) Find the average rate of change of the area of a circle with respect to its radius r as r changes from (i) 2 to 3 (ii) 2 to 2.5 (iii) 2 to 2.1 (b) Find the instantaneous rate of change when r − 2. (c) Show that the rate of change of the area of a circle with respect to its radius (at any r) is equal to the circumference of the circle. Try to explain geometrically why this is true by drawing a circle whose radius is increased by an amount Dr. How can you approximate the resulting change in area DA if Dr is small?

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 =

(i) 2 to 3:

ΔA / Δr = = 5π

(ii) 2 to 2.5

ΔA / Δr = = 4.5π

(iii) 2 to 2.1

ΔA / Δr = = 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:

= 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.

logv3(x+2)=1

rewrite in exponential form:

logb(x) = y is equal to b^y =x

b = 3, y = 1

now solve for x:

3^1 = x+2)

x = 3-2

x = 1

see the picture for the graph.