Calculus Applications - IB Mathematics Applications and Interpretation

For what values of $t$ is the particle moving in the positive direction?
Answer: $0 \le t < 2$ or $t > 4$ The particle moves in the positive direction when its velocity, $v(t)$ , is positive. First, find the velocity function by differentiating the displacement function: $v(t) = s'(t) = 3t^2 - 18t + 24$ . To find where $v(t) > 0$ , first find the roots of $v(t)=0$ : $3(t^2 - 6t + 8) = 0 \Rightarrow 3(t-2)(t-4) = 0$ . The roots are $t=2$ and $t=4$ . Since the velocity function is a parabola opening upwards, it is positive outside its roots. Therefore, $v(t) > 0$ for $t < 2$ and $t > 4$ . Considering the domain $t \ge 0$ , the particle moves in the positive direction for $0 \le t < 2$ or $t > 4$ .
The velocity, $v$ m/s, of a remote-controlled car is given by $v(t) = 5t - t^2$ , for $0 \le t \le 6$ seconds. What is the total distance travelled by the car in the first 6 seconds?
Answer: 23.7 m Total distance travelled requires integrating the absolute value of velocity, $\int_0^6 |v(t)| \,dt$ . First, find when the velocity is zero: $v(t) = t(5-t) = 0$ , which gives $t=0$ and $t=5$ . The car changes direction at $t=5$ s. The total distance is the sum of the distances travelled in the intervals $[0, 5]$ and $[5, 6]$ . Distance = $\int_0^5 (5t - t^2) \,dt + \int_5^6 -(5t - t^2) \,dt$ . $\int(5t-t^2)dt = \frac{5}{2}t^2 - \frac{1}{3}t^3$ . The first integral is $[\frac{5}{2}t^2 - \frac{1}{3}t^3]_0^5 = \frac{125}{6}$ . The second integral is $|[\frac{5}{2}t^2 - \frac{1}{3}t^3]_5^6| = |(90-72) - \frac{125}{6}| = |18 - \frac{125}{6}| = |-\frac{17}{6}| = \frac{17}{6}$ . Total distance = $\frac{125}{6} + \frac{17}{6} = \frac{142}{6} = \frac{71}{3} \approx 23.7$ m.
At what time is the volume of water in the reservoir at a minimum?
Answer: 5 hours This problem tests your understanding of optimization using calculus, specifically finding when a rate of change equals zero to locate minimum or maximum values. To find when the volume is minimized, you need to set up the volume function and find where its derivative equals zero. The net rate of water change is: inflow minus outflow = $(300 + 20t) - 400 = -100 + 20t$ litres/hour. The volume function is: $V(t) = 50000 + \int_0^t (-100 + 20s) ds = 50000 - 100t + 10t^2$ To find the minimum, take the derivative and set it equal to zero: $V'(t) = -100 + 20t = 0$ Solving: $20t = 100$ , so $t = 5$ hours. Since $V''(t) = 20 > 0$ , this confirms a minimum at $t = 5$ . Looking at the wrong answers: B) 0 hours represents the starting time, but the reservoir continues losing water initially since outflow exceeds inflow. C) 10 hours might tempt you if you mistakenly doubled the correct answer or made an algebraic error. D) 20 hours could result from confusing the coefficients in the rate equation or solving $t = 100/20$ incorrectly. A) 5 hours is correct because this is precisely when the net flow rate changes from negative (losing water) to positive (gaining water). Study tip: In reservoir problems, always identify the net rate first, then integrate to get volume. The minimum/maximum occurs when the net rate equals zero, not when individual rates are equal.
Find the rate, in litres per hour, at which water is leaking from the tank after 10 hours.
Answer: 121.5 The rate of change of volume is the derivative, $V'(t)$ . Using the chain rule: $V'(t) = 5000 \cdot 3(1 - 0.01t)^2 \cdot (-0.01) = -150(1 - 0.01t)^2$ . At $t=10$ , the rate of change is $V'(10) = -150(1 - 0.01 \cdot 10)^2 = -150(0.9)^2 = -150(0.81) = -121.5$ litres/hour. Since the question asks for the rate of leaking, it implies a positive value, which is 121.5 litres/hour.
What is the acceleration of the roller coaster car at $t=3$ seconds, to three significant figures?
Answer: -2.89 m/s $^2$ When you encounter a physics problem asking for acceleration given a position function, remember that acceleration is the second derivative of position with respect to time. This connects calculus directly to kinematics. Given the height function $h(t) = 10\sin(0.5t) - 0.2t^2 + 20$ , you need to find the second derivative and evaluate it at $t = 3$ . First, find the velocity by taking the first derivative: $h'(t) = 10 \cdot 0.5 \cos(0.5t) - 0.4t = 5\cos(0.5t) - 0.4t$ Then find acceleration by taking the second derivative: $h''(t) = 5 \cdot (-0.5)\sin(0.5t) - 0.4 = -2.5\sin(0.5t) - 0.4$ At $t = 3$ : $h''(3) = -2.5\sin(0.5 \cdot 3) - 0.4 = -2.5\sin(1.5) - 0.4$ Since $\sin(1.5) \approx 0.9975$ : $h''(3) = -2.5(0.9975) - 0.4 = -2.494 - 0.4 = -2.894$ To three significant figures, this is $-2.89$ m/s², which is answer A. Answer B (-10.4) likely comes from incorrectly using the coefficient 10 instead of properly applying the chain rule. Answer C (-0.847) might result from only considering the polynomial term's contribution. Answer D (28.2) appears to confuse signs or mix up derivative calculations entirely. Study tip: Always remember the kinematic hierarchy: position → velocity (first derivative) → acceleration (second derivative). Practice chain rule applications with trigonometric functions, as they appear frequently in oscillatory motion problems.
At the instant when the side length is 5 cm, what is the rate of change of the cube's volume?
Answer: -7.5 cm $^3$ /s This is a related rates problem involving calculus and the chain rule. When you see rates of change that depend on each other, you need to connect them through differentiation. Start with the volume formula for a cube: $V = x^3$ . To find how the volume changes with respect to time, differentiate both sides: $\frac{dV}{dt} = 3x^2 \cdot \frac{dx}{dt}$ . This uses the chain rule since volume depends on side length, which depends on time. You're given that $\frac{dx}{dt} = -0.1$ cm/s (negative because the side length is decreasing) and you need $\frac{dV}{dt}$ when $x = 5$ cm. Substituting: $\frac{dV}{dt} = 3(5)^2(-0.1) = 3(25)(-0.1) = -7.5$ cm³/s. Answer A (-125.0 cm³/s) likely comes from incorrectly calculating $5^3 \times (-0.1) = -12.5$ , then multiplying by 10. Answer B (75.0 cm³/s) makes the same calculation error but forgets the negative sign entirely. Answer C (7.5 cm³/s) gets the correct magnitude but misses that volume must be decreasing (positive sign) when the side length decreases. The key insight is recognizing the negative sign: if the side length decreases, the volume must also decrease, so $\frac{dV}{dt}$ should be negative. For related rates problems, always write down what you know ( $\frac{dx}{dt}$ ), what you want ( $\frac{dV}{dt}$ ), find the relationship between variables, then differentiate with respect to time using the chain rule.
Find the displacement of the particle at $t=\pi$ seconds.
Answer: 14.4 m First, find velocity by integrating acceleration: $v(t) = \int \cos(0.5t) \,dt = 2\sin(0.5t) + C$ . Given $v(0)=3$ , $3 = 2\sin(0) + C \Rightarrow C=3$ . So, $v(t) = 2\sin(0.5t) + 3$ . Next, find displacement by integrating velocity: $s(t) = \int (2\sin(0.5t) + 3) \,dt = -4\cos(0.5t) + 3t + D$ . Given $s(0)=1$ , $1 = -4\cos(0) + 0 + D \Rightarrow 1 = -4 + D \Rightarrow D=5$ . So, $s(t) = -4\cos(0.5t) + 3t + 5$ . At $t=\pi$ , $s(\pi) = -4\cos(0.5\pi) + 3\pi + 5 = -4(0) + 3\pi + 5 = 3\pi + 5 \approx 14.42$ m.
What is the company's total profit, to the nearest thousand dollars, after 6 months?
Answer: $103,000 The total profit after 6 months is the initial profit plus the accumulated profit over the 6 months. Accumulated profit = $\int_0^6 P'(t) \,dt = \int_0^6 (5\cos(0.2t) + 10) \,dt$ . We can solve this analytically or with a calculator. Analytically: $[\frac{5\sin(0.2t)}{0.2} + 10t]_0^6 = [25\sin(0.2t) + 10t]_0^6 = (25\sin(1.2) + 60) - 0 \approx 23.30 + 60 = 83.30$ . This is in thousands of dollars. The total profit is $P(6) = P(0) + \text{Accumulated Profit} = 20 + 83.30 = 103.30$ thousand dollars. To the nearest thousand, this is $103,000.
What is the net change in the fish population, to the nearest fish, during the first 10 months?
Answer: 132 The net change in the population is the definite integral of the rate of change from $t=0$ to $t=10$ . We need to calculate $\int_0^{10} 100(e^{-0.1t} - 0.5) \,dt$ . Using a GDC, the value of the integral is approximately 132.1. Analytically, the integral is $100[-\frac{e^{-0.1t}}{0.1} - 0.5t]_0^{10} = 100[-10e^{-0.1t} - 0.5t]_0^{10} = 100[(-10e^{-1} - 5) - (-10e^0 - 0)] = 100[-10/e - 5 + 10] = 100(5-10/e) \approx 132.1$ . To the nearest fish, the net change is 132.
What is the total amount of oil, in millions of barrels, predicted to be consumed from the start of 2025 to the start of 2030?
Answer: 8.75 The total amount consumed is the definite integral of the rate function $R(t)$ . The interval is from the start of 2025 ( $t=5$ ) to the start of 2030 ( $t=10$ ). We need to calculate $\int_5^{10} 1.2e^{0.05t} \,dt$ . Using a calculator, this integral is approximately 8.75. Analytically, the integral is $[\frac{1.2}{0.05}e^{0.05t}]_5^{10} = 24[e^{0.5} - e^{0.25}] \approx 8.75$ million barrels.
At $t=4$ , which of the following statements correctly describes the particle's speed?
Answer: The particle's speed is at a local maximum. First, evaluate velocity at $t=4$ : $v(4) = (4-3)e^{-4} = e^{-4} > 0$ . Since $v \neq 0$ , the particle is not stationary and not changing direction. To analyze the speed, we need the acceleration, $a(t) = v'(t)$ . Using the product rule, $a(t) = (1)e^{-t} + (t-3)(-e^{-t}) = e^{-t}(1 - (t-3)) = e^{-t}(4-t)$ . At $t=4$ , $a(4) = e^{-4}(4-4) = 0$ . Since $v(4) > 0$ and $a(4) = 0$ , the rate of change of speed is zero. To determine if it's a max or min, we check the sign of $a(t)$ around $t=4$ . For $t < 4$ , $a(t) > 0$ . For $t > 4$ , $a(t) < 0$ . Since acceleration changes from positive to negative, the velocity (and speed, since v is positive here) is at a local maximum.
What is the maximum height, to the nearest metre, reached by the object?
Answer: 127 m When you see motion problems with given acceleration functions, you need to work backwards through calculus to find velocity, then position, then determine when maximum height occurs. Starting with $a(t) = -9.8 - 0.1t$ , integrate to find velocity: $v(t) = -9.8t - 0.05t^2 + C$ . Using the initial velocity of 50 m/s, you get $v(t) = 50 - 9.8t - 0.05t^2$ . Integrate again for position: $s(t) = 50t - 4.9t^2 - \frac{0.05t^3}{3} + C$ . With initial height 2 m: $s(t) = 2 + 50t - 4.9t^2 - \frac{t^3}{60}$ . Maximum height occurs when velocity equals zero. Setting $50 - 9.8t - 0.05t^2 = 0$ and solving this quadratic equation gives $t ≈ 4.69$ seconds (taking the positive root since we want the upward journey). Substituting back: $s(4.69) = 2 + 50(4.69) - 4.9(4.69)^2 - \frac{(4.69)^3}{60} ≈ 127$ m. A) 127 m is correct - this accounts for the changing acceleration properly. B) 125 m likely results from computational errors in the quadratic solution or rounding too early. C) 130 m might come from ignoring the $-0.1t$ term in acceleration, treating it as constant $-9.8$ m/s². D) 247 m suggests a major conceptual error, possibly confusing the time calculation or using incorrect integration. Remember: in variable acceleration problems, always integrate step-by-step from acceleration to velocity to position, and find extremes by setting velocity to zero.
Find the time $t>0$ at which the car's instantaneous velocity is equal to its average velocity over the interval $[0, 6]$ .
Answer: 3.65 s This question tests the relationship between instantaneous velocity (derivative) and average velocity over an interval - a fundamental concept in calculus applications. To find when instantaneous velocity equals average velocity, you need both values. First, find the instantaneous velocity by taking the derivative: $v(t) = s'(t) = 3t^2 - 6t + 5$ . Next, calculate the average velocity over $[0, 6]$ using $\frac{s(6) - s(0)}{6 - 0}$ . You get $s(6) = 6^3 - 3(6^2) + 5(6) = 216 - 108 + 30 = 138$ and $s(0) = 0$ , so average velocity is $\frac{138}{6} = 23$ m/s. Now set instantaneous velocity equal to average velocity: $3t^2 - 6t + 5 = 23$ . Rearranging gives $3t^2 - 6t - 18 = 0$ , or $t^2 - 2t - 6 = 0$ . Using the quadratic formula: $t = \frac{2 \pm \sqrt{4 + 24}}{2} = \frac{2 \pm \sqrt{28}}{2} = 1 \pm \sqrt{7}$ . Since $t > 0$ , you want $t = 1 + \sqrt{7} \approx 3.65$ seconds. Choice A (1.00 s) likely comes from taking $1 - \sqrt{7}$ and using its absolute value. Choice B (3.00 s) might result from approximating $\sqrt{7} \approx 2$ . Choice C (23.0 s) confuses the time with the average velocity value itself. Strategy tip: When solving "instantaneous equals average" problems, always find the derivative first, then calculate the average rate of change, and finally solve the resulting equation. Watch for negative solutions that need to be rejected based on the domain.
How much chemical, to the nearest millilitre, leaks out of the container between $t=10$ and $t=20$ minutes?
Answer: 51 mL When you see a rate function and need to find the total amount over a time interval, you're looking at a definite integral problem. The rate $L(t) = \frac{100}{t+5}$ tells you how fast the chemical is leaking at any moment, but to find the total amount leaked between two times, you need to integrate. To find the chemical leaked between $t=10$ and $t=20$ , calculate: $\int_{10}^{20} \frac{100}{t+5} dt$ First, find the antiderivative. Since $\int \frac{1}{t+5} dt = \ln|t+5| + C$ , we have: $\int \frac{100}{t+5} dt = 100\ln|t+5| + C$ Now evaluate the definite integral: $100\ln|t+5|\Big|_{10}^{20} = 100[\ln(25) - \ln(15)] = 100\ln\left(\frac{25}{15}\right) = 100\ln\left(\frac{5}{3}\right)$ Using a calculator: $100\ln\left(\frac{5}{3}\right) \approx 100(0.511) = 51.1 \approx 51$ mL. Choice A (3 mL) likely comes from incorrectly evaluating just $\ln(5/3)$ without the factor of 100. Choice B (50 mL) might result from approximation errors or using $\ln(2)$ instead of $\ln(5/3)$ . Choice C (53 mL) could come from computational mistakes in the logarithm calculation. Remember: when you see a rate function, the total change is always the integral of that rate over the given interval. Don't confuse rate (the derivative) with total amount (the integral).
What is the maximum displacement of the buoy from the fixed point in the given interval?
Answer: 2.31 m To find the maximum displacement, we need to find the critical points by setting the derivative (velocity) to zero, and then test the displacement at the critical points and the endpoints. $v(t) = s'(t) = 1.5 \cdot \frac{\pi}{6}\cos(\frac{\pi}{6}t) - 0.5 = \frac{\pi}{4}\cos(\frac{\pi}{6}t) - 0.5$ . Set $v(t)=0$ : $\cos(\frac{\pi}{6}t) = \frac{0.5}{\pi/4} = \frac{2}{\pi}$ . Using a calculator, $\frac{\pi}{6}t = \arccos(\frac{2}{\pi}) \approx 0.8806$ , so $t \approx 1.682$ . We must also check the endpoints $t=0$ and $t=12$ . $s(0)=2$ . $s(12) = 1.5\sin(2\pi) - 0.5(12) + 2 = -4$ . $s(1.682) = 1.5\sin(\frac{\pi}{6}(1.682)) - 0.5(1.682) + 2 \approx 2.31$ . Comparing the values $s(0)=2$ , $s(12)=-4$ , and $s(1.682) \approx 2.31$ , the maximum displacement is 2.31 m.
What is the acceleration of the rocket at the time its velocity is at a maximum?
Answer: 0 m/s $^2$ The velocity is at a maximum when its rate of change is zero. The rate of change of velocity is acceleration, $a(t) = v'(t)$ . Therefore, at the time of maximum velocity, the acceleration must be 0. To verify, we can find the time of maximum velocity by setting $a(t) = v'(t) = -0.3t^2 + 4t + 5 = 0$ . Using a solver, this occurs at $t \approx 14.5$ s. The question asks for the acceleration at this time, which is by definition 0 m/s $^2$ .

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