题目

Problem

Figure 3

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=\frac{9x^2(5-\sqrt{x})}{5},\qquad x\geqslant0

The curve has a turning point at the point $M$ , as shown in Figure 3.

(a) Using calculus, find the coordinates of $M$ .

(5)

The curve crosses the $x$ -axis at the point $P$ , as shown in Figure 3.

(b) Use algebra to find the $x$ coordinate of $P$ .

(2)

The finite region $R$ , shown shaded in Figure 3, is bounded by the curve, the line through $M$ parallel to the $x$ -axis and the line through $P$ parallel to the $y$ -axis.

(5)

解答

(a)

解法一

思路

展开

先把函数展开成幂的形式，方便求导。turning point 要令 $\frac{dy}{dx}=0$ 。

答题过程

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First rewrite the curve:

\begin{align*} y=&\,\frac{9x^2(5-\sqrt{x})}{5}\\ =&\,9x^2-\frac95x^{5/2}. \end{align*}

Differentiate:

\begin{align*} \frac{dy}{dx} =&\,18x-\frac95\cdot\frac52x^{3/2}\\ =&\,18x-\frac92x^{3/2}. \end{align*}

At a turning point,

\begin{align*} 18x-\frac92x^{3/2}=&\,0\\ 18x=&\,\frac92x^{3/2}. \end{align*}

Since $M$ is not at the origin, divide by $x$ :

\begin{align*} 18=&\,\frac92x^{1/2}\\ 4=&\,\sqrt{x}\\ x=&\,16. \end{align*}

Now find $y$ :

\begin{align*} y=&\,\frac{9(16)^2(5-\sqrt{16})}{5}\\ =&\,\frac{9(256)(1)}{5}\\ =&\,\frac{2304}{5}\\ =&\,460.8. \end{align*}

Therefore,

\begin{align*} M=(16,460.8). \end{align*}

(b)

解法一

思路

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$P$ 是 $x$ -axis 上的交点，所以令 $y=0$ 。图中 $P$ 不是原点，因此使用 $5-\sqrt{x}=0$ 。

答题过程

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At the $x$ -axis,

\begin{align*} y=0. \end{align*}

\begin{align*} \frac{9x^2(5-\sqrt{x})}{5}=&\,0. \end{align*}

For point $P$ , $x\ne0$ , so

\begin{align*} 5-\sqrt{x}=&\,0\\ \sqrt{x}=&\,5\\ x=&\,25. \end{align*}

Therefore, the $x$ coordinate of $P$ is $25$ .

(c)

解法一

思路

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区域 $R$ 是水平线 $y=460.8$ 和曲线之间，从 $x=16$ 到 $x=25$ 的面积。因此可用“矩形面积减曲线下面积”。

答题过程

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The area of the rectangle is

\begin{align*} (25-16)(460.8)=&\,4147.2. \end{align*}

Now integrate the curve:

\begin{align*} \int\left(9x^2-\frac95x^{5/2}\right)\,\mathrm{d}x =&\,3x^3-\frac95\cdot\frac{2}{7}x^{7/2}\\[4mm] =&\,3x^3-\frac{18}{35}x^{7/2}. \end{align*}

So the area under the curve from $16$ to $25$ is

\begin{align*} \left[3x^3-\frac{18}{35}x^{7/2}\right]_{16}^{25}. \end{align*}

Therefore, the shaded area is

\begin{align*} R =&\,4147.2-\left[3x^3-\frac{18}{35}x^{7/2}\right]_{16}^{25}\\[4mm] =&\,1312.685\ldots . \end{align*}

Hence the area of $R$ is

\begin{align*} 1312.7 \end{align*}

to one decimal place.

解法二

思路

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单积分法。我们可以直接对「上方水平直线与下方曲线的差值」进行定积分估算： $\int_{16}^{25} (460.8 - y) dx$ 。这种方法将矩形面积和曲线下面积合并为单个积分式，计算步骤更紧凑。

答题过程

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The upper boundary of the region $R$ is the line $y = 460.8$ , and the lower boundary is the curve $y = 9x^2 - \frac95x^{5/2}$ .

The area of $R$ is given by the integral of the difference from $x = 16$ to $x = 25$ :

\begin{align*} R =&\,\, \int_{16}^{25} \left( 460.8 - \left(9x^2 - \frac95x^{5/2}\right) \right) \,\mathrm{d}x\\[4mm] =&\,\, \left[ 460.8x - 3x^3 + \frac{18}{35}x^{7/2} \right]_{16}^{25} \end{align*}

Evaluate the limits:

At $x = 25$ :

\begin{align*} & 460.8(25) - 3(25)^3 + \frac{18}{35}(25)^{7/2}\\[4mm] =&\,\, 11520 - 46875 + \frac{18}{35}(78125)\\[4mm] =&\,\, -35355 + \frac{281250}{7}\\[4mm] =&\,\, \frac{33765}{7} \approx 4823.571\dots \end{align*}

At $x = 16$ :

\begin{align*} & 460.8(16) - 3(16)^3 + \frac{18}{35}(16)^{7/2}\\[4mm] =&\,\, 7372.8 - 12288 + \frac{18}{35}(16384)\\[4mm] =&\,\, -4915.2 + \frac{294912}{35}\\[4mm] =&\,\, \frac{122880}{35} = \frac{12288}{7} \approx 3510.857\dots \end{align*}

Substitute back:

\begin{align*} R =&\,\, \frac{33765}{7} - \frac{12288}{7}\\[4mm] =&\,\, \frac{21477}{7}\\[4mm] \approx&\,\, 3068.14 \quad \text{--- wait, let me re-evaluate the calculations.} \end{align*}

Wait, let’s re-calculate: Let’s use the decimal values: $\text{Rectangle} = 4147.2$ Integral under curve: $\left[ 3x^3 - \frac{18}{35}x^{7/2} \right]_{16}^{25}$

At $x = 25$ : $3(15625) - \frac{18}{35}(78125) = 46875 - 40178.57 = 6696.43$
At $x = 16$ : $3(4096) - \frac{18}{35}(16384) = 12288 - 8426.06 = 3861.94$ So the integral under curve is $6696.43 - 3861.94 = 2834.49$ Shaded area = $4147.2 - 2834.49 = 1312.71 \approx 1312.7$

Let’s evaluate the direct integral formula again: $\left[ 460.8x - 3x^3 + \frac{18}{35}x^{7/2} \right]_{16}^{25}$ At $x = 25$ : $460.8(25) - 6696.43 = 11520 - 6696.43 = 4823.57$ At $x = 16$ : $460.8(16) - 3861.94 = 7372.8 - 3861.94 = 3510.86$ Difference: $4823.57 - 3510.86 = 1312.71 \approx 1312.7$

Perfect! The calculation matches. Let’s write the exact steps down clearly:

\begin{align*} R =&\,\, \left( 460.8(25) - 3(25)^3 + \frac{18}{35}(25)^{7/2} \right)\\[4mm] &\,\, - \left( 460.8(16) - 3(16)^3 + \frac{18}{35}(16)^{7/2} \right)\\[4mm] \approx&\,\, 4823.57 - 3510.86\\[4mm] =&\,\, 1312.71 \end{align*}

Hence the area of $R$ is $1312.7$ to one decimal place.

IAL 2025 Jan Q9

题目

解答

(a)

解法一

思路

答题过程

(b)

解法一

思路

答题过程

(c)

解法一

思路

答题过程

解法二

思路

答题过程