题目
Problem
In this question you must show all stages of your working.
Solutions relying entirely on calculator technology are not acceptable.
Figure 3 shows a sketch of part of the curve with equation
y = 9 x 2 ( 5 − x ) 5 , x ⩾ 0
y=\frac{9x^2(5-\sqrt{x})}{5},\qquad x\geqslant0 y = 5 9 x 2 ( 5 − x ) , x ⩾ 0 The curve has a turning point at the point M M M
(a) Using calculus, find the coordinates of M M M
(5)
The curve crosses the x x x P P P
(b) Use algebra to find the x x x P P P
(2)
The finite region R R R M M M x x x P P P y y y
(c) Use algebraic integration to find the area of R R R
(5)
题目中文翻译
本题必须展示所有计算步骤。完全依赖计算器技术的解答不可接受。
图3是下列曲线的一部分的示意图：
y = 9 x 2 ( 5 − x ) 5 , x ⩾ 0.
y=\frac{9x^2(5-\sqrt{x})}{5},\qquad x\geqslant0.
y = 5 9 x 2 ( 5 − x ) , x ⩾ 0. 曲线在点 M M M
(a) 用微积分求 M M M
曲线在点 P P P x x x
(b) 用代数方法求 P P P
图3中阴影所示的有限区域 R R R M M M x x x P P P y y y
(c) 用代数积分求 R R R
解答
(a)
解法一
思路
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先展开成幂函数求导。点 M M M x x x x x x
答题过程
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Expand the curve:
y = 9 x 2 − 9 5 x 5 / 2 , d y d x = 18 x − 9 2 x 3 / 2 .
\begin{align*}
y=&\,9x^2-\frac95x^{5/2},\\
\frac{\mathrm{d}y}{\mathrm{d}x}
=&\,18x-\frac92x^{3/2}.
\end{align*}
y = d x d y = 9 x 2 − 5 9 x 5/2 , 18 x − 2 9 x 3/2 . At M M M x > 0 x>0 x > 0
18 x − 9 2 x 3 / 2 = 0 , 18 = 9 2 x , x = 4 , x = 16.
\begin{align*}
18x-\frac92x^{3/2}=&\,0,\\
18=&\,\frac92\sqrt{x},\\
\sqrt{x}=&\,4,\\
x=&\,16.
\end{align*}
18 x − 2 9 x 3/2 = 18 = x = x = 0 , 2 9 x , 4 , 16. Substitute into the original curve:
y = 9 ( 16 ) 2 ( 5 − 4 ) 5 = 2304 5 = 460.8.
\begin{align*}
y=&\,\frac{9(16)^2(5-4)}5\\
=&\,\frac{2304}5\\
=&\,460.8.
\end{align*}
y = = = 5 9 ( 16 ) 2 ( 5 − 4 ) 5 2304 460.8. Therefore M = ( 16 , 460.8 ) M=(16,460.8) M = ( 16 , 460.8 )
解法二
思路
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保留原式为两个因子的乘积，用乘积法则。将导数中的 x x x M M M
答题过程
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For x > 0 x>0 x > 0
y = 9 5 x 2 ( 5 − x ) .
y=\frac95x^2(5-\sqrt{x}).
y = 5 9 x 2 ( 5 − x ) . Using the product rule,
d y d x = 9 5 [ 2 x ( 5 − x ) − 1 2 x 2 x − 1 / 2 ] = 9 5 ( 10 x − 5 2 x 3 / 2 ) = 9 2 x ( 4 − x ) .
\begin{align*}
\frac{\mathrm{d}y}{\mathrm{d}x}
=&\,\frac95\bigg[2x(5-\sqrt{x})\\
&\,\hspace{2pt}-\frac12x^2x^{-1/2}\bigg]\\
=&\,\frac95\bigg(10x-\frac52x^{3/2}\bigg)\\
=&\,\frac92x(4-\sqrt{x}).
\end{align*}
d x d y = = = 5 9 [ 2 x ( 5 − x ) − 2 1 x 2 x − 1/2 ] 5 9 ( 10 x − 2 5 x 3/2 ) 2 9 x ( 4 − x ) . At M M M x > 0 x>0 x > 0
4 − x = 0 , x = 4 , x = 16.
\begin{align*}
4-\sqrt{x}=&\,0,\\
\sqrt{x}=&\,4,\\
x=&\,16.
\end{align*}
4 − x = x = x = 0 , 4 , 16. Then
y = 9 ( 16 ) 2 ( 5 − 4 ) 5 = 460.8.
y=\frac{9(16)^2(5-4)}5=460.8.
y = 5 9 ( 16 ) 2 ( 5 − 4 ) = 460.8. Hence M = ( 16 , 460.8 ) M=(16,460.8) M = ( 16 , 460.8 )
(b)
解法一
思路
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令 y = 0 y=0 y = 0 P P P 5 − x 5-\sqrt{x} 5 − x x 2 x^2 x 2
答题过程
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At P P P y = 0 y=0 y = 0 x > 0 x>0 x > 0
9 x 2 ( 5 − x ) 5 = 0.
\frac{9x^2(5-\sqrt{x})}{5}=0.
5 9 x 2 ( 5 − x ) = 0. Since x 2 > 0 x^2>0 x 2 > 0
5 − x = 0 , x = 5 , x = 25.
\begin{align*}
5-\sqrt{x}=&\,0,\\
\sqrt{x}=&\,5,\\
x=&\,25.
\end{align*}
5 − x = x = x = 0 , 5 , 25. Thus the x x x P P P 25 25 25
(c)
解法一
思路
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用矩形面积减去曲线在 16 16 16 25 25 25
答题过程
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The bounding rectangle has area
( 25 − 16 ) 2304 5 = 20736 5 .
(25-16)\frac{2304}{5}=\frac{20736}{5}.
( 25 − 16 ) 5 2304 = 5 20736 . An antiderivative of the curve is
∫ ( 9 x 2 − 9 5 x 5 / 2 ) d x = 3 x 3 − 18 35 x 7 / 2 .
\begin{align*}
&\,\int\bigg(9x^2-\frac95x^{5/2}\bigg)\,\mathrm{d}x\\
=&\,3x^3-\frac{18}{35}x^{7/2}.
\end{align*}
= ∫ ( 9 x 2 − 5 9 x 5/2 ) d x 3 x 3 − 35 18 x 7/2 . At the limits,
F ( 25 ) = 3 ( 25 ) 3 − 18 35 ( 25 ) 7 / 2 = 46875 − 281250 7 = 46875 7 , F ( 16 ) = 3 ( 16 ) 3 − 18 35 ( 16 ) 7 / 2 = 12288 − 294912 35 = 135168 35 .
\begin{align*}
F(25)=&\,3(25)^3-\frac{18}{35}(25)^{7/2}\\
=&\,46875-\frac{281250}{7}\\
=&\,\frac{46875}{7},\\[4mm]
F(16)=&\,3(16)^3-\frac{18}{35}(16)^{7/2}\\
=&\,12288-\frac{294912}{35}\\
=&\,\frac{135168}{35}.
\end{align*}
F ( 25 ) = = = F ( 16 ) = = = 3 ( 25 ) 3 − 35 18 ( 25 ) 7/2 46875 − 7 281250 7 46875 , 3 ( 16 ) 3 − 35 18 ( 16 ) 7/2 12288 − 35 294912 35 135168 . Hence the area under the curve is
F ( 25 ) − F ( 16 ) = 234375 − 135168 35 = 99207 35 .
\begin{align*}
F(25)-F(16)
=&\,\frac{234375-135168}{35}\\
=&\,\frac{99207}{35}.
\end{align*}
F ( 25 ) − F ( 16 ) = = 35 234375 − 135168 35 99207 . Therefore the shaded area is
Area ( R ) = 20736 5 − 99207 35 = 145152 − 99207 35 = 9189 7 ≈ 1312.7 (1 d.p.) .
\begin{align*}
\operatorname{Area}(R)
=&\,\frac{20736}{5}-\frac{99207}{35}\\
=&\,\frac{145152-99207}{35}\\
=&\,\frac{9189}{7}\\
\approx&\,1312.7\quad\text{(1 d.p.)}.
\end{align*}
Area ( R ) = = = ≈ 5 20736 − 35 99207 35 145152 − 99207 7 9189 1312.7 (1 d.p.) .
解法二
思路
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直接对上方水平线与下方曲线的差积分。设积分原函数为 G G G G ( 25 ) G(25) G ( 25 ) G ( 16 ) G(16) G ( 16 )
答题过程
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The upper boundary is y = 2304 / 5 y=2304/5 y = 2304/5
Area ( R ) = ∫ 16 25 ( 2304 5 − 9 x 2 + 9 5 x 5 / 2 ) d x = [ 2304 5 x − 3 x 3 + 18 35 x 7 / 2 ] 16 25 .
\begin{align*}
\operatorname{Area}(R)
=&\,\int_{16}^{25}\bigg(\frac{2304}{5}-9x^2\\
&\,\hspace{2pt}+\frac95x^{5/2}\bigg)\,\mathrm{d}x\\
=&\,\Big[\frac{2304}{5}x-3x^3\\
&\,\hspace{2pt}+\frac{18}{35}x^{7/2}\Big]_{16}^{25}.
\end{align*}
Area ( R ) = = ∫ 16 25 ( 5 2304 − 9 x 2 + 5 9 x 5/2 ) d x [ 5 2304 x − 3 x 3 + 35 18 x 7/2 ] 16 25 . Write
G ( x ) = 2304 5 x − 3 x 3 + 18 35 x 7 / 2 .
\begin{align*}
G(x)=&\,\frac{2304}{5}x-3x^3\\
&\,\hspace{2pt}+\frac{18}{35}x^{7/2}.
\end{align*}
G ( x ) = 5 2304 x − 3 x 3 + 35 18 x 7/2 . Then
G ( 25 ) = 11520 − 46875 + 281250 7 = 33765 7 , G ( 16 ) = 36864 5 − 12288 + 294912 35 = 258048 − 430080 + 294912 35 = 122880 35 = 24576 7 .
\begin{align*}
G(25)=&\,11520-46875+\frac{281250}{7}\\
=&\,\frac{33765}{7},\\[4mm]
G(16)=&\,\frac{36864}{5}-12288\\
&\,\hspace{2pt}+\frac{294912}{35}\\
=&\,\frac{258048-430080+294912}{35}\\
=&\,\frac{122880}{35}\\
=&\,\frac{24576}{7}.
\end{align*}
G ( 25 ) = = G ( 16 ) = = = = 11520 − 46875 + 7 281250 7 33765 , 5 36864 − 12288 + 35 294912 35 258048 − 430080 + 294912 35 122880 7 24576 . Therefore
Area ( R ) = G ( 25 ) − G ( 16 ) = 33765 − 24576 7 = 9189 7 ≈ 1312.7 (1 d.p.) .
\begin{align*}
\operatorname{Area}(R)
=&\,G(25)-G(16)\\
=&\,\frac{33765-24576}{7}\\
=&\,\frac{9189}{7}\\
\approx&\,1312.7\quad\text{(1 d.p.)}.
\end{align*}
Area ( R ) = = = ≈ G ( 25 ) − G ( 16 ) 7 33765 − 24576 7 9189 1312.7 (1 d.p.) . 也可以使用辅助直线 M P MP M P L ( x ) = 256 5 ( 25 − x ) L(x)=\frac{256}{5}(25-x) L ( x ) = 5 256 ( 25 − x )
T = 1 2 ( 25 − 16 ) 2304 5 = 10368 5 .
T=\frac12(25-16)\frac{2304}{5}=\frac{10368}{5}.
T = 2 1 ( 25 − 16 ) 5 2304 = 5 10368 . 曲线与 M P MP M P
∫ 16 25 ( y − L ( x ) ) d x = 99207 35 − 10368 5 = 26631 35 .
\begin{align*}
&\,\int_{16}^{25}(y-L(x))\,\mathrm{d}x\\
=&\,\frac{99207}{35}-\frac{10368}{5}\\
=&\,\frac{26631}{35}.
\end{align*}
= = ∫ 16 25 ( y − L ( x )) d x 35 99207 − 5 10368 35 26631 . 于是同一区域为
T − 26631 35 = 9189 7 .
T-\frac{26631}{35}=\frac{9189}{7}.
T − 35 26631 = 7 9189 . 这只是相同面积的另一种分解，不必重复整套积分。
