The vertical displacement HHH, in millimetres, of a high-precision camera sensor at time t t\,t seconds is modeled by the function:
H(t)=sect+5t−7,0<t<π2 H(t) = \sec t + 5t - 7, \quad 0 < t < \frac{\pi}{2} H(t)=sect+5t−7,0<t<2πA calibration reset is triggered when the displacement is zero, at time t=αt = \alphat=α.
Show that 1.0<α<1.11.0 < \alpha < 1.11.0<α<1.1.
Show that α \alpha\,α is a solution of the equation
t=75−15cost t = \frac{7}{5} - \frac{1}{5\cos t} t=57−5cost1The iterative formula
tn+1=75−15costn t_{n+1} = \frac{7}{5} - \frac{1}{5\cos t_n} tn+1=57−5costn1is used to approximate the value of α\alphaα.
Starting with t1=1.05t_1 = 1.05t1=1.05 and using the iterative formula: (i) find, to 4 decimal places, the value of t2t_2t2. (ii) find, to 4 decimal places, the value of α\alphaα.
133 exam-style questions on OCR (MEI) A Level Maths 1.10 Numerical Methods (A-level only), covering 1.10.1 Locate roots by change of sign (A-level only), 1.10.2 When change of sign methods fail (A-level only), 1.10.3 Fixed point iteration (A-level only), 1.10.4 Newton-Raphson method (A-level only), 1.10.5 Convergence of iterations (A-level only), 1.10.6 Trapezium rule (A-level only), 1.10.7 Upper and lower bounds using rectangles (A-level only), and 1.10.8 Numerical methods to solve problems (A-level only). Each one has a worked solution and a mark scheme showing where the marks go.