AP Calculus BC : 7.8 Exponential Models with Differential  Equations- Exam Style questions with Answer- MCQ

Question

The number of bacteria in a culture is growing at a rate of\(3,000e^{2t/5}\) per unit of time t. At t = 0 , the number of bacteria present was 7,500. Find the number present at t = 5 .
(A)  \(1, 200e^{2}\)         (B)   \(3,000e^{2}\)          (C)  \(7,500e^{2}\)                    (D)\( 7,500e^{5}\)                    (E)\(\frac{15,000}{7}e^{7}\)

Answer/Explanation

Ans:C 

Question

\(f'(x) and f(1)=1\) ,then f(x)

(A)\(\frac{1}{2}e^{-2x+2}\)                             (B)\(e^{-x-1}\)                       (C)\(e^{1-x}\)                             (D)\(e^{-x} -e^{x}\)

Answer/Explanation

Ans:C

 

Question

 Plutonium has a half-life of 8,645 years. Using your calculator, if there are initially 20 g of plutonium, how many grams are left after 1,000 years?
(A) ≈ 45
(B) ≈ 18
(C) ≈ 35
(D) ≈ 15

Answer/Explanation

Ans:(B)

Recall the equation for exponential growth/decay is
\(y=y_{0}e^{kt}\)
Since the half-life is 8645 years and you start with 20 g
\(\Rightarrow 10=20e^{(k)(8645)}\)
\(\Rightarrow \frac{1}{2}=e^{8645k}\)
To solve the exponential equation for k you take the natural log of both sides.
\(\Rightarrow \ln \left ( \frac{1}{2} \right )=\ln (e^{8645k})\)
\(\Rightarrow \ln (1)-\ln (2)=(8645k)(ln(e))\)
Now recall that ln(1) = 0 and ln(e) = 1.
\(\Rightarrow -\ln (2)=8645k\Rightarrow k=-\frac{\ln (2)}{8645}\Rightarrow y=20e^{-\frac{\ln (2)}{8645}t}\)
Now to find how many grams are left after 1,000 years, you can plug t = 1,000 into the equation

Question

The amount of bacteria in a petri dish increases at a rate proportional to the amount of bacteria present. An initial amount of bacteria is placed in the dish, and shortly after they begin to multiply. Using your calculator, if there are 200 bacteria after one day in the dish, and 600 bacteria after 3 days in the dish, how many bacteria are in the petri dish after 7 days?
(A) 5,400
(B) 400
(C) 500
(D) 540

Answer/Explanation

Ans:(A)

Since the rate of increase is proportional to the amount of bacteria present, this is an exponential growth/decay model.
\(y(t)=y_{0}e^{kt}\)
\(y(1)=200,y(3)=600\Rightarrow 200=y_{0}e^{k(1)},600=y_{0}e^{k(3)}\)
Since you have 2 equations and 1 unknown variable,\(y_{0}\), you can use substitution to solve for\(y_{0}\) and k.
Solving the first equation for \(y_{0}\) you get: \(y_{0}=\frac{200}{e^{k}}\Rightarrow y_{0}=200e^{-k}\)
Now substituting your value of \(y_{0}\) into the second equation you get
\(600=(200e^{-k})e^{3k}\)
\(600=200e^{2k}\)
\(3=e^{2k}\)
\(\ln (3)=2k\)
\(k=\frac{\ln (3)}{2}\)
\(k=\ln (\sqrt{3})\)
Now that you have k, you can plug back into your first equation to solve for \(y_{0}\)
\(k=\ln (\sqrt{3}),y_{0}=200e^{-k}\Rightarrow y_{0}=200e^{-\ln (\sqrt{3})} \Rightarrow y_{0}\approx 115.47\)
To calculate how many bacteria are present after 1 week (7 days), you can use the following equation:
\(y(t)=115.47e^{\ln (\sqrt{3})(t)}\)

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