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A stone is dropped into a lake, creating a circular ripple that travels outward at a speed of 43 cm/s. Find the rate which the area within the circle is increasing after

a) 1 second, b) 3 seconds, and c) 5 seconds.
What can you conclude?

1 Answer

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The rate at which the area within the circle is increasing, obtained using the chain rule of differentiation, which indicates a linear relation with the duration of travel are;

a) 11617.61 cm²/s

b) 34852.83 cm²/s

c) 58088.05 cm²/s

There is a linear relationship between the rate of change of the area and the duration of travel of the ripple

What is a linear relation?

A linear relation is one in which the graph of the equation representing the relation is a straight line.

The chain rule of differentiation indicates that we get;

dA/dt = dA/dr × dr/dt

The area of a circle is; A = π·r²

dA/dr = 2·π·r

dr/dt = 43 cm/s

Therefore; dA/dt = 2·π·r × 43

dA/dt = 86·π·r

a) The radius of the ripple after 1 second is; r = 1 s × 43 cm/s = 43 cm

The radius after 1 second is r = 43 cm

The rate at which the area is increasing after 1 second is; 86 × π × 43 ≈ 11617.61 cm²/s

b) The radius, r, after 3 seconds is; 3 × 43 = 129 cm

The rate at which the area is increasing is; 86 × π × 129 ≈ 34852.83 cm²/s

c) The radius, r, after 5 seconds is; 5 × 43 = 215 cm

The rate at which the area is increasing is; 86 × π × 215 ≈ 58088.05 cm²/s

The first difference in the rate of change of the areas are;

34852.83 - 11617.61 = 23235.22

58088.05 - 34852.83 = 23235.22

The first rate of change in the area indicates a linear relationship, therefore, the area increases linearly with duration of travel of the circular ripple

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