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

(i)

Parametric curves

x=t+e−2tx=t+e^{-2t} and y=t−e−2ty=t-e^{-2t}. State the initial position of the particle.

Hint

  1. The motion starts at t=0t=0.

Answer · (i)

(1,−1)(1,-1).

(ii)

Long-term behaviour

Explain what would happen to the path of the particle after a long time (with no impact).

Explore this idea

Hint

  1. Let t→∞t\to\infty and see which terms survive.

Answer · (ii)

Original parametric path approaches y=xy=x as t→∞t\to\infty.

Explore the idea

The path hugs y = x as time goes on

Slide towards the right end: the tangent becomes almost parallel to y = x (gradient near 1).

Position along the curve(1, -1)t = 0
Try an example
Tangent gradient
-3

Start (1,−1)(1,-1): the gradient is 1+21−2=−3\frac{1+2}{1-2}=-3, and the particle begins by heading up and to the left before the curve turns.

For finite nonzero gradients, tangent gradient × normal gradient = −1. Both axes use equal scales.

  • Path
  • Y = x
  • Tangent
Key idea

For large tt, dydx→1\frac{dy}{dx}\to1 and the curve is asymptotic to y=xy=x.

Ideas to try
Slide to the far right: what is the gradient?
(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.).

(iv)

Axial intercepts of a composite path

After TT seconds (T>2T>2) the particle reaches the point AA on the xx-axis. Find AA, and sketch the path, indicating the coordinates of any axial intercepts.

Explore this idea

Hint

  1. AA is where the normal meets y=0y=0.

  2. The curve itself crosses the xx-axis where t=e−2tt=e^{-2t}.

  3. Sketch the curve for 0≤t≤20\le t\le2 followed by the straight segment.

Particle path and enclosed region
Particle path and enclosed regionScroll across to see the full diagram.

Answer · (iv)

A=(4.15070373…,0)A=(4.15070373\ldots,0); initial point (1,−1)(1,-1) and curve x-intercept (0.852606…,0)(0.852606\ldots,0).

No y-axis intersection; follows curve until t=2, then its normal to A.

Explore the idea

Curve until t = 2, then the normal to A

The fixed purple line is the normal. Jump to the key points.

Position along the curve(1, -1)t = 0
Try an example
Tangent gradient
-3

The tangent follows the curve at the selected point. Its normal is perpendicular to it.

For finite nonzero gradients, tangent gradient × normal gradient = −1. Both axes use equal scales.

  • Path for 0 ≤ t ≤ 2
  • Normal from the collision to A
  • Tangent at the moving point
Key idea

The path has two pieces: a curve up to the collision, then a straight line to AA.

Ideas to try
Find where the curved part crosses the xx-axis.
(v)

Area under a parametric path

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

Explore this idea

Hint

  1. The enclosed region begins where the curve crosses the xx-axis and ends at AA.

  2. Curved part: ∫y dxdt dt\int y\,\frac{dx}{dt}\,dt from t=0.4263t=0.4263 to 22.

  3. Straight part: ∫(−0.929x+3.857) dx\int(-0.929x+3.857)\,dx from x=2.02x=2.02 to 4.154.15.

Answer · (v)

3.563.56 square units (3 s.f.).

The enclosed region begins at the curve’s x-axis crossing, not the free launch endpoint.

Explore the idea

The area under the straight part is a slice of a trapezium

The normal from the collision point (2.018, 1.982) to A. The strips add up to the exact area.

Number of stripsn = 4
Reset the controls

Where does one strip’s area come from?

1 of 4

Click a strip in the graph to inspect it. The darker rectangle is this strip. Its red marks use the left edge at x* = 2.0183.

upperlowerwidth Δx
Zoomed-in rectangle: width and height are scaled independently. The sample sets its upper and lower values across the whole strip.
Sample x*
2.0183
Width Δx
0.5331
Upper value
1.9817
Lower value
0
Height: upper − lower
1.9817
Strip area
1.0564

0.5331 ×(1.9817 − 0)≈ 1.0564

Add this contribution to those of the other 3 strips to get the sum below. Displayed values are rounded.

Strip width
0.533
2.132 ÷ 4
Sum of strips
2.64108
Exact, ≈ 2.113
2.11286
Error
+0.52822
25.00% too big

How the sum improves as n grows

  • Normal
  • Area under it
Key idea

A straight line gives an exact trapezium: ≈2.113\approx2.113. Together with 1.4471.447 from the curve, the total is 3.563.56.

Ideas to try
Increase the number of strips: does the sum change much?