mensuration

A circular pipe has outer diameter 4 cm and thickness t cm.
a) Show that the area of the cross section, A cm², is given by A=pi(4t-t²)
b) Find the rate of increase of A with respect to t when t=1/4 and when t=1/2

Outer diameter = 4
outer radius = 2

inner diameter = (4 - 2t)
inner radius = (4 - 2t)/2 = 2 - t
cross sectional area = A = outer area - inner area = π{(2²) - (2- t)²} = π{4 - 4 + 4t – t²} = π(4t – t²)

Therefore, rate of change of area w.r.t. thickness t is

dA/dt = π (4 – 2t) = 2π (2 – t)

So, when t = ¼, we have dA/dt = 2π (2 – ¼ ) = 7π/2

Similarly, when t = ½ , we have dA/dt = 2π (2 – ½) = 3π