2004 Paper 2 Q8
Year: 2004
Paper: 2
Question Number: 8
Course: LFM Pure
Section: Differential equations
Difficulty: 1600.0
Banger: 1483.3
differential equation
first order
separation of variables
curve sketching
inequality
substitution
trigonometric
comparison of solutions
Problem
Let \(x\) satisfy the differential equation
$$
\frac {\d x}{\d t} = {\big( 1-x^n\big)\vphantom{\Big)}}^{\!1/n}
$$
and the condition \(x=0\) when \(t=0 \,\).
- Solve the equation in the case \(n=1\) and sketch the graph of the solution for \(t > 0 \,\).
- Prove that \(1-x < (1-x^2)^{1/2} \) for \(0 < x < 1 \,\). Use this result to sketch the graph of the solution in the case \(n=2\) for \(0 < t < \frac12 \pi \,\), using the same axes as your previous sketch. By setting \(x=\sin y\,\), solve the equation in this case.
- Use the result (which you need not prove) \[ (1-x^2)^{1/2} < (1-x^3)^{1/3} \text{ \ \ for \ \ } 0 < x < 1 \;, \] to sketch, without solving the equation, the graph of the solution of the equation in the case \(n=3\) using the same axes as your previous sketches. Use your sketch to show that \(x=1\) at a value of \(t\) less than \(\frac12 \pi \,\).
Solution
- \(\,\) \begin{align*}
&& \dot{x} &= (1-x) \\
\Rightarrow &&\int \frac{1}{1-x} \d x &= \int \d t \\
\Rightarrow && -\ln |1-x| &= t + C \\
t=0, x = 0: && -\ln 1 &= C \Rightarrow C = 0\\
\Rightarrow && -\ln|1-x| &= t \\
\Rightarrow && 1-x&= e^{-t} \\
\Rightarrow && x &= 1-e^{-t}
\end{align*}
- Notice that \((1-x^2)^{1/2} = (1-x)^{1/2}(1+x)^{1/2} > (1-x)^{1/2} > 1-x\)
\begin{align*} && \dot{x} &= \sqrt{1-x^2} \\ \Rightarrow && \int \frac{1}{\sqrt{1-x^2}} \d x &= t + C \\ x = \sin y, \d x = \cos y && \int \frac{\cos y}{\cos y} \d y &= t + C \\ \Rightarrow && y &= t + C \\ \Rightarrow && \sin^{-1} x &= t + C \\ t = 0, x = 0: && x &= \sin t \end{align*}
- \(\,\)
We know the gradient is steeper, so the solution must always be above \(\sin t\), which means it reaches \(1\) before \(\frac{\pi}{2}\)
Rating Information
Difficulty Rating: 1600.0
Difficulty Comparisons: 0
Banger Rating: 1483.3
Banger Comparisons: 1
Show LaTeX source
Problem source
Let $x$ satisfy the differential equation
$$
\frac {\d x}{\d t} = {\big( 1-x^n\big)\vphantom{\Big)}}^{\!1/n}
$$
and the condition $x=0$ when $t=0 \,$.
\begin{questionparts}
\item Solve the equation in the case $n=1$ and sketch the graph of the solution for $t > 0 \,$.
\item Prove that $1-x < (1-x^2)^{1/2} $ for $0 < x < 1 \,$.
Use this result to sketch the graph of the solution in the case $n=2$ for $0 < t < \frac12 \pi \,$, using the same axes as your previous sketch.
By setting $x=\sin y\,$, solve the equation in this case.
\item Use the result (which you need not prove)
\[ (1-x^2)^{1/2} < (1-x^3)^{1/3} \text{ \ \ for \ \ } 0 < x < 1 \;, \]
to sketch, without solving the equation, the graph of the solution of the equation in the case $n=3$ using the same axes as your previous sketches. Use your sketch to show that $x=1$ at a value of $t$ less than $\frac12 \pi \,$.
\end{questionparts}Solution source
\begin{questionparts}
\item $\,$ \begin{align*}
&& \dot{x} &= (1-x) \\
\Rightarrow &&\int \frac{1}{1-x} \d x &= \int \d t \\
\Rightarrow && -\ln |1-x| &= t + C \\
t=0, x = 0: && -\ln 1 &= C \Rightarrow C = 0\\
\Rightarrow && -\ln|1-x| &= t \\
\Rightarrow && 1-x&= e^{-t} \\
\Rightarrow && x &= 1-e^{-t}
\end{align*}
\begin{center}
\begin{tikzpicture}
\def\functionf(#1){1-exp(-(#1))};
\def\xl{-1};
\def\xu{8};
\def\yl{-.1};
\def\yu{1.2};
% Calculate scaling factors to make the plot square
\pgfmathsetmacro{\xrange}{\xu-\xl}
\pgfmathsetmacro{\yrange}{\yu-\yl}
\pgfmathsetmacro{\xscale}{10/\xrange}
\pgfmathsetmacro{\yscale}{10/\yrange}
% Define the styles for the axes and grid
\tikzset{
axis/.style={very thick, ->},
grid/.style={thin, gray!30},
x=\xscale cm,
y=\yscale cm
}
% Define the bounding region with clip
\begin{scope}
% You can modify these values to change your plotting region
\clip (\xl,\yl) rectangle (\xu,\yu);
% Draw a grid (optional)
% \draw[grid] (-5,-3) grid (5,3);
% \filldraw (-0.5, 0) circle (1.5pt) node[below]{$p$};
% \filldraw (1, 0) circle (1.5pt) node[below]{$q$};
% \filldraw (2.1, 0) circle (1.5pt) node[below]{$r$};
\draw[thick, blue, smooth, domain=0:\xu, samples=100]
plot (\x, {\functionf(\x)});
\draw[red, dashed] (0,1) -- (\xu, 1) node[pos=0.5, above] {$y = 1$};
\node[blue, above, rotate=5] at (6, {\functionf(6)}) {\tiny $x=1-e^{-t}$};
\end{scope}
% Set up axes
\draw[axis] (\xl,0) -- (\xu,0) node[right] {$t$};
\draw[axis] (0,\yl) -- (0,\yu) node[above] {$x$};
\end{tikzpicture}
\end{center}
\item Notice that $(1-x^2)^{1/2} = (1-x)^{1/2}(1+x)^{1/2} > (1-x)^{1/2} > 1-x$
\begin{center}
\begin{tikzpicture}
\def\functionf(#1){1-exp(-(#1))};
\def\functiong(#1){sin(180*#1/pi)};
\def\xl{-1};
\def\xu{5*pi/8};
\def\yl{-.1};
\def\yu{1.2};
% Calculate scaling factors to make the plot square
\pgfmathsetmacro{\xrange}{\xu-\xl}
\pgfmathsetmacro{\yrange}{\yu-\yl}
\pgfmathsetmacro{\xscale}{10/\xrange}
\pgfmathsetmacro{\yscale}{10/\yrange}
% Define the styles for the axes and grid
\tikzset{
axis/.style={very thick, ->},
grid/.style={thin, gray!30},
x=\xscale cm,
y=\yscale cm
}
% Define the bounding region with clip
\begin{scope}
% You can modify these values to change your plotting region
\clip (\xl,\yl) rectangle (\xu,\yu);
% Draw a grid (optional)
% \draw[grid] (-5,-3) grid (5,3);
% \filldraw (-0.5, 0) circle (1.5pt) node[below]{$p$};
% \filldraw (1, 0) circle (1.5pt) node[below]{$q$};
% \filldraw (2.1, 0) circle (1.5pt) node[below]{$r$};
\draw[thick, red, smooth, domain=0:{pi/2}, samples=100]
plot (\x, {\functionf(\x)});
\draw[thick, blue, smooth, domain=0:{pi/2}, samples=100]
plot (\x, {\functiong(\x)});
\draw[red, dashed] (0,1) -- (\xu, 1) node[pos=0.5, above] {$y = 1$};
\node[blue, above, rotate=50] at ({pi/4}, {\functiong({pi/4})}) {\tiny $x$};
\node[red, below, rotate=40] at ({pi/4}, {\functionf({pi/4})}) {\tiny $x=1-e^{-t}$};
\end{scope}
% Set up axes
\draw[axis] (\xl,0) -- (\xu,0) node[right] {$t$};
\draw[axis] (0,\yl) -- (0,\yu) node[above] {$x$};
\end{tikzpicture}
\end{center}
\begin{align*}
&& \dot{x} &= \sqrt{1-x^2} \\
\Rightarrow && \int \frac{1}{\sqrt{1-x^2}} \d x &= t + C \\
x = \sin y, \d x = \cos y && \int \frac{\cos y}{\cos y} \d y &= t + C \\
\Rightarrow && y &= t + C \\
\Rightarrow && \sin^{-1} x &= t + C \\
t = 0, x = 0: && x &= \sin t
\end{align*}
\item $\,$
\begin{center}
\begin{tikzpicture}
\def\functionf(#1){1-exp(-(#1))};
\def\functiong(#1){sin(180*#1/pi)};
\def\functionh(#1){(#1)*(2-(#1))};
\def\xl{-1};
\def\xu{5*pi/8};
\def\yl{-.1};
\def\yu{1.2};
% Calculate scaling factors to make the plot square
\pgfmathsetmacro{\xrange}{\xu-\xl}
\pgfmathsetmacro{\yrange}{\yu-\yl}
\pgfmathsetmacro{\xscale}{10/\xrange}
\pgfmathsetmacro{\yscale}{10/\yrange}
% Define the styles for the axes and grid
\tikzset{
axis/.style={very thick, ->},
grid/.style={thin, gray!30},
x=\xscale cm,
y=\yscale cm
}
% Define the bounding region with clip
\begin{scope}
% You can modify these values to change your plotting region
\clip (\xl,\yl) rectangle (\xu,\yu);
% Draw a grid (optional)
% \draw[grid] (-5,-3) grid (5,3);
% \filldraw (-0.5, 0) circle (1.5pt) node[below]{$p$};
% \filldraw (1, 0) circle (1.5pt) node[below]{$q$};
% \filldraw (2.1, 0) circle (1.5pt) node[below]{$r$};
\draw[thick, red, smooth, domain=0:{pi/2}, samples=100]
plot (\x, {\functionf(\x)});
\draw[thick, red, smooth, domain=0:{pi/2}, samples=100]
plot (\x, {\functiong(\x)});
\draw[thick, blue, smooth, domain=0:{1}, samples=100]
plot (\x, {\functionh(\x)});
\draw[red, dashed] (0,1) -- (\xu, 1) node[pos=0.5, above] {$y = 1$};
\node[red, above, rotate=50] at ({pi/4}, {\functiong({pi/4})}) {\tiny $x=\sin t$};
\node[red, below, rotate=40] at ({pi/4}, {\functionf({pi/4})}) {\tiny $x=1-e^{-t}$};
\end{scope}
% Set up axes
\draw[axis] (\xl,0) -- (\xu,0) node[right] {$t$};
\draw[axis] (0,\yl) -- (0,\yu) node[above] {$x$};
\end{tikzpicture}
\end{center}
We know the gradient is steeper, so the solution must always be above $\sin t$, which means it reaches $1$ before $\frac{\pi}{2}$
\end{questionparts}