CalcGospel 國際數學圖譜
返回

IAL 2023 May Q7

A Level / Edexcel / P2

IAL 2023 May Paper · Question 7

题目

Problem

The height of a river above a fixed point on the riverbed was monitored over a 7-day period.

The height of the river, HH metres, tt days after monitoring began, was given by

H=120(20+6t−t2)t+170⩽t⩽7H = \frac{1}{20}(20 + 6t - t^2)\sqrt{t} + 17 \qquad 0 \leqslant t \leqslant 7

Given that HH has a stationary value at t=αt = \alpha,

(a) use calculus to show that α\alpha satisfies the equation

5α2−18α−20=05\alpha^2 - 18\alpha - 20 = 0

(b) Hence find the value of α\alpha, giving your answer to 3 decimal places.

(c) Use further calculus to prove that HH is a maximum at this value of α\alpha.

(8)
题目中文翻译

某河流在固定点处的水深在7天内被监测。

水深 HH（米）与监测开始后 tt 天的关系为 H=120(20+6t−t2)t+17H=\dfrac{1}{20}(20+6t-t^2)\sqrt{t}+17，0⩽t⩽70\leqslant t\leqslant 7。

已知 HH 在 t=αt=\alpha 处有驻值，

(a) 用微积分证明 α\alpha 满足 5α2−18α−20=05\alpha^2-18\alpha-20=0

(b) 求 α\alpha 的值（保留3位小数）

(c) 用进一步的微积分证明 HH 在该 α\alpha 处取极大值。

（8分）

解答

解法一

思路

展开 HH 中含 t\sqrt{t} 的项，逐项求导，令 dHdt=0\frac{\mathrm{d}H}{\mathrm{d}t}=0，两边乘以适当的 tt 的幂次消去分数指数，整理得二次方程。二阶导数小于0则为极大值。

答题过程

(a) The height is H=120(20+6t−t2)t+17H = \dfrac{1}{20}(20 + 6t - t^2)\sqrt{t} + 17, i.e.

H=120(20t1/2+6t3/2−t5/2)+17H = \frac{1}{20}\left(20t^{1/2} + 6t^{3/2} - t^{5/2}\right) + 17

dHdt=120(10t−1/2+9t1/2−52t3/2)\frac{\mathrm{d}H}{\mathrm{d}t} = \frac{1}{20}\left(10t^{-1/2} + 9t^{1/2} - \frac{5}{2}t^{3/2}\right)

At the stationary value t=αt = \alpha, dHdt=0\dfrac{\mathrm{d}H}{\mathrm{d}t} = 0:

10α−1/2+9α1/2−52α3/2=010\alpha^{-1/2} + 9\alpha^{1/2} - \frac{5}{2}\alpha^{3/2} = 0

Multiply through by 2α1/2\dfrac{2}{\alpha^{1/2}}:

20α+18−5α=0\frac{20}{\alpha} + 18 - 5\alpha = 0

Multiply by α\alpha:

20 + 18\alpha - 5\alpha^2 =&\, 0 \\[4mm] 5\alpha^2 - 18\alpha - 20 =&\, 0 \end{align*}$$ $$\boxed{5\alpha^2 - 18\alpha - 20 = 0}$$ **(b)** Using the quadratic formula: $$\begin{align*} \alpha =&\, \frac{18 \pm \sqrt{(-18)^2 - 4(5)(-20)}}{2(5)} \\[4mm] =&\, \frac{18 \pm \sqrt{324 + 400}}{10} \\[4mm] =&\, \frac{18 \pm \sqrt{724}}{10} \\[4mm] =&\, \frac{18 \pm 2\sqrt{181}}{10} \\[4mm] =&\, \frac{9 \pm \sqrt{181}}{5} \end{align*}$$ Since $0 \leqslant \alpha \leqslant 7$, take the positive root: $$\alpha = \frac{9 + \sqrt{181}}{5} \approx \boxed{4.491}$$ **(c)** The second derivative: $$\frac{\mathrm{d}^2H}{\mathrm{d}t^2} = \frac{1}{20}\left(-5t^{-3/2} + \frac{9}{2}t^{-1/2} - \frac{15}{4}t^{1/2}\right)$$ At $t = \alpha \approx 4.491$: $$\frac{\mathrm{d}^2H}{\mathrm{d}t^2} \approx \frac{1}{20}\left(-5(4.491)^{-3/2} + \frac{9}{2}(4.491)^{-1/2} - \frac{15}{4}(4.491)^{1/2}\right)$$ $$\approx \frac{1}{20}(-0.523 + 2.127 - 7.942) \approx \frac{1}{20}(-6.338) \approx -0.317 < 0$$ Since $\dfrac{\mathrm{d}^2H}{\mathrm{d}t^2} < 0$ at $t = \alpha$, $H$ has a **maximum** at this value of $\alpha$. $\quad\blacksquare$ ## 解法二 #### 思路 (a) 用乘积法则求导：将 $H$ 看成 $\frac{1}{20}(20+6t-t^2)$ 与 $t^{1/2}$ 的乘积。(c) 不代入数值，而是证明二阶导数在 $t\in(0,7)$ 上恒负。 #### 答题过程 **(a)** Write $H = \dfrac{1}{20}(20 + 6t - t^2) \cdot t^{1/2} + 17$. Using the product rule with $u = \dfrac{1}{20}(20 + 6t - t^2)$ and $v = t^{1/2}$: $$u' = \frac{1}{20}(6 - 2t), \quad v' = \frac{1}{2}t^{-1/2}$$ $$\begin{align*} \frac{\mathrm{d}H}{\mathrm{d}t} =&\, u'v + uv' \\[4mm] =&\, \frac{6 - 2t}{20} \cdot t^{1/2} + \frac{20 + 6t - t^2}{20} \cdot \frac{1}{2}t^{-1/2} \\[4mm] =&\, \frac{1}{20}\left[(6 - 2t)t^{1/2} + \frac{20 + 6t - t^2}{2}t^{-1/2}\right] \end{align*}$$ Setting $\dfrac{\mathrm{d}H}{\mathrm{d}t} = 0$ and multiplying through by $20t^{1/2}$: $$(6 - 2t)t + \frac{20 + 6t - t^2}{2} = 0$$ $$12t - 4t^2 + 20 + 6t - t^2 = 0$$ $$-5t^2 + 18t + 20 = 0$$ $$5t^2 - 18t - 20 = 0$$ At $t = \alpha$: $5\alpha^2 - 18\alpha - 20 = 0$. $\quad\blacksquare$ **(c)** The second derivative can be written as: $$\frac{\mathrm{d}^2H}{\mathrm{d}t^2} = -\frac{1}{80}t^{-3/2}\left(20t^2 - 18t + 15\right)$$ For the quadratic $20t^2 - 18t + 15$, the discriminant is: $$\Delta = (-18)^2 - 4(20)(15) = 324 - 1200 = -876 < 0$$ Since the leading coefficient $20 > 0$ and $\Delta < 0$, we have $20t^2 - 18t + 15 > 0$ for all $t$. Also, $t^{-3/2} > 0$ for $t > 0$. Therefore $\dfrac{\mathrm{d}^2H}{\mathrm{d}t^2} < 0$ for all $t \in (0, 7)$, so $H$ has a **maximum** at $t = \alpha$. $\quad\blacksquare$