(\frac{10}{3-2e^{-0.5x}}\) ( x ≥ 0 ).
Find an expression for \(\frac{dy}{dx}\) . [3]
At the point on the curve where x = 4 , it is given that \(\frac{dy}{dt}\)= – 0.1 m s–1.
Find the value of \(\frac{dy}{dt}\) at this exact same instant. [3]
▶️Answer/Explanation
Ans
(a) valid attempt to use chain rule or quotient rule
Question
The function f is defined by \(f(x) = x\sqrt {9 – {x^2}} + 2\arcsin \left( {\frac{x}{3}} \right)\).
(a) Write down the largest possible domain, for each of the two terms of the function, f , and hence state the largest possible domain, D , for f .
(b) Find the volume generated when the region bounded by the curve y = f(x) , the x-axis, the y-axis and the line x = 2.8 is rotated through \(2\pi \) radians about the x-axis.
(c) Find \(f'(x)\) in simplified form.
(d) Hence show that \(\int_{ – p}^p {\frac{{11 – 2{x^2}}}{{\sqrt {9 – {x^2}} }}} {\text{d}}x = 2p\sqrt {9 – {p^2}} + 4\arcsin \left( {\frac{p}{3}} \right)\), where \(p \in D\) .
(e) Find the value of p which maximises the value of the integral in (d).
(f) (i) Show that \(f”(x) = \frac{{x(2{x^2} – 25)}}{{{{(9 – {x^2})}^{\frac{3}{2}}}}}\).
(ii) Hence justify that f(x) has a point of inflexion at x = 0 , but not at \(x = \pm \sqrt {\frac{{25}}{2}} \) .
▶️Answer/Explanation
Markscheme
(a) For \(x\sqrt {9 – {x^2}} \), \( – 3 \leqslant x \leqslant 3\) and for \(2\arcsin \left( {\frac{x}{3}} \right)\), \( – 3 \leqslant x \leqslant 3\) A1
\( \Rightarrow D{\text{ is }} – 3 \leqslant x \leqslant 3\) A1
[2 marks]
(b) \(V = \pi \int_0^{2.8} {{{\left( {x\sqrt {9 – {x^2}} = 2\arcsin \frac{x}{3}} \right)}^2}{\text{d}}x} \) M1A1
= 181 A1
[3 marks]
(c) \(\frac{{{\text{d}}y}}{{{\text{d}}x}} = {(9 – {x^2})^{\frac{1}{2}}} – \frac{{{x^2}}}{{{{(9 – {x^2})}^{\frac{1}{2}}}}} + \frac{{\frac{2}{3}}}{{\sqrt {1 – \frac{{{x^2}}}{9}} }}\) M1A1
\( = {(9 – {x^2})^{\frac{1}{2}}} – \frac{{{x^2}}}{{{{(9 – {x^2})}^{\frac{1}{2}}}}} + \frac{2}{{{{(9 – {x^2})}^{\frac{1}{2}}}}}\) A1
\( = \frac{{9 – {x^2} – {x^2} + 2}}{{{{(9 – {x^2})}^{\frac{1}{2}}}}}\) A1
\( = \frac{{11 – 2{x^2}}}{{\sqrt {9 – {x^2}} }}\) A1
[5 marks]
(d) \(\int_{ – p}^p {\frac{{11 – 2{x^2}}}{{\sqrt {9 – {x^2}} }}{\text{d}}x = \left[ {x\sqrt {9 – {x^2}} + 2\arcsin \frac{x}{3}} \right]_{ – p}^p} \) M1
\( = p\sqrt {9 – {p^2}} + 2\arcsin \frac{p}{3} + p\sqrt {9 – {p^2}} + 2\arcsin \frac{p}{3}\) A1
\( = 2p\sqrt {9 – {p^2}} + 4\arcsin \left( {\frac{p}{3}} \right)\) AG
[2 marks]
(e) \(11 – 2{p^2} = 0\) M1
\(p = 2.35\,\,\,\,\,\left( {\sqrt {\frac{{11}}{2}} } \right)\) A1
Note: Award A0 for \(p = \pm 2.35\) .
[2 marks]
(f) (i) \(f”(x) = \frac{{{{(9 – {x^2})}^{\frac{1}{2}}}( – 4x) + x(11 – 2{x^2}){{(9 – {x^2})}^{ – \frac{1}{2}}}}}{{9 – {x^2}}}\) M1A1
\( = \frac{{ – 4x(9 – {x^2}) + x(11 – 2{x^2})}}{{{{(9 – {x^2})}^{\frac{3}{2}}}}}\) A1
\( = \frac{{ – 36x + 4{x^3} + 11x – 2{x^3}}}{{{{(9 – {x^2})}^{\frac{3}{2}}}}}\) A1
\( = \frac{{x(2{x^2} – 25)}}{{{{(9 – {x^2})}^{\frac{3}{2}}}}}\) AG
(ii) EITHER
When \(0 < x < 3\), \(f”(x) < 0\). When \( – 3 < x < 0\), \(f”(x) > 0\). A1
OR
\(f”(0) = 0\) A1
THEN
Hence \(f”(x)\) changes sign through x = 0 , giving a point of inflexion. R1
EITHER
\(x = \pm \sqrt {\frac{{25}}{2}} \) is outside the domain of f. R1
OR
\(x = \pm \sqrt {\frac{{25}}{2}} \) is not a root of \(f”(x) = 0\) . R1
[7 marks]
Total [21 marks]
Examiners report
It was disappointing to note that some candidates did not know the domain for arcsin. Most candidates knew what to do in (b) but sometimes the wrong answer was obtained due to the calculator being in the wrong mode. In (c), the differentiation was often disappointing with \(\arcsin \left( {\frac{x}{3}} \right)\) causing problems. In (f)(i), some candidates who failed to do (c) guessed the correct form of \(f'(x)\) (presumably from (d)) and then went on to find \(f”(x)\) correctly. In (f)(ii), the justification of a point of inflexion at x = 0 was sometimes incorrect – for example, some candidates showed simply that \(f'(x)\) is positive on either side of the origin which is not a valid reason.
Question
A water trough which is 10 metres long has a uniform cross-section in the shape of a semicircle with radius 0.5 metres. It is partly filled with water as shown in the following diagram of the cross-section. The centre of the circle is O and the angle KOL is \(\theta \) radians.
The volume of water is increasing at a constant rate of \(0.0008{\text{ }}{{\text{m}}^3}{{\text{s}}^{ – 1}}\).
a.Find an expression for the volume of water \(V{\text{ }}({{\text{m}}^3})\) in the trough in terms of \(\theta \).[3]
b.Calculate \(\frac{{{\text{d}}\theta }}{{{\text{d}}t}}\) when \(\theta = \frac{\pi }{3}\).[4]
▶️Answer/Explanation
Markscheme
area of segment \( = \frac{1}{2} \times {0.5^2} \times (\theta – \sin \theta )\) M1A1
\(V = {\text{area of segment}} \times 10\)
\(V = \frac{5}{4}(\theta – \sin \theta )\) A1
[3 marks]
METHOD 1
\(\frac{{{\text{d}}V}}{{{\text{d}}t}} = \frac{5}{4}(1 – \cos \theta )\frac{{{\text{d}}\theta }}{{{\text{d}}t}}\) M1A1
\(0.0008 = \frac{5}{4}\left( {1 – \cos \frac{\pi }{3}} \right)\frac{{{\text{d}}\theta }}{{{\text{d}}t}}\) (M1)
\(\frac{{{\text{d}}\theta }}{{{\text{d}}t}} = 0.00128{\text{ }}({\text{rad}}\,{s^{ – 1}})\) A1
METHOD 2
\(\frac{{{\text{d}}\theta }}{{{\text{d}}t}} = \frac{{{\text{d}}\theta }}{{{\text{d}}V}} \times \frac{{{\text{d}}V}}{{{\text{d}}t}}\) (M1)
\(\frac{{{\text{d}}V}}{{{\text{d}}\theta }} = \frac{5}{4}(1 – \cos \theta )\) A1
\(\frac{{{\text{d}}\theta }}{{{\text{d}}t}} = \frac{{4 \times 0.0008}}{{5\left( {1 – \cos \frac{\pi }{3}} \right)}}\) (M1)
\(\frac{{{\text{d}}\theta }}{{{\text{d}}t}} = 0.00128\left( {\frac{4}{{3125}}} \right)({\text{rad }}{s^{ – 1}})\) A1
[4 marks]
Question
An earth satellite moves in a path that can be described by the curve \(72.5{x^2} + 71.5{y^2} = 1\) where \(x = x(t)\) and \(y = y(t)\) are in thousands of kilometres and \(t\) is time in seconds.
Given that \(\frac{{{\text{d}}x}}{{{\text{d}}t}} = 7.75 \times {10^{ – 5}}\) when \(x = 3.2 \times {10^{ – 3}}\), find the possible values of \(\frac{{{\text{d}}y}}{{{\text{d}}t}}\).
Give your answers in standard form.
▶️Answer/Explanation
Markscheme
METHOD 1
substituting for \(x\) and attempting to solve for \(y\) (or vice versa) (M1)
\(y = ( \pm )0.11821 \ldots \) (A1)
EITHER
\(145x + 143y\frac{{{\text{d}}y}}{{{\text{d}}x}} = 0{\text{ }}\left( {\frac{{{\text{d}}y}}{{{\text{d}}x}} = – \frac{{145x}}{{143y}}} \right)\) M1A1
OR
\(145x\frac{{{\text{d}}x}}{{{\text{d}}t}} + 143y\frac{{{\text{d}}y}}{{{\text{d}}t}} = 0\) M1A1
THEN
attempting to find \(\frac{{{\text{d}}x}}{{{\text{d}}t}}{\text{ }}\left( {\frac{{{\text{d}}y}}{{{\text{d}}t}} = – \frac{{145(3.2 \times {{10}^{ – 3}})}}{{143\left( {( \pm )0.11821 \ldots } \right)}} \times (7.75 \times {{10}^{ – 5}})} \right)\) (M1)
\(\frac{{{\text{d}}y}}{{{\text{d}}t}} = \pm 2.13 \times {10^{ – 6}}\) A1
Note: Award all marks except the final A1 to candidates who do not consider ±.
METHOD 2
\(y = ( \pm )\sqrt {\frac{{1 – 72.5{x^2}}}{{71.5}}} \) M1A1
\(\frac{{{\text{d}}y}}{{{\text{d}}x}} = ( \pm )0.0274 \ldots \) (M1)(A1)
\(\frac{{{\text{d}}y}}{{{\text{d}}t}} = ( \pm )0.0274 \ldots \times 7.75 \times {10^{ – 5}}\) (M1)
\(\frac{{{\text{d}}y}}{{{\text{d}}t}} = \pm 2.13 \times {10^{ – 6}}\) A1
Note: Award all marks except the final A1 to candidates who do not consider ±.
[6 marks]
Examiners report
Question
(a) Differentiate \(f(x) = \arcsin x + 2\sqrt {1 – {x^2}} \) , \(x \in [ – 1, 1]\) .
(b) Find the coordinates of the point on the graph of \(y = f (x)\) in \([ – 1, 1]\), where the gradient of the tangent to the curve is zero.
▶️Answer/Explanation
Markscheme
(a) \(f'(x) = \frac{1}{{\sqrt {1 – {x^2}} }} – \frac{{2x}}{{\sqrt {1 – {x^2}} }}\) \(\left( { = \frac{{1 – 2x}}{{\sqrt {1 – {x^2}} }}} \right)\) M1A1A1
Note: Award A1 for first term,
M1A1 for second term (M1 for attempting chain rule).
(b) \(f'(x) = 0\) (M1)
\(x = 0.5\) , \(y = 2.26\) or \(\frac{\pi }{6} + \sqrt 3 \) (accept (\(0.500\), \(2.26\)) A1A1 N3
[6 marks]
Examiners report
Most candidates scored well on this question, showing competence at non-trivial differentiation. The follow through rules allowed candidates to recover from minor errors in part (a). Some candidates demonstrated their resourcefulness in using their GDC to answer part (b) even when they had been unable to gain full marks on part (a).
Question
A function \(f\) is defined by \(f(x) = {x^3} + {{\text{e}}^x} + 1,{\text{ }}x \in \mathbb{R}\). By considering \(f'(x)\) determine whether \(f\) is a one-to-one or a many-to-one function.
▶️Answer/Explanation
Markscheme
\(f'(x) = 3{x^2} + {{\text{e}}^x}\) A1
Note: Accept labelled diagram showing the graph \(y = f'(x)\) above the x-axis;
do not accept unlabelled graphs nor graph of \(y = f(x)\).
EITHER
this is always \( > 0\) R1
so the function is (strictly) increasing R1
and thus \(1 – 1\) A1
OR
this is always \( > 0\;\;\;{\text{(accept }} \ne 0{\text{)}}\) R1
so there are no turning points R1
and thus \(1 – 1\) A1
Note: A1 is dependent on the first R1.
[4 marks]
Examiners report
The differentiation was normally completed correctly, but then a large number did not realise what was required to determine the type of the original function. Most candidates scored 1/4 and wrote explanations that showed little or no understanding of the relation between first derivative and the given function. For example, it was common to see comments about horizontal and vertical line tests but applied to the incorrect function.In term of mathematical language, it was noted that candidates used many terms incorrectly showing no knowledge of the meaning of terms like ‘parabola’, ‘even’ or ‘odd’ ( or no idea about these concepts).