Comprehensive assessment covering all four lessons: optimisation with exponential functions, growth and decay models, half-life and doubling time, continuously compounded interest, motion applications, displacement versus distance travelled, and volumes of revolution.
Assessment
Select the best answer for each question. 1 mark each.
The differential equation that models exponential growth and decay is:
The general solution of $\frac{dN}{dt} = kN$ is:
The derivative of $f(x) = xe^{-x}$ is:
The maximum value of $f(x) = xe^{-x}$ for $x \geq 0$ is:
For a decaying quantity $N = N_0 e^{kt}$, the half-life is given by:
A quantity follows $N = N_0 e^{kt}$ with $k = -0.05$ per year. The quantity is:
After exactly $3$ half-lives, the fraction of a radioactive substance remaining is:
For a particle moving in a straight line, velocity is the derivative of:
A particle is momentarily at rest when:
The total distance travelled by a particle from $t = t_1$ to $t = t_2$ is:
A particle has velocity $v(t) = t^2 - 4t + 3$ m/s. It is momentarily at rest when:
A colony grows so that $\frac{dN}{dt} = kN$ with $k = \ln 2$ per hour. Its doubling time is:
Integration by substitution is the reverse of:
The region under $y = x$ from $x = 0$ to $x = 3$ is rotated about the x-axis. The volume of the solid generated is:
When evaluating a definite integral such as $\int_1^2 3x^2\, dx$, the constant of integration $C$ should be:
Short Answer
Find the maximum value of $f(x) = 2xe^{-x}$ for $x \geq 0$. Show all working.
An amount of $\$5000$ is invested at $6\%$ per annum compounded continuously, so that $\frac{dA}{dt} = 0.06A$. Find the value of the investment after $8$ years, correct to the nearest dollar. Show all working.
A radioactive substance has a half-life of $12$ years. Find the decay constant $k$, then find the mass remaining from an initial $200$ g after $30$ years. Show all working.
A town's population was $10\,000$ in $2010$ and $12\,000$ in $2015$. Assuming the population grows exponentially as $P = P_0 e^{kt}$, find $k$ and hence the year in which the population first reaches $20\,000$. Show all working.
A particle moves in a straight line with acceleration $a(t) = 6t - 4$ m/s$^2$. Initially it is at rest at the origin. Find expressions for $v(t)$ and $x(t)$, and hence find its position at $t = 2$. Show all working.
A particle moves with velocity $v(t) = t^2 - 4t + 3$ m/s. Find the total distance travelled from $t = 0$ to $t = 3$. Show all working.
A particle has constant acceleration $a(t) = 2$ m/s$^2$. At $t = 0$, its velocity is $5$ m/s and its position is $10$ m. Find $x(t)$, and hence its position at $t = 3$. Show all working.
The region bounded by $y = \sqrt{x}$, the x-axis, and the line $x = 4$ is rotated about the x-axis. Find the exact volume of the solid of revolution. Show all working.
Q1: B, the model assumes the rate of change is proportional to the current amount: $\frac{dN}{dt} = kN$.
Q2: C, separating variables gives $\ln|N| = kt + C$, so $N = N_0 e^{kt}$ where $N_0 = N(0)$.
Q3: A, product rule: $f'(x) = (1)e^{-x} + x(-e^{-x}) = e^{-x}(1 - x)$.
Q4: D, $f'(x) = e^{-x}(1 - x) = 0$ gives $x = 1$, and $f(1) = 1 \cdot e^{-1} = \frac{1}{e}$.
Q5: B, setting $N = \frac{N_0}{2}$ gives $\frac{1}{2} = e^{kt_{1/2}}$, so $t_{1/2} = \frac{\ln 2}{|k|}$.
Q6: D, a negative value of $k$ means $N = N_0 e^{kt}$ decreases, so the quantity is decaying exponentially.
Q7: C, after $3$ half-lives the fraction remaining is $\left(\frac{1}{2}\right)^3 = \frac{1}{8}$.
Q8: A, velocity is the rate of change of position: $v = \frac{dx}{dt}$.
Q9: B, the particle is momentarily at rest when its velocity is zero, $v(t) = 0$.
Q10: D, total distance uses the absolute value of velocity, $\int_{t_1}^{t_2} |v(t)|\, dt$, so reversals still add to the total.
Q11: C, $v(t) = t^2 - 4t + 3 = (t - 1)(t - 3) = 0$ gives $t = 1$ and $t = 3$.
Q12: A, doubling time is $t_{double} = \frac{\ln 2}{k} = \frac{\ln 2}{\ln 2} = 1$ hour.
Q13: B, integration by substitution reverses the chain rule.
Q14: C, $V = \pi \int_0^3 y^2\, dx = \pi \int_0^3 x^2\, dx = \pi \left[\frac{x^3}{3}\right]_0^3 = 9\pi$.
Q15: D, a definite integral evaluates to a number; the constant $C$ cancels in $F(b) - F(a)$, so it is omitted.
Q16 (3 marks): Product rule: $f'(x) = 2e^{-x} + 2x(-e^{-x}) = 2e^{-x}(1 - x)$ [1]. Since $2e^{-x} > 0$ for all $x$, $f'(x) = 0$ only when $1 - x = 0$, so $x = 1$ [1]. Then $f(1) = 2(1)e^{-1} = \frac{2}{e} \approx 0.736$. Sign analysis ($f' > 0$ for $x < 1$, $f' < 0$ for $x > 1$) confirms this is a maximum, so the maximum value is $\frac{2}{e}$ [1].
Q17 (3 marks): The solution of $\frac{dA}{dt} = 0.06A$ is $A(t) = 5000e^{0.06t}$ [1]. At $t = 8$: $A(8) = 5000e^{0.06 \times 8} = 5000e^{0.48}$ [1]. Evaluating, $e^{0.48} \approx 1.61607$, so $A(8) \approx 5000 \times 1.61607 \approx \$8080$ [1].
Q18 (3 marks): From $t_{1/2} = \frac{\ln 2}{|k|}$ with $t_{1/2} = 12$: $k = -\frac{\ln 2}{12} \approx -0.0578$ per year (negative for decay) [1]. Then $N(t) = 200e^{kt}$, so $N(30) = 200e^{-0.0578 \times 30} = 200e^{-1.733}$ [1]. Evaluating, $N(30) \approx 200 \times 0.1768 \approx 35.4$ g [1]. (Check: $30 \div 12 = 2.5$ half-lives, so $200 \times (\tfrac{1}{2})^{2.5} \approx 35.4$ g.)
Q19 (4 marks): Let $t = 0$ correspond to $2010$, so $P_0 = 10\,000$ and $P(t) = 10\,000e^{kt}$. At $t = 5$: $12\,000 = 10\,000e^{5k}$, so $e^{5k} = 1.2$ [1]. Thus $k = \frac{\ln 1.2}{5} \approx 0.03646$ per year [1]. Set $20\,000 = 10\,000e^{kt}$, so $e^{kt} = 2$ and $t = \frac{\ln 2}{k} = \frac{\ln 2}{0.03646} \approx 19.0$ years [1]. Since $2010 + 19 = 2029$, the population first reaches $20\,000$ during $2029$ [1].
Q20 (4 marks): $v(t) = \int (6t - 4)\, dt = 3t^2 - 4t + C_1$; since $v(0) = 0$, $C_1 = 0$, so $v(t) = 3t^2 - 4t$ [1.5]. $x(t) = \int (3t^2 - 4t)\, dt = t^3 - 2t^2 + C_2$; since $x(0) = 0$, $C_2 = 0$, so $x(t) = t^3 - 2t^2$ [1.5]. At $t = 2$: $x(2) = 8 - 8 = 0$ m, so the particle has returned to the origin [1].
Q21 (4 marks): $v(t) = (t - 1)(t - 3)$, zero at $t = 1$ and $t = 3$. On $[0, 1]$, $v \geq 0$; on $[1, 3]$, $v \leq 0$ [1]. Distance $= \int_0^1 v\, dt - \int_1^3 v\, dt$. With antiderivative $\frac{t^3}{3} - 2t^2 + 3t$: on $[0, 1]$ the value is $\frac{1}{3} - 2 + 3 = \frac{4}{3}$ [1]. On $[1, 3]$ the signed integral is $(9 - 18 + 9) - (\frac{4}{3}) = 0 - \frac{4}{3} = -\frac{4}{3}$, so its magnitude is $\frac{4}{3}$ [1]. Total distance $= \frac{4}{3} + \frac{4}{3} = \frac{8}{3}$ m [1].
Q22 (4 marks): $v(t) = \int 2\, dt = 2t + C_1$; since $v(0) = 5$, $C_1 = 5$, so $v(t) = 2t + 5$ [1.5]. $x(t) = \int (2t + 5)\, dt = t^2 + 5t + C_2$; since $x(0) = 10$, $C_2 = 10$, so $x(t) = t^2 + 5t + 10$ [1.5]. At $t = 3$: $x(3) = 9 + 15 + 10 = 34$ m [1].
Q23 (4 marks): Using the disc method about the x-axis, $V = \pi \int_0^4 y^2\, dx$ [1]. Here $y^2 = (\sqrt{x})^2 = x$, so $V = \pi \int_0^4 x\, dx$ [1] $= \pi \left[\frac{x^2}{2}\right]_0^4 = \pi \left(\frac{16}{2} - 0\right)$ [1] $= 8\pi$ cubic units [1].