The potential energy UUU (in millijoules) of a micro-mechanical sensor is modeled by the function U(y)=(2+32y)4U(y) = \left(2 + \frac{3}{2}y\right)^4U(y)=(2+23y)4, where yyy is the displacement in micrometers. The expansion of this function is given by:
U(y)=P+Qy+54y2+Ry3+5.0625y4 U(y) = P + Qy + 54y^2 + Ry^3 + 5.0625y^4 U(y)=P+Qy+54y2+Ry3+5.0625y4Find the value of PPP and the value of QQQ.
The net energy flux Φ\PhiΦ in the sensor is defined by the difference between two opposing displacement states:
Φ=(2+32y)4−(2−32y)4 \Phi = \left(2 + \frac{3}{2}y\right)^4 - \left(2 - \frac{3}{2}y\right)^4 Φ=(2+23y)4−(2−23y)4Show that Φ=Cy+Dy3\Phi = Cy + Dy^3Φ=Cy+Dy3, where CCC and DDD are constants to be found.
Hence, or otherwise, calculate:
∫((2+32y)4−(2−32y)4)dy \int \left( \left(2 + \frac{3}{2}y\right)^4 - \left(2 - \frac{3}{2}y\right)^4 \right) dy ∫((2+23y)4−(2−23y)4)dy144 exam-style questions on AQA A Level Maths 1.7.1 Binomial expansion. Each one has a worked solution and a mark scheme showing where the marks go.