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.
717 exam-style questions on Edexcel A Level Maths Differentiation, 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. Each one has a worked solution and a mark scheme showing where the marks go.