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]
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]