Question

A circle is inside a square. The radius of the circle is increasing at a rate of 5 meters per day and the sides of the square are increasing at a rate of 4 meters per day. When the radius is 3 meters, and the sides are 21 meters, then how fast is the AREA outside the circle but inside the square changing

Answer 1

Answer:

Area of the portion outside the circle but inside the square is reduced by 114 m² after the third day, reduced by  239.14284 in the fourth day.

Step-by-step explanation:

First day

Radius of the circle = 3 meters

Length of sides of the square = 21 meters

Area of the circle = π R2   = 22/7 x 32  = 28.285714 m²

Area of the square = L x B = 21 x 21 = 441 m²

Area of area outside the circle but inside the square = 412.71429 m²

Second day

Radius of the circle = 8 meters

Length of sides of the square = 25 meters

Area of the circle = π R2   = 22/7 x 82  = 201.14286 m²  

Area of the square = L x B = 25 x 25 = 625 m²

Area of area outside the circle but inside the square = 423.85714 m²

Third day

Radius of the circle =13 meters

Length of sides of the square = 29 meters

Area of the circle = π R2   = 22/7 x 132  = 531.14286 m²

Area of the square = L x B = 29 x 29 = 841 m²

Area of area outside the circle but inside the square = 309.85714 m²

Fourth day

Radius of the circle =18 meters

Length of sides of the square = 33 meters

Area of the circle = π R²   = 22/7 x 182  =1018.2857 m²  

Area of the square = L x B = 33 x 33 = 1089 m²

Area of area outside the circle but inside the square = 70.7143 m²

 

From the second and third day, it can be noticed that the area of the portion outside the circle but inside the square is reducing changing by half.  

Answer 2

Answer:

The area outside the circle but inside the square is changing at the speed of (168 - 30π)m^2/day or 73.8m^2 per day

Step-by-step explanation:

Let us denote the radius of the circle by "r" and the length of a side of the square by "L".

Then given that the rate of change of the radius of the circle is 5 metres per day;

then dr/dt = 5

Again, given that the rate of change of a side of the square is 4 metres per day;

then dL/dt = 4

Let the area outside the circle but inside the square be "A".

Therefore, A = (area of the square) - (area of circle)

= L^2 - πr^2

We are now required to calculate dA/dt If the initial radius of the circle was 3 metres and each side of the square was 21 metres initially.

i.e find dA/dt

When r = 3 and L = 21

dA/dt = 2L(dL/dt) - 2πr(dr/dt)

= [2×(21)×(4)] - [2π×(3)×(5)]

= (168 - 30π)m^2/day

OR

If π = 3.14

Then: 168 - (30 × 3.14)

=168 - 94.2

= 73.8m^2 per day

The area outside the circle but inside the square is changing at the speed of 73.8m^2 per day