UG Applied Mathematics Booster Test 3 - Applications of Differential Equations -

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UG Applied Mathematics Booster Test 3 - Applications of Differential Equations

⏱️ 30 Minutes

❓ 20 Questions

📌 Answers are locked once submitted — results and explanations appear at the end.

UG Applied Mathematics Booster Test 3 - Applications of Differential Equations

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QUESTION 1 OF 20

The cake cools from 185°F to 150°F in 30 minutes in a 75°F room. This establishes
\(e^{-30k}=\frac{75}{110}.\)
To find the temperature at 45 minutes, the equation
\(T-75=110e^{-45k}\)
is mathematically restructured as:

Answer Review

1 The cake cools from 185°F to 150°F in 30 minutes in a 75°F room. This establishes
\(e^{-30k}=\frac{75}{110}.\)
To find the temperature at 45 minutes, the equation
\(T-75=110e^{-45k}\)
is mathematically restructured as:

A \(T=75+110{\left(e^{-30k}\right)}^{1.5}\)

B \(T=75+110{\left(e^{-30k}\right)}^{2}\)

C \(T=75+110{\left(e^{-30k}\right)}^{0.5}\)

D \(T=110{\left(e^{-30k}\right)}^{1.5}\)

💡 Short Explanation

Since \(45=1.5\times 30,\) we have \(e^{-45k}=e^{-30k\times 1.5}={\left(e^{-30k}\right)}^{1.5}.\) Therefore, \(T=75+110{\left(e^{-30k}\right)}^{1.5}.\)

✅ Correct Answer Explanation

Newton's Law of Cooling gives \(T-75=110e^{-45k}.\) Using the exponent law, \(e^{ab}=(e^{a})^{b},\) we write \(e^{-45k}=e^{-30k\left(\frac{45}{30}\right)}=e^{-30k(1.5)}={\left(e^{-30k}\right)}^{1.5}.\) Substituting, \(T-75=110{\left(e^{-30k}\right)}^{1.5}.\) Adding 75 to both sides, \(T=75+110{\left(e^{-30k}\right)}^{1.5}.\) Hence, Option A is correct.

❓ Why Other Options Are Incorrect

Option A)
\(T=75+110{\left(e^{-30k}\right)}^{1.5}\)
Correct.
Since \(45=1.5\times 30\), the exponent becomes \(1.5\).
Option B)
\(T=75+110{\left(e^{-30k}\right)}^{2}\)
Incorrect because squaring corresponds to 60 minutes, not 45 minutes.
Option C)
\(T=75+110{\left(e^{-30k}\right)}^{0.5}\)
Incorrect because the exponent \(0.5\) corresponds to 15 minutes, not 45 minutes.
Option D)
\(T=110{\left(e^{-30k}\right)}^{1.5}\)
Incorrect because it omits the surrounding temperature term 75, which must be added back.

🎯 Strategy

Used

Apply the Exponent Law

Application:

1. Write the cooling equation:
\(T-75=110e^{-45k}.\)
1. Express 45 as
\(45=1.5\times 30.\)
1. Use
\(e^{ab}=(e^{a})^{b}.\)
1. Rearrange to obtain the required expression for \(T\).

🧠 Memory Trick

Mnemonic: "Multiply Time → Raise Exponential."

📚 Class XII – NCERT Mathematics Part II – Chapter: Differential Equations – Topic: Newton's Law of Cooling

2 Match the specific derived constant formulas to the applied mathematical models detailed in the text.

List I	List II
1. Carbon-14 decay constant \(\left(k_{1}\right)\)	a. \(\frac{\ln\,2}{k}\)
2. Doubling time \(\left(t\right)\)	b. \(2250\times 2^{1/5}\)
3. Nembutal initial dose \(\left(x_{0}\right)\)	c. \(e^{-30k}=0.6818\)
4. Cake cooling constant structure at 30 min	d. \(\frac{\ln\,2}{5700}\)

A 1-a, 2-d, 3-b, 4-c

B 1-d, 2-a, 3-b, 4-c

C 1-d, 2-b, 3-a, 4-c

D 1-c, 2-a, 3-b, 4-d

💡 Short Explanation

The Carbon-14 decay constant is derived using the radioactive decay formula and its half-life of 5700 years. The doubling time formula is obtained from the exponential growth model. The Nembutal problem requires calculating the initial dose using the exponential growth relationship. The cake cooling model uses the exponential cooling equation to determine the cooling constant.

✅ Correct Answer Explanation

The correct matching is: 1. Carbon-14 decay constant \(\left(k_{1}\right)\)→ d. \(\frac{\ln\,2}{5700}\) 2. Doubling time \(\left(t\right)\)→ a. \(\frac{\ln\,2}{k}\) 3. Nembutal initial dose \(\left(x_{0}\right)\)→ b. \(2250\times 2^{1/5}\) 4. Cake cooling constant structure at 30 min → c. \(e^{-30k}=0.6818\) Thus, the correct sequence is: 1-d, 2-a, 3-b, 4-c Hence, Option B is the correct answer.

❓ Why Other Options Are Incorrect

Option A. 1-a, 2-d, 3-b, 4-c → Incorrect because the formulas for the Carbon-14 decay constant and doubling time are interchanged. The Carbon-14 decay constant specifically uses the half-life formula \(\frac{\ln\,2}{5700}\), whereas the doubling time is \(\frac{\ln\,2}{k}\).
Option C. 1-d, 2-b, 3-a, 4-c → Incorrect because the doubling time is not \(2250\times 2^{1/5}\), and the Nembutal initial dose is not \(\frac{\ln\,2}{k}\).
Option D. 1-c, 2-a, 3-b, 4-d → Incorrect because the cake cooling equation and Carbon-14 decay constant are incorrectly interchanged.

🎯 Strategy

Formula Identification

Application:

Identify the model first and then recall its standard derived formula:
Radioactive decay → \(\frac{\ln\,2}{Half-life}\)
Exponential growth → Doubling time \(=\frac{\ln\,2}{k}\)
Drug concentration problem → Initial value calculation
Newton's Law of Cooling → Exponential cooling equation

Final Logic:

Only Option B correctly matches each mathematical model with its corresponding derived formula.

🧠 Memory Trick

Cake → \(e^{-30k}=0.6818\)

📚 Class XII – Mathematics – Chapter: Differential Equations – Topic: Applications of Differential Equations (Exponential Growth and Decay Models, Newton's Law of Cooling, Radioactive Decay and Biological Models).

3 To calculate the doubling time of Ms. Rajni's 10,000 deposit at 4% continuous interest, which mathematical evaluations explicitly occurred?
1. The equation simplified to 0.04t = ln 2
2. The value ln 2 ≈ 0.6931 was substituted
3. The calculation t = 0.6931 / 0.04 yielded 17.32 years
4. The 10,000 initial principal canceled out on both sides

A 1,2,4

B 1,2

C 1,2,3

D 1,2,3,4

💡 Short Explanation

All listed steps are used Log transformation applied Numerical substitution performed

✅ Correct Answer Explanation

All steps occur in derivation: 1→ rt = ln2 2 → ln2 ≈ 0.6931 3 → t computed 4 → principal cancels during division

❓ Why Other Options Are Incorrect

Option A → incomplete
Option B → missing computation steps
Option C → excludes cancellation logic

🎯 Strategy

Elimination

Final Logic: full derivation includes all steps

🧠 Memory Trick

"Doubling = log + substitution + cancellation"

📚 Class XII – NCERT Mathematics Part II – Chapter: Differential Equations – Topic: Continuous Compound Interest

4 Identify the incorrect mathematical step in the derivation of the continuous compound interest differential equation from discrete intervals.

A The interest over a short time \(\Delta t\) is \(\Delta A=rA(t)\Delta t.\)

B Dividing by \(\Delta t\) yields \(\frac{\Delta A}{\Delta t}=rA.\)

C Taking the limit as \(\Delta t\) approaches \(0\) yields \(\frac{dA}{dt}=rA.\)

D Taking the limit as \(\Delta t\) approaches \(\infty\) yields \(\frac{dA}{dt}=rA.\)

💡 Short Explanation

A derivative is defined by taking the limit as \(\Delta t\rightarrow 0,\) not as \(\Delta t\rightarrow \infty .\) Hence, Option D is the incorrect step.

✅ Correct Answer Explanation

The derivation of the continuous compound interest model begins with \(\Delta A=rA(t)\Delta t.\) Dividing both sides by \(\Delta t\), \(\frac{\Delta A}{\Delta t}=rA.\) To obtain the instantaneous rate of change, we take the limit as \(\Delta t\rightarrow 0.\) This gives \(\frac{dA}{dt}=rA.\) Taking the limit as \(\Delta t\rightarrow \infty\) does not define a derivative and is mathematically incorrect. Therefore, Option D is correct.

❓ Why Other Options Are Incorrect

Option A)
Incorrect as a choice because this is the correct approximation for the interest earned over a small time interval.
Option B)
Incorrect as a choice because dividing by \(\Delta t\) correctly forms the average rate of change.
Option C)
Incorrect as a choice because the derivative is correctly obtained by taking the limit as
\(\Delta t\rightarrow 0.\)
Option D)
Correct.
The derivative is never defined using
\(\Delta t\rightarrow \infty .\)
The correct limit is
\(\Delta t\rightarrow 0.\)

🎯 Strategy

Used

Recall the Definition of a Derivative

Application:

1. Start with the finite difference equation.
2. Divide by the time interval.
3. Apply the limit
\(\Delta t\rightarrow 0.\)
1. Obtain the differential equation.

🧠 Memory Trick

Mnemonic: "Zero for Slope, Infinity is Nope!"

📚 Bottom of Form

5 When deriving the general solution
\(P(t)=Ae^{kt}\)
for population, the constant of integration \(c\) is converted into the coefficient \(A\). What is the exact algebraic step showing this relationship?

A \(e^{c}=A\)

B \(ln⁡c=A\)

C \(c^{e}=A\)

D \(A=e^{-c}\)

💡 Short Explanation

After integration, \(ln⁡P=kt+c.\) Exponentiating both sides gives \(P=e^{kt+c}=e^{c}e^{kt}.\) Since \(e^{c}\)is also a constant, we write \(e^{c}=A.\)

✅ Correct Answer Explanation

Starting with the differential equation \(\frac{dP}{dt}=kP,\) separate the variables: \(\frac{dP}{P}=k dt.\) Integrating, \(ln⁡P=kt+c.\) Taking the exponential of both sides, \(P=e^{kt+c}.\) Using the exponent law, \(e^{a+b}=e^{a}e^{b},\) we obtain \(P=e^{c}e^{kt}.\) Since \(e^{c}\)is a constant, let \(A=e^{c}.\) Hence, \(P(t)=Ae^{kt}.\) Therefore, Option A is correct.

❓ Why Other Options Are Incorrect

Option A)
\(e^{c}=A\)
Correct.
This is the standard substitution used after exponentiating the integrated equation.
Option B)
\(ln⁡c=A\)
Incorrect because the exponential of the integration constant is taken, not its logarithm.
Option C)
\(c^{e}=A\)
Incorrect because the correct expression is
\(e^{c},\)
not \(c^{e}\).
Option D)
\(A=e^{-c}\)
Incorrect because exponentiating
\(ln⁡P=kt+c\)
produces
\(e^{c},\)
not \(e^{-c}\).

🎯 Strategy

Used

Exponentiate After Integration

Application:

1. Integrate to obtain
\(ln⁡P=kt+c.\)
1. Apply the exponential function.
2. Use
\(e^{a+b}=e^{a}e^{b}.\)
1. Replace the constant
\(e^{c}\)
by a new constant \(A\).

🧠 Memory Trick

Mnemonic: "Exponentiate, Then Rename the Constant."

📚 Chapter: Differential Equations

6 After finding
\(e^{k}=2,\)
the initial population \(\lambda\) is found using
\(\lambda (2)^{3}=10,000.\)
What is the precise numerical calculation to solve for \(\lambda\)?

A \(\lambda =\frac{10000}{6}=1666.6\)

B \(\lambda =\frac{10000}{8}=1250\)

C \(\lambda =\frac{10000}{4}=2500\)

D \(\lambda =\frac{10000}{2}=5000\)

💡 Short Explanation

Since \(e^{k}=2,\) we have \(e^{3k}=(e^{k})^{3}=2^{3}=8.\) Thus, \(\lambda =\frac{10000}{8}=1250.\)

✅ Correct Answer Explanation

The population model is \(P(t)=\lambda e^{kt}.\) Given, \(P(3)=10,000,\) so \(\lambda e^{3k}=10,000.\) Since \(e^{k}=2,\) we obtain \(e^{3k}=(e^{k})^{3}=2^{3}=8.\) Therefore, \(\lambda (8)=10,000.\) Solving, \(\lambda =\frac{10,000}{8}=1250.\) Hence, Option B is correct.

❓ Why Other Options Are Incorrect

Option A)
\(\lambda =\frac{10000}{6}=1666.6\)
Incorrect because
\(2^{3}=8,\)
not 6.
Option B)
\(\lambda =\frac{10000}{8}=1250\)
Correct.
Since
\(e^{3k}=8,\)
this gives the correct initial population.
Option C)
\(\lambda =\frac{10000}{4}=2500\)
Incorrect because
\(2^{3}\neq 4.\)
Option D)
\(\lambda =\frac{10000}{2}=5000\)
Incorrect because the exponent is 3, so the exponential factor is 8, not 2.

🎯 Strategy

Used

Substitute the Known Exponential Value

Application:

1. Use
\(e^{k}=2.\)
1. Compute
\(e^{3k}=(e^{k})^{3}=8.\)
1. Substitute into
\(\lambda e^{3k}=10,000.\)
1. Solve for \(\lambda\).

🧠 Memory Trick

Mnemonic: "Raise the Power, Then Divide."

📚 Chapter: Differential Equations

7 To estimate the age of charcoal containing \(\frac{1}{4}\)of its original Carbon-14, the text demonstrates substituting
\(x(t)=\frac{x_{0}}{4}\)
into
\(\frac{x_{0}}{4}=x_{0}e^{-k_{1}t}.\)
Removing \(x_{0}\)and taking natural logarithms gives exactly:

A \(ln⁡\left(\frac{1}{4}\right)=k_{1}t\)

B \(-ln⁡4=-k_{1}t\)

C \(ln⁡4=k_{1}t\)

D \(-ln⁡2=k_{1}t\)

💡 Short Explanation

After cancelling \(x_{0}\), \(\frac{1}{4}=e^{-k_{1}t}.\) Taking natural logarithms, \(ln⁡\left(\frac{1}{4}\right)=-k_{1}t.\) Since \(ln⁡\left(\frac{1}{4}\right)=-ln⁡4,\) the equation becomes \(-ln⁡4=-k_{1}t.\)

✅ Correct Answer Explanation

The Carbon-14 decay model is \(x=x_{0}e^{-k_{1}t}.\) For a sample containing one-fourth of its original Carbon-14, \(\frac{x_{0}}{4}=x_{0}e^{-k_{1}t}.\) Cancel \(x_{0}\): \(\frac{1}{4}=e^{-k_{1}t}.\) Taking the natural logarithm of both sides, \(ln⁡\left(\frac{1}{4}\right)=ln⁡\left(e^{-k_{1}t}\right).\) Therefore, \(ln⁡\left(\frac{1}{4}\right)=-k_{1}t.\) Using the logarithmic identity, \(ln⁡\left(\frac{1}{4}\right)=-ln⁡4,\) we obtain \(-ln⁡4=-k_{1}t.\) Hence, Option B is correct.

❓ Why Other Options Are Incorrect

Option A)
\(ln⁡\left(\frac{1}{4}\right)=k_{1}t\)
Incorrect because the exponent is negative, so the right-hand side must be \(-k_{1}t\).
Option B)
\(-ln⁡4=-k_{1}t\)
Correct.
This follows from
\(ln⁡\left(\frac{1}{4}\right)=-ln⁡4.\)
Option C)
\(ln⁡4=k_{1}t\)
Incorrect because this is obtained only after multiplying both sides of Option B by \(-1\). It is not the equation obtained immediately after taking logarithms.
Option D)
\(-ln⁡2=k_{1}t\)
Incorrect because the remaining fraction is
\(\frac{1}{4},\)
which corresponds to \(\ln\,4\), not \(\ln\,2\).

🎯 Strategy

Used

Apply Natural Logarithms

Application:

1. Substitute the remaining fraction into the decay model.
2. Cancel the initial amount \(x_{0}\).
3. Take the natural logarithm of both sides.
4. Use the identity
\(ln⁡\left(\frac{1}{a}\right)=-ln⁡a.\)

🧠 Memory Trick

Mnemonic: "Fraction Inside Log → Negative Outside Log."

📚 Chapter: Differential Equations

8 Radium disappears at a rate proportional to its amount and half vanishes in 1600 years. To find the percentage lost in 100 years, one must evaluate
\(A(100)=A_{0}e^{-100k},\)
where
\(k=\frac{\ln\,2}{1600}.\)
What is the exponential fraction representing this remaining amount?

A \(A(100)=A_{0}e^{-\frac{\ln\,2}{16}}\)

B \(A(100)=A_{0}e^{-\frac{\ln\,2}{100}}\)

C \(A(100)=A_{0}e^{\frac{\ln\,2}{16}}\)

D \(A(100)=A_{0}e^{-\frac{1600}{100}}\)

💡 Short Explanation

Substitute \(k=\frac{\ln\,2}{1600}\) into \(A(100)=A_{0}e^{-100k}.\) Then, \(A(100)=A_{0}e^{-100\left(\frac{\ln\,2}{1600}\right)}=A_{0}e^{-\frac{\ln\,2}{16}}.\)

✅ Correct Answer Explanation

The radioactive decay model is \(A(t)=A_{0}e^{-kt}.\) Given the half-life is 1600 years, \(k=\frac{\ln\,2}{1600}.\) To find the amount remaining after 100 years, \(A(100)=A_{0}e^{-100k}.\) Substitute the value of \(k\): \(A(100)=A_{0}e^{-100\left(\frac{\ln\,2}{1600}\right)}.\) Since \(\frac{100}{1600}=\frac{1}{16},\) we obtain \(A(100)=A_{0}e^{-\frac{\ln\,2}{16}}.\) Hence, Option A is correct.

❓ Why Other Options Are Incorrect

Option A)
\(A(100)=A_{0}e^{-\frac{\ln\,2}{16}}\)
Correct.
This is obtained by directly substituting \(k=\frac{\ln\,2}{1600}\)into the decay model.
Option B)
\(A(100)=A_{0}e^{-\frac{\ln\,2}{100}}\)
Incorrect because the denominator should be 16, not 100.
Option C)
\(A(100)=A_{0}e^{\frac{\ln\,2}{16}}\)
Incorrect because radioactive decay requires a negative exponent.
Option D)
\(A(100)=A_{0}e^{-\frac{1600}{100}}\)
Incorrect because the exponent should involve the decay constant \(\ln\,2\), not the ratio \(\frac{1600}{100}\).

🎯 Strategy

Used

Substitute the Decay Constant

Application:

1. Write the decay model:
\(A(t)=A_{0}e^{-kt}.\)
1. Substitute
\(k=\frac{\ln\,2}{1600}.\)
1. Replace \(t\) with 100.
2. Simplify the fraction in the exponent.

🧠 Memory Trick

Mnemonic: "Substitute \(k\), Reduce the Fraction."

📚 Chapter: Differential Equations

9 From the exact solution steps in Example 16 for estimating the \(\frac{1}{4}\)Carbon-14 charcoal age, the equation evaluates to
\(t=\frac{-5700(-ln⁡4)}{\ln\,2}.\)
Since
\(ln⁡4=2ln⁡2,\)
what is the mathematically simplified result?

A 5700×2

B 5700/2

C 5700×4

D 5700/4

💡 Short Explanation

ln4 = 2ln2 Cancellation occurs Final multiplication

✅ Correct Answer Explanation

t = 5700 × (2ln2)/ln2 = 5700 × 2.

❓ Why Other Options Are Incorrect

B → inverse error
C → wrong scaling
D → incorrect reduction

🎯 Strategy

Substitution

Final Logic: log identity simplifies ratio

🧠 Memory Trick

"ln4 = 2ln2"

📚 Class XII – NCERT Mathematics Part II – Chapter: Differential Equations – Topic: Carbon Dating

10 If a bone is observed to contain 75% of Carbon-14 (Exercise 5, Q8), the equation to find its antiquity \(t\) is
\(0.75=e^{-\left(\frac{\ln\,2}{5700}\right)t}.\)
Isolating \(t\) directly produces the analytical expression:

A \(t=-\frac{5700 ln⁡(0.75)}{\ln\,2}\)

B \(t=\frac{5700 ln⁡(0.75)}{\ln\,2}\)

C \(t=-\frac{5700 ln⁡(1.33)}{\ln\,2}\)

D \(t=\frac{5700 ln⁡(0.25)}{\ln\,2}\)

💡 Short Explanation

Taking natural logarithms of \(0.75=e^{-\left(\frac{\ln\,2}{5700}\right)t}\) and solving for \(t\) gives \(t=-\frac{5700 ln⁡(0.75)}{\ln\,2}.\)

✅ Correct Answer Explanation

The Carbon-14 decay model is \(x=x_{0}e^{-k_{1}t},\) where \(k_{1}=\frac{\ln\,2}{5700}.\) Since the bone contains 75% of its original Carbon-14, \(0.75=e^{-\left(\frac{\ln\,2}{5700}\right)t}.\) Taking natural logarithms, \(ln⁡(0.75)=-\left(\frac{\ln\,2}{5700}\right)t.\) Multiplying both sides by \(-\frac{5700}{\ln\,2},\) we obtain \(t=-\frac{5700 ln⁡(0.75)}{\ln\,2}.\) Hence, Option A is correct.

❓ Why Other Options Are Incorrect

Option A)
\(t=-\frac{5700 ln⁡(0.75)}{\ln\,2}\)
Correct.
It is obtained by correctly isolating \(t\) after taking natural logarithms.
Option B)
\(t=\frac{5700 ln⁡(0.75)}{\ln\,2}\)
Incorrect because the negative sign is omitted.
Since \(ln⁡(0.75)<0\), the negative sign is required to obtain a positive age.
Option C)
\(t=-\frac{5700 ln⁡(1.33)}{\ln\,2}\)
Incorrect because the equation starts with the remaining fraction 0.75, not its reciprocal.
Although \(ln⁡(1.33)\approx -ln⁡(0.75)\), this is not the direct expression obtained from the given equation.
Option D)
\(t=\frac{5700 ln⁡(0.25)}{\ln\,2}\)
Incorrect because 0.25 (25%) corresponds to one-fourth remaining, whereas the problem states 75% remaining.

🎯 Strategy

Used

Apply Natural Logarithms and Isolate the Variable

Application:

1. Substitute the remaining fraction into the decay model.
2. Take the natural logarithm of both sides.
3. Rearrange algebraically to isolate \(t\).
4. Simplify the expression.

🧠 Memory Trick

Mnemonic: "Remaining Fraction → Log → Negative Makes Age Positive."

📚 Chapter: Differential Equations

11 In the drug assimilation model, the amount of drug initially in the GI-tract is \(x_{0}\). Solving
\(\frac{dx}{dt}=-k_{1}x\)
gives
\(x(t)=ce^{-k_{1}t}.\)
At \(t=0\),
\(x(0)=x_{0}.\)
This boundary condition proves mathematically that:

A \(c=0\)

B \(c=x_{0}\)

C \(c=k_{1}\)

D \(c=e\)

💡 Short Explanation

Substituting \(t=0\) into \(x(t)=ce^{-k_{1}t}\) gives \(x_{0}=ce^{0}=c.\) Hence, \(c=x_{0}.\)

✅ Correct Answer Explanation

The solution of the differential equation \(\frac{dx}{dt}=-k_{1}x\) is \(x(t)=ce^{-k_{1}t},\) where \(c\) is the constant of integration. Using the initial condition \(x(0)=x_{0},\) we substitute \(t=0\): \(x_{0}=ce^{-k_{1}(0)}.\) Since \(e^{0}=1,\) we obtain \(x_{0}=c.\) Therefore, \(c=x_{0}.\) Hence, Option B is correct.

❓ Why Other Options Are Incorrect

Option A)
\(c=0\)
Incorrect because if \(c=0\), then
\(x(t)=0\)
for all \(t\), which contradicts the given initial amount \(x_{0}\).
Option B)
\(c=x_{0}\)
Correct.
Applying the initial condition directly gives
\(c=x_{0}.\)
Option C)
\(c=k_{1}\)
Incorrect because \(k_{1}\)is the decay constant, whereas \(c\) is determined by the initial condition.
Option D)
\(c=e\)
Incorrect because \(e\) is Euler's number and is unrelated to the integration constant in this problem.

🎯 Strategy

Used

Apply the Initial Condition

Application:

1. Write the general solution:
\(x(t)=ce^{-k_{1}t}.\)
1. Substitute the given initial condition:
\(t=0,x=x_{0}.\)
1. Use
\(e^{0}=1.\)
1. Solve for the integration constant.

🧠 Memory Trick

Mnemonic: "Time Zero Reveals the Constant."

📚 Chapter: Differential Equations

12 If the drug enters the blood at rate \(k_{1}x\) and leaves at \(k_{2}y\), the differential equation
\(\frac{dy}{dt}=k_{1}x-k_{2}y\)
is formed. Substituting the known GI-tract solution
\(x=x_{0}e^{-k_{1}t}\)
into this equation yields a linear first-order differential equation:

A \(\frac{dy}{dt}+k_{2}y=k_{1}x_{0}e^{-k_{1}t}\)

B \(\frac{dy}{dt}-k_{2}y=k_{1}x_{0}e^{-k_{1}t}\)

C \(\frac{dy}{dt}+k_{1}y=k_{2}x_{0}e^{-k_{1}t}\)

D \(\frac{dy}{dt}=k_{1}x_{0}e^{-k_{2}t}\)

💡 Short Explanation

Substitute \(x=x_{0}e^{-k_{1}t}\) into \(\frac{dy}{dt}=k_{1}x-k_{2}y,\) and rearrange to the standard linear form: \(\frac{dy}{dt}+k_{2}y=k_{1}x_{0}e^{-k_{1}t}.\)

✅ Correct Answer Explanation

The blood compartment satisfies \(\frac{dy}{dt}=k_{1}x-k_{2}y.\) The GI-tract solution is \(x=x_{0}e^{-k_{1}t}.\) Substituting this into the equation, \(\frac{dy}{dt}=k_{1}x_{0}e^{-k_{1}t}-k_{2}y.\) Moving the term \(-k_{2}y\) to the left-hand side, \(\frac{dy}{dt}+k_{2}y=k_{1}x_{0}e^{-k_{1}t}.\) This is the standard form of a first-order linear differential equation. Hence, Option A is correct.

❓ Why Other Options Are Incorrect

Option A)
\(\frac{dy}{dt}+k_{2}y=k_{1}x_{0}e^{-k_{1}t}\)
Correct.
It is obtained by direct substitution and rearrangement into the standard linear form.
Option B)
\(\frac{dy}{dt}-k_{2}y=k_{1}x_{0}e^{-k_{1}t}\)
Incorrect because moving \(-k_{2}y\) to the left changes its sign to +\(k_{2}y\).
Option C)
\(\frac{dy}{dt}+k_{1}y=k_{2}x_{0}e^{-k_{1}t}\)
Incorrect because the constants \(k_{1}\)and \(k_{2}\)are incorrectly interchanged.
Option D)
\(\frac{dy}{dt}=k_{1}x_{0}e^{-k_{2}t}\)
Incorrect because the exponential term depends on the GI-tract decay constant \(k_{1}\), and the elimination term \(-k_{2}y\) has been omitted.

🎯 Strategy

Used

Substitution into a Coupled Differential Equation

Application:

1. Write the blood-compartment equation.
2. Substitute the known expression for \(x(t)\).
3. Rearrange all \(y\)-terms to the left-hand side.
4. Obtain the standard linear form
\(\frac{dy}{dt}+P(t)y=Q(t).\)

🧠 Memory Trick

Mnemonic: "Known \(x\), Move \(y\), Linear DE Ready!"

📚 Subtopic: Formation and Solution of First-Order Linear Differential Equations in Drug Absorption (Compartment Models)

13 The necessary Nembutal dose for a 50 kg dog is calculated using
\(x_{0}=2250(e^{k}).\)
With the half-life being 5 hours
\(e^{-5k}=\frac{1}{2},\)
the equation
\(e^{5k}=2\)
leads to
\(e^{k}=2^{1/5}.\)
Evaluating
\(2250\times 2^{1/5}\)
yields approximately:

A 2250 mg

B 2585 mg

C 4500 mg

D 1125 mg

💡 Short Explanation

Since \(e^{k}=2^{1/5},\) the initial dose is \(x_{0}=2250\times 2^{1/5}.\) Using \(2^{1/5}\approx 1.1487,\) we get \(x_{0}\approx 2250\times 1.1487\approx 2585 mg.\)

✅ Correct Answer Explanation

The given equation is \(x_{0}=2250e^{k}.\) From the half-life condition, \(e^{-5k}=\frac{1}{2}.\) Taking reciprocals, \(e^{5k}=2.\) Taking the fifth root, \(e^{k}=2^{1/5}.\) Therefore, \(x_{0}=2250\times 2^{1/5}.\) Using the approximation \(2^{1/5}\approx 1.1487,\) we obtain \(x_{0}\approx 2250\times 1.1487\approx 2585 mg.\) Hence, Option B is correct.

❓ Why Other Options Are Incorrect

Option A) 2250 mg
Incorrect because it ignores the multiplying factor
\(2^{1/5}>1.\)
Option B) 2585 mg
Correct.
This is the approximate value of
\(2250\times 2^{1/5}.\)
Option C) 4500 mg
Incorrect because this assumes multiplication by 2 instead of by
\(2^{1/5}.\)
Option D) 1125 mg
Incorrect because this is half of 2250 mg and is unrelated to the required initial dose calculation.

🎯 Strategy

Used

Substitute the Derived Exponential Value

Application:

1. Use the half-life equation to determine
\(e^{k}.\)
1. Substitute into
\(x_{0}=2250e^{k}.\)
1. Evaluate the numerical value.

🧠 Memory Trick

Mnemonic: "Reciprocal → Root → Multiply."

📚 Chapter: Differential Equations

14 The mathematical model for a sedative clearing from the bloodstream simply states
\(\frac{dx}{dt}=-kx.\)
If the initial dose \(x_{0}\)was instantly infused directly into the bloodstream at \(t=0\), the equation
\(x(t)=x_{0}e^{-kt}\)
applies. What happens to the rate of decay as \(t\) approaches infinity?

A The rate becomes constant

B The rate approaches infinity

C The rate approaches zero

D The rate equals the initial dose

💡 Short Explanation

As time increases, \(e^{-kt}\rightarrow 0.\) Therefore, \(\frac{dx}{dt}=-kx_{0}e^{-kt}\rightarrow 0.\) The rate of decay gradually becomes zero.

✅ Correct Answer Explanation

The decay model is \(x(t)=x_{0}e^{-kt},\) where \(k>0\). Differentiating, \(\frac{dx}{dt}=-kx_{0}e^{-kt}.\) As \(t\rightarrow \infty ,\) the exponential term satisfies \(e^{-kt}\rightarrow 0.\) Hence, \(\frac{dx}{dt}=-kx_{0}e^{-kt}\rightarrow 0.\) This means the amount of drug remaining becomes extremely small, so its rate of decrease also becomes extremely small. Therefore, Option C is correct.

❓ Why Other Options Are Incorrect

Option A) The rate becomes constant
Incorrect because
\(\frac{dx}{dt}=-kx_{0}e^{-kt}\)
continues to decrease in magnitude and is not constant.
Option B) The rate approaches infinity
Incorrect because the exponential term approaches zero, not infinity.
Option C) The rate aproaches zero
Correct.
As the drug amount approaches zero, the decay rate also approaches zero.
Option D) The rate equals the initial dose
Incorrect because \(x_{0}\)represents the initial amount, not the rate of decay.

🎯 Strategy

Used

Evaluate the Limit of the Exponential Function

Application:

1. Differentiate the exponential decay model.
2. Use the limit
\(e^{-kt}\rightarrow 0ast\rightarrow \infty .\)
1. Conclude the limiting value of the decay rate.

🧠 Memory Trick

Mnemonic: "Amount Vanishes → Rate Vanishes."

📚 Chapter: Differential Equations

15 the satellite power function
\(y=50e^{-0.004t}\)
is given to find the power after 90 days. Using the approximation
\(e^{-0.36}=0.6976,\)
what is the exact evaluated power output?

A \(50\times 0.6976=34.88 watts\)

B \(50\times 0.36=18 watts\)

C \(\frac{50}{0.6976}=71.67 watts\)

D \(50-0.36=49.64 watts\)

💡 Short Explanation

Substitute \(t=90\) into the model: \(y=50e^{-0.36}.\) Using \(e^{-0.36}=0.6976,\) we get \(y=50(0.6976)=34.88 watts.\)

✅ Correct Answer Explanation

The satellite power model is \(y=50e^{-0.004t}.\) For \(t=90\) days, \(-0.004\times 90=-0.36.\) Thus, \(y=50e^{-0.36}.\) Using the given approximation, \(e^{-0.36}=0.6976,\) we obtain \(y=50\times 0.6976=34.88 watts.\) Therefore, \(y=34.88 watts.\) Hence, Option A is correct.

❓ Why Other Options Are Incorrect

Option A)
\(50\times 0.6976=34.88 watts\)
Correct.
This is the direct evaluation of the exponential decay model.
Option B)
\(50\times 0.36=18 watts\)
Incorrect because 0.36 is the exponent, not the value of the exponential function.
Option C)
\(\frac{50}{0.6976}=71.67 watts\)
Incorrect because the model requires multiplication by \(e^{-0.36}\), not division.
Option D)
\(50-0.36=49.64 watts\)
Incorrect because exponential decay is not evaluated by simple subtraction.

🎯 Strategy

Used

Substitute and Evaluate

Application:

1. Substitute the given time into the exponent.
2. Use the provided approximation for the exponential term.
3. Multiply by the initial power.

🧠 Memory Trick

Mnemonic: "Substitute → Exponent → Multiply."

📚 Subtopic: Exponential Decay Model and Numerical Evaluation of Exponential Functions

16 To analytically find the half-life of the satellite energy output from
\(y=50e^{-0.004t},\)
the required equation is
\(25=50e^{-0.004t},\)
which simplifies to
\(-0.004t=-ln⁡2.\)
The exact evaluation of \(t\) is:

A \(\frac{0.6931}{0.004}=173.27 days\)

B \(0.6931\times 0.004=0.0027 days\)

C \(\frac{2}{0.004}=500 days\)

D \(\frac{0.3010}{0.004}=75.25 days\)

💡 Short Explanation

At half-life, \(25=50e^{-0.004t}.\) Taking natural logarithms gives \(-0.004t=-ln⁡2.\) Thus, \(t=\frac{\ln\,2}{0.004}=\frac{0.6931}{0.004}\approx 173.27 days.\)2. Correct Answer Explanation At half-life, \(y=25.\) Substitute into the model: \(25=50e^{-0.004t}.\) Divide both sides by 50: \(\frac{1}{2}=e^{-0.004t}.\) Taking the natural logarithm, \(ln⁡\left(\frac{1}{2}\right)=-0.004t.\) Since \(ln⁡\left(\frac{1}{2}\right)=-ln⁡2,\) we obtain \(-0.004t=-ln⁡2.\) Hence, \(t=\frac{\ln\,2}{0.004}=\frac{0.6931}{0.004}\approx 173.27 days.\) Therefore, Option A is correct.

❓ Why Other Options Are Incorrect

Option A)
\(\frac{0.6931}{0.004}=173.27 days\)
Correct.
It correctly evaluates the half-life formula.
Option B)
\(0.6931\times 0.004=0.0027 days\)
Incorrect because the equation requires division, not multiplication.
Option C)
\(\frac{2}{0.004}=500 days\)
Incorrect because the numerator should be \(ln⁡2\approx 0.6931\), not 2.
Option D)
\(\frac{0.3010}{0.004}=75.25 days\)
Incorrect because 0.3010 is the common logarithm (\({log⁡}_{10}2\)), whereas the exponential model requires the natural logarithm \(\ln\,2\).

🎯 Strategy

Used

Apply the Half-Life Formula

Application:

1. Set the remaining amount equal to half the initial amount.
2. Cancel the initial value.
3. Take the natural logarithm.
4. Solve for time using
\(t=\frac{\ln\,2}{k}.\)

🧠 Memory Trick

Mnemonic: "Log 2 Above, Decay Constant Below."

📚 Subtopic: Exponential Decay Model, Half-Life and Continuous Decay Applications

17 A bacterial culture increases 5 times in 10 hours. The growth model is
\(y=y_{0}e^{kt}.\)
This translates to
\(5y_{0}=y_{0}e^{10k}.\)
The constant \(k\) is isolated exactly as:

A \(k=\frac{\ln\,5}{10}\)

B \(k=\frac{10}{\ln\,5}\)

C \(k=\frac{\ln\,10}{5}\)

D \(k=\frac{5}{\ln\,10}\)

💡 Short Explanation

Since the population becomes 5 times its initial value in 10 hours, \(5=e^{10k}.\) Taking natural logarithms, \(ln⁡5=10k,\) so \(k=\frac{\ln\,5}{10}.\)

✅ Correct Answer Explanation

The exponential growth model is \(y=y_{0}e^{kt},\) where \(y_{0}\)= initial population, \(k\)= growth constant. Given that after 10 hours the population becomes 5 times the initial population, \(5y_{0}=y_{0}e^{10k}.\) Dividing both sides by \(y_{0}\), \(5=e^{10k}.\) Taking the natural logarithm, \(ln⁡5=10k.\) Therefore, \(k=\frac{\ln\,5}{10}.\) Hence, Option A is correct.

❓ Why Other Options Are Incorrect

Option A)
\(k=\frac{\ln\,5}{10}\)
Correct.
It follows directly from
\(5=e^{10k}.\)
Option B)
\(k=\frac{10}{\ln\,5}\)
Incorrect because the numerator and denominator are interchanged.
Option C)
\(k=\frac{\ln\,10}{5}\)
Incorrect because the logarithm should be taken of the growth factor (5), not the time (10).
Option D)
\(k=\frac{5}{\ln\,10}\)
Incorrect because neither the numerator nor the denominator follows from the exponential growth equation.

🎯 Strategy

Used

Use the Exponential Growth Formula

Application:

1. Write the growth model:
\(y=y_{0}e^{kt}.\)
1. Substitute the given growth factor and time:
\(5y_{0}=y_{0}e^{10k}.\)
1. Cancel \(y_{0}\).
2. Take the natural logarithm.
3. Solve for \(k\).

🧠 Memory Trick

Mnemonic: "Log of the Factor, Divide by Time."

📚 Subtopic: Exponential Growth Model, Separation of Variables and Determination of Growth Constant

18 An oil well's output decreases at a continuous rate of 10% per year, meaning
\(\frac{dy}{dt}=-0.10y.\)
To mathematically find when the output falls to one-fourth, i.e.,
\(0.25y_{0}=y_{0}e^{-0.10t},\)
the time \(t\) evaluates exactly to:

A \(t=\frac{\ln\,4}{0.10}\)

B \(t=\frac{\ln\,0.25}{0.10}\)

C \(t=\frac{0.10}{\ln\,4}\)

D \(t=\frac{4}{0.10}\)

💡 Short Explanation

Starting with \(0.25=e^{-0.10t},\) take natural logarithms: \(ln⁡(0.25)=-0.10t.\) Since \(ln⁡(0.25)=ln⁡\left(\frac{1}{4}\right)=-ln⁡4,\) we get \(-ln⁡4=-0.10t,\) so \(t=\frac{\ln\,4}{0.10}.\)

✅ Correct Answer Explanation

The exponential decay model is \(y=y_{0}e^{-0.10t}.\) When the output becomes one-fourth of the initial output, \(0.25y_{0}=y_{0}e^{-0.10t}.\) Cancel \(y_{0}\): \(0.25=e^{-0.10t}.\) Taking natural logarithms, \(ln⁡(0.25)=-0.10t.\) Using the identity \(ln⁡\left(\frac{1}{4}\right)=-ln⁡4,\) we obtain \(-ln⁡4=-0.10t.\) Dividing both sides by \(-0.10\), \(t=\frac{\ln\,4}{0.10}.\) Hence, Option A is correct.

❓ Why Other Options Are Incorrect

Option A)
\(t=\frac{\ln\,4}{0.10}\)
Correct.
It follows directly after applying logarithms and simplifying.
Option B)
\(t=\frac{\ln\,0.25}{0.10}\)
Incorrect because \(ln⁡(0.25)\)is negative, giving a negative time. The simplified positive expression is \(\frac{\ln\,4}{0.10}\).
Option C)
\(t=\frac{0.10}{\ln\,4}\)
Incorrect because the numerator and denominator are reversed.
Option D)
\(t=\frac{4}{0.10}\)
Incorrect because the exponential model requires the natural logarithm, not the number 4 itself.

🎯 Strategy

Used

Solve an Exponential Decay Equation

Application:

1. Substitute the remaining fraction into the decay model.
2. Cancel the initial quantity.
3. Take the natural logarithm.
4. Use the identity
\(ln⁡\left(\frac{1}{a}\right)=-ln⁡a.\)
1. Solve for \(t\).

🧠 Memory Trick

Mnemonic: "One-Fourth → Log Four."

📚 Subtopic: Exponential Decay, Continuous Decay Models and Solving Exponential Equations Using Natural Logarithms

📄 Passage

For space vehicles, the power supply often comes from radioactive isotopes. The output \(y\)(in watts) of a radioactive power supply for a certain satellite decays exponentially according to the function
\(y=50e^{-0.004t},\)
where \(t\) is the time in days.

Based on the passage and mathematical modeling principles, what is the implicit differential equation that governs this specific satellite's power decay?

A \(\frac{dy}{dt}=-0.004y\)

B \(\frac{dy}{dt}=50y\)

C \(\frac{dy}{dt}=-0.004t\)

D \(\frac{dy}{dt}=e^{-0.004t}\)

💡 Short Explanation

The function \(y=50e^{-0.004t}\) is an exponential decay model. Differentiating gives \(\frac{dy}{dt}=-0.004(50e^{-0.004t})=-0.004y.\)

✅ Correct Answer Explanation

Given \(y=50e^{-0.004t},\) differentiate with respect to \(t\): \(\frac{dy}{dt}=50(-0.004)e^{-0.004t}.\) Since \(50e^{-0.004t}=y,\) the equation becomes \(\frac{dy}{dt}=-0.004y.\) Hence, Option A is correct.

❓ Why Other Options Are Incorrect

Option A)
Correct.
It is obtained by differentiating the exponential decay function.
Option B)
Incorrect because 50 is the initial value, not the decay constant.
Option C)
Incorrect because the rate of change is proportional to \(y\), not time \(t\).
Option D)
Incorrect because the derivative must include the factor \(-0.004\times 50\) and be proportional to \(y\).

🎯 Strategy

Used

Differentiate the Exponential Function

Application:

1. Differentiate \(Ae^{kt}\).
2. Factor out the original function.
3. Replace \(Ae^{kt}\)with \(y\).

🧠 Memory Trick

Mnemonic: "Differentiate → Multiply by \(k\)."

📚 Subtopic: Exponential Growth and Decay Models

📄 Passage

Using the satellite model passage, the analytical step to find when the power is reduced to exactly half is taking the natural logarithm of both sides of
\(0.5=e^{-0.004t}.\)
What is the mathematical result of this specific step?

A \(ln⁡(0.5)=-0.004t\)

B \(ln⁡2=-0.004t\)

C \(0.5=-0.004ln⁡(t)\)

D \(ln⁡(-0.004)=0.5t\)

💡 Short Explanation

Taking the natural logarithm of both sides gives \(ln⁡(0.5)=-0.004t.\)

✅ Correct Answer Explanation

Starting with \(0.5=e^{-0.004t},\) take the natural logarithm of both sides: \(ln⁡(0.5)=ln⁡\left(e^{-0.004t}\right).\) Since \(ln⁡(e^{x})=x,\) we obtain \(ln⁡(0.5)=-0.004t.\) This is the immediate analytical step before solving for \(t\). Hence, Option A is correct.

❓ Why Other Options Are Incorrect

Option A)
Correct.
This is the direct result of taking natural logarithms on both sides.
Option B)
Incorrect because
\(ln⁡(0.5)=-ln⁡2,\)
not \(\ln\,2\).
Option C)
Incorrect because the logarithm is applied to the constant 0.5, not to the variable \(t\).
Option D)
Incorrect because \(ln⁡(-0.004)\)is undefined in the real number system and is unrelated to the derivation.

🎯 Strategy

Used

Take Natural Logarithms

Application:

1. Start with the exponential equation.
2. Apply \(ln⁡\)to both sides.
3. Use the identity
\(ln⁡(e^{x})=x.\)
1. Simplify to obtain the linear equation in \(t\).

🧠 Memory Trick

Mnemonic: "Take Log, Remove the Exponential."

📚 Subtopic: Exponential Decay Models and Solving Exponential Equations Using Natural Logarithms

UG Applied Mathematics Booster Test 3 - Applications of Differential Equations

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