The Workbook

1. Differentiate term by term: \(\dfrac{dy}{dx} = 6x + 2\).

2. At \(x = 2\): \(6 \times 2 + 2 = 14\).

3. \(6x + 2 = 0\) gives \(x = -\dfrac{1}{3}\). The tangent is horizontal there.

Use \(\dfrac{d}{dx}x^n = nx^{n-1}\).

1. \(5x^4\).

2. \(x^{-2}\) gives \(-2x^{-3} = -\dfrac{2}{x^3}\).

3. \(x^{1/2}\) gives \(\dfrac{1}{2}x^{-1/2} = \dfrac{1}{2\sqrt{x}}\).

4. A constant does not change, so its derivative is \(0\).

1. \(v = \dfrac{dx}{dt} = 15t^2 - 2\).

2. \(a = \dfrac{dv}{dt} = 30t\).

3. At \(t = 1\) s: \(v = 15 - 2 = 13\) m/s and \(a = 30\) m/s\(^2\).

1. \(\cos x\).

2. \(-\sin x\).

3. \(e^{x}\); this function equals its own derivative.

4. \(\dfrac{1}{x}\).

The slope of \(\sin x\) is \(\cos x\), which is first zero at \(x = \dfrac{\pi}{2}\). That is the top of the sine curve.

1. At \(t = 2\) s the curve is at \(x = 2^2 = 4\) m.

2. Slope \(= \dfrac{8 - 0}{3 - 1} = 4\) m/s.

3. \(\dfrac{dx}{dt} = 2t = 4\) m/s at \(t = 2\) s. The derivative equals the slope of the tangent, as it must.

1. The line passes through \((0, 1)\) and \((3, 7)\). Slope \(= \dfrac{7 - 1}{3 - 0} = 2\); intercept \(= 1\). So \(y = 2x + 1\).

2. \(\dfrac{dy}{dx} = 2\) at every \(x\), so it is \(2\) at \(x = 1\) and at \(x = 2\). A straight line has the same slope everywhere.

1. At \(P\) the curve rises, so the slope is positive. At \(Q\) the curve is at its top, so the slope is zero. At \(R\) the curve falls, so the slope is negative.

2. \(\dfrac{dy}{dx} = 4 - 2x\). At \(x = 1\): \(+2\). At \(x = 2\): \(0\). At \(x = 3\): \(-2\). The signs agree.

3. \(y = 4 \times 2 - 2^2 = 4\).

1. \(v = 20\) m/s at \(t = 0\) and \(v = 0\) at \(t = 4\) s.

2. Slope \(= \dfrac{0 - 20}{4 - 0} = -5\) m/s\(^2\). The slope of a \(v\)-\(t\) graph is the acceleration.

3. The slope is negative, so the velocity decreases. The car slows down and stops at \(t = 4\) s.

Chain rule: differentiate the outer function, then multiply by the derivative of the inner one.

1. \(3(2x + 1)^2 \times 2 = 6(2x + 1)^2\).

2. \(\cos 3t \times 3 = 3\cos 3t\).

3. \(e^{-2t} \times (-2) = -2e^{-2t}\).

4. \(-\sin(\omega t + \phi) \times \omega = -\omega\sin(\omega t + \phi)\).

Product rule: \((uv)' = u'v + uv'\).

1. \(2x\sin x + x^2\cos x\).

2. \(1 \times e^{t} + t\,e^{t} = e^{t}(1 + t)\).

Quotient rule: \(\left(\dfrac{u}{v}\right)' = \dfrac{u'v - uv'}{v^2}\).

1. \(\dfrac{1 \times (x + 1) - x \times 1}{(x + 1)^2} = \dfrac{1}{(x + 1)^2}\).

2. \(\dfrac{x\cos x - \sin x}{x^2}\).

1. \(y = (1 + x^2)^{1/2}\), so \(\dfrac{dy}{dx} = \dfrac{1}{2}(1 + x^2)^{-1/2} \times 2x = \dfrac{x}{\sqrt{1 + x^2}}\).

2. At \(x = \dfrac{3}{4}\): \(\sqrt{1 + \dfrac{9}{16}} = \dfrac{5}{4}\), so the slope is \(\dfrac{3/4}{5/4} = \dfrac{3}{5}\).

\(\dfrac{dV}{dt} = \dfrac{dV}{dr}\dfrac{dr}{dt} = 4\pi r^2\dfrac{dr}{dt}\).

\(= 4\pi \times 100 \times 0.2 = 80\pi \approx 251\) cm\(^3\)/s.

1. \(i = \dfrac{dq}{dt} = 4t + 3\). At \(t = 2\) s: \(i = 11\) A.

2. \(\dfrac{di}{dt} = 4\) A/s, the same at every instant.

1. The lamp, the top of the head and the tip of the shadow lie on one line. The two right triangles are similar: \(\dfrac{2}{s} = \dfrac{6}{x + s}\). So \(2x + 2s = 6s\), which gives \(s = \dfrac{x}{2}\).

2. \(\dfrac{ds}{dt} = \dfrac{1}{2}\dfrac{dx}{dt} = \dfrac{1}{2} \times 2 = 1\) m/s.

3. The tip is at \(x + s = \dfrac{3x}{2}\) from the post. Its speed is \(\dfrac{3}{2} \times 2 = 3\) m/s.

\(A = x^2\), so \(\dfrac{dA}{dt} = 2x\dfrac{dx}{dt} = 2 \times 4 \times 0.5 = 4\) cm\(^2\)/s.

1. \(v = \dfrac{dx}{dt} = -8\sin 2t\) and \(a = \dfrac{dv}{dt} = -16\cos 2t\).

2. \(a = -16\cos 2t = -4(4\cos 2t) = -4x\). The acceleration is proportional to \(x\) and opposite to it, which is simple harmonic motion.

3. \(|v| = 8|\sin 2t|\) is largest when \(\sin 2t = \pm 1\): \(v_{\max} = 8\) m/s.

1. \(K = \dfrac{1}{2}mv^2 = \dfrac{1}{2} \times 4 \times 4t^2 = 8t^2\) J.

2. \(P = \dfrac{dK}{dt} = 16t = 48\) W at \(t = 3\) s.

1. \(\dfrac{dy}{dx} = 2x - 6 = 0\) gives \(x = 3\).

2. \(\dfrac{d^2y}{dx^2} = 2 > 0\), so the curve bends upward there. It is a minimum.

3. \(y_{\min} = 9 - 18 + 5 = -4\).

\(\dfrac{dy}{dx} = 12 - 3x^2 = 0\) gives \(x = \pm 2\).

\(\dfrac{d^2y}{dx^2} = -6x\). At \(x = 2\) this is \(-12 < 0\): a maximum, \(y = 24 - 8 = 16\). At \(x = -2\) it is \(+12 > 0\): a minimum, \(y = -24 + 8 = -16\).

1. \(F = -\dfrac{dU}{dx} = -(2x - 4) = 4 - 2x\). \(F = 0\) at \(x = 2\) m, the bottom of the curve.

2. \(\dfrac{d^2U}{dx^2} = 2 > 0\), so \(U\) is a minimum at \(x = 2\) m. A small push either way raises \(U\), so the force pushes the particle back. The equilibrium is stable.

3. The graph bottoms at \(U = 1\) J. Check: \(U(2) = 4 - 8 + 5 = 1\) J.

1. \(\dfrac{dR}{d\theta} = 40 \times 2\cos 2\theta = 80\cos 2\theta\).

2. \(\cos 2\theta = 0\) gives \(2\theta = 90^\circ\), so \(\theta = 45^\circ\). Also \(\dfrac{d^2R}{d\theta^2} = -160\sin 2\theta < 0\) there, so it is a maximum.

3. \(R_{\max} = 40\sin 90^\circ = 40\) m. This is \(\dfrac{u^2}{g} = \dfrac{400}{10}\).

1. \(\dfrac{dh}{dt} = 20 - 10t = 0\) gives \(t = 2\) s. The velocity is zero at the top.

2. \(h_{\max} = 20 \times 2 - 5 \times 4 = 20\) m.

3. \(\dfrac{d^2h}{dt^2} = -10\) m/s\(^2\): the acceleration due to gravity, downward. It is negative, so \(t = 2\) s gives a maximum.

1. \(A = x(20 - x) = 20x - x^2\).

2. \(\dfrac{dA}{dx} = 20 - 2x = 0\) gives \(x = 10\) cm. \(\dfrac{d^2A}{dx^2} = -2 < 0\), a maximum. The rectangle is a square.

3. \(A_{\max} = 10 \times 10 = 100\) cm\(^2\).