(Growth of Cells) Suppose that after t hrs, there are p (t) cells present in a culture, where P(t) = 5000e*^2t
a) How many cells were present initially?
b) Give a differential equation satisfied by P(t)
c) When will the population double?
d) When will 20,000 cells be present?
a) To find the initial number of cells present, we need to know the value of P(0). Substitute t=0 into the equation:
P(0) = 5000e^(2*0)
P(0) = 5000e^0
P(0) = 5000 * 1
P(0) = 5000
So, the initial number of cells present is 5000.
b) To find a differential equation satisfied by P(t), we differentiate P(t) with respect to t:
dP(t)/dt = d(5000e^(2t))/dt
Using the chain rule, the derivative of e^(2t) is 2e^(2t):
dP(t)/dt = 5000 * 2e^(2t)
dP(t)/dt = 10,000e^(2t)
Therefore, the differential equation satisfied by P(t) is dP(t)/dt = 10,000e^(2t).
c) To find when the population doubles, we need to find the value of t for which P(t) is twice the initial value:
2 * P(0) = P(t)
Substituting P(0) = 5000:
2 * 5000 = 5000e^(2t)
Divide both sides by 5000:
2 = e^(2t)
Taking the natural logarithm of both sides to solve for t:
ln(2) = 2t
Divide by 2:
t = ln(2) / 2
Calculate the value of t to find when the population will double.
d) To find when 20,000 cells will be present, we need to find the value of t for which P(t) is equal to 20,000:
P(t) = 20,000
Substituting the equation for P(t):
5000e^(2t) = 20,000
Divide both sides by 5000:
e^(2t) = 4
Take the natural logarithm of both sides to solve for t:
ln(e^(2t)) = ln(4)
Using the logarithmic property, the exponent comes down:
2t * ln(e) = ln(4)
Since ln(e) equals 1, we have:
2t = ln(4)
Divide by 2:
t = ln(4) / 2
Calculate the value of t to find when 20,000 cells will be present.
a) To find the initial number of cells, we need to find P(0). Substituting t = 0 into the equation, we have:
P(0) = 5000e^(2*0)
P(0) = 5000e^0
P(0) = 5000 * 1
P(0) = 5000
So, the initial number of cells is 5000.
b) To find the differential equation satisfied by P(t), we can differentiate P(t) with respect to t.
dP/dt = d/dt(5000e^(2t))
dP/dt = 5000 * d/dt(e^(2t))
dP/dt = 5000 * 2e^(2t)
Therefore, the differential equation satisfied by P(t) is dP/dt = 10000e^(2t).
c) To find when the population doubles, we need to solve the equation P(t) = 2P(0).
2P(0) = 5000e^(2t)
2 * 5000 = 5000e^(2t)
10000 = 5000e^(2t)
Divide both sides by 5000:
2 = e^(2t)
Take the natural logarithm of both sides:
ln(2) = ln(e^(2t))
Use the property of logarithms:
ln(2) = 2t * ln(e)
Since ln(e) = 1, the equation simplifies to:
ln(2) = 2t
Solve for t:
t = ln(2)/2
Using a calculator, t ≈ 0.34657359.
So, the population will double approximately after 0.3466 hours.
d) To find when 20,000 cells will be present, we need to solve the equation P(t) = 20,000.
20,000 = 5000e^(2t)
Divide both sides by 5000:
4 = e^(2t)
Take the natural logarithm of both sides:
ln(4) = ln(e^(2t))
Use the property of logarithms:
ln(4) = 2t * ln(e)
Again, since ln(e) = 1, the equation simplifies to:
ln(4) = 2t
Solve for t:
t = ln(4)/2
Using a calculator, t ≈ 0.69314718.
So, approximately after 0.6931 hours, 20,000 cells will be present.
a. "initially" means time = 0
P(0) = 5000e^2(0) = 5000e^0 = 5000
b. P(t) = 5000*e^(2t)
[d/du]e^u = e^u du
Therefore, P'(t) = 5000*2e^(2t)
P'(t) = 10000e^(2t)
c. Set e^(2t) = 2.
2t = ln2
t = (ln2)/2
The population will double every (ln2)/2 hours.
d. set P = 20000 and solve for t.
20000 = 5000e^(2t)
4 = e^(2t)
ln4 = 2t
2ln2 = 2t
ln2 = t