题目
Problem
A scientist is studying a population of fish in a lake. The number of fish, N N N t t t
N = 600 e 0.3 t 2 + e 0.3 t t ≥ 0 N=\frac{600e^{0.3t}}{2+e^{0.3t}}\qquad t\ge 0 N = 2 + e 0.3 t 600 e 0.3 t t ≥ 0 Use the equation of the model to answer parts (a), (b), (c), (d) and (e).
(a) Find the number of fish in the lake at the start of the study.
(1)
(b) Find the upper limit to the number of fish in the lake.
(1)
(c) Find the time, after the start of the study, when there are predicted to be 500 fish in the lake. Give your answer in years and months to the nearest month.
(4)
(d) Show that
d N d t = A e 0.3 t ( 2 + e 0.3 t ) 2 \frac{\mathrm{d}N}{\mathrm{d}t}
=\frac{Ae^{0.3t}}{\big(2+e^{0.3t}\big)^2} d t d N = ( 2 + e 0.3 t ) 2 A e 0.3 t where A A A
(3)
Given that when t = T t=T t = T d N d t = 8 \dfrac{\mathrm{d}N}{\mathrm{d}t}=8 d t d N = 8
(e) find the value of T T T
(Solutions relying entirely on calculator technology are not acceptable.)
(4)
题目中文翻译
一位科学家正在研究湖中鱼类种群。研究开始后 t t t N N N
N = 600 e 0.3 t 2 + e 0.3 t t ≥ 0 N=\frac{600e^{0.3t}}{2+e^{0.3t}}\qquad t\ge 0 N = 2 + e 0.3 t 600 e 0.3 t t ≥ 0 利用该模型方程回答(a)、(b)、(c)、(d) 和 (e)。
(a) 求研究开始时湖中的鱼数。
(b) 求湖中鱼数的上限。
(c) 求研究开始后湖中鱼数预计达到 500 条的时间。答案用“几年几个月”表示，并取最近的整月。
(d) 证明
d N d t = A e 0.3 t ( 2 + e 0.3 t ) 2 \frac{\mathrm{d}N}{\mathrm{d}t}
=\frac{Ae^{0.3t}}{\big(2+e^{0.3t}\big)^2} d t d N = ( 2 + e 0.3 t ) 2 A e 0.3 t 其中 A A A
已知当 t = T t=T t = T d N d t = 8 \dfrac{\mathrm{d}N}{\mathrm{d}t}=8 d t d N = 8
(e) 求 T T T
（不接受完全依赖计算器技术的解法。）
解答
(a)
解法一
思路
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研究开始时 t = 0 t=0 t = 0 e 0 = 1 e^0=1 e 0 = 1
答题过程
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At t = 0 t=0 t = 0
N = 600 e 0 2 + e 0 = 600 3 = 200. N=\frac{600e^0}{2+e^0}
=\frac{600}{3}
=200. N = 2 + e 0 600 e 0 = 3 600 = 200. Therefore the number of fish at the start of the study is
200 . \boxed{200}. 200 .
(b)
解法一
思路
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把分子、分母同除以 e 0.3 t e^{0.3t} e 0.3 t t t t e − 0.3 t → 0 e^{-0.3t}\to0 e − 0.3 t → 0
答题过程
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Rewrite the model as
N = 600 2 e − 0.3 t + 1 . N=\frac{600}{2e^{-0.3t}+1}. N = 2 e − 0.3 t + 1 600 . As t → ∞ t\to\infty t → ∞ e − 0.3 t → 0 e^{-0.3t}\to0 e − 0.3 t → 0
N → 600 1 = 600. N\to\frac{600}{1}=600. N → 1 600 = 600. Therefore the upper limit is
600 . \boxed{600}. 600 .
(c)
解法一
思路
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令 N = 500 N=500 N = 500 e 0.3 t e^{0.3t} e 0.3 t t t t 12 12 12
答题过程
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Set N = 500 N=500 N = 500
500 = 600 e 0.3 t 2 + e 0.3 t . 500=\frac{600e^{0.3t}}{2+e^{0.3t}}. 500 = 2 + e 0.3 t 600 e 0.3 t . Then
500 ( 2 + e 0.3 t ) = 600 e 0.3 t , e 0.3 t = 10. \begin{align*}
500\left(2+e^{0.3t}\right)
=&\,600e^{0.3t},\\[2mm]
e^{0.3t}=&\,10.
\end{align*} 500 ( 2 + e 0.3 t ) = e 0.3 t = 600 e 0.3 t , 10. Therefore
t = ln 10 0.3 = 7.675 … t=\frac{\ln10}{0.3}
=7.675\ldots t = 0.3 ln 10 = 7.675 … The fractional part corresponds to
0.675 … × 12 = 8.10 … 0.675\ldots\times12=8.10\ldots 0.675 … × 12 = 8.10 … months. Hence, to the nearest month, the time is
7 years 8 months . \boxed{7\text{ years }8\text{ months}}. 7 years 8 months .
(d)
解法一
思路
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把分子与分母分别设为函数并使用商法则。展开分子后，两个含 e 0.6 t e^{0.6t} e 0.6 t
答题过程
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Let
u = 600 e 0.3 t and v = 2 + e 0.3 t . u=600e^{0.3t}
\quad\text{and}\quad
v=2+e^{0.3t}. u = 600 e 0.3 t and v = 2 + e 0.3 t . Then
u ′ = 180 e 0.3 t and v ′ = 0.3 e 0.3 t . u'=180e^{0.3t}
\quad\text{and}\quad
v'=0.3e^{0.3t}. u ′ = 180 e 0.3 t and v ′ = 0.3 e 0.3 t . By the quotient rule,
d N d t = v u ′ − u v ′ v 2 . \frac{\mathrm{d}N}{\mathrm{d}t}
=\frac{vu'-uv'}{v^2}. d t d N = v 2 v u ′ − u v ′ . Thus
d N d t = ( 2 + e 0.3 t ) ( 180 e 0.3 t ) − ( 600 e 0.3 t ) ( 0.3 e 0.3 t ) ( 2 + e 0.3 t ) 2 . \frac{\mathrm{d}N}{\mathrm{d}t}
=
\frac{
\left(2+e^{0.3t}\right)\left(180e^{0.3t}\right)
-\left(600e^{0.3t}\right)\left(0.3e^{0.3t}\right)
}{
\left(2+e^{0.3t}\right)^2
}. d t d N = ( 2 + e 0.3 t ) 2 ( 2 + e 0.3 t ) ( 180 e 0.3 t ) − ( 600 e 0.3 t ) ( 0.3 e 0.3 t ) . The numerator simplifies as follows:
( 2 + e 0.3 t ) ( 180 e 0.3 t ) − ( 600 e 0.3 t ) ( 0.3 e 0.3 t ) = 360 e 0.3 t + 180 e 0.6 t − 180 e 0.6 t = 360 e 0.3 t . \begin{align*}
&\left(2+e^{0.3t}\right)\left(180e^{0.3t}\right)
-\left(600e^{0.3t}\right)\left(0.3e^{0.3t}\right)\\[2mm]
=&\,360e^{0.3t}
+180e^{0.6t}
-180e^{0.6t}\\[2mm]
=&\,360e^{0.3t}.
\end{align*} = = ( 2 + e 0.3 t ) ( 180 e 0.3 t ) − ( 600 e 0.3 t ) ( 0.3 e 0.3 t ) 360 e 0.3 t + 180 e 0.6 t − 180 e 0.6 t 360 e 0.3 t . Hence
d N d t = 360 e 0.3 t ( 2 + e 0.3 t ) 2 , \frac{\mathrm{d}N}{\mathrm{d}t}
=\frac{360e^{0.3t}}{\left(2+e^{0.3t}\right)^2}, d t d N = ( 2 + e 0.3 t ) 2 360 e 0.3 t , so
A = 360 . \boxed{A=360}. A = 360 .
解法二
思路
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官方评分资料可先把模型拆成常数减去一个简单分式，再使用链式法则。这样能直接避开商法则中的抵消步骤。
答题过程
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Rewrite the model:
N = 600 ( 1 − 2 2 + e 0.3 t ) = 600 − 1200 2 + e 0.3 t . \begin{align*}
N
=&\,600\left(
1-\frac{2}{2+e^{0.3t}}
\right)\\[2mm]
=&\,600-\frac{1200}{2+e^{0.3t}}.
\end{align*} N = = 600 ( 1 − 2 + e 0.3 t 2 ) 600 − 2 + e 0.3 t 1200 . Differentiating,
d N d t = − 1200 ( − 1 ) ( 2 + e 0.3 t ) − 2 ( 0.3 e 0.3 t ) = 360 e 0.3 t ( 2 + e 0.3 t ) 2 . \begin{align*}
\frac{\mathrm{d}N}{\mathrm{d}t}
=&\,-1200\left(-1\right)
\left(2+e^{0.3t}\right)^{-2}
\left(0.3e^{0.3t}\right)\\[2mm]
=&\,\frac{360e^{0.3t}}
{\left(2+e^{0.3t}\right)^2}.
\end{align*} d t d N = = − 1200 ( − 1 ) ( 2 + e 0.3 t ) − 2 ( 0.3 e 0.3 t ) ( 2 + e 0.3 t ) 2 360 e 0.3 t . Therefore
A = 360 . \boxed{A=360}. A = 360 .
解法三
思路
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官方评分资料还接受把分母写成负指数幂，对两个因子的乘积使用乘积法则，再合并同分母。
答题过程
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Write
N = 600 e 0.3 t ( 2 + e 0.3 t ) − 1 . N=600e^{0.3t}\left(2+e^{0.3t}\right)^{-1}. N = 600 e 0.3 t ( 2 + e 0.3 t ) − 1 . Using the product and chain rules,
d N d t = 180 e 0.3 t ( 2 + e 0.3 t ) − 1 − 180 e 0.6 t ( 2 + e 0.3 t ) − 2 = 180 e 0.3 t ( 2 + e 0.3 t ) − 180 e 0.6 t ( 2 + e 0.3 t ) 2 = 360 e 0.3 t ( 2 + e 0.3 t ) 2 . \begin{align*}
\frac{\mathrm{d}N}{\mathrm{d}t}
=&\,180e^{0.3t}
\left(2+e^{0.3t}\right)^{-1}\\
&-180e^{0.6t}
\left(2+e^{0.3t}\right)^{-2}\\[2mm]
=&\,\frac{
180e^{0.3t}\left(2+e^{0.3t}\right)
-180e^{0.6t}
}{
\left(2+e^{0.3t}\right)^2
}\\[2mm]
=&\,\frac{360e^{0.3t}}
{\left(2+e^{0.3t}\right)^2}.
\end{align*} d t d N = = = 180 e 0.3 t ( 2 + e 0.3 t ) − 1 − 180 e 0.6 t ( 2 + e 0.3 t ) − 2 ( 2 + e 0.3 t ) 2 180 e 0.3 t ( 2 + e 0.3 t ) − 180 e 0.6 t ( 2 + e 0.3 t ) 2 360 e 0.3 t . Hence
A = 360 . \boxed{A=360}. A = 360 .
(e)
解法一
思路
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代入 d N d t = 8 \frac{\mathrm{d}N}{\mathrm{d}t}=8 d t d N = 8 u = e 0.3 T u=e^{0.3T} u = e 0.3 T T ≥ 0 T\geq0 T ≥ 0 u ≥ 1 u\geq1 u ≥ 1
答题过程
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Using part (d),
8 = 360 e 0.3 T ( 2 + e 0.3 T ) 2 . 8=\frac{360e^{0.3T}}
{\left(2+e^{0.3T}\right)^2}. 8 = ( 2 + e 0.3 T ) 2 360 e 0.3 T . Let u = e 0.3 T u=e^{0.3T} u = e 0.3 T
8 = 360 u ( 2 + u ) 2 , 8=\frac{360u}{(2+u)^2}, 8 = ( 2 + u ) 2 360 u , so
8 ( 2 + u ) 2 = 360 u , u 2 − 41 u + 4 = 0. \begin{align*}
8(2+u)^2=&\,360u,\\[2mm]
u^2-41u+4=&\,0.
\end{align*} 8 ( 2 + u ) 2 = u 2 − 41 u + 4 = 360 u , 0. Therefore
u = 41 ± 1665 2 . u=\frac{41\pm\sqrt{1665}}{2}. u = 2 41 ± 1665 . Since T ≥ 0 T\geq0 T ≥ 0 u = e 0.3 T ≥ 1 u=e^{0.3T}\geq1 u = e 0.3 T ≥ 1
u = 41 + 1665 2 . u=\frac{41+\sqrt{1665}}{2}. u = 2 41 + 1665 . Thus
T = 1 0.3 ln ( 41 + 1665 2 ) = 12.3706 … T=\frac{1}{0.3}
\ln\left(
\frac{41+\sqrt{1665}}{2}
\right)
=12.3706\ldots T = 0.3 1 ln ( 2 41 + 1665 ) = 12.3706 … Therefore, to one decimal place,
T = 12.4 . \boxed{T=12.4}. T = 12.4 . 