Jordan is attempting to use differentiation from first principles to prove that the rate of change of the displacement of a pendulum, given by s(t)=sints(t) = \sin ts(t)=sint, is −1-1−1 at the instant where t=πt = \pit=π.
Jordan's teacher points out that mistakes were made starting in Step 4 of the derivation. The working is shown below.
Step 1: Gradient of chord PQ=sin(π+h)−sin(π)hPQ = \frac{\sin(\pi + h) - \sin(\pi)}{h}PQ=hsin(π+h)−sin(π)
Step 2: =sin(π)cos(h)+cos(π)sin(h)−sin(π)h= \frac{\sin(\pi)\cos(h) + \cos(\pi)\sin(h) - \sin(\pi)}{h}=hsin(π)cos(h)+cos(π)sin(h)−sin(π)
Step 3: =sin(π)(cos(h)−1h)+cos(π)(sin(h)h)= \sin(\pi)\left(\frac{\cos(h) - 1}{h}\right) + \cos(\pi)\left(\frac{\sin(h)}{h}\right)=sin(π)(hcos(h)−1)+cos(π)(hsin(h))
Step 4: For the rate of change at t=πt = \pit=π, let h=0h = 0h=0 then
cos(h)−1h=1 and sin(h)h=0 \frac{\cos(h) - 1}{h} = 1 \text{ and } \frac{\sin(h)}{h} = 0 hcos(h)−1=1 and hsin(h)=0Step 5: Hence the rate of change is given by
sin(π)×1+cos(π)×0=0 \sin(\pi) \times 1 + \cos(\pi) \times 0 = 0 sin(π)×1+cos(π)×0=0Complete Steps 4 and 5 of Jordan's working to correct the proof.
Practise AQA A Level Maths 1.10 G: Differentiation with exam-style questions for A Level Maths. 321 questions covering 1.10.1 The derivative and second derivative, 1.10.2 Differentiating standard functions, 1.10.3 Applications of differentiation, 1.10.4 Product, quotient and chain rules (A-level only), 1.10.5 Implicit and parametric differentiation (A-level only), and 1.10.6 Constructing differential equations (A-level only), matched to the AQA A Level Maths (7357) specification and written in Paper 1, Paper 2 and Paper 3 style. Every question includes a full worked solution and mark scheme, so you can see where marks are awarded rather than just whether you got the answer right.
