Unit 25 Review Exercises
Applications of Differentiation
These questions mix the ideas from the whole unit. Paper 1 means no calculator; Paper 2 means a GDC is allowed. Questions marked AHL are for HL students only.
1 Paper 1 Find the equations of the tangent and the normal to the curve \(y = x^3 - 2x^2 + 1\) at the point where \(x = 2\).
Point \((2, 1)\). \(\frac{dy}{dx} = 3x^2 - 4x = 4\) at \(x = 2\).
Tangent: \(y - 1 = 4(x - 2)\), so \(y = 4x - 7\). Normal: \(y - 1 = -\frac{1}{4}(x - 2)\), so \(y = -\frac{1}{4}x + \frac{3}{2}\).
See 25.1.
2 Paper 1 Find the values of \(x\) for which \(f(x) = x^3 - 3x^2 - 24x + 2\) is decreasing.
\(f'(x) = 3x^2 - 6x - 24 = 3(x - 4)(x + 2)\). Negative between the roots: \(-2 < x < 4\).
See 25.2.
3 Paper 1 Find the coordinates of the stationary points of \(y = 2x^3 - 3x^2 - 12x + 4\) and determine their nature.
\(\frac{dy}{dx} = 6x^2 - 6x - 12 = 6(x - 2)(x + 1) = 0 \Rightarrow x = -1, 2\). \(\frac{d^2y}{dx^2} = 12x - 6\).
\(x = -1\): \(y = -2 - 3 + 12 + 4 = 11\), \(\frac{d^2y}{dx^2} = -18 < 0\): \((-1, 11)\) local maximum.
\(x = 2\): \(y = 16 - 12 - 24 + 4 = -16\), \(\frac{d^2y}{dx^2} = 18 > 0\): \((2, -16)\) local minimum.
See 25.3.
4 Paper 1 Let \(f(x) = x^4 - 2x^3\).
(a) Find the stationary points of \(f\) and classify them.
(b) Find the points of inflexion of \(f\), and say whether each is stationary or non-stationary.
(a) \(f'(x) = 4x^3 - 6x^2 = 2x^2(2x - 3) = 0 \Rightarrow x = 0\) or \(x = \frac{3}{2}\).
Sign of \(f'\): \(-\) for \(x < 0\), \(-\) for \(0 < x < \frac{3}{2}\), \(+\) for \(x > \frac{3}{2}\).
\((0, 0)\) is a stationary point of inflexion; \(\left(\frac{3}{2}, -\frac{27}{16}\right)\) is a local minimum.
(b) \(f''(x) = 12x^2 - 12x = 12x(x - 1)\): zero and changing sign at \(x = 0\) and \(x = 1\).
\((0, 0)\): stationary inflexion. \((1, -1)\): \(f'(1) = -2 \neq 0\), non-stationary inflexion.
See 25.3.
5 Paper 1 The graph of \(y = f'(x)\) is the parabola \(y = -(x + 1)(x - 3)\). Find the \(x\)-coordinates of the local maximum, the local minimum and the point of inflexion of \(f\), and state where \(f\) is concave up.
Sign of \(f'\): \(-\) for \(x < -1\), \(+\) for \(-1 < x < 3\), \(-\) for \(x > 3\). Local minimum at \(x = -1\); local maximum at \(x = 3\).
The vertex of \(f'\) is at \(x = 1\), so \(f\) has a point of inflexion at \(x = 1\). \(f'\) is increasing for \(x < 1\), so \(f\) is concave up for \(x < 1\).
See 25.4.
6 Paper 1 For the curve \(y = x\mathrm{e}^{2x}\), find the exact coordinates of (a) the stationary point, and show that it is a minimum, (b) the point of inflexion.
(a) \(\frac{dy}{dx} = \mathrm{e}^{2x} + 2x\mathrm{e}^{2x} = \mathrm{e}^{2x}(1 + 2x) = 0 \Rightarrow x = -\frac{1}{2}\), \(y = -\frac{1}{2}\mathrm{e}^{-1} = -\frac{1}{2\mathrm{e}}\).
\(\frac{d^2y}{dx^2} = 2\mathrm{e}^{2x}(1 + 2x) + 2\mathrm{e}^{2x} = \mathrm{e}^{2x}(4 + 4x)\). At \(x = -\frac{1}{2}\): \(2\mathrm{e}^{-1} > 0\), a minimum.
(b) \(\frac{d^2y}{dx^2} = 0\) at \(x = -1\) and it changes sign there: \(\left(-1, -\mathrm{e}^{-2}\right)\).
See 25.3.
7 Paper 1 AHL The area of a circle increases at a constant rate of \(12\) cm\(^2\) s\(^{-1}\). Find the rate of increase of the radius when the radius is \(3\) cm.
\(A = \pi r^2\), \(\frac{dA}{dt} = 2\pi r\frac{dr}{dt}\). \(12 = 2\pi(3)\frac{dr}{dt}\), so \(\frac{dr}{dt} = \frac{2}{\pi} \approx 0.637\) cm s\(^{-1}\).
See 25.5.
8 Paper 2 The tangent to the curve \(y = \ln(x^2 + 1)\) at the point where \(x = 2\) meets the \(x\)-axis at \(Q\). Find the \(x\)-coordinate of \(Q\).
Point \((2, \ln 5)\). \(\frac{dy}{dx} = \frac{2x}{x^2 + 1} = \frac{4}{5}\) at \(x = 2\).
Tangent: \(y - \ln 5 = \frac{4}{5}(x - 2)\). Set \(y = 0\): \(x = 2 - \frac{5}{4}\ln 5 \approx -0.0118\).
See 25.1.
9 Paper 2 AHL Water is poured into a hemispherical bowl of radius \(10\) cm at \(20\) cm\(^3\) s\(^{-1}\). When the water is \(h\) cm deep, its volume is \(V = \frac{\pi}{3}h^2(30 - h)\) cm\(^3\). Find the rate at which the depth is increasing when \(h = 4\).
\(V = 10\pi h^2 - \frac{\pi}{3}h^3\), so \(\frac{dV}{dh} = 20\pi h - \pi h^2 = \pi(80 - 16) = 64\pi\) at \(h = 4\).
\(\frac{dh}{dt} = \frac{20}{64\pi} = \frac{5}{16\pi} \approx 0.0995\) cm s\(^{-1}\).
See 25.5.
10 Paper 1 An open-topped cylindrical can (it has a base but no lid) must hold \(27\pi\) cm\(^3\). Find the radius that makes the area of metal used as small as possible, and find this area.
\(\pi r^2 h = 27\pi \Rightarrow h = \frac{27}{r^2}\). \(S = \pi r^2 + 2\pi r h = \pi r^2 + \frac{54\pi}{r}\).
\(\frac{dS}{dr} = 2\pi r - \frac{54\pi}{r^2} = 0 \Rightarrow r^3 = 27 \Rightarrow r = 3\) (then \(h = 3\)).
\(\frac{d^2S}{dr^2} = 2\pi + \frac{108\pi}{r^3} > 0\): minimum. \(S = 9\pi + 18\pi = 27\pi \approx 84.8\) cm\(^2\).
See 25.6.