An exponential height law makes the rod never land · source reference

Printed question

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

(i) State an equation relating $x$ and $y$. [1]

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

(ii) Based on Jesse’s proposal, show that $\frac{dx}{dt}=\frac{k(36-x^2)}x$, where $k$ is a constant. [3]

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

(iii) Find an expression for $x$ in terms of $t$ 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. [5]

(iv) Sketch the graph of $x$ against $t$. Hence, comment on whether Jesse’s proposed model is appropriate. Justify your answer. [3]

[2012/HCI/I/8]

Printed working

$x^2+y^2=36$.

$dy/dt=-ky$, $k>0$. Differentiate $x^2+y^2=36$: $2x\,dx/dt+2y\,dy/dt=0$.

$dx/dt=-(y/x)dy/dt=-(y/x)(-ky)=ky^2/x=k(36-x^2)/x$.

With $k=2$, $\int x/(36-x^2)dx=\int2dt$.

$-\frac12\int(-2x)/(36-x^2)dx=\int2dt$; $-\frac12\ln(36-x^2)=2t+C$ since $36-x^2>0$.

$\ln(36-x^2)=-4t+C'$; $36-x^2=Ae^{-4t}$, $A=e^{C'}$; $x^2=36-Ae^{-4t}$; $x=\sqrt{36-Ae^{-4t}}$ since $x>0$.

At 0, $x=4$: $4=\sqrt{36-A}\Rightarrow A=20$. Thus $x=\sqrt{36-20e^{-4t}}$.

For $y=3$, $x=\sqrt{36-9}=\sqrt{27}$; $27=36-20e^{-4t}\Rightarrow e^{-4t}=9/20$; $t=-\frac14\ln(9/20)=0.2$ s (1 d.p.).

The source sketch starts at $x=4$, approaches asymptote $x=6$ for $t\ge0$. Based on Jesse’s model, $x$ will never be 6: the rod would never fall flat. Thus the model is not appropriate.

Printed answer key

(i) $x^2+y^2=36$. (iii) $x=\sqrt{36-Ae^{-4t}}$; 0.2 s. (iv) No; native sketch below.

Teacher note 1

Since $y$ decreases as $t$ increases, $dy/dt<0$. Since $y>0$, the constant of proportionality is negative for $dy/dt<0$.

Teacher note 2

Given a DE of form $dx/dt=f(x)$, rearrange before integrating with respect to $t$: $\frac1{f(x)}\frac{dx}{dt}=1$, hence $\int\frac1{f(x)}dx=\int1dt$.

Teacher note 3

Note $t\ge0$; do not sketch any curve for $t<0$.

Diagram 10-10-rod

X is left of O, Y above O; right-angle square and x,y,6 labels retained.

Diagram 10-10-solution

Mathematical review

(i) $x^2+y^2=36$. (ii) Differentiation gives $xx'+yy'=0$, hence $x'=ky^2/x=k(36-x^2)/x$.

(iii) With k=2, $\int x/(36-x^2)\,dx=2t+C$, so $36-x^2=Ae^{-4t}$. Initial x=4 gives A=20. Thus $x=\sqrt{36-20e^{-4t}}$ for t≥0, with $y=\sqrt{20}e^{-2t}$. At y=3, $e^{-4t}=9/20$, so $t=\tfrac14\ln(20/9)\approx0.1996$, or 0.2 s.

(iv) x starts at 4, increases concavely toward the asymptote x=6. Since y never reaches zero at finite t, this model predicts the rod never lies flat, a limitation for an actual slipping rod.

Provenance

TJC JC2 2025 H2 Maths (9758) MSM. Native transcription checked against original visible question, working and sidebars. Question page(s): 3. Working page(s): 13. Original records remain private for QA; this native article has no PDF or raster worksheet dependency.