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2001 Paper 1 Q7
D: 1500.0 B: 1516.0

In a cosmological model, the radius \(\rm R\) of the universe is a function of the age \(t\) of the universe. The function \(\rm R\) satisfies the three conditions: $$ \mbox{\({\rm R}(0)=0\)}, \ \ \ \ \ \ \ \ \ \mbox{\({\rm R'}(t)>0\) for \(t>0\)}, \ \ \ \ \ \ \ \ \ \ \mbox{\({\rm R''}(t)<0\) for \(t>0\)}, \tag{*} $$ where \({\rm R''}\) denotes the second derivative of \(\rm R\). The function \({\rm H}\) is defined by \[ {\rm H} (t)= \frac{{\rm R}'(t)}{{\rm R}( t)}\;. \]

  1. Sketch a graph of \({\rm R} (t)\). By considering a tangent to the graph, show that \(t<1/{\rm H}(t)\).
  2. Observations reveal that \({\rm H}(t) = a/t\), where \(a\) is constant. Derive an expression for \({\rm R}(t)\). What range of values of \(a\) is consistent with the three conditions \((*)\)?
  3. Suppose, instead, that observations reveal that \({\rm H}(t)= b t^{-2}\), where \(b\) is constant. Show that this is not consistent with conditions \((*)\) for any value of \(b\).


Solution:

  1. \(\,\)
    TikZ diagram
    Notice the tangent must hit the \(y\)-axis above the origin, ie \begin{align*} && 0 &< R'(t)(0-t) + R(t) \\ \Rightarrow && R'(t) t &< R(t) \\ \Rightarrow && t &< \frac{R(t)}{R'(t)} = \frac{1}{H(t)} \end{align*}
  2. Suppose \(H(t) = a/t\) then \begin{align*} && \frac{R'}{R} &= \frac{a}{t} \\ \Rightarrow && \int \frac{1}{R} \d R &= \int \frac{a}{t} \d t \\ \Rightarrow && \ln R &= a \ln t + C \tag{t, R > 0} \\ \Rightarrow && R &= Kt^a \end{align*} Since we need \(R(t) > 0\), \(K > 0\), since \(R'(t) > 0\) we need \(a > 0\), since \(R''(t) < 0\) we need \(a(a-1) < 0\) ie \(0 < a < 1\)
  3. Suppose instead \(H(t) = bt^{-2}\) then \begin{align*} && \frac{R'}{R} &= \frac{b}{t^2} \\ \Rightarrow && \int \frac{1}{R} \d R &= \int \frac{b}{t^2} \d t \\ \Rightarrow && \ln R &= -bt^{-1} + C \tag{R > 0} \\ \Rightarrow && R &= Ke^{-b/t} \end{align*} Since \(R > 0\) we must have \(K > 0\). \begin{align*} R' > 0: && R' &= K(b/t^2)e^{-b/t} > 0 \\ \Rightarrow && b &> 0 \\ R'' < 0: && R'' &= K(b^2/t^4)e^{-b/t} -K2b/t^3 e^{-b/t} \\ &&&= Kb/t^4 (b-2t)e^{-b/t} < 0 \\ \Rightarrow && b &< 2t\\ \Rightarrow && b &< 2t \end{align*} which cannot be true for all \(t\), ie there is no \(b\) which satisfies this.