Standard +0.3 This is a straightforward separable variables question with clear setup: translate the word problem into dx/dt = -k/√x, use given conditions to find k, separate and integrate (standard √x integration), then solve for the required time. While it requires multiple steps and careful handling of initial conditions, it follows a completely standard template with no conceptual surprises, making it slightly easier than average.
12 A patch of disease on a leaf is being chemically treated. At time \(t\) days after treatment starts, its length is \(x \mathrm {~cm}\) and the rate of decrease of its length is observed to be inversely proportional to the square root of its length. At time \(t = 3 , x = 4\) and, at this instant, the length is decreasing at 0.05 cm per day.
Write down and solve a differential equation to model this situation. Hence find the time it takes for the length to decrease to 0.01 cm . [0pt]
[10]
Substitute \(x = 4\) and \(\frac{\mathrm{d}x}{\mathrm{d}t} = \pm 0.05\) to find \(k\) M1\*
Obtain \(k = 0.1\) A1
Substitute \(t = 3\) and \(x = 4\) to find \(C\) (dependent on a value for \(k\) obtained from using \(x = 4\) and \(\frac{\mathrm{d}x}{\mathrm{d}t} = \pm 0.05\)) depM1
Obtain \(C = 5.63(3333...)\) or \(\frac{169}{30}\) or \(\frac{169}{3}\) from \(\frac{20}{3}x^{\frac{3}{2}} = -t + C\) or \(-\frac{169}{30}\) if \(c\) is placed on LHS A1
Substitute \(x = 0.01\) into their solution provided of form \(px^{\frac{3}{2}} = \pm mt + C\) to find \(t\) M1
Obtain \(t = 56.3\) or 56 days A1
[10]
SR if \(\frac{\mathrm{d}x}{\mathrm{d}t} = k\sqrt{x}\) award a maximum of B1 M3
SR if \(-\frac{k}{\sqrt{x}}\) stated then \(k = -0.1\) leads to final correct answer deduct A1 for \(k\) and A1 for the final answer \(= 8/10\)
State $\frac{\mathrm{d}x}{\mathrm{d}t}$ **B1**
State $-\frac{k}{\sqrt{x}}$ (award B1 for $\frac{k}{\sqrt{x}}$ if $k = -0.1$) **B1**
Separate variables and integrate both sides, raising the powers by 1 **M1**
Obtain $\frac{2}{3}x^{\frac{3}{2}} = -kt + C$ **A1**
Substitute $x = 4$ and $\frac{\mathrm{d}x}{\mathrm{d}t} = \pm 0.05$ to find $k$ **M1\***
Obtain $k = 0.1$ **A1**
Substitute $t = 3$ and $x = 4$ to find $C$ (dependent on a value for $k$ obtained from using $x = 4$ and $\frac{\mathrm{d}x}{\mathrm{d}t} = \pm 0.05$) **depM1**
Obtain $C = 5.63(3333...)$ or $\frac{169}{30}$ or $\frac{169}{3}$ from $\frac{20}{3}x^{\frac{3}{2}} = -t + C$ or $-\frac{169}{30}$ if $c$ is placed on LHS **A1**
Substitute $x = 0.01$ into their solution provided of form $px^{\frac{3}{2}} = \pm mt + C$ to find $t$ **M1**
Obtain $t = 56.3$ or 56 days **A1**
**[10]**
**SR** if $\frac{\mathrm{d}x}{\mathrm{d}t} = k\sqrt{x}$ award a maximum of B1 M3
**SR** if $-\frac{k}{\sqrt{x}}$ stated then $k = -0.1$ leads to final correct answer deduct A1 for $k$ and A1 for the final answer $= 8/10$
12 A patch of disease on a leaf is being chemically treated. At time $t$ days after treatment starts, its length is $x \mathrm {~cm}$ and the rate of decrease of its length is observed to be inversely proportional to the square root of its length. At time $t = 3 , x = 4$ and, at this instant, the length is decreasing at 0.05 cm per day.
Write down and solve a differential equation to model this situation. Hence find the time it takes for the length to decrease to 0.01 cm .\\[0pt]
[10]
\hfill \mbox{\textit{Pre-U Pre-U 9794/1 2016 Q12 [10]}}