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Date May 2017 Marks available 6 Reference code 17M.2.hl.TZ2.2
Level HL only Paper 2 Time zone TZ2
Command term Find Question number 2 Adapted from N/A

Question

Consider the curve defined by the equation 4x2+y2=7.

Find the equation of the normal to the curve at the point (1, 3).

[6]
a.

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

[3]
b.

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.

Examiners report

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a.
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b.

Syllabus sections

Topic 6 - Core: Calculus » 6.1 » Informal ideas of limit, continuity and convergence.

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