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).
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.
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]
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]