A particle switches from a curve to its normal · source reference

Printed question

A particle is moving along a curve, $C$, such that its position at time $t$ seconds after it is set into motion is given by the parametric equations $x=t+e^{-2t}$, $y=t-e^{-2t}$.

(i) State the coordinates of the initial position of the article. [1]

(ii) Explain what would happen to the path of the particle after a long time. [1]

At the time of 2 seconds after the particle was set into motion, an external force struck the particle resulting in the particle moving in a straight line along the normal to the path at the point of collision.

(iii) Find an equation for the normal to the curve $C$ at the point $t=2$, leaving your answer correct to 3 decimal places. [3]

After $T$ seconds, where $T>2$, the particle reaches point $A$, which lies on the $x$-axis, and stops moving.

(iv) Find the coordinates of the point $A$. Hence, give a sketch of the path traced by the particle, indicating the coordinates of any axial intercepts. [4]

(v) Find the total area bounded by the path of the particle in the first $T$ seconds and the positive $x$-axis. [4]

[2017/SRJC/Prelim/I/10]

Printed working

(i) Original position $t=0$: $x=0+e^0=1$, $y=0-e^0=-1$, so $(1,-1)$.

(ii) As $t\to\infty$, $e^{-2t}\to0$, so the source writes $x\to t$, $y\to t$: the unforced path approaches $y=x$ because $y-x=-2e^{-2t}\to0$.

(iii) $dx/dt=1-2e^{-2t}$, $dy/dt=1+2e^{-2t}$, so $dy/dx=(1+2e^{-2t})/(1-2e^{-2t})$.

At $t=2$, $x=2+e^{-4}=2.01832$, $y=2-e^{-4}=1.98168$; normal gradient $=(2e^{-4}-1)/(1+2e^{-4})=-0.92933$.

$$y-1.98168=-0.92933(x-2.01832)\Rightarrow y=-0.92933x+3.85737\Rightarrow y=-0.929x+3.857$$

(to 3 d.p.).

(iv) At $A$, $y=0$: $0=-0.929x+3.857\Rightarrow x=4.15178$, so $A=(4.15,0)$. The source sketch shows initial $(1,-1)$, collision $(2.02,1.98)$, normal segment to $A(4.15,0)$, and curved portion crossing the $x$-axis.

(v) On $C$, $y=0\Rightarrow t-e^{-2t}=0\Rightarrow t=e^{-2t}$. GC gives $t=0.4263$, $x=0.85261$.

$$\begin{aligned}\text{Required area}&=\int_{0.852}^{2.02}y\,dx+\int_{2.02}^{4.15}(-0.929x+3.857)\,dx\\ &=\int_{0.4263}^{2}(t-e^{-2t})(1-2e^{-2t})\,dt+\int_{2.02}^{4.15}(-0.929x+3.857)\,dx\\ &=3.56\text{ units}^2\quad(3\text{ s.f.}).\end{aligned}$$

This is the closed region above the positive axis after the curved path crosses it. The initial arc below the axis alone does not enclose a region with that axis. Numerical work should keep unrounded collision/normal values internally.

Teacher notes

“After a long time” means $t\to\infty$.

Gradient of normal is $-1/\text{gradient of tangent}=-1/(dy/dx)$.

Curve $C$ has $x=t+e^{-2t}$, $y=t-e^{-2t}$. Note: (1) $dx$ changes to $(dx/dt)\,dt$; (2) remember to change the $x$ limits to $t$ limits.

Answer key

(i) $(1,-1)$. (ii) Approaches $y=x$ (unforced extrapolation). (iii) $y=-0.929x+3.857$. (iv) $(4.15,0)$ and the native path sketch. (v) $3.56$ units$^2$.

Native diagrams