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
I don't know if you are the same person, but this question is a cut-and-past of
http://www.jiskha.com/display.cgi?id=1422453360
areas are measured in cm^2 , not cm^3
I cannot understand 2 cm2 ´ s -1
we know that
dA/dt = 2
and A can be written as pi*r^2
inital value of A, meaning t=0, is 1
from that, we get the equation
A = 2t+1
example:
t = 2
A = 2(2) + 1 = 5
1 + 2 + 2 = 5
we can then write
pi*r^2 = 2t + 1
r^2 = 2t + 1/pi
r = sqrt. 2t+1/pi
To determine 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 increases at a constant rate of 2 cm^2/s, we can express this relationship as:
dA/dt = 2
Where dA/dt represents the rate of change of the area with respect to time t.
By differentiating the equation for the area of a circle with respect to time, we get:
dA/dt = 2πr(dr/dt)
Since dA/dt = 2, we can rewrite the equation as:
2 = 2πr(dr/dt)
Now, let's solve for dr/dt:
(dr/dt) = 2/(2πr)
(dr/dt) = 1/(πr)
To find the radius at time t, we need to integrate both sides with respect to t:
∫(dr/dt) dt = ∫[1/(πr)] dt
Integrating both sides gives:
r(t) = ∫[1/(πr)] dt
To evaluate this integral, we need to determine the bounds. We know that at time t=0 (the initial time), the area A is 1 cm^2. Substituting this into the area formula:
1 = πr(0)^2
1 = πr(0)
r(0) = 1/π
Now, we can solve for r(t):
r(t) = ∫[1/(πr)] dt + r(0)
Integrating 1/(πr) with respect to t gives:
r(t) = (1/π)∫(1/r) dt + r(0)
r(t) = (1/π)ln(r) + r(0)
To find the value of r, we can rearrange the equation:
r(t) - r(0) = (1/π)ln(r)
Substituting r(0) = 1/π and rearranging, we get:
r(t) - 1/π = (1/π)ln(r)
Now, let's solve for r:
r(t) - 1/π = (1/π)ln(r)
π(r(t) - 1/π) = ln(r)
e^(π(r(t) - 1/π)) = r
Finally, let's simplify the expression:
e^(π(r(t) - 1/π)) = r
e^(πr(t) - 1) = r
e^(πr(t)) = r * e
Taking the square root of both sides:
e^(πr(t)/2) = sqrt(r * e)
Now we have the equation:
e^(πr(t)/2) = sqrt(r * e)
Simplifying the equation further:
πr(t)/2 = ln(sqrt(r * e))
πr(t)/2 = (1/2)ln(r * e)
Multiplying both sides by 2 and dividing by π:
r(t) = (1/π)ln(r * e)
Since ln(r * e) = ln(r) + ln(e) = ln(r) + 1, we can rewrite the equation:
r(t) = (1/π)ln(r) + (1/π)
Therefore, the radius r at time t is given by:
r(t) = (1/π)ln(r) + (1/π) = sqrt((2t + 1)/π)
Note that the square root symbol should be applied to the entire expression (2t + 1) in the numerator and π in the denominator.