Exponential Models with Differential Equations - AP Calculus AB

1. For $\frac{dy}{dt} = ky$, determine $y(t)$ if $y_0 = 12$.

   $y(t) = 12e^{kt}$. General solution with specified initial value $y_0=12$.

2. Determine $y(t)$ for $\frac{dy}{dt} = -3y$ and $y(0) = 7$.

   $y(t) = 7e^{-3t}$. Apply $k=-3$ and initial value $y_0=7$ to get the solution.

3. What is the initial value in $N(t) = N_0 e^{-\lambda t}$?

   $N_0$. The initial amount or concentration at time $t=0$.

4. For $\frac{dy}{dt} = -2y$, what is $y(t)$ if $y(0) = 4$?

   $y(t) = 4e^{-2t}$. Use $k=-2$ and $y_0=4$ in the exponential decay formula.

5. What is the general solution form for a differential equation $\frac{dy}{dt} = ky$?

   $y(t) = y_0 e^{kt}$. This is the general exponential form where $k$ determines growth/decay rate.

6. Identify $y(t)$ for $\frac{dy}{dt} = 0.1y$ with $y(1) = 3$.

   $y(t) = 3e^{0.1(t-1)}$. Apply initial condition at $t=1$ with $k=0.1$ and $y_0=3$.

7. Find $y(t)$ if $\frac{dy}{dt} = 4y$ and $y(2) = 16$.

   $y(t) = 16e^{4(t-2)}$. With condition at $t=2$, substitute to get $y=16e^{4(t-2)}$.

8. Which function describes a quantity halving every 5 years?

   $y(t) = y_0 \left(\frac{1}{2}\right)^{\frac{t}{5}}$. Half-life of 5 years means the exponent is $t/5$ with base $1/2$.

9. Find the $y$-intercept for $y(t) = 5e^{3t}$.

   $y$-intercept is 5. At $t=0$, $y(0) = 5e^0 = 5 \cdot 1 = 5$.

10. Determine the constant $k$ for the doubling time formula $P(t) = P_0 2^{\frac{t}{d}}$.

    $k = \frac{\ln(2)}{d}$. Since $2^{t/d} = e^{kt}$, we have $k = \ln(2)/d$.

11. What is the $k$ value if a population triples in 6 years?

    $k = \frac{\ln(3)}{6}$. Tripling means $3 = e^{6k}$, so $k = \ln(3)/6$.

12. State the exponential growth rate if $k = 0.05$.

    5%. The value $k=0.05$ corresponds to a 5% growth rate per unit time.

13. What is the exponential decay formula for a substance?

    $N(t) = N_0 e^{-\lambda t}$. Standard decay model where $\lambda$ is the decay constant.

14. What is the half-life formula for exponential decay?

    $y(t) = y_0 \left(\frac{1}{2}\right)^{\frac{t}{h}}$. The quantity reduces by half every $h$ time units using base $\frac{1}{2}$.

15. What does $y_0$ represent in the solution $y(t) = y_0 e^{kt}$?

    Initial value of $y$. The starting value or amount at time $t=0$.

16. What is the natural base in the exponential model $y(t) = y_0 e^{kt}$?

    $e$. Euler's number $e \approx 2.718$ is the natural exponential base.

17. What is the solution for $\frac{dy}{dt} = ky$ if $y(0) = y_0$?

    $y(t) = y_0 e^{kt}$. Separating variables and integrating yields this exponential solution.

18. What is the specific solution if $\frac{dy}{dt} = 5y$ and $y(1) = 10$?

    $y(t) = 10e^{5(t-1)}$. Use initial condition at $t=1$ to find $C=10e^{-5}$, so $y=10e^{5(t-1)}$.

19. What is the solution to $\frac{dy}{dt} = -y$ given $y(0) = 15$?

    $y(t) = 15e^{-t}$. Apply $k=-1$ and initial condition $y_0=15$ to the general form.

20. Determine $y(t)$ for $\frac{dy}{dt} = -0.3y$ and $y(0) = 9$.

    $y(t) = 9e^{-0.3t}$. Use decay constant $k=-0.3$ with initial value $y_0=9$.

21. What is the exponential model used for population growth?

    $P(t) = P_0 e^{kt}$. Standard exponential growth model where $P_0$ is initial population.

22. Identify the $y$-intercept in $y(t) = 10e^{-t}$.

    $y$-intercept is 10. The $y$-intercept occurs when $t=0$, giving $y(0)=10e^0=10$.

23. Which equation represents a population doubling every period?

    $P(t) = P_0 2^{\frac{t}{d}}$. The base 2 with time divided by doubling period $d$ models doubling.

24. Determine $y(t)$ for $\frac{dy}{dt} = 2y$ with $y(0) = 8$.

    $y(t) = 8e^{2t}$. Substitute $k=2$ and $y_0=8$ into the general exponential solution.

25. What initial condition is used for solving $\frac{dy}{dt} = ky$?

    $y(0) = y_0$. This boundary condition determines the constant in the general solution.

26. Calculate the decay constant $k$ if the half-life is 10 years.

    $k = \frac{\ln(2)}{10}$. For half-life decay, $k = \ln(2)/t_{1/2} = \ln(2)/10$.

27. Find $y(t)$ for $\frac{dy}{dt} = 0.5y$ if $y(0) = 1$.

    $y(t) = e^{0.5t}$. With $k=0.5$ and $y_0=1$, the solution simplifies to $e^{0.5t}$.

28. What is the $k$ value for a quantity decreasing by 15% annually?

    $k = -0.15$. A 15% annual decrease corresponds to $k=-0.15$ in the model.

29. State the differential equation model for continuous exponential growth.

    $\frac{dy}{dt} = ky$. The rate of change is proportional to the current value with constant $k$.

30. Identify the differential equation for exponential decay.

    $\frac{dy}{dt} = -ky$. Negative $k$ indicates the quantity decreases over time exponentially.

31. Find the solution to $\frac{dy}{dt} = 3y$ given $y(0) = 5$.

    $y(t) = 5e^{3t}$. Substitute $k=3$ and $y_0=5$ into the general solution $y=y_0e^{kt}$.

32. Solve for $k$ in $y = y_0 e^{kt}$ if $y(2) = 8$ and $y_0 = 2$.

    $k = \frac{\ln(4)}{2}$. From $8 = 2e^{2k}$, get $4 = e^{2k}$, so $k = \ln(4)/2$.

33. Calculate $y(t)$ for $\frac{dy}{dt} = -\frac{1}{2}y$ and $y(0) = 20$.

    $y(t) = 20e^{-\frac{1}{2}t}$. Use $k=-1/2$ and $y_0=20$ in the exponential solution form.

34. State the condition for exponential growth.

    $k > 0$ in $\frac{dy}{dt} = ky$. Positive $k$ means the derivative and function have the same sign.

35. What is the $y$-intercept for $P(t) = 2e^{0.3t}$?

    $y$-intercept is 2. At $t=0$, $P(0) = 2e^0 = 2 \cdot 1 = 2$.

36. Express $y(t)$ in terms of $k$ if $\frac{dy}{dt} = ky$ and $y(0) = 6$.

    $y(t) = 6e^{kt}$. General form with initial condition $y_0=6$ and growth constant $k$.

37. Identify the exponential model for radioactive decay.

    $N(t) = N_0 e^{-kt}$. Standard form for radioactive decay with decay constant $k$.

38. State the condition for exponential decay.

    $k < 0$ in $\frac{dy}{dt} = ky$. Negative $k$ causes the quantity to decrease exponentially over time.

39. Calculate $y(t)$ for $\frac{dy}{dt} = 0.2y$ and $y(0) = 7$.

    $y(t) = 7e^{0.2t}$. Substitute $k=0.2$ and $y_0=7$ into the exponential growth formula.

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