Show that $5cos⁡2t−5cos⁡6t≈80t25\cos 2t - 5\cos 6t \approx 80t^2$ for small values of $t$ .

The output signal of a high-frequency sensor is modeled by the function $\sqrt{\frac{1}{20}(5\cos 2t - 5\cos 6t)}$ . Show that for small positive values of $t$ , the area enclosed by the curve $y = R (t)$ , the $t$ -axis, and the line $t = 0.4$ can be approximated by $Area≈2m×5n\text{Area} \approx 2^m \times 5^n$ where $m$ and $n$ are integers to be found.

Explain why $dt\int_{12.6}^{12.7} 2t \, dt$ is not a suitable approximation for $dt\int_{12.6}^{12.7} R(t) \, dt$ .

Explain how $dt\int_{12.6}^{12.7} R(t) \, dt$ may be approximated by $dt\int_{a}^{b} 2t \, dt$ for suitable values of $a$ and $b$ .

Show that $5cos⁡2t−5cos⁡6t≈80t25\cos 2t - 5\cos 6t \approx 80t^2$ for small values of $t$ .

Explain why $dt\int_{12.6}^{12.7} 2t \, dt$ is not a suitable approximation for $dt\int_{12.6}^{12.7} R(t) \, dt$ .

Explain how $dt\int_{12.6}^{12.7} R(t) \, dt$ may be approximated by $dt\int_{a}^{b} 2t \, dt$ for suitable values of $a$ and $b$ .

Radians

Question 63

Markscheme

Radians

Question 63

Markscheme

Radians

Question 63

Radians Questions

Radians

Question 63

Radians Questions