题目

Problem

Figure 1

Figure 1 is a sketch of the curve $C$ with equation

y=2x^{\frac32}(4-x) \qquad x\geq0

The point $P$ is the stationary point of $C$ .

(a) Find, using calculus, the $x$ coordinate of $P$ .

(4)

The region $R_1$ , shown shaded in Figure 1, is bounded by $C$ and the $x$ -axis.

The region $R_2$ , also shown shaded in Figure 1, is bounded by $C$ , the $x$ -axis and the line with equation $x=k$ , where $k$ is a constant.

Given that the area of $R_1$ is equal to the area of $R_2$

(b) find, using calculus, the exact value of $k$ .

(4)

解答

展开

先把函数展开成幂函数之和，再求导。驻点满足 $\frac{\mathrm{d}y}{\mathrm{d}x}=0$ 。

展开

Rewrite the curve:

\begin{align*} y=&\,2x^{\frac32}(4-x)\\ =&\,8x^{\frac32}-2x^{\frac52}. \end{align*}

Differentiate:

\begin{align*} \frac{\mathrm{d}y}{\mathrm{d}x} =&\,8\left(\frac32\right)x^{\frac12} -2\left(\frac52\right)x^{\frac32}\\ =&\,12x^{\frac12}-5x^{\frac32}. \end{align*}

At a stationary point,

\begin{align*} \frac{\mathrm{d}y}{\mathrm{d}x}=0. \end{align*}

\begin{align*} 12x^{\frac12}-5x^{\frac32}=&\,0\\ x^{\frac12}(12-5x)=&\,0. \end{align*}

The stationary point $P$ is not at $x=0$ , so

\begin{align*} 12-5x=&\,0\\ x=&\,\frac{12}{5}. \end{align*}

展开

$R_1$ 在 $x$ -axis 上方， $R_2$ 在 $x$ -axis 下方。两块面积相等，表示从 $0$ 到 $k$ 的有向面积为 $0$ 。

展开

Integrate the curve:

\begin{align*} \int \left(8x^{\frac32}-2x^{\frac52}\right)\,\mathrm{d}x =&\,\frac{16}{5}x^{\frac52} -\frac47x^{\frac72}. \end{align*}

Since the area above the $x$ -axis equals the area below the $x$ -axis,

\begin{align*} \int_0^k y\,\mathrm{d}x=0. \end{align*}

\begin{align*} \left[ \frac{16}{5}x^{\frac52} -\frac47x^{\frac72} \right]_0^k=&\,0\\ \frac{16}{5}k^{\frac52} -\frac47k^{\frac72}=&\,0. \end{align*}

Factorise:

\begin{align*} k^{\frac52}\left(\frac{16}{5}-\frac47k\right)=&\,0. \end{align*}

Since $k\neq0$ ,

\begin{align*} \frac{16}{5}-\frac47k=&\,0\\ \frac47k=&\,\frac{16}{5}\\ k=&\,\frac{16}{5}\cdot\frac74\\ =&\,\frac{28}{5}. \end{align*}