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 Edexcel A Level Maths Differentiation with exam-style questions for A Level Maths. 311 questions covering 9.1 Differentiating sin x and cos x, 9.2 Differentiating exponentials and logarithms, 9.3 The Chain Rule, 9.4 The Product Rule, 9.5 The Quotient Rule, 9.6 Differentiating Trigonometric Functions, 9.7 Parametric Differentiation, 9.8 Implicit Differentiation, 9.9 Using Second Derivatives, and 9.10 Rates of Change, matched to the Edexcel A Level Maths (9MA0) 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.
