A chemical plant food loses effectiveness at a rate proportional to the amount
present in the soil. The amount grams of plant food effective after days
satisfies the differential equation

(a) Find the general solution for in terms of where the initial amount of
plant food is grams.
(b) Find the value of if, after 30 days, only half the initial amount of plant
food is effective.
© What percentage of the original amount of plant food is effective after 35
days?
(d) The plant food has to be renewed when its effectiveness falls below .
Is the manufacturer of the plant food justified in calling its product “sixty
day super food”?

Ask by Gray Hammond. in the United Kingdom

Mar 26,2025

Question

A chemical plant food loses effectiveness at a rate proportional to the amount
present in the soil. The amount grams of plant food effective after days
satisfies the differential equation

(a) Find the general solution for in terms of where the initial amount of
plant food is grams.
(b) Find the value of if, after 30 days, only half the initial amount of plant
food is effective.
© What percentage of the original amount of plant food is effective after 35
days?
(d) The plant food has to be renewed when its effectiveness falls below .
Is the manufacturer of the plant food justified in calling its product “sixty
day super food”?

Ask by Gray Hammond. in the United Kingdom

Mar 26,2025

UpStudy AI · Accepted Answer

Let's solve the problem step by step.

### (a) Find the general solution for $$ M $$ in terms of $$ t $$

The differential equation given is:

$$
\frac{d M}{d t} = k M
$$

This is a separable differential equation. We can separate the variables and integrate:

1. Rearranging gives:

$$
\frac{d M}{M} = k \, dt
$$

2. Integrating both sides:

$$
\int \frac{d M}{M} = \int k \, dt
$$

This results in:

$$
\ln |M| = kt + C
$$

3. Exponentiating both sides to solve for $$ M $$:

$$
M = e^{kt + C} = e^C e^{kt}
$$

Let $$ M_0 = e^C $$, which represents the initial amount of plant food. Thus, the general solution is:

$$
M(t) = M_0 e^{kt}
$$

### (b) Find the value of $$ k $$ if, after 30 days, only half the initial amount of plant food is effective.

We know that after 30 days, $$ M(30) = \frac{M_0}{2} $$. Using the general solution:

$$
M(30) = M_0 e^{30k} = \frac{M_0}{2}
$$

Dividing both sides by $$ M_0 $$:

$$
e^{30k} = \frac{1}{2}
$$

Taking the natural logarithm of both sides:

$$
30k = \ln\left(\frac{1}{2}\right)
$$

Thus, we can solve for $$ k $$:

$$
k = \frac{\ln\left(\frac{1}{2}\right)}{30}
$$

### (c) What percentage of the original amount of plant food is effective after 35 days?

Using the value of $$ k $$ found in part (b), we can find $$ M(35) $$:

$$
M(35) = M_0 e^{35k}
$$

Substituting $$ k $$:

$$
M(35) = M_0 e^{35 \cdot \frac{\ln\left(\frac{1}{2}\right)}{30}} = M_0 e^{\frac{35}{30} \ln\left(\frac{1}{2}\right)}
$$

This simplifies to:

$$
M(35) = M_0 \left(\frac{1}{2}\right)^{\frac{35}{30}} = M_0 \left(\frac{1}{2}\right)^{\frac{7}{6}}
$$

To find the percentage of the original amount:

$$
\text{Percentage} = \left(\frac{M(35)}{M_0}\right) \times 100 = \left(\frac{1}{2}\right)^{\frac{7}{6}} \times 100
$$

### (d) The plant food has to be renewed when its effectiveness falls below $$ 25\% $$. Is the manufacturer justified in calling its product "sixty day super food"?

We need to find out when $$ M(t) < 0.25 M_0 $$:

$$
M_0 e^{kt} < 0.25 M_0
$$

Dividing both sides by $$ M_0 $$:

$$
e^{kt} < 0.25
$$

Taking the natural logarithm:

$$
kt < \ln(0.25)
$$

Substituting $$ k $$:

$$
t < \frac{\ln(0.25)}{k}
$$

Now substituting $$ k = \frac{\ln\left(\frac{1}{2}\right)}{30} $$:

$$
t < \frac{\ln(0.25)}{\frac{\ln\left(\frac{1}{2}\right)}{30}} = 30 \cdot \frac{\ln(0.25)}{\ln\left(\frac{1}{2}\right)}
$$

Calculating $$ \ln(0.25) $$ and $$ \ln\left(\frac{1}{2}\right) $$:

$$
\ln(0.25) = \ln\left(\frac{1}{2^2}\right) = 2 \ln\left(\frac{1}{2}\right)
$$

Thus:

$$
t < 30 \cdot 2 = 60
$$

This means the effectiveness falls below $$ 25\% $$ before 60 days, so the manufacturer is not justified in calling it "sixty day super food".

Now, let's calculate the percentage of the original amount of plant food effective after 35 days.

We will compute $$ \left(\frac{1}{2}\right)^{\frac{7}{6}} $$ to find the percentage.
Calculate the value by following steps:
- step0: Calculate:
  $$\left(\frac{1}{2}\right)^{\frac{7}{6}}$$
- step1: Use the properties of exponents:
  $$\frac{1^{\frac{7}{6}}}{2^{\frac{7}{6}}}$$
- step2: Evaluate the power:
  $$\frac{1}{2^{\frac{7}{6}}}$$
- step3: Rewrite the expression:
  $$\frac{1}{2\sqrt[6]{2}}$$
- step4: Multiply by the Conjugate:
  $$\frac{\sqrt[6]{2^{5}}}{2\sqrt[6]{2}\times \sqrt[6]{2^{5}}}$$
- step5: Simplify:
  $$\frac{\sqrt[6]{32}}{2\sqrt[6]{2}\times \sqrt[6]{2^{5}}}$$
- step6: Multiply the numbers:
  $$\frac{\sqrt[6]{32}}{4}$$
The calculation of $$ \left(\frac{1}{2}\right)^{\frac{7}{6}} $$ gives approximately $$ 0.445449 $$.

To find the percentage of the original amount of plant food that is effective after 35 days:

$$
\text{Percentage} \approx 0.445449 \times 100 \approx 44.54\%
$$

### Summary of Results:
- **(a)** The general solution is $$ M(t) = M_0 e^{kt} $$.
- **(b)** The value of $$ k $$ is $$ \frac{\ln\left(\frac{1}{2}\right)}{30} $$.
- **(c)** Approximately $$ 44.54\% $$ of the original amount of plant food is effective after 35 days.
- **(d)** The manufacturer is not justified in calling it "sixty day super food" since the effectiveness falls below $$ 25\% $$ before 60 days.
(a) The general solution is $$ M(t) = M_0 e^{kt} $$. (b) The value of $$ k $$ is $$ \frac{\ln\left(\frac{1}{2}\right)}{30} $$. (c) After 35 days, approximately $$ 44.54\% $$ of the original plant food remains effective. (d) The manufacturer is not justified in calling the product "sixty day super food" because the effectiveness falls below $$ 25\% $$ before 60 days.

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