The area, A cm3 , of a circle increases at a constant rate of 2 cm2 ´ s -1 . If the
initial area of A is 1 cm2 , show that the radius of the circle at time t is given by
r=sqrt( 2t+1/ pi)
sqrt on both denominator and numerator
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To find the radius of the circle at time t, we can use the formula for the area of a circle:
A = πr^2
Given that the area of the circle increases at a constant rate of 2 cm^2/s, we can write the rate of change of the area as:
dA/dt = 2 cm^2/s
Since we are interested in finding the radius at time t, we need to express the rate of change of the area in terms of the radius, by using the chain rule:
dA/dt = dA/dr * dr/dt,
Where dA/dr represents the change in area with respect to radius, and dr/dt represents the change in radius with respect to time.
From the formula for the area of a circle, we can calculate dA/dr:
dA/dr = 2πr.
Now, we can rewrite the equation dA/dt = dA/dr * dr/dt, using the values we have:
2 = 2πr * dr/dt.
We are given that the initial area A is 1 cm^2, so when t = 0, the area is 1 cm^2:
A = πr^2 = 1 cm^2.
Solving for r in terms of A:
r^2 = 1/π,
r = sqrt(1/π) = 1/sqrt(π).
Now, let's solve for dr/dt:
2 = 2πr * dr/dt.
Substitute the value of r:
2 = 2π * (1/sqrt(π)) * dr/dt,
1 = sqrt(π) * dr/dt,
dr/dt = 1 / sqrt(π).
Now, we can integrate dr/dt with respect to t to find the radius as a function of time:
∫dr/dt dt = ∫(1 / sqrt(π)) dt,
r = ∫(1 / sqrt(π)) dt.
Integrating both sides:
r = (1 / sqrt(π)) * t + C.
Since the initial radius r_0 at t = 0 is 1 / sqrt(π):
1 / sqrt(π) = (1 / sqrt(π)) * 0 + C,
C = 1 / sqrt(π).
Therefore, the expression for the radius as a function of time t is:
r = (1 / sqrt(π)) * t + 1 / sqrt(π).
Simplifying this expression:
r = (t + 1) / sqrt(π).
Finally, if we want to express the radius in terms of (2t + 1) and π, we substitute t + 1 for 2t + 1:
r = ((2t + 1) / 2) / sqrt(π),
r = (2t + 1) / (2 sqrt(π)),
r = sqrt((2t + 1) / π).
Therefore, the radius of the circle at time t is given by:
r = sqrt((2t + 1) / π).