A-Level Differential equations: worked solution

Question

The duckweed patch satisfies r=√(100 − 99 e^{−kt}) with k>0, starting at r=1 m. Sketch r against t. [3]

A sketch earns its marks by showing the three features you can prove: the starting point, the direction and bend of the curve, and the long-term behaviour. Work each one out before drawing. 1. Starting point (t = 0). r = √(100 − 99e⁰) = √(100 − 99) = √1 = 1 So the curve passes through (0, 1), which matches the stated starting radius of 1 m. 2. Long-term behaviour (t → ∞). Since k > 0, e^{−kt} → 0 as t → ∞, so r → √(100 − 0) = 10 Also 99e^{−kt} > 0 for every t, so 100 − 99e^{−kt} < 100 and r < 10 always: the curve approaches r = 10 from below and never reaches it. So r = 10 is a horizontal asymptote. 3. The curve is increasing. Square first, then differentiate implicitly (easier than differentiating the root): r² = 100 − 99e^{−kt} 2r (dr/dt) = 99k e^{−kt} dr/dt = 99k e^{−kt} / (2r) k > 0, e^{−kt} > 0 and r > 0, so dr/dt > 0 for all t: r increases throughout. 4. The curve is concave down. As t increases, e^{−kt} decreases and r increases, so the numerator of dr/dt falls while the denominator rises. Hence dr/dt is always decreasing, i.e. d²r/dt² < 0, and the curve bends downwards everywhere. There is no point of inflexion and no maximum. Sketch (t ≥ 0 only, since t is time): - Begin at the point (0, 1), clearly labelled. - Rise steeply at first, then flatten out, with no turning point and no crossing. - Draw a dashed horizontal line at r = 10 and label it, with the curve creeping up to it from below. - Label the axes t and r (r in metres).