Show that 5cos2t−5cos6t≈80t25\cos 2t - 5\cos 6t \approx 80t^25cos2t−5cos6t≈80t2 for small values of ttt.
The output signal of a high-frequency sensor is modeled by the function R(t)=120(5cos2t−5cos6t)R(t) = \sqrt{\frac{1}{20}(5\cos 2t - 5\cos 6t)}R(t)=201(5cos2t−5cos6t). Show that for small positive values of ttt, the area enclosed by the curve y=R(t)y = R(t)y=R(t), the ttt-axis, and the line t=0.4t = 0.4t=0.4 can be approximated by Area≈2m×5n\text{Area} \approx 2^m \times 5^nArea≈2m×5n where mmm and nnn are integers to be found.
Explain why ∫12.612.72t dt\int_{12.6}^{12.7} 2t \, dt∫12.612.72tdt is not a suitable approximation for ∫12.612.7R(t) dt\int_{12.6}^{12.7} R(t) \, dt∫12.612.7R(t)dt.
Explain how ∫12.612.7R(t) dt\int_{12.6}^{12.7} R(t) \, dt∫12.612.7R(t)dt may be approximated by ∫ab2t dt\int_{a}^{b} 2t \, dt∫ab2tdt for suitable values of aaa and bbb.
Practise Edexcel A Level Maths 5.5 Small Angle Approximations with exam-style questions for A Level Maths. 25 questions, 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.