# IB DP Maths Topic 6.1 The derivative interpreted as a gradient function and as a rate of change HL Paper 2

## Question

Given that the graph of $$y = {x^3} – 6{x^2} + kx – 4$$ has exactly one point at which the gradient is zero, find the value of k .

## Markscheme

$$\frac{{{\text{d}}y}}{{{\text{d}}x}} = 3{x^2} – 12x + k$$     M1A1

For use of discriminant $${b^2} – 4ac = 0$$ or completing the square $$3{(x – 2)^2} + k – 12$$     (M1)

$$144 – 12k = 0$$     (A1)

Note: Accept trial and error, sketches of parabolas with vertex (2,0) or use of second derivative.

$$k = 12$$     A1

[5 marks]

## Examiners report

Generally candidates answer this question well using a diversity of methods. Surprisingly, a small number of candidates were successful in answering this question using the discriminant of the quadratic and in many cases reverted to trial and error to obtain the correct answer.

## Question

a.Sketch the curve $$y = \frac{{\cos x}}{{\sqrt {{x^2} + 1} }},{\text{ }} – 4 \leqslant x \leqslant 4$$ showing clearly the coordinates of the x-intercepts, any maximum points and any minimum points.[4]

b.Write down the gradient of the curve at x = 1 .[1]

c.Find the equation of the normal to the curve at x = 1 .[3]

## Markscheme

A1A1A1A1

Note: Award A1 for correct shape. Do not penalise if too large a domain is used,

A1 for correct x-intercepts,

A1 for correct coordinates of two minimum points,

A1 for correct coordinates of maximum point.

Accept answers which correctly indicate the position of the intercepts, maximum point and minimum points.

[4 marks]

a.

gradient at x = 1 is –0.786     A1

[1 mark]

b.

gradient of normal is $$\frac{{ – 1}}{{ – 0.786}}( = 1.272…)$$     (A1)

when x = 1, y = 0.3820…     (A1)

Equation of normal is y – 0.382 = 1.27(x – 1)     A1

$$( \Rightarrow y = 1.27x – 0.890)$$

[3 marks]

## Question

Consider the curve, $$C$$ defined by the equation $${y^2} – 2xy = 5 – {{\text{e}}^x}$$. The point A lies on $$C$$ and has coordinates $$(0,{\text{ }}a),{\text{ }}a > 0$$.

a.Find the value of $$a$$.[2]

b.Show that $$\frac{{{\text{d}}y}}{{{\text{d}}x}} = \frac{{2y – {{\text{e}}^x}}}{{2(y – x)}}$$.[4]

c.Find the equation of the normal to $$C$$ at the point A.[3]

d.Find the coordinates of the second point at which the normal found in part (c) intersects $$C$$.[4]

e.Given that $$v = {y^3},{\text{ }}y > 0$$, find $$\frac{{{\text{d}}v}}{{{\text{d}}x}}$$ at $$x = 0$$.[3]

$${a^2} = 5 – 1$$     (M1)

$$a = 2$$     A1

[2 marks]

a.

$$2y\frac{{{\text{d}}y}}{{{\text{d}}x}} – \left( {2x\frac{{{\text{d}}y}}{{{\text{d}}x}} + 2y} \right) = – {{\text{e}}^x}$$     M1A1A1A1

Note:     Award M1 for an attempt at implicit differentiation, A1 for each part.

$$\frac{{{\text{d}}y}}{{{\text{d}}x}} = \frac{{2y – {{\text{e}}^x}}}{{2(y – x)}}$$     AG

[4 marks]

b.

at $$x = 0,{\text{ }}\frac{{{\text{d}}y}}{{{\text{d}}x}} = \frac{3}{4}$$     (A1)

finding the negative reciprocal of a number     (M1)

gradient of normal is $$– \frac{4}{3}$$

$$y = – \frac{4}{3}x + 2$$     A1

[3 marks]

c.

substituting linear expression     (M1)

$${\left( { – \frac{4}{3}x + 2} \right)^2} – 2x\left( { – \frac{4}{3}x + 2} \right) + {{\text{e}}^x} – 5 = 0$$ or equivalent

$$x = 1.56$$     (M1)A1

$$y = – 0.0779$$     A1

$$(1.56,{\text{ }} – 0.0779)$$

[4 marks]

d.

$$\frac{{{\text{d}}v}}{{{\text{d}}x}} = 3{y^2}\frac{{{\text{d}}y}}{{{\text{d}}x}}$$    M1A1

$$\frac{{{\text{d}}v}}{{{\text{d}}x}} = 3 \times 4 \times \frac{3}{4} = 9$$    A1

[3 marks]

## Question

The curve $$C$$ is defined by equation $$xy – \ln y = 1,{\text{ }}y > 0$$.

a.Find $$\frac{{{\text{d}}y}}{{{\text{d}}x}}$$ in terms of $$x$$ and $$y$$.[4]

b.Determine the equation of the tangent to $$C$$ at the point $$\left( {\frac{2}{{\text{e}}},{\text{ e}}} \right)$$[3]

## Markscheme

$$y + x\frac{{{\text{d}}y}}{{{\text{d}}x}} – \frac{1}{y}\frac{{{\text{d}}y}}{{{\text{d}}x}} = 0$$     M1A1A1

Note:     Award A1 for the first two terms, A1 for the third term and the 0.

$$\frac{{{\text{d}}y}}{{{\text{d}}x}} = \frac{{{y^2}}}{{1 – xy}}$$     A1

Note:     Accept $$\frac{{ – {y^2}}}{{\ln y}}$$.

Note:     Accept $$\frac{{ – y}}{{x – \frac{1}{y}}}$$.

[4 marks]

a.

$${m_T} = \frac{{{{\text{e}}^2}}}{{1 – {\text{e}} \times \frac{2}{{\text{e}}}}}$$     (M1)

$${m_T} = – {{\text{e}}^2}$$     (A1)

$$y – {\text{e}} = – {{\text{e}}^2}x + 2{\text{e}}$$

$$– {{\text{e}}^2}x – y + 3{\text{e}} = 0$$ or equivalent     A1

Note:     Accept $$y = – 7.39x + 8.15$$.

[3 marks]

## Question

Consider the curve defined by the equation $$4{x^2} + {y^2} = 7$$.

a.Find the equation of the normal to the curve at the point $$\left( {1,{\text{ }}\sqrt 3 } \right)$$.[6]

b.Find the volume of the solid formed when the region bounded by the curve, the $$x$$-axis for $$x \geqslant 0$$ and the $$y$$-axis for $$y \geqslant 0$$ is rotated through $$2\pi$$ about the $$x$$-axis.[3]

## Markscheme

METHOD 1

$$4{x^2} + {y^2} = 7$$

$$8x + 2y\frac{{{\text{d}}y}}{{{\text{d}}x}} = 0$$     (M1)(A1)

$$\frac{{{\text{d}}y}}{{{\text{d}}x}} = – \frac{{4x}}{y}$$

Note:     Award M1A1 for finding $$\frac{{{\text{d}}y}}{{{\text{d}}x}} = – 2.309 \ldots$$ using any alternative method.

hence gradient of normal $$= \frac{y}{{4x}}$$     (M1)

hence gradient of normal at $$\left( {1,{\text{ }}\sqrt 3 } \right)$$ is $$\frac{{\sqrt 3 }}{4}\,\,\,( = 0.433)$$     (A1)

hence equation of normal is $$y – \sqrt 3 = \frac{{\sqrt 3 }}{4}(x – 1)$$     (M1)A1

$$\left( {y = \frac{{\sqrt 3 }}{4}x + \frac{{3\sqrt 3 }}{4}} \right)\,\,\,(y = 0.433x + 1.30)$$

METHOD 2

$$4{x^2} + {y^2} = 7$$

$$y = \sqrt {7 – 4{x^2}}$$     (M1)

$$\frac{{{\text{d}}y}}{{{\text{d}}x}} = – \frac{{4x}}{{\sqrt {7 – 4{x^2}} }}$$     (A1)

Note:     Award M1A1 for finding $$\frac{{{\text{d}}y}}{{{\text{d}}x}} = – 2.309 \ldots$$ using any alternative method.

hence gradient of normal $$= \frac{{\sqrt {7 – 4{x^2}} }}{{4x}}$$     (M1)

hence gradient of normal at $$\left( {1,{\text{ }}\sqrt 3 } \right)$$ is $$\frac{{\sqrt 3 }}{4}\,\,\,( = 0.433)$$     (A1)

hence equation of normal is $$y – \sqrt 3 = \frac{{\sqrt 3 }}{4}(x – 1)$$     (M1)A1

$$\left( {y = \frac{{\sqrt 3 }}{4}x + \frac{{3\sqrt 3 }}{4}} \right)\,\,\,(y = 0.433x + 1.30)$$

[6 marks]

a.

Use of $$V = \pi \int\limits_0^{\frac{{\sqrt 7 }}{2}} {{y^2}{\text{d}}x}$$

$$V = \pi \int\limits_0^{\frac{{\sqrt 7 }}{2}} {\left( {7 – 4{x^2}} \right){\text{d}}x}$$    (M1)(A1)

Note:     Condone absence of limits or incorrect limits for M mark.

Do not condone absence of or multiples of $$\pi$$.

$$= 19.4\,\,\,\left( { = \frac{{7\sqrt 7 \pi }}{3}} \right)$$     A1

[3 marks]

b.

[N/A]

a.

[N/A]

b.

## Question

Consider the function $$f(x) = \frac{{\sqrt x }}{{\sin x}},{\text{ }}0 < x < \pi$$.

Consider the region bounded by the curve $$y = f(x)$$, the $$x$$-axis and the lines $$x = \frac{\pi }{6},{\text{ }}x = \frac{\pi }{3}$$.

a.i.Show that the $$x$$-coordinate of the minimum point on the curve $$y = f(x)$$ satisfies the equation $$\tan x = 2x$$.[5]

a.ii.Determine the values of $$x$$ for which $$f(x)$$ is a decreasing function.[2]

b.Sketch the graph of $$y = f(x)$$ showing clearly the minimum point and any asymptotic behaviour.[3]

c.Find the coordinates of the point on the graph of $$f$$ where the normal to the graph is parallel to the line $$y = – x$$.[4]

d.This region is now rotated through $$2\pi$$ radians about the $$x$$-axis. Find the volume of revolution.[3]

## Markscheme

attempt to use quotient rule or product rule     M1

$$f’(x) = \frac{{\sin x\left( {\frac{1}{2}{x^{ – \frac{1}{2}}}} \right) – \sqrt x \cos x}}{{{{\sin }^2}x}}{\text{ }}\left( { = \frac{1}{{2\sqrt x \sin x}} – \frac{{\sqrt x \cos x}}{{{{\sin }^2}x}}} \right)$$     A1A1

Note:     Award A1 for $$\frac{1}{{2\sqrt x \sin x}}$$ or equivalent and A1 for $$– \frac{{\sqrt x \cos x}}{{{{\sin }^2}x}}$$ or equivalent.

setting $$f’(x) = 0$$     M1

$$\frac{{\sin x}}{{2\sqrt x }} – \sqrt x \cos x = 0$$

$$\frac{{\sin x}}{{2\sqrt x }} = \sqrt x \cos x$$ or equivalent     A1

$$\tan x = 2x$$     AG

[5 marks]

a.i.

$$x = 1.17$$

$$0 < x \leqslant 1.17$$     A1A1

Note:     Award A1 for $$0 < x$$ and A1 for $$x \leqslant 1.17$$. Accept $$x < 1.17$$.

[2 marks]

a.ii.

concave up curve over correct domain with one minimum point above the $$x$$-axis.     A1

approaches $$x = 0$$ asymptotically     A1

approaches $$x = \pi$$ asymptotically     A1

Note:     For the final A1 an asymptote must be seen, and $$\pi$$ must be seen on the $$x$$-axis or in an equation.

[3 marks]

b.

$$f’(x){\text{ }}\left( { = \frac{{\sin x\left( {\frac{1}{2}{x^{ – \frac{1}{2}}}} \right) – \sqrt x \cos x}}{{{{\sin }^2}x}}} \right) = 1$$     (A1)

attempt to solve for $$x$$     (M1)

$$x = 1.96$$     A1

$$y = f(1.96 \ldots )$$

$$= 1.51$$     A1

[4 marks]

c.

$$V = \pi \int_{\frac{\pi }{6}}^{\frac{\pi }{3}} {\frac{{x{\text{d}}x}}{{{{\sin }^2}x}}}$$     (M1)(A1)

Note:     M1 is for an integral of the correct squared function (with or without limits and/or $$\pi$$).

$$= 2.68{\text{ }}( = 0.852\pi )$$     A1

## Question

The following graph shows the two parts of the curve defined by the equation $${x^2}y = 5 – {y^4}$$, and the normal to the curve at the point P(2 , 1).

a.Show that there are exactly two points on the curve where the gradient is zero.[7]

b.Find the equation of the normal to the curve at the point P.[5]

c.The normal at P cuts the curve again at the point Q. Find the $$x$$-coordinate of Q.[3]

d.The shaded region is rotated by 2$$\pi$$ about the $$y$$-axis. Find the volume of the solid formed.[7]

## Markscheme

differentiating implicitly:       M1

$$2xy + {x^2}\frac{{{\text{d}}y}}{{{\text{d}}x}} = – 4{y^3}\frac{{{\text{d}}y}}{{{\text{d}}x}}$$     A1A1

Note: Award A1 for each side.

if $$\frac{{{\text{d}}y}}{{{\text{d}}x}} = 0$$ then either $$x = 0$$ or $$y = 0$$       M1A1

$$x = 0 \Rightarrow$$ two solutions for $$y\left( {y = \pm \sqrt[4]{5}} \right)$$      R1

$$y = 0$$ not possible (as 0 ≠ 5)     R1

hence exactly two points      AG

Note: For a solution that only refers to the graph giving two solutions at  $$x = 0$$ and no solutions for $$y = 0$$ award R1 only.

[7 marks]

a.

at (2, 1)  $$4 + 4\frac{{{\text{d}}y}}{{{\text{d}}x}} = – 4\frac{{{\text{d}}y}}{{{\text{d}}x}}$$     M1

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

gradient of normal is 2       M1

1 = 4 + c       (M1)

equation of normal is $$y = 2x – 3$$     A1

[5 marks]

b.

substituting      (M1)

$${x^2}\left( {2x – 3} \right) = 5 – {\left( {2x – 3} \right)^4}$$ or $${\left( {\frac{{y + 3}}{2}} \right)^2}\,y = 5 – {y^4}$$       (A1)

$$x = 0.724$$      A1

[3 marks]

c.

recognition of two volumes      (M1)

volume $$1 = \pi \int_1^{\sqrt[4]{5}} {\frac{{5 – {y^4}}}{y}} {\text{d}}y\left( { = 101\pi = 3.178 \ldots } \right)$$      M1A1A1

Note: Award M1 for attempt to use $$\pi \int {{x^2}} {\text{d}}y$$, A1 for limits, A1 for $${\frac{{5 – {y^4}}}{y}}$$ Condone omission of $$\pi$$ at this stage.

volume 2

EITHER

$$= \frac{1}{3}\pi \times {2^2} \times 4\left( { = 16.75 \ldots } \right)$$     (M1)(A1)

OR

$$= \pi \int_{ – 3}^1 {{{\left( {\frac{{y + 3}}{2}} \right)}^2}} {\text{d}}y\left( { = \frac{{16\pi }}{3} = 16.75 \ldots } \right)$$     (M1)(A1)

THEN

total volume = 19.9      A1

[7 marks]

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