Problems

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Solution: \begin{align*} && \frac{\d y}{\d x} &= \frac{\frac{\d y}{\d t}}{\frac{\d x}{\d t}} \\ &&&= \frac{6t^2}{6t} = t \\ \Rightarrow && \frac{y-2p^3}{x - 3p^2} &= p \\ \Rightarrow && y &= px-3p^3+2p^3 \\ && y &= px - p^3 \end{align*} The two lines will be \begin{align*} && y &= px - p^3 \\ && y &= qx - q^3 \\ \Rightarrow && p^3-q^3 &= (p-q)x \\ \Rightarrow && x &= p^2+pq+q^2 \\ && y &= p(p^2+pq+q^2)-p^3 \\ &&&= pq(p+q) \\ && (x,y) &= (p^2+pq+q^2,pq(p+q)) \\ \end{align*} If the tangents are \(\perp\) then \(pq=-1\), so we have \begin{align*} && (x,y) &= (p^2+2pq+q^2-pq, pq(p+q)) \\ &&&= ((p+q)^2-1, -(p+q)) \\ &&&= (u^2-1, -u) \end{align*} We have \(x = y^2+1\) and \(\left ( \frac{x}{3} \right)^3 = \left ( \frac{y}{2}\right)^2 \Rightarrow y^2 = \frac{4}{27}x^3\) so \begin{align*} && 0 &= \frac{4}{27}x^3-x+1 \\ &&0&=4x^3-27x+27 \\ &&&= (x+3)(2x-3)^2 \end{align*} So we have the points \((x,y) = \left (\frac32, \pm\frac{1}{\sqrt{2}}\right)\)

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Solution: \(\angle POQ = 90^\circ\) means that if \(P(p^2,p^3)\) and \(Q(q^2,q^3)\) are our points then \(OP^2+OQ^2 = PQ^2\), so \begin{align*} && p^4+p^6+q^4+q^6 &= (p^2-q^2)^2+(p^3-q^3)^2 \\ &&&= p^4+q^4-2p^2q^2+p^6+q^6-2p^3q^3 \\ \Rightarrow && 0 &= 2p^2q^2(1+pq) \\ \Rightarrow && pq &= -1 \\ \\ && \frac{\d y}{ \d x} &= \frac{\frac{\d y }{\d t}}{\frac{\d x}{\d t}} \\ &&&= \frac{3t^2}{2t} = \tfrac32t \\ \Rightarrow && \frac{y-p^3}{x-p^2} &= \tfrac32p \\ \Rightarrow && 2(y-p^3) &=3p(x-p^2) \\ && 2(y-q^3) &=3q(x-q^2) \\ \Rightarrow && 2(q^3-p^3) &= (3p-3q)x+3(q^3-p^3) \\ && p^3-q^3 &= 3(p-q)x \\ \Rightarrow && x &= \tfrac13(p^2+q^2+pq) \\ && 2y &= 3p(\tfrac13(p^2+q^2+pq)-p^2)+2p^3 \\ &&&= p(p^2+q^2+pq)-p^3 \\ &&&= pq^2+p^2q \\ &&&= -p-q \\ &&y&= -\frac{p+q}{2} \\ \\ && 4y^2 &= p^2+q^2 \\ && 3x-1 &= p^2+q^2 \\ \end{align*} To check if they meet, try \(4t^6=3t^2 - 1\). Consider \(y = 4x^3-3x+1\) \(y(0) = 1\) and \(y' = 12x^2-3 = 3(4x^2-1)\) which has roots at \(\pm \tfrac12\), therefore we need to test \(y(\tfrac12) = \tfrac12-\tfrac32 + 1 = 0\), so there is a one intersection at \(x = \tfrac1{2}, y = \tfrac1{2\sqrt{2}}\)

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Solution:

1. 

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   Clearly the only solution is \(x = 1\)
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   There is clearly only one solution, with \(x \approx -2\) \begin{align*} && 4(1-x) &= 6+2\sqrt{9-x^2} \\ && -2x-1 &=\sqrt{9-x^2} \\ \Rightarrow && 4x^2+4x+1 &= 9-x^2 \\ \Rightarrow && 0 &= 5x^2+4x-8 \\ &&x&= \frac{-2\pm 2\sqrt{11}}{5} \\ \Rightarrow && x &= -\left ( \frac{2+2\sqrt{11}}{5} \right) \end{align*}

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Solution: \begin{align*} && y &= ax^3 - 6ax^2+ \left( 12a + 12 \right)x - \left( 8a + 16 \right) \\ && y(2) &= 8a-24a+24a+24-8a-16 \\ &&&= 8 \\ && y'(x) &= 3ax^2-12ax+(12a+12) \\ && y'(0) &= 12a-24a+12a+12 \\ &&&= 12 \end{align*} Therefore since our curve has the same value and gradient at \((2,8)\) as \(y = x^3\) they must touch at this point. Therefore \begin{align*} && ax^3 - 6ax^2+ \left( 12a + 12 \right)x - \left( 8a + 16 \right) - x^3 &= (x-2)^2((a-1)x-(2a+4)) \end{align*} Therefore if \(a \neq 1\), they touch again when \(x = \frac{2a+4}{a-1}\).

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2. 

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3. 

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Solution: \begin{align*} && y &= \frac{2x(x^2-5)}{x^2-4} \\ &&&= 2x(x^2-5)(-\tfrac14)(1-\tfrac14x^2)^{-1} \\ &&&= \tfrac52x + \cdots \\ &&&= \frac{2x(x^2-4)-2x}{x^2-4} \\ &&&= 2x - \frac{2x}{x^2-4} \end{align*}

1. We are looking for the intersections of \(y = \frac23(x+3)\) and \(y = f(x)\)
   

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   Therefore 3 real roots.
2. We are looking for intersections of \(y = \frac12(5x-2)\) and \(y = f(x)\)
   

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   so one solution.
3. We are looking for intersections of \(y = f(x)^2\) and \(y = x^2+1\), or \(y = \sqrt{x^2+1}\) and \(y = f(x)\) where \(f(x) \geq 0\)
   

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   So \(3\) solutions.

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Solution:

Suppose the circle has centre \((c,d)\), then \begin{align*} && c^2+d^2 &= (t^2-c)^2+(t^3-d)^2 \\ \Rightarrow && 0 &= t^4-2ct^2+t^6-2t^3d \\ \Rightarrow && 0 &= t^4+t^2-2td-2c \end{align*} So by setting \(a = -2d\) and \(b = -2c\) we have the desired equation. By matching the coefficients of \(t^4, t^3, t^2\) we must have: \begin{align*} && 0 &= (t^2-(r+s)t+rs)(t^2+t(r+s)-rs+(r+s)^2+1) \\ \Rightarrow && 0 &= t^2+(r+s)t-rs+(r+s)^2+1 \\ && \Delta &= (r+s)^2 -4(1-rs+(r+s)^2) \\ &&&= -4+4rs-3(r+s)^2 \\ &&&=-4-2(r+s)^2-(r-s)^2 < 0 \end{align*} Therefore there are no further (real) solutions. Hence \(O, R, S\) are the only solutions.

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Solution:

Clearly we can achieve \(0\), \(1\), and \(2\) intersections by never entering the range, entering too flat, or entering before hitting the second branch. To achieve \(3\) we can go at a flat enough slope that we hit somewhere near the top of the second branch, and since the gradient there will be \(\approx 0\), and our gradient is positive, we must intersect before that point as well, ie \(3\) intersections. Clearly we cannot intersect the second branch \(3\) times or the first branch twice, therefore there are at most \(3\) intersections. To intersect the graph only once, we need to:

• be below \(((4n+1)\tfrac{\pi}{2}, 1)\) and
• not touch the second gradient

The first condition means that \(k (4n+1)\tfrac{\pi}{2} < 1 \Rightarrow k < \frac{2}{(4n+1)\pi}\). For the second condition, consider a point on the curve \(\sin x\) whose tangent line goes through the origin, ie \(\frac{y - \sin t}{x - t} = \cos t \Rightarrow y = (\cos t)x - t \cos t+\sin t\) ie \(\sin t = t \cos t\). For this point \(t\) to be in the required interval, we need \((4n+5) \tfrac{\pi}{2} -t \in (0, \frac{\pi}{2})\), so let's call this value \(\delta\). Then our result is: The gradient needs to be steeper than \(\cos t = \cos \left ( (4n+5) \tfrac{\pi}{2} - \delta \right) = \sin \delta\) and \(\cos \delta =\left ( (4n+5) \tfrac{\pi}{2} - \delta \right) \sin \delta \). If \(n\) is large, then, \begin{align*} && 1 &\approx \left ( (4n+5) \tfrac{\pi}{2} - \delta \right) \delta \\ \Rightarrow && 1 &\approx (4n+5) \tfrac{\pi}{2} \delta \\ \Rightarrow && \delta &\approx \frac{2}{(4n+5)\pi} \end{align*}. To higher order: \begin{align*} && 1-\frac12 \delta^2 &\approx \left ( (4n+5) \tfrac{\pi}{2} - \delta \right) \delta \\ \Rightarrow && 1-\frac12 \delta^2 &\approx (4n+5) \tfrac{\pi}{2} \delta - \delta^2 \\ \Rightarrow && 0 &\approx 1 - (4n + 5)\tfrac{\pi}{2} \delta + \frac12 \delta^2 \\ \Rightarrow && \delta &\approx (4n+5) \tfrac{\pi}{2} - \sqrt{(4n+5)^2 \frac{\pi^2}{4} - 2} \\ &&&= \frac{2}{(4n+5) \tfrac{\pi}{2} + \sqrt{(4n+5)^2 \frac{\pi^2}{4} - 2}} \end{align*}.

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Solution:

The gradient at the origin is \(\frac{5}{2}\) which we can observe by looking at the taylor series

1. \begin{align*} && 4x^{2}(x^{2}-5)&=(5x-2)(x^{2}-4) \\ && \frac{2x(x^2-5)}{x^2-4} &= \frac{5x-2}{2x} = \frac52 -\frac1x \end{align*} Therefore it will have two roots as the hyperbola will intersect our graph in two places. (In the upper left and lower right quadrants).
2. \begin{align*} && 4x^{2}(x^{2}-5)^{2}&=(x^{2}-4)^{2}(x^{2}+1) \\ && \frac{2x(x^2-5)}{x^2-4} &= \frac{(x^2-4)(x^2+1)}{2x} \end{align*}
   

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   No solutions \begin{align*} && 4z^{2}(z-5)^{2}&=(z-4)^{2}(z+1) \\ && \frac{2z(z^2-5)}{z^2-4} &= \frac{(z+5)(z-4)(z+1)}{(z-5)(z+4)} \end{align*}
   

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   5 solutions