To prepare for the General Election, a political party studies the effectiveness of spreading information on Instantglam, a social media platform, by uploading a particular video. The number of Instantglam users who have viewed the video increases at a rate that is directly proportional to the number of Instantglam users who have yet to view the video. The number of Instantglam users who have viewed the video tt hours after the video is uploaded is xx. The total number of Instantglam users is PP, assumed to be fixed.

  1. (i)

    Write down a differential equation and show that x=P(1−e−kt)x = P(1 - \mathrm{e}^{-kt}), where kk is a positive constant.

  2. (ii)

    It is known that 12 hours after the video is uploaded, half the total number of Instantglam users have viewed the video.

    Estimate to the nearest hour, the time needed for 80% of the total number of Instantglam users to have viewed the video.

  3. (iii)

    Sketch the graph of xx against tt.

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Worked solution

1. 根据题意列出微分方程并分离变量积分

根据题意,观看视频的人数 xx 增加的速率 dxdt\dfrac{\mathrm{d}x}{\mathrm{d}t} 与尚未观看视频的人数 (P−x)(P - x) 成正比:
dxdt=k(P−x)(k>0)\dfrac{\mathrm{d}x}{\mathrm{d}t} = k(P - x) \quad (k > 0)
分离变量得:
∫1P−x dx=∫k dt\int \dfrac{1}{P - x} \,\mathrm{d}x = \int k \,\mathrm{d}t
−ln⁡(P−x)=kt+C(因 x<P,故 P−x>0)-\ln(P - x) = kt + C \quad (\text{因 } x < P \text{,故 } P - x > 0)
ln⁡(P−x)=−kt−C\ln(P - x) = -kt - C
P−x=Ae−kt(其中 A=e−C>0)P - x = A\mathrm{e}^{-kt} \quad (\text{其中 } A = \mathrm{e}^{-C} > 0)

2. 代入初始条件确定常数,推导目标关系式

在视频刚上传时(t=0t = 0),已观看人数 x=0x = 0。
代入上式:
P−0=Ae0  ⟹  A=PP - 0 = A \mathrm{e}^{0} \implies A = P
代演可得:
P−x=Pe−ktP - x = P \mathrm{e}^{-kt}
x=P−Pe−kt=P(1−e−kt)x = P - P \mathrm{e}^{-kt} = P(1 - \mathrm{e}^{-kt})(证毕)

3. 根据 $t = 12$ 时的观看人数占比求出常数 $k$

已知当 t=12t = 12 时,x=0.5Px = 0.5P:
0.5P=P(1−e−12k)0.5P = P(1 - \mathrm{e}^{-12k})
0.5=1−e−12k0.5 = 1 - \mathrm{e}^{-12k}
e−12k=0.5\mathrm{e}^{-12k} = 0.5
−12k=ln⁡(0.5)=−ln⁡2-12k = \ln(0.5) = -\ln 2
k=ln⁡212≈0.05776k = \dfrac{\ln 2}{12} \approx 0.05776

4. 计算 80% 用户观看视频所需的时间 $t$

当 80% 的用户观看视频时,x=0.8Px = 0.8P:
0.8P=P(1−e−kt)0.8P = P(1 - \mathrm{e}^{-kt})
0.8=1−e−kt0.8 = 1 - \mathrm{e}^{-kt}
e−kt=0.2\mathrm{e}^{-kt} = 0.2
−kt=ln⁡(0.2)=−ln⁡5-kt = \ln(0.2) = -\ln 5
t=ln⁡5k=ln⁡5ln⁡212=12ln⁡5ln⁡2≈27.86 小时t = \dfrac{\ln 5}{k} = \dfrac{\ln 5}{\frac{\ln 2}{12}} = \dfrac{12\ln 5}{\ln 2} \approx 27.86 \text{ 小时}
精确到最接近的小时,时间为 2828 小时。

5. 绘制 $x$ 关于 $t$ 的函数图像

分析函数 x=P(1−e−kt)x = P(1 - \mathrm{e}^{-kt}) 的图像特征:

  1. 当 t=0t = 0 时,x=0x = 0,图像过原点 (0,0)(0, 0);
  2. 当 t>0t > 0 时,dxdt=kPe−kt>0\dfrac{\mathrm{d}x}{\mathrm{d}t} = kP\mathrm{e}^{-kt} > 0,函数单调递增;
  3. d2xdt2=−k2Pe−kt<0\dfrac{\mathrm{d}^2 x}{\mathrm{d}t^2} = -k^2 P\mathrm{e}^{-kt} < 0,曲线凹向下方(增长速率逐渐减慢);
  4. 当 t→∞t \to \infty 时,x→Px \to P,存在水平渐近线 x=Px = P;
  5. 图像经过点 (12,0.5P)(12, 0.5P)。

绘图要点:在第一象限画一条从原点 (0,0)(0,0) 开始上凸递增、并无限趋近于虚线渐近线 x=Px = P 的平滑曲线,标出横轴 tt、纵轴 xx 以及渐近线 x=Px = P。

Answer
  1. 微分方程为 dxdt=k(P−x)\dfrac{\mathrm{d}x}{\mathrm{d}t} = k(P - x),证明过程见解题步骤。
  2. 达到 80% 观看率所需时间约为 2828 小时。
  3. 图象从原点 (0,0)(0,0) 出发,单调递增且凹向下方,当 t→∞t \to \infty 时无限趋近于水平渐近线 x=Px = P,并经过点 (12,0.5P)(12, 0.5P)。

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