An exponential height law makes the rod never land

Slipping rod XYO; X; Y; x; y; 6. X is left of O, Y above O; right-angle square and x,y,6 labels retained.Slipping rod XYOXYxy6
Question diagram 1
Slipping rod XYO; X; Y; x; y; 6. X is left of O, Y above O; right-angle square and x,y,6 labels retained.Slipping rod XYOXYxy6

Given 6 metre rod XY: X lies on the horizontal surface to the left of O, Y on the vertical surface above O, OX=x and OY=y, with a right angle at O.

Jesse leans a 6 metre rod XY in the position as shown below. At time tt seconds, the two ends X and Y are xx and yy metres from O respectively.

(i) State an equation relating xx and yy. mark

As the rod slips, Jesse proposes that yy decreases at a rate proportional to yy.

(ii) Based on Jesse’s proposal, show that dxdt=k(36−x2)x\frac{dx}{dt}=\frac{k(36-x^2)}x, where kk is a constant. marks

Initially, X is 4 metres from O. Take k=2k=2.

(iii) Find an expression for xx in terms of tt and find the time taken for Y to be 3 metres from O from the instance Jesse lets go of the rod, giving your answer to 1 decimal place. marks

(iv) Sketch the graph of xx against tt. Hence, comment on whether Jesse’s proposed model is appropriate. Justify your answer. marks

[2012/HCI/I/8]

Total marks: 12

Solution

(i)

Pythagoras

State an equation relating xx and yy for a 6 m rod.

Hint

  1. Right-angled triangle with hypotenuse 6.

Answer · (i)

x2+y2=36x^2+y^2=36.

(ii)

Related rates

If dydt=−ky\frac{dy}{dt}=-ky, show dxdt=k(36−x2)x\dfrac{dx}{dt}=\dfrac{k(36-x^2)}x.

Hint

  1. Differentiate x2+y2=36x^2+y^2=36 with respect to tt.

Answer · (ii)

x′=k(36−x2)/xx\prime=k(36-x^2)/x.

(iii)

Separation of variables

With k=2k=2 and x=4x=4 initially, find xx in terms of tt and the time for yy to reach 3 m.

Explore this idea

Hint

  1. Separate: x36−x2dx=2 dt\frac x{36-x^2}dx=2\,dt.

  2. y=3y=3 means x2=27x^2=27.

Answer · (iii)

x=36−20e−4tx=\sqrt{36-20e^{-4t}}; at y=3y=3, t=14ln⁡(20/9)≈0.2t=\frac14\ln(20/9)\approx0.2 s.

Explore the idea

The foot of the rod

Each dash shows the slope the differential equation gives at that point. Change the starting value and the solution follows the dashes.

Starting value of x at t = 04
Read the solution at t0.25
Try an example
x at t = 0.25
5.352
Rate dx/dt there
2.751
  • Solution that starts at the purple dot
  • Closed-form answer
Key idea

The curve rises towards 6 but never reaches it.

Ideas to try
  1. Slide the starting value close to 6 and see how the curve approaches it.
  2. Read xx at t=0.25t=0.25 s.
(iv)

Evaluating the model

Sketch xx against tt and comment on whether the model is appropriate.

Hint

  1. As t→∞t\to\infty, e−4t→0e^{-4t}\to0.

Position of the lower rod end
Position of the lower rod endScroll across to see the full diagram.

Answer · (iv)

Rises concavely from 44 toward asymptote 66; inappropriate for a rod expected to fall flat in finite time.