题目
Problem
A cone, shown in Figure 2, has
fixed height 5 5 5
base radius r r r
slant height l l l
(a) Find an expression for l l l r r r
(1)
Given that the base radius is increasing at a constant rate of 3 3 3
(b) find the rate at which the total surface area of the cone is changing when the radius of the cone is 1.5 1.5 1.5 2 ^2 2
[The total surface area, S S S S = π r 2 + π r l S=\pi r^2+\pi rl S = π r 2 + π r l
(4)
题目中文翻译
图 2 所示圆锥满足：
高固定为 5 5 5
底面半径为 r r r
母线长为 l l l
(a) 用 r r r l l l
已知底面半径以每分钟 3 3 3
(b) 求当圆锥半径为 1.5 1.5 1.5 2 ^2 2
[圆锥总表面积 S S S S = π r 2 + π r l S=\pi r^2+\pi rl S = π r 2 + π r l
解答
(a)
解法一
思路
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圆锥的高、半径和母线长组成直角三角形。高固定为 5 5 5 r r r l l l
答题过程
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By Pythagoras’ theorem,
l 2 = r 2 + 5 2 . l^2=r^2+5^2. l 2 = r 2 + 5 2 . Since l > 0 l>0 l > 0
l = r 2 + 25 . \boxed{l=\sqrt{r^2+25}}. l = r 2 + 25 .
(b)
解法一
思路
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先把总表面积 S S S r r r d S d r \frac{\mathrm{d}S}{\mathrm{d}r} d r d S d r d t = 3 \frac{\mathrm{d}r}{\mathrm{d}t}=3 d t d r = 3 d S d t = d S d r d r d t \frac{\mathrm{d}S}{\mathrm{d}t}=\frac{\mathrm{d}S}{\mathrm{d}r}\frac{\mathrm{d}r}{\mathrm{d}t} d t d S = d r d S d t d r
答题过程
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The total surface area is
S = π r 2 + π r l . S=\pi r^2+\pi rl. S = π r 2 + π r l . Using l = r 2 + 25 l=\sqrt{r^2+25} l = r 2 + 25
S = π r 2 + π r r 2 + 25 . S=\pi r^2+\pi r\sqrt{r^2+25}. S = π r 2 + π r r 2 + 25 . Differentiate with respect to r r r
d S d r = 2 π r + π r 2 + 25 + π r ⋅ 1 2 ( r 2 + 25 ) − 1 2 ⋅ 2 r = 2 π r + π r 2 + 25 + π r 2 r 2 + 25 . \begin{align*}
\frac{\mathrm{d}S}{\mathrm{d}r}
=&\, 2\pi r
+\pi\sqrt{r^2+25}
+\pi r\cdot\frac{1}{2}(r^2+25)^{-\frac12}\cdot 2r \\[2mm]
=&\, 2\pi r
+\pi\sqrt{r^2+25}
+\frac{\pi r^2}{\sqrt{r^2+25}}.
\end{align*} d r d S = = 2 π r + π r 2 + 25 + π r ⋅ 2 1 ( r 2 + 25 ) − 2 1 ⋅ 2 r 2 π r + π r 2 + 25 + r 2 + 25 π r 2 . Given that
d r d t = 3 , \frac{\mathrm{d}r}{\mathrm{d}t}=3, d t d r = 3 , we have
d S d t = d S d r d r d t . \frac{\mathrm{d}S}{\mathrm{d}t}
=\frac{\mathrm{d}S}{\mathrm{d}r}\frac{\mathrm{d}r}{\mathrm{d}t}. d t d S = d r d S d t d r . When r = 1.5 r=1.5 r = 1.5
r 2 + 25 = 1.5 2 + 25 = 27.25 = 109 2 . \sqrt{r^2+25}
=\sqrt{1.5^2+25}
=\sqrt{27.25}
=\frac{\sqrt{109}}{2}. r 2 + 25 = 1. 5 2 + 25 = 27.25 = 2 109 . Therefore
d S d t = 3 ( 2 π ( 1.5 ) + π ⋅ 109 2 + π ( 1.5 ) 2 109 2 ) = 3 ( 3 π + π 109 2 + 9 π 2 109 ) = 81.5354 … . \begin{align*}
\frac{\mathrm{d}S}{\mathrm{d}t}
=&\, 3\left(
2\pi(1.5)
+\pi\cdot\frac{\sqrt{109}}{2}
+\frac{\pi(1.5)^2}{\frac{\sqrt{109}}{2}}
\right) \\[2mm]
=&\, 3\left(
3\pi+\frac{\pi\sqrt{109}}{2}
+\frac{9\pi}{2\sqrt{109}}
\right) \\[2mm]
=&\, 81.5354\ldots.
\end{align*} d t d S = = = 3 ( 2 π ( 1.5 ) + π ⋅ 2 109 + 2 109 π ( 1.5 ) 2 ) 3 ( 3 π + 2 π 109 + 2 109 9 π ) 81.5354 … . So the rate of change of the total surface area is
81.5 cm 2 per minute . \boxed{81.5\text{ cm}^2\text{ per minute}}. 81.5 cm 2 per minute .
解法二
思路
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另一种做法是保留 S = π r 2 + π r l S=\pi r^2+\pi rl S = π r 2 + π r l t t t l = r 2 + 25 l=\sqrt{r^2+25} l = r 2 + 25 d l d t \frac{\mathrm{d}l}{\mathrm{d}t} d t d l d S d t \frac{\mathrm{d}S}{\mathrm{d}t} d t d S
答题过程
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From
l = r 2 + 25 , l=\sqrt{r^2+25}, l = r 2 + 25 , we get
d l d r = r r 2 + 25 . \frac{\mathrm{d}l}{\mathrm{d}r}
=\frac{r}{\sqrt{r^2+25}}. d r d l = r 2 + 25 r . Since
d r d t = 3 , \frac{\mathrm{d}r}{\mathrm{d}t}=3, d t d r = 3 , it follows that
d l d t = d l d r d r d t = 3 r r 2 + 25 . \frac{\mathrm{d}l}{\mathrm{d}t}
=\frac{\mathrm{d}l}{\mathrm{d}r}\frac{\mathrm{d}r}{\mathrm{d}t}
=\frac{3r}{\sqrt{r^2+25}}. d t d l = d r d l d t d r = r 2 + 25 3 r . Now
S = π r 2 + π r l . S=\pi r^2+\pi rl. S = π r 2 + π r l . Differentiate with respect to t t t
d S d t = 2 π r d r d t + π l d r d t + π r d l d t . \frac{\mathrm{d}S}{\mathrm{d}t}
=2\pi r\frac{\mathrm{d}r}{\mathrm{d}t}
+\pi l\frac{\mathrm{d}r}{\mathrm{d}t}
+\pi r\frac{\mathrm{d}l}{\mathrm{d}t}. d t d S = 2 π r d t d r + π l d t d r + π r d t d l . When r = 1.5 r=1.5 r = 1.5
l = 109 2 , d l d t = 3 ( 1.5 ) 109 2 = 9 109 . l=\frac{\sqrt{109}}{2},
\qquad
\frac{\mathrm{d}l}{\mathrm{d}t}
=\frac{3(1.5)}{\frac{\sqrt{109}}{2}}
=\frac{9}{\sqrt{109}}. l = 2 109 , d t d l = 2 109 3 ( 1.5 ) = 109 9 . Therefore
d S d t = 2 π ( 1.5 ) ( 3 ) + π ( 109 2 ) ( 3 ) + π ( 1.5 ) ( 9 109 ) = 9 π + 3 π 109 2 + 27 π 2 109 = 81.5354 … . \begin{align*}
\frac{\mathrm{d}S}{\mathrm{d}t}
=&\, 2\pi(1.5)(3)
+\pi\left(\frac{\sqrt{109}}{2}\right)(3)
+\pi(1.5)\left(\frac{9}{\sqrt{109}}\right) \\[2mm]
=&\, 9\pi+\frac{3\pi\sqrt{109}}{2}
+\frac{27\pi}{2\sqrt{109}} \\[2mm]
=&\, 81.5354\ldots.
\end{align*} d t d S = = = 2 π ( 1.5 ) ( 3 ) + π ( 2 109 ) ( 3 ) + π ( 1.5 ) ( 109 9 ) 9 π + 2 3 π 109 + 2 109 27 π 81.5354 … . Thus
81.5 cm 2 per minute . \boxed{81.5\text{ cm}^2\text{ per minute}}. 81.5 cm 2 per minute . 
