Problems
Filters
1 problem found
2019 Paper 3 Q7
D: 1500.0
B: 1500.0
curve sketching
implicit curve
quartic
tangent-normal
quadratic in disguise
symmetry
asymptotic behaviour
Devil's Curve
The Devil's Curve is given by $$y^2(y^2 - b^2) = x^2(x^2 - a^2),$$ where \(a\) and \(b\) are positive constants.
- In the case \(a = b\), sketch the Devil's Curve.
- Now consider the case \(a = 2\) and \(b = \sqrt{5}\), and \(x \geq 0\), \(y \geq 0\).
- Show by considering a quadratic equation in \(x^2\) that either \(0 \leq y \leq 1\) or \(y \geq 2\).
- Describe the curve very close to and very far from the origin.
- Find the points at which the tangent to the curve is parallel to the \(x\)-axis and the point at which the tangent to the curve is parallel to the \(y\)-axis.
- Sketch the Devil's Curve in the case \(a = 2\) and \(b = \sqrt{5}\) again, but with \(-\infty < x < \infty\) and \(-\infty < y < \infty\).
Solution:
- Suppose \(a=b\), ie
\begin{align*}
&& y^2(y^2-a^2) &= x^2(x^2-a^2) \\
\Rightarrow && 0 &= x^4-y^4-a^2(x^2-y^2) \\
&&&= (x^2-y^2)(x^2+y^2-a^2)
\end{align*}
Therefore we have the lines \(y = \pm x\) and a circle radius \(a\).
- Since \(x^4 - 4x^2 - y^2(y^2-5)= 0\), we must have \(0 \leq \Delta = 16 + 4y^2(y^2-5) \Rightarrow y^4-5y^2+4 = (y^2-4)(y^2-1) \geq 0\), therefore \(0 \leq y \leq 1\) or \(y \geq 2\) (since we are only considering positive values of \(y\)).
- When \((x, y) \approx 0\) the equation is more like \(4x^2 \approx 5y^2\) or \(y \approx \frac{2}{\sqrt{5}}x\) If \(|x|, |y|\) are very large, it is more like \(x^4 \approx y^4\), ie \(y \approx x\)
- \(\,\) \begin{align*} && (2y(y^2-5)+y^2(2y))y' &= 2x(x^2-4)+2x^3 \\ \Rightarrow && (4y^3-10y)y' &= 4x^3-8x \end{align*} Therefore the gradient is parallel to the \(x\)-axis when \(x = 0, x = \sqrt{2}\). We need \(x = 0, y \neq 0\), ie \(y = \sqrt{5}\), so \((0, \sqrt{5})\) and \((\sqrt{2}, 0)\) It is parallel to the \(y\)-axis when \(y = 0\) or \(y = \sqrt{\frac52}\), ie \((2, 0)\)
- \(\,\)
