Question
Consider the curve with equation \({x^2} + xy + {y^2} = 3\).
(a) Find in terms of k, the gradient of the curve at the point (−1, k).
(b) Given that the tangent to the curve is parallel to the x-axis at this point, find the value of k.
▶️Answer/Explanation
Markscheme
(a) Attempting implicit differentiation M1
\(2x + y + x\frac{{{\text{d}}y}}{{{\text{d}}x}} + 2y\frac{{{\text{d}}y}}{{{\text{d}}x}} = 0\) A1
EITHER
Substituting \(x = – 1,{\text{ }}y = k\,\,\,\,\,\)e.g. \( – 2 + k – \frac{{{\text{d}}y}}{{{\text{d}}x}} + 2k\frac{{{\text{d}}y}}{{{\text{d}}x}} = 0\) M1
Attempting to make \(\frac{{{\text{d}}y}}{{{\text{d}}x}}\) the subject M1
OR
Attempting to make \(\frac{{{\text{d}}y}}{{{\text{d}}x}}\) the subject e.g. \(\frac{{{\text{d}}y}}{{{\text{d}}x}} = \frac{{ – (2x + y)}}{{x + 2y}}\) M1
Substituting \(x = – 1,{\text{ }}y = k{\text{ into }}\frac{{{\text{d}}y}}{{{\text{d}}x}}\) M1
THEN
\(\frac{{{\text{d}}y}}{{{\text{d}}x}} = \frac{{2 – k}}{{2k – 1}}\) A1 N1
(b) Solving \(\frac{{{\text{d}}y}}{{{\text{d}}x}} = 0{\text{ for }}k{\text{ gives }}k = 2\) A1
[6 marks]
Question
Consider the part of the curve \(4{x^2} + {y^2} = 4\) shown in the diagram below.
(a) Find an expression for \(\frac{{{\text{d}}y}}{{{\text{d
}}x}}\) in terms of x and y .
(b) Find the gradient of the tangent at the point \(\left( {\frac{2}{{\sqrt 5 }},\frac{2}{{\sqrt 5 }}} \right)\).
(c) A bowl is formed by rotating this curve through \(2\pi \) radians about the x-axis.
Calculate the volume of this bowl.
▶️Answer/Explanation
Markscheme
(a) \(8x + 2y\frac{{{\text{d}}y}}{{{\text{d}}x}} = 0\) M1A1
Note: Award M1A0 for \(8x + 2y\frac{{{\text{d}}y}}{{{\text{d}}x}} = 4\) .
\(\frac{{{\text{d}}y}}{{{\text{d}}x}} = – \frac{{4x}}{y}\) A1
(b) – 4 A1
(c) \(V = \int {\pi {y^2}{\text{d}}x} \) or equivalent M1
\(V = \pi \int_0^1 {(4 – 4{x^2}){\text{d}}x} \) A1
\( = \pi \left[ {4x – \frac{4}{3}{x^3}} \right]_0^1\) A1
\( = \frac{{8\pi }}{3}\) A1
Note: If it is correct except for the omission of \(\pi \) , award 2 marks.
[8 marks]
Question
Show that the points (0, 0) and (\(\sqrt {2\pi } \) , \( – \sqrt {2\pi } \) ) on the curve \({{\text{e}}^{\left( {x + y} \right)}} = \cos \left( {xy} \right)\) have a common tangent.
▶️Answer/Explanation
Markscheme
attempt at implicit differentiation M1
\({{\text{e}}^{\left( {x + y} \right)}}\left( {1 + \frac{{{\text{d}}y}}{{{\text{d}}x}}} \right) = – \sin \left( {xy} \right)\left( {x\frac{{{\text{d}}y}}{{{\text{d}}x}} + y} \right)\) A1A1
let \(x = 0\), \(y = 0\) M1
\({{\text{e}}^0}\left( {1 + \frac{{{\text{d}}y}}{{{\text{d}}x}}} \right) = 0\)
\(\frac{{{\text{d}}y}}{{{\text{d}}x}} = – 1\) A1
let \(x = \sqrt {2\pi } \) , \(y = – \sqrt {2\pi } \)
\({{\text{e}}^0}\left( {1 + \frac{{{\text{d}}y}}{{{\text{d}}x}}} \right) = – \sin \left( {- 2\pi } \right)\left( {x\frac{{{\text{d}}y}}{{{\text{d}}x}} + y} \right) = 0\)
so \(\frac{{{\text{d}}y}}{{{\text{d}}x}} = – 1\) A1
since both points lie on the line \(y = – x\) this is a common tangent R1
Note: \(y = – x\) must be seen for the final R1. It is not sufficient to note that the gradients are equal.
[7 marks]
Question
The curve C has equation \(2{x^2} + {y^2} = 18\). Determine the coordinates of the four points on C at which the normal passes through the point (1, 0) .
▶️Answer/Explanation
Markscheme
\(4x + 2y\frac{{{\text{d}}y}}{{{\text{d}}x}} = 0 \Rightarrow \frac{{{\text{d}}y}}{{{\text{d}}x}} = -\frac{{2x}}{y}\) M1A1
Note: Allow follow through on incorrect \(\frac{{{\text{d}}y}}{{{\text{d}}x}}\) from this point.
gradient of normal at (a, b) is \(\frac{b}{{2a}}\)
Note: No further A marks are available if a general point is not used
equation of normal at (a, b) is \(y – b = \frac{b}{{2a}}(x – a)\left( { \Rightarrow y = \frac{b}{{2a}}x + \frac{b}{2}} \right)\) M1A1
substituting (1, 0) M1
\(b = 0\) or \(a = -1\) A1A1
four points are \((3,{\text{ }}0),{\text{ }}( – 3,{\text{ 0}}),{\text{ }}( – 1,{\text{ }}4),{\text{ }}( – 1,{\text{ }} – 4)\) A1A1
Note: Award A1A0 for any two points correct.
[9 marks]
Question
Let \({x^3}y = a\sin nx\) . Using implicit differentiation, show that
\[{x^3}\frac{{{{\text{d}}^2}y}}{{{\text{d}}{x^2}}} + 6{x^2}\frac{{{\text{d}}y}}{{{\text{d}}x}} + ({n^2}{x^2} + 6)xy = 0\] .
▶️Answer/Explanation
Markscheme
\({x^3}y = a\sin nx\)
attempt to differentiate implicitly M1
\( \Rightarrow 3{x^2}y + {x^3}\frac{{{\text{d}}y}}{{{\text{d}}x}} = an\cos nx\) A2
Note: Award A1 for two out of three correct, A0 otherwise.
\( \Rightarrow 6xy + 3{x^2}\frac{{{\text{d}}y}}{{{\text{d}}x}} + 3{x^2}\frac{{{\text{d}}y}}{{{\text{d}}x}} + {x^3}\frac{{{{\text{d}}^2}y}}{{{\text{d}}{x^2}}} = -a{n^2}\sin nx\) A2
Note: Award A1 for three or four out of five correct, A0 otherwise.
\( \Rightarrow 6xy + 6{x^2}\frac{{{\text{d}}y}}{{{\text{d}}x}} + {x^3}\frac{{{{\text{d}}^2}y}}{{{\text{d}}{x^2}}} = -a{n^2}\sin nx\)
\( \Rightarrow {x^3}\frac{{{{\text{d}}^2}y}}{{{\text{d}}{x^2}}} + 6{x^2}\frac{{{\text{d}}y}}{{{\text{d}}x}} + 6xy + {n^2}{x^3}y = 0\) A1
\( \Rightarrow {x^3}\frac{{{{\text{d}}^2}y}}{{{\text{d}}{x^2}}} + 6{x^2}\frac{{{\text{d}}y}}{{{\text{d}}x}} + ({n^2}{x^2} + 6)xy = 0\) AG
[6 marks]
Question
Consider the curve defined by the equation \({x^2} + \sin y – xy = 0\) .
a.Find the gradient of the tangent to the curve at the point \((\pi ,{\text{ }}\pi )\) .[6]
b.Hence, show that \(\tan \theta = \frac{1}{{1 + 2\pi }}\), where \(\theta \) is the acute angle between the tangent to the curve at \((\pi ,{\text{ }}\pi )\) and the line y = x .[3]
▶️Answer/Explanation
Markscheme
a.attempt to differentiate implicitly M1
\(2x + \cos y\frac{{{\text{d}}y}}{{{\text{d}}x}} – y – x\frac{{{\text{d}}y}}{{{\text{d}}x}} = 0\) A1A1
Note: A1 for differentiating \({x^2}\) and sin y ; A1 for differentiating xy.
substitute x and y by \(\pi \) M1
\(2\pi – \frac{{{\text{d}}y}}{{{\text{d}}x}} – \pi – \pi \frac{{{\text{d}}y}}{{{\text{d}}x}} = 0 \Rightarrow \frac{{{\text{d}}y}}{{{\text{d}}x}} = \frac{\pi }{{1 + \pi }}\) M1A1
Note: M1 for attempt to make dy/dx the subject. This could be seen earlier.
[6 marks]
\(\theta = \frac{\pi }{4} – \arctan \frac{\pi }{{1 + \pi }}\) (or seen the other way) M1
\(\tan \theta = \tan \left( {\frac{\pi }{4} – \arctan \frac{\pi }{{1 + \pi }}} \right) = \frac{{1 – \frac{\pi }{{1 + \pi }}}}{{1 + \frac{\pi }{{1 + \pi }}}}\) M1A1
\(\tan \theta = \frac{1}{{1 + 2\pi }}\) AG
[3 marks]
Question
A curve is defined by the equation \(8y\ln x – 2{x^2} + 4{y^2} = 7\). Find the equation of the tangent to the curve at the point where x = 1 and \(y > 0\).
▶️Answer/Explanation
Markscheme
\(8y \times \frac{1}{x} + 8\frac{{{\text{d}}y}}{{{\text{d}}x}}\ln x – 4x + 8y\frac{{{\text{d}}y}}{{{\text{d}}x}} = 0\) M1A1A1
Note: M1 for attempt at implicit differentiation. A1 for differentiating \(8y\ln x\), A1 for differentiating the rest.
when \(x = 1,{\text{ }}8y \times 0 – 2 \times 1 + 4{y^2} = 7\) (M1)
\({y^2} = \frac{9}{4} \Rightarrow y = \frac{3}{2}{\text{ (as }}y > 0)\) A1
at \(\left( {1,\frac{3}{2}} \right)\frac{{{\text{d}}y}}{{{\text{d}}x}} = – \frac{2}{3}\) A1
\(y – \frac{3}{2} = – \frac{2}{3}(x – 1)\) or \(y = – \frac{2}{3}x + \frac{{13}}{6}\) A1
[7 marks]
Question
The curve C is given implicitly by the equation \(\frac{{{x^2}}}{y} – 2x = \ln y\) for \(y > 0\).
a.Express \(\frac{{{\text{d}}y}}{{{\text{d}}x}}\) in terms of x and y.[4]
b.Find the value of \(\frac{{{\text{d}}y}}{{{\text{d}}x}}\) at the point on C where y = 1 and \(x > 0\).[2]
▶️Answer/Explanation
Markscheme
attempt at implicit differentiation M1
EITHER
\(\frac{{2x}}{y} – \frac{{{x^2}}}{{{y^2}}}\frac{{{\text{d}}y}}{{{\text{d}}x}} – 2 = \frac{1}{y}\frac{{{\text{d}}y}}{{{\text{d}}x}}\) A1A1
Note: Award A1 for each side.
a.\(\frac{{{\text{d}}y}}{{{\text{d}}x}} = \frac{{\frac{{2x}}{y} – 2}}{{\frac{1}{y} + \frac{{{x^2}}}{{{y^2}}}}}{\text{ }}\left( { = \frac{{2xy – 2{y^2}}}{{{x^2} + y}}} \right)\) A1
OR
after multiplication by y
\(2x – 2y – 2x\frac{{{\text{d}}y}}{{{\text{d}}x}} = \frac{{{\text{d}}y}}{{{\text{d}}x}}\ln y + y\frac{1}{y}\frac{{{\text{d}}y}}{{{\text{d}}x}}\) A1A1
Note: Award A1 for each side.
\(\frac{{{\text{d}}y}}{{{\text{d}}x}} = \frac{{2(x – y)}}{{1 + 2x + \ln y}}\) A1
[4 marks]
for \(y = 1,{\text{ }}{x^2} – 2x = 0\)
\(x = (0{\text{ or) 2}}\) A1
for \(x = 2\), \(\frac{{{\text{d}}y}}{{{\text{d}}x}} = \frac{2}{5}\) A1
[2 marks]
Question
The curve C is given implicitly by the equation \(\frac{{{x^2}}}{y} – 2x = \ln y\) for \(y > 0\).
a.Express \(\frac{{{\text{d}}y}}{{{\text{d}}x}}\) in terms of x and y.[4]
b.Find the value of \(\frac{{{\text{d}}y}}{{{\text{d}}x}}\) at the point on C where y = 1 and \(x > 0\).[2]
▶️Answer/Explanation
Markscheme
attempt at implicit differentiation M1
EITHER
\(\frac{{2x}}{y} – \frac{{{x^2}}}{{{y^2}}}\frac{{{\text{d}}y}}{{{\text{d}}x}} – 2 = \frac{1}{y}\frac{{{\text{d}}y}}{{{\text{d}}x}}\) A1A1
Note: Award A1 for each side.
\(\frac{{{\text{d}}y}}{{{\text{d}}x}} = \frac{{\frac{{2x}}{y} – 2}}{{\frac{1}{y} + \frac{{{x^2}}}{{{y^2}}}}}{\text{ }}\left( { = \frac{{2xy – 2{y^2}}}{{{x^2} + y}}} \right)\) A1
OR
after multiplication by y
\(2x – 2y – 2x\frac{{{\text{d}}y}}{{{\text{d}}x}} = \frac{{{\text{d}}y}}{{{\text{d}}x}}\ln y + y\frac{1}{y}\frac{{{\text{d}}y}}{{{\text{d}}x}}\) A1A1
Note: Award A1 for each side.
a.
\(\frac{{{\text{d}}y}}{{{\text{d}}x}} = \frac{{2(x – y)}}{{1 + 2x + \ln y}}\) A1
[4 marks]
for \(y = 1,{\text{ }}{x^2} – 2x = 0\)
\(x = (0{\text{ or) 2}}\) A1
for \(x = 2\), \(\frac{{{\text{d}}y}}{{{\text{d}}x}} = \frac{2}{5}\) A1
[2 marks]
Question
A curve has equation \({x^3}{y^2} + {x^3} – {y^3} + 9y = 0\). Find the coordinates of the three points on the curve where \(\frac{{{\text{d}}y}}{{{\text{d}}x}} = 0\).
▶️Answer/Explanation
Markscheme
\(3{x^2}{y^2} + 2{x^3}y\frac{{{\text{d}}y}}{{{\text{d}}x}} + 3{x^2} – 3{y^2}\frac{{{\text{d}}y}}{{{\text{d}}x}} + 9\frac{{{\text{d}}y}}{{{\text{d}}x}} = 0\) M1M1A1
Note: First M1 for attempt at implicit differentiation, second M1 for use of product rule.
\(\left( {\frac{{{\text{d}}y}}{{{\text{d}}x}} = \frac{{3{x^2}{y^2} + 3{x^2}}}{{3{y^2} – 2{x^3}y – 9}}} \right)\)
\( \Rightarrow 3{x^2} + 3{x^2}{y^2} = 0\) (A1)
\( \Rightarrow 3{x^2}\left( {1 + {y^2}} \right) = 0\)
\(x = 0\) A1
Note: Do not award A1 if extra solutions given eg \(y = \pm 1\).
substituting \(x = 0\) into original equation (M1)
\({y^3} – 9y = 0\)
\(y(y + 3)(y – 3) = 0\)
\(y = 0,{\text{ }}y = \pm 3\)
coordinates \((0, 0), (0, 3), (0, – 3)\) A1
[7 marks]
Examiners report
The majority of candidates were able to apply implicit differentiation and the product rule correctly to obtain \(3{x^2}\left( {1 + {y^2}} \right) = 0\). The better then recognised that \(x = 0\) was the only possible solution. Such candidates usually went on to obtain full marks. A number decided that \(y = \pm 1\) though then made no further progress. The solution set \(x = 0\) and \(y = \pm i\) was also occasionally seen. A small minority found the correct x and y values for the three co-ordinates but then surprisingly expressed them as \({\text{(0, 0), (3, 0)}}\) and \((- 3, 0)\).
Question
A curve has equation \(\arctan {x^2} + \arctan {y^2} = \frac{\pi }{4}\).
(a) Find \(\frac{{{\text{d}}y}}{{{\text{d}}x}}\) in terms of x and y.
(b) Find the gradient of the curve at the point where \(x = \frac{1}{{\sqrt 2 }}\) and \(y < 0\).
▶️Answer/Explanation
Markscheme
(a) METHOD 1
\(\frac{{2x}}{{1 + {x^4}}} + \frac{{2y}}{{1 + {y^4}}}\frac{{{\text{d}}y}}{{{\text{d}}x}} = 0\) M1A1A1
Note: Award M1 for implicit differentiation, A1 for LHS and A1 for RHS.
\(\frac{{{\text{d}}y}}{{{\text{d}}x}} = – \frac{{x\left( {1 + {y^4}} \right)}}{{y\left( {1 + {x^4}} \right)}}\) A1
METHOD 2
\({y^2} = \tan \left( {\frac{\pi }{4} – \arctan {x^2}} \right)\)
\( = \frac{{\tan \frac{\pi }{4} – \tan \left( {\arctan {x^2}} \right)}}{{1 + \left( {\tan \frac{\pi }{4}} \right)\left( {\tan \left( {\arctan {x^2}} \right)} \right)}}\) (M1)
\( = \frac{{1 – {x^2}}}{{1 + {x^2}}}\) A1
\(2y\frac{{{\text{d}}y}}{{{\text{d}}x}} = \frac{{ – 2x\left( {1 + {x^2}} \right) – 2x\left( {1 – {x^2}} \right)}}{{{{\left( {1 + {x^2}} \right)}^2}}}\) M1
\(2y\frac{{{\text{d}}y}}{{{\text{d}}x}} = \frac{{ – 4x}}{{{{\left( {1 + {x^2}} \right)}^2}}}\)
\(\frac{{{\text{d}}y}}{{{\text{d}}x}} = – \frac{{2x}}{{y{{\left( {1 + {x^2}} \right)}^2}}}\) A1
\(\left( { = \frac{{2x\sqrt {1 + {x^2}} }}{{\sqrt {1 – {x^2}} {{\left( {1 + {x^2}} \right)}^2}}}} \right)\)
[4 marks]
(b) \({y^2} = \tan \left( {\frac{\pi }{4} – \arctan \frac{1}{2}} \right)\) (M1)
\( = \frac{{\tan \frac{\pi }{4} – \tan \left( {\arctan \frac{1}{2}} \right)}}{{1 + \left( {\tan \frac{\pi }{4}} \right)\left( {\tan \left( {\arctan \frac{1}{2}} \right)} \right)}}\) (M1)
Note: The two M1s may be awarded for working in part (a).
\( = \frac{{1 – \frac{1}{2}}}{{1 + \frac{1}{2}}} = \frac{1}{3}\) A1
\(y = – \frac{1}{{\sqrt 3 }}\) A1
substitution into \(\frac{{{\text{d}}y}}{{{\text{d}}x}}\)
\( = \frac{{4\sqrt 6 }}{9}\) A1
Note: Accept \(\frac{{8\sqrt 3 }}{{9\sqrt 2 }}\) etc.
[5 marks]
Total [9 marks]
Examiners report
Question
A curve is defined by \(xy = {y^2} + 4\).
a.Show that there is no point where the tangent to the curve is horizontal.[4]
b.Find the coordinates of the points where the tangent to the curve is vertical.[4]
▶️Answer/Explanation
Markscheme
a.\(x\frac{{{\text{d}}y}}{{{\text{d}}x}} + y = 2y\frac{{{\text{d}}y}}{{{\text{d}}x}}\) M1A1
a horizontal tangent occurs if \(\frac{{{\text{d}}y}}{{{\text{d}}x}} = 0\) so \(y = 0\) M1
we can see from the equation of the curve that this solution is not possible \((0 = 4)\) and so there is not a horizontal tangent R1
[4 marks]
\(\frac{{{\text{d}}y}}{{{\text{d}}x}} = \frac{y}{{2y – x}}\) or equivalent with \(\frac{{{\text{d}}x}}{{{\text{d}}y}}\)
the tangent is vertical when \(2y = x\) M1
substitute into the equation to give \(2{y^2} = {y^2} + 4\) M1
\(y = \pm 2\) A1
coordinates are \((4,{\text{ }}2),{\text{ }}( – 4,{\text{ }} – 2)\) A1
[4 marks]
Total [8 marks]
Question
A curve has equation \(3x – 2{y^2}{{\text{e}}^{x – 1}} = 2\).
a. Find an expression for \(\frac{{{\text{d}}y}}{{{\text{d}}x}}\) in terms of \(x\) and \(y\).[5]
b.Find the equations of the tangents to this curve at the points where the curve intersects the line \(x = 1\).[4]
▶️Answer/Explanation
Markscheme
attempt to differentiate implicitly M1
a.\(3 – \left( {4y\frac{{{\text{d}}y}}{{{\text{d}}x}} + 2{y^2}} \right){{\text{e}}^{x – 1}} = 0\) A1A1A1
Note: Award A1 for correctly differentiating each term.
\(\frac{{{\text{d}}y}}{{{\text{d}}x}} = \frac{{3 \bullet {{\text{e}}^{1 – x}} – 2{y^2}}}{{4y}}\) A1
Note: This final answer may be expressed in a number of different ways.
[5 marks]
\(3 – 2{y^2} = 2 \Rightarrow {y^2} = \frac{1}{2} \Rightarrow y = \pm \sqrt {\frac{1}{2}} \) A1
\(\frac{{{\text{d}}y}}{{{\text{d}}x}} = \frac{{3 – 2 \bullet \frac{1}{2}}}{{ \pm 4\sqrt {\frac{1}{2}} }} = \pm \frac{{\sqrt 2 }}{2}\) M1
at \(\left( {1,{\text{ }}\sqrt {\frac{1}{2}} } \right)\) the tangent is \(y – \sqrt {\frac{1}{2}} = \frac{{\sqrt 2 }}{2}(x – 1)\) and A1
at \(\left( {1,{\text{ }} – \sqrt {\frac{1}{2}} } \right)\) the tangent is \(y + \sqrt {\frac{1}{2}} = – \frac{{\sqrt 2 }}{2}(x – 1)\) A1
Note: These equations simplify to \(y = \pm \frac{{\sqrt 2 }}{2}x\).
Note: Award A0M1A1A0 if just the positive value of \(y\) is considered and just one tangent is found.
[4 marks]
Question
The folium of Descartes is a curve defined by the equation \({x^3} + {y^3} – 3xy = 0\), shown in the following diagram.
Determine the exact coordinates of the point P on the curve where the tangent line is parallel to the \(y\)-axis.
▶️Answer/Explanation
Markscheme
\({x^3} + {y^3} – 3xy = 0\)
\(3{x^2} + 3{y^2}\frac{{{\text{d}}y}}{{{\text{d}}x}} – 3x\frac{{{\text{d}}y}}{{{\text{d}}x}} – 3y = 0\) M1A1
Note: Differentiation wrt \(y\) is also acceptable.
\(\frac{{{\text{d}}y}}{{{\text{d}}x}} = \frac{{3y – 3{x^2}}}{{3{y^2} – 3x}}{\text{ }}\left( { = \frac{{y – {x^2}}}{{{y^2} – x}}} \right)\) (A1)
Note: All following marks may be awarded if the denominator is correct, but the numerator incorrect.
\({y^2} – x = 0\) M1
EITHER
\(x = {y^2}\)
\({y^6} + {y^3} – 3{y^3} = 0\) M1A1
\({y^6} – 2{y^3} = 0\)
\({y^3}({y^3} – 2) = 0\)
\((y \ne 0)\therefore y = \sqrt[3]{2}\) A1
\(x = {\left( {\sqrt[3]{2}} \right)^2}{\text{ }}\left( { = \sqrt[3]{4}} \right)\) A1
OR
\({x^3} + xy – 3xy = 0\) M1
\(x({x^2} – 2y) = 0\)
\(x \ne 0 \Rightarrow y = \frac{{{x^2}}}{2}\) A1
\({y^2} = \frac{{{x^4}}}{4}\)
\(x = \frac{{{x^4}}}{4}\)
\(x({x^3} – 4) = 0\)
\((x \ne 0)\therefore x = \sqrt[3]{4}\) A1
\(y = \frac{{{{\left( {\sqrt[3]{4}} \right)}^2}}}{2} = \sqrt[3]{2}\) A1
[8 marks]