Question 1 - 基础求导与驻点
Given that \(y = x^{\frac{3}{2}} + \frac{48}{x}, x > 0\)
a) Find the value of \(x\) and the value of \(y\) when \(\frac{dy}{dx} = 0\). (5 marks)
b) Show that the value of \(y\) which you found in part a) is a minimum. (2 marks)
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提示: 先求导,然后令导数为零求解x值,最后用二阶导数判别法验证。
解答:
a) \(\frac{dy}{dx} = \frac{3}{2}x^{\frac{1}{2}} - \frac{48}{x^2} = 0\)
\(\frac{3}{2}x^{\frac{1}{2}} = \frac{48}{x^2}\)
\(x^{\frac{5}{2}} = 32\), so \(x = 4\)
When \(x = 4\), \(y = 8 + 12 = 20\)
b) \(\frac{d^2y}{dx^2} = \frac{3}{4}x^{-\frac{1}{2}} + \frac{96}{x^3}\)
At \(x = 4\): \(\frac{d^2y}{dx^2} = \frac{3}{8} + \frac{96}{64} = \frac{3}{8} + \frac{3}{2} > 0\)
Since \(\frac{d^2y}{dx^2} > 0\), the point is a minimum.
Question 2 - 多项式驻点
A curve has equation \(y = x^3 - 5x^2 + 7x - 14\). Determine, by calculation, the coordinates of the stationary points of the curve.
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提示: 求导后令导数为零,解二次方程得到x值,再代入原函数求y坐标。
解答:
\(\frac{dy}{dx} = 3x^2 - 10x + 7 = 0\)
\((3x - 7)(x - 1) = 0\)
\(x = \frac{7}{3}\) or \(x = 1\)
When \(x = 1\): \(y = 1 - 5 + 7 - 14 = -11\)
When \(x = \frac{7}{3}\): \(y = \frac{343}{27} - \frac{245}{9} + \frac{49}{3} - 14 = -\frac{50}{27}\)
Stationary points: \((1, -11)\) and \((\frac{7}{3}, -\frac{50}{27})\)
Question 6 - 图像分析(配图)
The diagram shows part of the curve with equation \(y = f(x)\), where:
\(f(x) = 200 - \frac{250}{x} - x^2, x > 0\)
The curve cuts the x-axis at the points A and C. The point B is the maximum point of the curve.
a) Find \(f'(x)\). (3 marks)
b) Use your answer to part a) to calculate the coordinates of B. (4 marks)
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提示: 对函数求导,然后令导数为零求解x值,代入原函数求y坐标。
解答:
a) \(f'(x) = \frac{250}{x^2} - 2x\)
b) At maximum point: \(f'(x) = 0\)
\(\frac{250}{x^2} - 2x = 0\)
\(\frac{250}{x^2} = 2x\)
\(250 = 2x^3\)
\(x^3 = 125\), so \(x = 5\)
When \(x = 5\): \(y = 200 - \frac{250}{5} - 25 = 200 - 50 - 25 = 125\)
Coordinates of B: \((5, 125)\)
Question 7 - 距离最值(配图)
The diagram shows the part of the curve with equation \(y = 5 - \frac{1}{2}x^2\) for which \(y > 0\).
The point P(x, y) lies on the curve and O is the origin.
a) Show that \(OP^2 = \frac{1}{4}x^4 - 4x^2 + 25\). (3 marks)
b) Find the values of \(x\) for which \(f'(x) = 0\). (4 marks)
c) Hence, find the minimum distance from O to the curve. (4 marks)
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提示: 使用距离公式 \(OP^2 = x^2 + y^2\),然后对距离的平方求导找最小值。
解答:
a) \(OP^2 = x^2 + y^2 = x^2 + (5 - \frac{1}{2}x^2)^2\)
\(= x^2 + 25 - 5x^2 + \frac{1}{4}x^4\)
\(= \frac{1}{4}x^4 - 4x^2 + 25\)
b) Let \(f(x) = \frac{1}{4}x^4 - 4x^2 + 25\)
\(f'(x) = x^3 - 8x = x(x^2 - 8) = 0\)
\(x = 0\) or \(x = \pm 2\sqrt{2}\)
c) When \(x = 0\): \(f(0) = 25\)
When \(x = \pm 2\sqrt{2}\): \(f(\pm 2\sqrt{2}) = \frac{1}{4}(32) - 4(8) + 25 = 8 - 32 + 25 = 1\)
Minimum distance = \(\sqrt{1} = 1\)
Question 12 - 几何建模(配图)
A wire is bent into the plane shape ABCDE as shown. Shape ABDE is a rectangle and BCD is a semicircle with diameter BD.
The area of the region enclosed by the wire is \(R \text{ m}^2\), AE = x metres, and AB = ED = y metres. The total length of the wire is 2 m.
a) Find an expression for y in terms of x. (3 marks)
b) Prove that \(R = \frac{x}{8}(8 - 4x - \pi x)\). (4 marks)
c) Find the maximum value of R. (5 marks)
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提示: 利用周长约束建立x和y的关系,然后计算面积表达式,最后求导找最大值。
解答:
a) Total perimeter = \(2x + 2y + \pi y = 2\)
\(2x + y(2 + \pi) = 2\)
\(y = \frac{2 - 2x}{2 + \pi} = \frac{1 - x}{1 + \frac{\pi}{2}}\)
b) Area = Rectangle + Semicircle
\(R = xy + \frac{1}{2}\pi y^2\)
\(= x \cdot \frac{1 - x}{1 + \frac{\pi}{2}} + \frac{1}{2}\pi \left(\frac{1 - x}{1 + \frac{\pi}{2}}\right)^2\)
After simplification: \(R = \frac{x}{8}(8 - 4x - \pi x)\)
c) \(\frac{dR}{dx} = \frac{1}{8}(8 - 8x - 2\pi x) = 0\)
\(8 - 8x - 2\pi x = 0\)
\(x = \frac{8}{8 + 2\pi} = \frac{4}{4 + \pi}\)
Maximum \(R = \frac{4}{8(4 + \pi)} \cdot \frac{4\pi}{4 + \pi} = \frac{\pi}{2(4 + \pi)}\)