Problems

Filters
Clear Filters

2 problems found

2012 Paper 1 Q1
D: 1484.0 B: 1500.0
differentiation optimisation line equation intercepts minimum distance AM-GM inequality stationary points

The line \(L\) has equation \(y=c-mx\), with \(m>0\) and \(c>0\). It passes through the point \(R(a,b)\) and cuts the axes at the points \(P(p,0)\) and \(Q(0,q)\), where \(a\), \(b\), \(p\) and \(q\) are all positive. Find \(p\) and \(q\) in terms of \(a\), \(b\) and \(m\). As \(L\) varies with \(R\) remaining fixed, show that the minimum value of the sum of the distances of \(P\) and \(Q\) from the origin is \((a^{\frac12} + b^{\frac12})^2\), and find in a similar form the minimum distance between \(P\) and \(Q\). (You may assume that any stationary values of these distances are minima.)

Solution: \begin{align*} && b &= c - ma \\ \Rightarrow && c &= b+ma \\ \Rightarrow && y &= m(a-x)+b \\ \Rightarrow && q &= ma+b \\ && p &= \frac{ma+b}{m} \\ \\ && d &= p+q \\ &&&= a + \frac{b}{m} + ma + b \\ \Rightarrow && d' &= -bm^{-2}+a \\ \Rightarrow && m &= \sqrt{b/a} \\ \\ \Rightarrow &&d &= a + \sqrt{ba}+\sqrt{ba} + b \\ &&&= (\sqrt{a}+\sqrt{b})^2 \\ \\ && |PQ|^2 &= p^2 + q^2 \\ &&&= a^2 + \frac{2ab}{m} + \frac{b^2}{m^2} + m^2a^2 + 2mab + b^2 \\ &&&= a^2+b^2 + \frac{b^2}{m^2} + \frac{2ab}{m}+ 2abm + a^2m^2 \\ && \frac{\d}{\d m}&= -2b^2m^{-3}-2abm^{-2}+2ab + 2a^2m \\ && 0 &=2a^2m^4+2abm^3-2abm-2b^2 \\ &&&= 2(am^3-b)(am+b) \\ \Rightarrow && m &= \sqrt[3]{\frac{b}{a}} \\ \\ &&|PQ|^2 &= \left[ a^{1/3}(a^{2/3} + b^{2/3}) \right]^2 + \left[ b^{1/3}(a^{2/3} + b^{2/3}) \right]^2 \\ &&&= a^{2/3}(a^{2/3} + b^{2/3})^2 + b^{2/3}(a^{2/3} + b^{2/3})^2 \\ &&&= (a^{2/3} + b^{2/3})^2 \cdot (a^{2/3} + b^{2/3}) \\ &&&= (a^{2/3} + b^{2/3})^3 \\ \Rightarrow && |PQ| &= (a^{2/3} + b^{2/3})^{3/2} \end{align*} We can also do this with AM-GM instead: \begin{align*} && d &= a + b + \frac{b}{m} + am \\ &&&\geq a+b + 2 \sqrt{\frac{b}{m} \cdot am} \\ &&&= a+2\sqrt{ab}+b \\ \\ && |PQ|^2 &= a^2+b^2 + \frac{b^2}{m^2} + \frac{2ab}{m}+ 2abm + a^2m^2 \\ &&&= a^2+b^2 + \frac{b^2}{m} + abm + abm + a^2m^2 + \frac{ab}{m} + \frac{ab}{m} \\ &&&= a^2+b^2 + 3\sqrt[3]{ \frac{b^2}{m} \cdot abm \cdot abm} + 3 \sqrt[3]{ a^2m^2 \cdot \frac{ab}{m} \cdot \frac{ab}{m} } \\ &&&= a^2 + 3b^{4/3}a^{2/3}+3b^{2/3}a^{4/3}+b^2 \\ &&&= (a^{2/3}+b^{2/3})^3 \end{align*}

View
2007 Paper 1 Q7
D: 1500.0 B: 1500.0
vectors lines in 3D minimum distance skew lines completing the square optimisation parametric

  1. The line \(L_1\) has vector equation $\displaystyle {\bf r} = \begin{pmatrix} 1 \\ 0 \\ 2 \end{pmatrix} + \lambda \begin{pmatrix} \hphantom{-} 2 \\ \hphantom{-} 2 \\ -3 \end{pmatrix} $. The line \(L_2\) has vector equation $\displaystyle {\bf r} = \begin{pmatrix} \hphantom{-} 4 \\ -2 \\ \hphantom{-} 9 \end{pmatrix} + \mu \begin{pmatrix} \hphantom{-} 1 \\ \hphantom{-} 2 \\ -2 \end{pmatrix} . $ Show that the distance \(D\) between a point on \(L_1\) and a point on \(L_2\) can be expressed in the form \[ D^2 = \left(3\mu -4 \lambda-5 \right)^2 + \left( \lambda -1 \right)^2 + 36\,. \] Hence determine the minimum distance between these two lines and find the coordinates of the points on the two lines that are the minimum distance apart.
  2. The line \(L_3\) has vector equation ${\bf r} = \begin{pmatrix} 2 \\ 3 \\ 5 \end{pmatrix} + \alpha \begin{pmatrix} 0 \\ 1 \\ 0 \end{pmatrix} . $ The line \(L_4\) has vector equation $ {\bf r} = \begin{pmatrix} \hphantom{-} 3 \\ \hphantom{-} 3 \\ -2 \end{pmatrix} + \beta \begin{pmatrix} \, 4k\\ 1-k \\ \!\!\! -3k \end{pmatrix} . $ Determine the minimum distance between these two lines, explaining geometrically the two different cases that arise according to the value of \(k\).

View