[半径为 r 的球体体积为 43πr3\dfrac{4}{3}\pi r^3,其表面积为 4πr24\pi r^2。] 下图显示了一个容量为 360π cm3360\pi\ \text{cm}^3 的珍珠奶茶杯。它由一个底面半径为 rr cm、高为 hh cm 的圆柱体杯身和一个半球形杯盖组成。设杯子的总表面积为 S cm2S\ \text{cm}^2。

  1. (a)

    证明 S=π(53r2+720r)S = \pi\left(\dfrac{5}{3}r^2 + \dfrac{720}{r}\right)。

  2. (b)

    已知 rr 是可变的,求 SS 的驻值并确定其性质。

先不看答案,让 AI 诊断你卡在哪里?

它不会直接告诉你答案,而是问你卡在哪一步看看 AI 是怎么陪你一步步弄懂这道题的

完整解法

8(a)

1. Formulate an equation for the total volume of the cup.

Volume of cylinder + Volume of hemisphere = 360π360\pi
πr2h+23πr3=360π\pi r^2 h + \frac{2}{3}\pi r^3 = 360\pi

2. Express the height $h$ in terms of $r$.

r2h+23r3=360r^2 h + \frac{2}{3}r^3 = 360
r2h=360−23r3r^2 h = 360 - \frac{2}{3}r^3
h=360r2−23rh = \frac{360}{r^2} - \frac{2}{3}r

3. Formulate an expression for the total surface area $S$.

The surface area consists of the circular base, the curved surface of the cylinder, and the curved surface of the hemisphere.
S=πr2+2πrh+2πr2S = \pi r^2 + 2\pi r h + 2\pi r^2
S=3πr2+2πrhS = 3\pi r^2 + 2\pi r h

4. Substitute $h$ into the surface area expression and simplify to show the given result.

S=3πr2+2πr(360r2−23r)S = 3\pi r^2 + 2\pi r \left( \frac{360}{r^2} - \frac{2}{3}r \right)
S=3πr2+720πr−43πr2S = 3\pi r^2 + \frac{720\pi}{r} - \frac{4}{3}\pi r^2
S=(3−43)πr2+720πrS = \left(3 - \frac{4}{3}\right)\pi r^2 + \frac{720\pi}{r}
S=53πr2+720πrS = \frac{5}{3}\pi r^2 + \frac{720\pi}{r}
S=π(53r2+720r)S = \pi\left(\frac{5}{3}r^2 + \frac{720}{r}\right)

答案

S=π(53r2+720r)S = \pi\left(\frac{5}{3}r^2 + \frac{720}{r}\right) (Shown)

8(b)

1. Differentiate $S$ with respect to $r$.

S=π(53r2+720r−1)S = \pi\left(\frac{5}{3}r^2 + 720r^{-1}\right)

dSdr=π(103r−720r−2)\frac{\text{d}S}{\text{d}r} = \pi\left(\frac{10}{3}r - 720r^{-2}\right)

2. Set the derivative to zero to find the value of $r$ at the stationary point.

For a stationary value, dSdr=0\frac{\text{d}S}{\text{d}r} = 0.

π(103r−720r2)=0\pi\left(\frac{10}{3}r - \frac{720}{r^2}\right) = 0

103r=720r2\frac{10}{3}r = \frac{720}{r^2}

r3=720×310=216r^3 = 720 \times \frac{3}{10} = 216

r=2163=6r = \sqrt[3]{216} = 6

3. Calculate the stationary value of $S$.

Substitute r=6r = 6 into SS:

S=π(53(6)2+7206)S = \pi\left(\frac{5}{3}(6)^2 + \frac{720}{6}\right)

S=π(60+120)=180πS = \pi(60 + 120) = 180\pi

4. Determine the nature of the stationary value using the second derivative test.

d2Sdr2=π(103+1440r−3)\frac{\text{d}^2S}{\text{d}r^2} = \pi\left(\frac{10}{3} + 1440r^{-3}\right)

When r=6r = 6,
d2Sdr2=π(103+144063)=π(103+1440216)=10π\frac{\text{d}^2S}{\text{d}r^2} = \pi\left(\frac{10}{3} + \frac{1440}{6^3}\right) = \pi\left(\frac{10}{3} + \frac{1440}{216}\right) = 10\pi

Since d2Sdr2>0\frac{\text{d}^2S}{\text{d}r^2} > 0, the stationary value is a minimum.

答案

Stationary value is 180π180\pi, and it is a minimum.

AI 分析,可能出错

用 AI 批改,圈出写错的那一步,并自动将你的薄弱点存入错题本,形成你的错因库。

同一知识点的其他题

  1. 1如图所示,一个建在矩形底面上的温室。建造拱形屋顶和两个完全相同的半圆形侧面共使用了 140 m2140\text{ m}^2 的聚乙烯薄膜。每个半圆的半径为 r mr\text{ m},温室的长度为 y my\text{ m}。 证明温室的体积 V m3V\text{ m}^3…
  2. 2在三角形 ABCABC 中,AB=x cmAB = x\ \text{cm},AC=3x cmAC = 3x\ \text{cm},且角 BAC=2π3BAC = \dfrac{2\pi}{3}。如果 xx 以 0.5 cm/s0.5\ \text{cm/s} 的速率增加,求当 x=4x = 4…
  3. 3已知一条曲线的方程为 y=2x−33x+4y = \frac{2x - 3}{3x + 4},其中 x≠−43x \ne -\frac{4}{3}。 曲线在点 PP(其中 x>0x > 0)处的法线平行于直线 y=−2517x+2y = -\frac{25}{17}x + 2。…
  4. 4求曲线 y=2x2−ln⁡x+cy = 2x^2 - \ln x + c 的导函数。 曲线在点 TT 处与正 xx 轴相切。求 TT 的坐标。 证明 c=−ln⁡2ec = -\ln 2\sqrt{e}。 求在点 TT 处 d2ydx2\frac{\text{d}^2y}{\text{d}x^2}…
  5. 5曲线 y=2tan⁡3x−1y = 2\tan 3x - 1 在点 P(π3,−1)P\left(\frac{\pi}{3}, -1\right) 处的法线交 yy 轴于点 AA,交 xx 轴于点 BB。求三角形 AOBAOB 的精确面积。