Answer and Step-by-step explanation:
The rate of change of an area of a circle in respect to tis radius is given by:
ΔA / Δr = [tex]\frac{\pi(r_{1}^{2} - r_{2}^{2}) }{r_{1} - r_{2}}[/tex]
(i) 2 to 3:
ΔA / Δr = [tex]\frac{\pi(3^{2} - 2^{2}) }{3-2}[/tex] = 5π
(ii) 2 to 2.5
ΔA / Δr = [tex]\frac{\pi(2.5^{2} - 2^{2}) }{2.5-2}[/tex] = 4.5π
(iii) 2 to 2.1
ΔA / Δr = [tex]\frac{\pi(2.1^{2}-2^{2})}{2.1-2}[/tex] = 4.1π
Instantaneous rate of change is the change of a rate at a specific value. In this case, it wants at r = 2, so take the derivative of area and determine the area with the specific radius:
[tex]\frac{dA}{dr}[/tex] = π.r²
A'(r) = 2.π.r (1)
A'(2) = 4π
Note that expression (1) is the circumference of a circle, so it is shown that to determine the instantaneous rate of change of the are of a circle, is the circumference of any given radius.
The circles in the attachment shows a circle with radius r (light green) and a circle with radius r + Δr (darker green).
The rate of change between the two circles is the area shown by the arrow:
ΔA = π,[r² - (r+Δr)²]
ΔA = π.[r² - r² + 2rΔr + Δr²]
ΔA = π.(2rΔr + Δr²)
If Δr is too small, Δr² will be approximately zero, so
Δr²≈0
ΔA = π.(2rΔr)
The rate of change with respect to its radius will be:
ΔA / Δr = π.(2rΔr) / Δr
ΔA / Δr = 2.π.r
which is the circumference of a circle, so, this is the geometrical proof.
