# IB DP Math AA Topic: SL 5.7 The second derivative: IB Style Questions HL Paper 2

## Question

The following diagram shows a vertical cross section of a building. The cross section of the roof of the building can be modelled by the curve $$f(x) = 30{{\text{e}}^{ – \frac{{{x^2}}}{{400}}}}$$, where $$– 20 \le x \le 20$$.

Ground level is represented by the $$x$$-axis.

a.Find $$f”(x)$$.[4]

b.Show that the gradient of the roof function is greatest when $$x = – \sqrt {200}$$.[3]

c.The cross section of the living space under the roof can be modelled by a rectangle $$CDEF$$ with points $${\text{C}}( – a,{\text{ }}0)$$ and $${\text{D}}(a,{\text{ }}0)$$, where $$0 < a \le 20$$.

Show that the maximum area $$A$$ of the rectangle $$CDEF$$ is $$600\sqrt 2 {{\text{e}}^{ – \frac{1}{2}}}$$.[5]

d.A function $$I$$ is known as the Insulation Factor of $$CDEF$$. The function is defined as $$I(a) = \frac{{P(a)}}{{A(a)}}$$ where $${\text{P}} = {\text{Perimeter}}$$ and $${\text{A}} = {\text{Area of the rectangle}}$$.

(i)     Find an expression for $$P$$ in terms of $$a$$.

(ii)     Find the value of $$a$$ which minimizes $$I$$.

(iii)     Using the value of $$a$$ found in part (ii) calculate the percentage of the cross sectional area under the whole roof that is not included in the cross section of the living space.[9]

▶️Answer/Explanation

## Markscheme

$$f'(x) = 30{{\text{e}}^{ – \frac{{{x^2}}}{{400}}}} \bullet – \frac{{2x}}{{400}}\;\;\;\left( { = – \frac{{3x}}{{20}}{{\text{e}}^{ – \frac{{{x^2}}}{{400}}}}} \right)$$     M1A1

Note:     Award M1 for attempting to use the chain rule.

$$f”(x) = – \frac{3}{{20}}{{\text{e}}^{ – \frac{{{x^2}}}{{400}}}} + \frac{{3{x^2}}}{{4000}}{{\text{e}}^{ – \frac{{{x^2}}}{{400}}}}\;\;\;\left( { = \frac{3}{{20}}{{\text{e}}^{ – \frac{{{x^2}}}{{400}}}}\left( {\frac{{{x^2}}}{{200}} – 1} \right)} \right)$$     M1A1

Note:     Award M1 for attempting to use the product rule.

[4 marks]

a.

the roof function has maximum gradient when $$f”(x) = 0$$     (M1)

Note:     Award (M1) for attempting to find $$f”\left( { – \sqrt {200} } \right)$$.

EITHER

$$= 0$$     A1

OR

$$f”(x) = 0 \Rightarrow x = \pm \sqrt {200}$$     A1

THEN

valid argument for maximum such as reference to an appropriate graph or change in the sign of $$f”(x)$$ eg $$f”( – 15) = 0.010 \ldots ( > 0)$$ and $$f”( – 14) = – 0.001 \ldots ( < 0)$$     R1

$$\Rightarrow x = – \sqrt {200}$$     AG

[3 marks]

b.

$$A = 2a \bullet 30{{\text{e}}^{ – \frac{{{a^2}}}{{400}}}}\;\;\;\left( { = 60a{{\text{e}}^{ – \frac{{{a^2}}}{{400}}}} = – 400g'(a)} \right)$$     (M1)(A1)

EITHER

$$\frac{{{\text{d}}A}}{{{\text{d}}a}} = 60a{{\text{e}}^{ – \frac{{{a^2}}}{{400}}}} \bullet – \frac{a}{{200}} + 60{{\text{e}}^{ – \frac{{{a^2}}}{{400}}}} = 0 \Rightarrow a = \sqrt {200} {\text{ }}\left( { – 400f”(a) = 0 \Rightarrow a = \sqrt {200} } \right)$$     M1A1

OR

by symmetry eg $$a = – \sqrt {200}$$ found in (b) or $${A_{{\text{max}}}}$$ coincides with $$f”(a) = 0$$     R1

$$\Rightarrow a = \sqrt {200}$$     A1

Note:     Award A0(M1)(A1)M0M1 for candidates who start with $$a = \sqrt {200}$$ and do not provide any justification for the maximum area. Condone use of $$x$$.

THEN

$${A_{{\text{max}}}} = 60 \bullet \sqrt {200} {{\text{e}}^{ – \frac{{200}}{{400}}}}$$     M1

$$= 600\sqrt 2 {{\text{e}}^{ – \frac{1}{2}}}$$     AG

[5 marks]

c.

(i)     perimeter $$= 4a + 60{{\text{e}}^{ – \frac{{{a^2}}}{{400}}}}$$     A1A1

Note:     Condone use of $$x$$.

(ii)     $$I(a) = \frac{{4a + 60{{\text{e}}^{ – \frac{{{a^2}}}{{400}}}}}}{{60a{{\text{e}}^{ – \frac{{{a^2}}}{{400}}}}}}$$     (A1)

graphing $$I(a)$$ or other valid method to find the minimum     (M1)

$$a = 12.6$$     A1

(iii)     area under roof $$= \int_{ – 20}^{20} {30{{\text{e}}^{ – \frac{{{x^2}}}{{400}}}}} {\text{d}}x$$     M1

$$= 896.18 \ldots$$     (A1)

area of living space $$= 60 \cdot (12.6…) \cdot e – {\frac{{(12.6…)}}{{400}}^2} = 508.56…$$

percentage of empty space $$= 43.3\%$$     A1

[9 marks]

Total [21 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]

▶️Answer/Explanation

## 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]

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