A particle moves in a straight line so that its distance, \(x\) metres, from a fixed origin \(O\) after time \(t\) seconds is given by the differential equation \(\dfrac{d x}{d t}=\dfrac{2 e^{-x}}{1+4 t^2}\), where \(x=0\) when \(t=0\).
- i. Express the differential equation in the form \(\displaystyle \int g(x)dx=\int f(t)dt\). (1 mark)
- ii. Hence, show that \(x=\log _e\left(\tan ^{-1}(2 t)+1\right)\). (2 marks)
- The graph of \(x=\log _e\left(\tan ^{-1}(2 t)+1\right)\) has a horizontal asymptote.
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- Write down the equation of this asymptote. (1 mark)
- Sketch the graph of \(x=\log _e\left(\tan ^{-1}(2 t)+1\right)\) and the horizontal asymptote on the axes below. Using coordinates, plot and label the point where \(t=10\), giving the value of \(x\) correct to two decimal places. (2 marks)
- Find the speed of the particle when \(t=3\). Give your answer in metres per second, correct to two decimal places. (1 mark)
Two seconds after the first particle passed through \(O\), a second particle passes through \(O\).
Its distance \(x\) metres from \(O, t\) seconds after the first particle passed through \(O\), is given by \(x=\log _e\left(\tan ^{-1}(3 t-6)+1\right).\)
- Verify that the particles are the same distance from \(O\) when \(t=6\). (1 mark)
- Find the ratio of the speed of the first particle to the speed of the second particle when the particles are at the same distance from \(O\). Give your answer as \(\dfrac{a}{b}\) in simplest form, where \(a\) and \(b\) are positive integers. (2 marks)
a.i. \(\displaystyle\int e^x d x=\int \frac{2}{1+4 t^2}\,dt\)
a.ii. \(x=\log _e\left(\tan ^{-1}(2 t)+1\right) \)
b.i. \(\text{Asymptote: }\ x=\log _e\left(\dfrac{\pi}{2}+1\right)\)
b.ii.
c. \(0.02\)
d. \(\text {Particle positions at}\ \ t=6:\)
\(x_1=\log _e\left(\tan ^{-1}(2 \times 6)+1\right)=\log _e\left(\tan ^{-1}(12)+1\right)\)
\(x_2=\log _e\left(\tan ^{-1}(3 \times 6-6)+1\right)=\log _e\left(\tan ^{-1}(12)+1\right)=x_1\)
e. \(\dfrac{v_1}{v_2}=\dfrac{2}{3}\)
a.i. \(\dfrac{d x}{d t}=\dfrac{2 e^{-x}}{1+4 t^2}\)
\(\displaystyle\int e^x d x=\int \frac{2}{1+4 t^2}\,dt\)
a.ii. \(e^x=\tan ^{-1}(2 t)+c\)
\(\text {When}\ \ t=0, \ x=0 \ \Rightarrow \ c=1\)
\begin{aligned}
e^x& =\tan ^{-1}(2 t)+1 \\
x&=\log _e\left(\tan ^{-1}(2 t)+1\right)
\end{aligned}
b.i. \(\text {As}\ \ t \rightarrow \infty, \ \tan ^{-1}(2 t) \rightarrow \dfrac{\pi}{2}\)
\(x \rightarrow \log _e\left(\dfrac{\pi}{2}+1\right)\)
\(\therefore \text{Asymptote: }\ x=\log _e\left(\dfrac{\pi}{2}+1\right)\)
b.ii. \(\log _e\left(\dfrac{\pi}{2}+1\right) \approx 0.944\)
\(\text{When}\ \ t=10 :\)
\(x=\log _e\left( \tan ^{-1}(20)+1\right)=0.92\ \text{(2 d.p.)}\)
c. \(\text {At}\ \ t=3, x=\log _e\left(\tan ^{-1}(6)+1\right)\)
\(\text {Substitute } t \text { and } x \text { into } \dfrac{d x}{d t}:\)
\(\dfrac{d x}{d t}=\dfrac{2 e^{-x}}{1+4 t^2}=0.02\, \text{ms} ^{-1}\ \text{(2 d.p.)}\)
d. \(\text {Particle positions at}\ \ t=6:\)
\(x_1=\log _e\left(\tan ^{-1}(2 \times 6)+1\right)=\log _e\left(\tan ^{-1}(12)+1\right)\)
\(x_2=\log _e\left(\tan ^{-1}(3 \times 6-6)+1\right)=\log _e\left(\tan ^{-1}(12)+1\right)=x_1\)
e. \(e^x=e^{\log _e\left(\tan ^{-1}(12)+1\right)}=\tan ^{-1}(12)+1\)
\(\text {At}\ \ t=6:\)
\(\dfrac{d x_1}{d t}=\dfrac{2}{1+4 t^2} \times \dfrac{1}{\tan ^{-1}(2 t)+1}=\dfrac{2}{145\left(\tan ^{-1}(12)+1\right)}\)
\(\dfrac{d x^2}{d t}=\dfrac{3}{1+(3 t-6)^2} \times \dfrac{1}{\tan ^{-1}(3 t-6)-1}=\dfrac{3}{145\left(\tan ^{-1}(12)+1\right)}\)
\(\therefore \dfrac{v_1}{v_2}=\dfrac{2}{3}\)