A circle with a radius of 5cm has it radius increasing at the rate of 0.2cm/s. What will be the corresponding increase in the area?
rate of area increase = circumference * rate of radius increase = 2π cm^2/s
Its good
r=5cm; dr/dt=0.2cm/s
Area of Circle, A=πr²
Required to find: Increase in area=dA/dt
dA/dt= (dA/dr) x (dr/dt)
But dA/dr=2πr
Hence,
dA/dt=(2πr) x (0.2)
=0.4πr
Since r=5cm,
dA/dt=0.4π x 5
=2π cm²/s
To find the corresponding increase in the area of a circle when the radius is increasing at a certain rate, you need to use the formula for the area of a circle: A = πr^2, where A is the area and r is the radius.
Given that the radius is increasing at a rate of 0.2cm/s, we can calculate the increase in the radius over a certain time period by multiplying the rate (0.2cm/s) by the time (in seconds).
Let's say we want to find the corresponding increase in the area over a time period of t seconds. The increase in the radius would be 0.2t cm.
Now, we need to find the area of the circle with the initial radius (5cm) and the area of the circle with the increased radius (5 + 0.2t cm).
Let's calculate the two areas:
Initial Area (A1) = π(5cm)^2
Increased Area (A2) = π(5 + 0.2t cm)^2
The corresponding increase in the area is the difference between the increased area and the initial area:
Corresponding Increase in Area = A2 - A1
Now, let's substitute the values and calculate the increase in the area:
Corresponding Increase in Area = π(5 + 0.2t cm)^2 - π(5cm)^2
Simplifying further using algebraic identities:
Corresponding Increase in Area = π[(25 + 2(0.2t) + (0.2t)^2) - 25]
= π[25 + 0.4t + 0.04t^2 - 25]
= π(0.4t + 0.04t^2)
Therefore, the corresponding increase in the area of the circle is given by π(0.4t + 0.04t^2) square centimeters.