# IB DP Math AA Topic: AHL 5.12 Definition of derivative from first principles HL Paper 2

## Question

Consider the curve with equation $${x^3} + {y^3} = 4xy$$.

The tangent to this curve is parallel to the $$x$$-axis at the point where $$x = k,{\text{ }}k > 0$$.

a.Use implicit differentiation to show that $$\frac{{{\text{d}}y}}{{{\text{d}}x}} = \frac{{4y – 3{x^2}}}{{3{y^2} – 4x}}$$.[3]

b.Find the value of $$k$$.[5]

## Markscheme

$$3{x^2} + 3{y^2}\frac{{{\text{d}}y}}{{{\text{d}}x}} = 4\left( {y + x\frac{{{\text{d}}y}}{{{\text{d}}x}}} \right)$$    M1A1

$$(3{y^2} – 4x)\frac{{{\text{d}}y}}{{{\text{d}}x}} = 4y – 3{x^2}$$    A1

$$\frac{{{\text{d}}y}}{{{\text{d}}x}} = \frac{{4y – 3{x^2}}}{{3{y^2} – 4x}}$$    AG

[3 marks]

a.

$$\frac{{{\text{d}}y}}{{{\text{d}}x}} = 0 \Rightarrow 4y – 3{x^2} = 0$$    (M1)

substituting $$x = k$$ and $$y = \frac{3}{4}{k^2}$$ into $${x^3} + {y^3} = 4xy$$     M1

$${k^3} + \frac{{27}}{{64}}{k^6} = 3{k^3}$$    A1

attempting to solve $${k^3} + \frac{{27}}{{64}}{k^6} = 3{k^3}$$ for $$k$$     (M1)

$$k = 1.68{\text{ }}\left( { = \frac{4}{3}\sqrt[3]{2}} \right)$$    A1

Note:     Condone substituting $$y = \frac{3}{4}{x^2}$$ into $${x^3} + {y^3} = 4xy$$ and solving for $$x$$.

## 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]

## 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

[3 marks]

## Question

A curve C is given by the implicit equation $$x + y – {\text{cos}}\left( {xy} \right) = 0$$.

The curve $$xy = – \frac{\pi }{2}$$ intersects C at P and Q.

a.Show that $$\frac{{{\text{d}}y}}{{{\text{d}}x}} = – \left( {\frac{{1 + y\,{\text{sin}}\left( {xy} \right)}}{{1 + x\,{\text{sin}}\left( {xy} \right)}}} \right)$$.[5]

b.i.Find the coordinates of P and Q.[4]

b.ii.Given that the gradients of the tangents to C at P and Q are m1 and m2 respectively, show that m1 × m2 = 1.[3]

c.Find the coordinates of the three points on C, nearest the origin, where the tangent is parallel to the line $$y = – x$$.[7]

## Markscheme

attempt at implicit differentiation      M1

$$1 + \frac{{{\text{d}}y}}{{{\text{d}}x}} + \left( {y + x\frac{{{\text{d}}y}}{{{\text{d}}x}}} \right){\text{sin}}\left( {xy} \right) = 0$$     A1M1A1

Note: Award A1 for first two terms. Award M1 for an attempt at chain rule A1 for last term.

$$\left( {1 + x\,{\text{sin}}\left( {xy} \right)} \right)\frac{{{\text{d}}y}}{{{\text{d}}x}} = – 1 – y\,{\text{sin}}\left( {xy} \right)$$     A1

$$\frac{{{\text{d}}y}}{{{\text{d}}x}} = – \left( {\frac{{1 + y\,{\text{sin}}\left( {xy} \right)}}{{1 + x\,{\text{sin}}\left( {xy} \right)}}} \right)$$     AG

[5 marks]

a.

EITHER

when $$xy = – \frac{\pi }{2},\,\,{\text{cos}}\,xy = 0$$     M1

$$\Rightarrow x + y = 0$$    (A1)

OR

$$x – \frac{\pi }{{2x}} – {\text{cos}}\left( {\frac{{ – \pi }}{2}} \right) = 0$$ or equivalent      M1

$$x – \frac{\pi }{{2x}} = 0$$     (A1)

THEN

therefore $${x^2} = \frac{\pi }{2}\left( {x = \pm \sqrt {\frac{\pi }{2}} } \right)\left( {x = \pm 1.25} \right)$$     A1

$${\text{P}}\left( {\sqrt {\frac{\pi }{2}} ,\, – \sqrt {\frac{\pi }{2}} } \right),\,\,{\text{Q}}\left( { – \sqrt {\frac{\pi }{2}} ,\,\sqrt {\frac{\pi }{2}} } \right)$$ or $$P\left( {1.25,\, – 1.25} \right),\,Q\left( { – 1.25,\,1.25} \right)$$     A1

[4 marks]

b.i.

m1 = $$– \left( {\frac{{1 – \sqrt {\frac{\pi }{2}} \times – 1}}{{1 + \sqrt {\frac{\pi }{2}} \times – 1}}} \right)$$     M1A1

m= $$– \left( {\frac{{1 + \sqrt {\frac{\pi }{2}} \times – 1}}{{1 – \sqrt {\frac{\pi }{2}} \times – 1}}} \right)$$     A1

mm= 1     AG

Note: Award M1A0A0 if decimal approximations are used.
Note: No FT applies.

[3 marks]

b.ii.irst case, attempt to solve $$2x = {\text{cos}}\left( {{x^2}} \right)$$     M1

(0.486,0.48

equate derivative to −1    M1

$$\left( {y – x} \right){\text{sin}}\left( {xy} \right) = 0$$     (A1)

$$y = x,\,{\text{sin}}\left( {xy} \right) = 0$$     R1

in the f6)      A1

in the second case, $${\text{sin}}\left( {xy} \right) = 0 \Rightarrow xy = 0$$ and $$x + y = 1$$     (M1)

(0,1), (1,0)      A1

[7 marks]

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