If in culture of yeast the rate of growth y'(t) is proportional to the amount y(t ) present at time t, and if y(t) doubles in 1 day, how much can be expected after 3 days at the same rate of the growth? After 1 week

 
 (Hint: Continuous growth model)
 

Since we are "doubling" it is easy to use a base of 2

amount = a(2)^(t/1)
= a (2)^t , where t ≥ 0, and t=0 ---> now

after 1 day we have 2a
after 3 days , we have a(2)^3 or 8a
after 1 week , we have a(2)^7 = 128a

normally we would use base e
here we have
amount = a e^(kt), where k is a constant
so after 1 day:
2a = a e^k
2 = e^k
k = ln2

amount = a e^(ln2 t)
checking with our previous answer:
if t = 7
amount = a e(7ln2) = 128a

Well, let me first put on my comedy cap and say that yeast must be quite the overachiever if it's doubling its growth in just one day! It's like the Usain Bolt of microorganisms!

Now, to answer your question on a more serious note, we can use the continuous growth model to figure this out. Since the rate of growth, y'(t), is proportional to the amount y(t) at any given time, we can express this relationship as:

y'(t) = ky(t)

Here, k is the constant of proportionality. Now, if y(t) doubles in 1 day, we can say that:

y'(t) = ky(t) = 2y(t)

To solve this equation, we can simply divide both sides by y(t):

ky'(t) / y(t) = 2

Now, integrating both sides, we get:

∫(1/y(t)) dy(t) = ∫2 dt

This simplifies to:

ln(y(t)) = 2t + C

Here, C is the constant of integration. Exponentiating both sides, we have:

y(t) = e^(2t+C) = Ce^(2t)

Now, we need to find the value of C. Since we know that y(0) = y_0, where y_0 is the initial amount of yeast, we substitute these values into the equation:

y(0) = Ce^(2*0) = C

So, C = y_0. Thus, the equation becomes:

y(t) = y_0 * e^(2t)

Now we're ready to tackle your question! After 3 days, the quantity of yeast would be:

y(3) = y_0 * e^(2*3) = y_0 * e^6

And after 1 week (7 days), we have:

y(7) = y_0 * e^(2*7) = y_0 * e^14

So, after 3 days, the amount of yeast can be expected to be y_0 * e^6, and after 1 week, it would be y_0 * e^14.

Now, if you'll excuse me, I think I've worked up quite an appetite talking about yeast. Time for some bread and butter jokes!

In a continuous growth model, the rate of growth is proportional to the amount present at any given time. Let's denote the rate of growth as y'(t) and the amount at time t as y(t).

Given that y'(t) is proportional to y(t), we can write this relationship as y'(t) = ky(t), where k is the constant proportionality factor.

We are also given that y(t) doubles in 1 day. This means that after 1 day, the amount becomes 2y(t), which can be written as y(t+1) = 2y(t).

To solve for k, we can differentiate both sides of the equation y(t+1) = 2y(t) with respect to t:

(y(t+1))' = (2y(t))'

Taking the derivative of y(t+1) with respect to t, we get:

y'(t+1) = 2y'(t)

Since y'(t) = ky(t), we can substitute this into the equation:

ky'(t+1) = 2ky(t)

Dividing both sides by k, we get:

y'(t+1) = 2y(t)

Notice that this equation is the same as the original equation y'(t) = ky(t). This implies that k is a constant and remains the same over time.

Now that we know k is a constant, we can solve for it using given information. We know that if y(t) doubles after 1 day, then y(t+1) = 2y(t):

y(t+1) = 2y(t)
ky'(t+1) = 2ky(t)
k(2y(t)) = 2ky(t)
2k = 2k

This equation holds true, which confirms that k is a constant.

Since we have the continuous growth model with the rate of growth equal to y'(t) = ky(t), we can use this model to find the amount after a given time period.

After 3 days, we want to find y(3):

y'(t) = ky(t)
y'(t) = k * y(t)

Since y'(t) = dy(t)/dt, we can rewrite the equation as:

dy(t)/dt = k * y(t)

Separating variables, we get:

1/y(t) * dy(t) = k * dt

Integrating both sides, we have:

∫[1/y(t)] dy(t) = k ∫dt

ln|y(t)| = kt + C

Exponentiating both sides, we get:

|y(t)| = e^(kt + C)

Considering the initial condition that y(0) = y0, we can substitute y(0) for t=0:

|y0| = e^(k * 0 + C)
|y0| = e^C
C = ln|y0|

Therefore, the equation becomes:

|y(t)| = e^(kt + ln|y0|)
|y(t)| = e^(kt) * e^(ln|y0|)
|y(t)| = e^(kt) * |y0|

Since we are dealing with quantities, the absolute value can be disregarded:

y(t) = e^(kt) * y0

Now we can use this equation to find the amount after 3 days:

y(3) = e^(k*3) * y0

We are not given the value of k or y0, so we cannot determine the exact amount without more information. However, this equation gives us the general formula to calculate the amount after a given time using the continuous growth model.

To find the amount after 1 week (7 days), you can use the same equation:

y(7) = e^(k*7) * y0

Similarly, without the values of k and y0, we cannot determine the exact amount.

To solve this problem, we can use the concept of continuous growth model, where the rate of growth y'(t) is proportional to the amount y(t) present at time t.

Let's denote the constant of proportionality as k. Therefore, we have the following differential equation:

y'(t) = k * y(t)

We are given that y(t) doubles in 1 day. This means that if we have y(0) amount of yeast at time t = 0, then after 1 day (t = 1), y(1) = 2 * y(0).

To find the value of k, we can substitute these values into the differential equation:

k * y(0) = 2 * y(0)

This simplifies to:

k = 2

Now, we can solve the differential equation to find y(t). First, let's rewrite it:

y'(t) = 2 * y(t)

Separate the variables:

1 / y(t) * dy = 2 * dt

Integrate both sides:

∫(1 / y(t) * dy) = ∫(2 * dt)

ln|y(t)| = 2t + C

where C is the constant of integration.

To find the value of C, we can use the initial condition y(0) = y(0). Substitute these values in:

ln|y(0)| = 2 * 0 + C

ln|y(0)| = C

Now, we can rewrite the equation as:

ln|y(t)| = 2t + ln|y(0)|

Exponentiate both sides:

|y(t)| = e^(2t + ln|y(0)|)

Since y(t) represents the amount of yeast, it cannot be negative. Therefore, we can drop the absolute value:

y(t) = e^(2t + ln|y(0)|)

Finally, we can find the amount of yeast after 3 days and after 1 week by substituting the given time values into this equation:

For 3 days (t = 3):

y(3) = e^(2 * 3 + ln|y(0)|)

For 1 week (t = 7):

y(7) = e^(2 * 7 + ln|y(0)|)

Please note that the exact values for y(3) and y(7) will depend on the initial amount of yeast y(0).