A particle switches from a curve to its normal

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

(i) State the coordinates of the initial position of the article. mark

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

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 CC at the point t=2t=2, leaving your answer correct to 3 decimal places. marks

After TT seconds, where T>2T>2, the particle reaches point AA, which lies on the xx-axis, and stops moving.

(iv) Find the coordinates of the point AA. Hence, give a sketch of the path traced by the particle, indicating the coordinates of any axial intercepts. marks

(v) Find the total area bounded by the path of the particle in the first TT seconds and the positive xx-axis. marks

[2017/SRJC/Prelim/I/10]

Total marks: 13

Solution

If this part uses an earlier result, revisit (ii) Long-term behaviour or open all working together.

(iii)

Normals to parametric curves

At t=2t=2 the particle is struck and moves along the normal. Find an equation for the normal to CC at t=2t=2, correct to 3 decimal places.

Hint

  1. dydx=1+2e−2t1−2e−2t\frac{dy}{dx}=\frac{1+2e^{-2t}}{1-2e^{-2t}}.

  2. Normal gradient =−1−2e−2t1+2e−2t=-\frac{1-2e^{-2t}}{1+2e^{-2t}}.

  3. Use the point (2+e−4, 2−e−4)(2+e^{-4},\,2-e^{-4}).

Answer · (iii)

y=−0.929x+3.857y=-0.929x+3.857 (3 d.p.).