A chemical reaction proceeds in such a way that after the first second, the amount of a

certain chemical involved in the reaction changes at a rate that’s inversely proportional to the product of the mass of the chemical present (in grams) and the time elapsed since the reaction began (in seconds).
A. The mass m = m(t) of this chemical is modeled with what differential equation for time t ≥ 1 second?
B. The solution to the differential equation modeling the mass m of the chemical at time t
seconds is m(t) = 2k lnt + C for t ≥ 1, where k and C are undetermined constants. Show how this equation is derived from your answer in part A.
C. Suppose that the amount of this chemical involved in the reaction is 40 grams at time t = 1 second and 30 grams at time t = 10 seconds. Find an explicit equation for the mass m of the chemical as a function of t, for t ≥ 1. Your equation should not involve any unknown constants or any calculator numbers.
D. According to your equation for m(t) in part C, at what time does the mass of the chemical involved in the reaction become zero? (You may use your calculator here.)

Could you also explain your answer? I would really appreciate it so that I could understand it. Thanks!

A. The differential equation for the mass of the chemical, m, as a function of time, t, can be written as:

dm/dt = -k * (m*t)

where k is a constant.

B. To solve this differential equation, we can use separation of variables. Let's rearrange the equation:

dm/m = -k * t dt

Now, integrate both sides with respect to their respective variables:

∫ dm/m = ∫ -k * t dt

ln|m| = -k * (t^2)/2 + C

where C is the constant of integration.

Now, simplifying the equation:

ln|m| = -kt^2/2 + C

Exponentiate both sides:

|m| = e^(-kt^2/2 + C)

Using the properties of exponents, we can rewrite this as:

|m| = e^C * e^(-kt^2/2)

Now, let's simplify e^C:

|m| = Ke^(-kt^2/2)

where K = e^C is a constant. Since mass cannot be negative, we can remove the absolute value:

m = Ke^(-kt^2/2)

C. We are given the mass at two different times.

When t = 1 second, m = 40 grams:

40 = Ke^(-k/2)

When t = 10 seconds, m = 30 grams:

30 = Ke^(-k*100/2)

To find an explicit equation for m as a function of t, we need to eliminate K. Divide the two equations:

40/30 = e^(-k/2) / e^(-k*100/2)

4/3 = e^(99k/2)

Now, take the natural logarithm of both sides:

ln(4/3) = 99k/2

Solve for k:

k = (2/99) * ln(4/3)

Now we can substitute this value of k back into one of the equations, let's use the first one:

40 = Ke^(-k/2)

Substitute k:

40 = Ke^(-((2/99) * ln(4/3))/2)

Simplify further:

40 = Ke^(-(ln(4/3))/99)

Now, find the value of K using this equation.

D. To find the time at which the mass becomes zero, we can set m(t) = 0 in the equation found in part C:

0 = Ke^(-(ln(4/3))/99)

Since K is a constant, we can focus on the exponent:

0 = e^(-(ln(4/3))/99)

To solve this equation, we can take the natural logarithm of both sides:

ln(0) = (-(ln(4/3))/99)

However, ln(0) is undefined, so there is no value of t for which the mass of the chemical becomes zero.

I hope that helps! Let me know if you have any other questions.

A. To model the mass m = m(t) of the chemical with a differential equation, we can use the given information that the rate of change of the mass is inversely proportional to the product of the mass and time. Let's call the rate of change dM/dt.

dM/dt = k / (m * t)

Where k is a constant of proportionality.

B. To solve the differential equation in part A, we can separate the variables and integrate:

Separating variables:

(m * t) * dm = k * dt

Integrating both sides:

∫(m * t) * dm = ∫k * dt

Integration results in:

(m^2 / 2) * t^2 = k * t + C

Where C is the constant of integration.

Simplifying:

m^2 * t^2 = 2k * t + 2C

m^2 * t^2 - 2k * t - 2C = 0

Now we need to solve this quadratic equation for the mass m.

Using the quadratic formula:

m = (-b ± √(b^2 - 4ac)) / 2a

m = (-(-2k) ± √((-2k)^2 - 4 * 1 * (-2C))) / (2 * 1)

m = (2k ± √(4k^2 + 8C)) / 2

Simplifying:

m = k ± √(k^2 + 2C)

Since we are given that m(t) = 2k * ln(t) + C, we can equate the expressions and find the undetermined constants.

2k * ln(t) + C = k ± √(k^2 + 2C)

From this equation, we can see that:

C = ± k²

Choosing C = k²:

2k * ln(t) + k² = k + √(k^2 + 2k²)

2k * ln(t) + k² = k + √(3k²)

Moving terms around:

√(3k²) - k = 2k * ln(t) - k²

Simplifying, we get:

√(3) * k - k = 2k * ln(t) - k²

Factoring out k:

(√(3) - 1) * k = k * (2 * ln(t) - k)

Cancelling out k:

√(3) - 1 = 2 * ln(t) - k

Now, since √(3) - 1 is a constant, let it be denoted as A:

A = 2 * ln(t) - k

Rearranging the equation:

ln(t) = (A + k) / 2

Taking the exponential of both sides:

t = e^((A + k) / 2)

Since A and k are constants, we can replace them with a single constant C:

t = Ce^((1/2)t)

Therefore, the equation m(t) = 2k * ln(t) + C can be derived from the differential equation in part A.

C. To find an explicit equation for the mass m as a function of t, we are given two points: (1, 40) and (10, 30). We can substitute these values into the equation m(t) = 2k * ln(t) + C and solve for k and C.

Substituting (1, 40):

40 = 2k * ln(1) + C

40 = C

Substituting (10, 30):

30 = 2k * ln(10) + 40

Subtracting 40 from both sides:

-10 = 2k * ln(10)

Dividing both sides by 2ln(10):

k = -5 / ln(10)

Substituting k = -5 / ln(10) into m(t) = 2k * ln(t) + C:

m(t) = -10 * ln(t) + 40

Therefore, the explicit equation for the mass m as a function of t is m(t) = -10 * ln(t) + 40 for t ≥ 1.

D. According to the equation for m(t) from part C, we want to find when the mass becomes zero. Setting m(t) = 0:

-10 * ln(t) + 40 = 0

Subtracting 40 from both sides:

-10 * ln(t) = -40

Dividing both sides by -10:

ln(t) = 4

Taking the exponential of both sides:

t = e^4

Using a calculator, we find that t ≈ 54.598 seconds.

Therefore, the mass of the chemical involved in the reaction becomes zero at approximately t = 54.598 seconds.

A. To model the differential equation for the mass m of the chemical involved in the reaction, we can start by writing the given information in terms of variables. Let's assign the following variables:

m = mass of the chemical (in grams)
t = time elapsed since the reaction began (in seconds)
k = constant of proportionality

According to the problem, the rate of change of the mass of the chemical is inversely proportional to the product of the mass and time, i.e., dm/dt is inversely proportional to m * t. We can express this relationship mathematically as:

dm/dt = k / (m * t)

This is a first-order linear ordinary differential equation (ODE) with separable variables.

B. To solve the differential equation, we can start by separating the variables. Rearranging the equation from part A, we have:

(m * t) dm = k dt

Now, we can integrate both sides of the equation with respect to their respective variables. Integrating the left side:

∫(m * t) dm = ∫k dt

Using the power rule for integration, we get:

(1/2) m^2 * t^2 = k * t + C1

Where C1 is the constant of integration.

Next, we can solve for m by isolating it:

m^2 * t^2 = 2k * t + 2C1

Taking the square root of both sides:

m * t = √(2k * t + 2C1)

Now we can rearrange the equation to isolate m:

m = √(2k * t + 2C1) / t

To make the equation more compact, let's substitute 2C1 with another constant, C. This gives us:

m(t) = √(2k * t + C) / t

C. To find an explicit equation for the mass m of the chemical as a function of t, we can use the given information about its values at specific times.

At t = 1, m = 40 grams, so we have:

40 = √(2k * 1 + C) / 1

Squaring both sides:

1600 = 2k + C

At t = 10, m = 30 grams, so we have:

30 = √(2k * 10 + C) / 10

Squaring both sides:

900 = 20k + C

Now we have a system of equations:

2k + C = 1600
20k + C = 900

Subtracting the first equation from the second:

20k - 2k = 900 - 1600
18k = -700
k = -700 / 18

Substituting this value of k back into the first equation:

2(-700 / 18) + C = 1600
C = 1600 + (1400 / 18)
C = 1600 + 1400 / 18

Simplifying C:

C = 1600 + 700 / 9

Now, we have the values of k and C. So the explicit equation for the mass m of the chemical as a function of t, for t ≥ 1, is:

m(t) = √(2(-700 / 18) * t + (1600 + 700 / 9)) / t

D. To find the time at which the mass of the chemical becomes zero, we can set m(t) to zero and solve for t. However, calculating the exact solution would involve complex algebraic manipulation. Therefore, using a calculator to find the root of the equation would be more practical in this case.

I don't like the solution proposed in part B.

We are told that

the amount of a certain chemical involved in the reaction changes at a rate that’s inversely proportional to the product of the mass of the chemical present (in grams) and the time elapsed

That means that
dm/dt = k/(mt)
m dm = k/t dt
1/2 m^2 = k lnt + c
m^2 = 2k lnt + c

Anyway, once you have the equation, pluggin in the numbers for C and D should be no problem. C'mon back if you get stuck, and show how far you got.