Limits and continuity #2 - Questions and Answers

Question 1 A tank initially contains 120 L of pure water. A mixture containing a concentration of y g/L of salt enters the tank at a rate of 2 L/min, and the well-stirred mixture leaves the tank at the same rate. Find an expression in terms of y for the amount of salt in the tank at any time t. Also find the limiting amount of salt in the tank as t→∞ Answer 1 Step 1 Let Q(t) salt content of tank at time t, be the initial salt content is Q(0) = 0 The differential equation that describes this issue is dQ/dt = rate_in - rate_out rate_in = 2γ rate_out = Q(t)/120 * 2 Consequently, the problem of beginning value is dQ/dt = 2γ - Q/60 Q(0) = 0 Step 2 Division with differential equation Q - 120γ to obtain the standard separable equation. This is acceptable for everyone. Q ≠ 120 dQ/(Q - 120γ) = -1/60 dt Integrate both parties in relation to t Keeping in mind the chain rule, you should be aware that the integration of the left side may be expressed as simply "integration with respect to Q." ∫ 1/(Q - 120γ) dQ = ∫ -1/60 dt ln|Q - 120γ| + C₁ = -t/60 + C₂ ln|Q - 120γ| = -t/60 + C₂ - C₁ |Q - 120γ| = e^(-t/60 + C₂ - C₁) Since C₁ and C₂ are constants e^(C₂ - C₁) is a positive constant. Call it C The right side of the equation is positive for all t, as you can see. Notice that Q < 120γ (because 120γ is max. amount of salt in the tank) 120γ - Q = Ce^(-t/60) Q = 120γ - Ce^(-t/60) Step 3 Calculate constant C using the starting conditions. Q(0) = 0 120γ - Ce^0 = 0 C = 120γ Consequently, the salt content Q(t) each moment t is in the tank Q(t) = 120γ - 120γe^(-t/60) Additionally, the tank's maximum salt content is lim t→∞ Q(t) = lim t→∞ (120γ - 120γe^(-t/60)) = 120γ Answer 2 At time t=0, the tank contains 120 L of pure water, which means there is no salt in the tank initially. Therefore, S(0) = 0 g. Now, let's consider the rate at which salt enters the tank. The mixture entering the tank has a concentration of y g/L of salt, and the rate of flow is 2 L/min. This means that the amount of salt entering the tank per minute is 2y g/min . Since the mixture entering the tank is well-stirred and leaves the tank at the same rate, the amount of salt leaving the tank per minute is also 2yg/min. Therefore, the rate of change of the amount of salt in the tank is given by: dS/dt = (rate of salt entering the tank) - (rate of salt leaving the tank) dS/dt = 2γ - 2γ = 0 We obtained 0 as the derivative of S(t) with respect to t. This implies that the amount of salt in the tank is constant over time. Therefore, the expression for the amount of salt in the tank at any time t is: S(t) = S(0) + ∫(dS/dt)dt S(t) = 0 + ∫0 dt S(t) = 0 Hence, the amount of salt in the tank at any time t is 0 g. Next, let's find the limiting amount of salt in the tank as t approaches infinity (t→∞). In this case, we can evaluate the integral from 0 to infinity: S(∞) = S(0) + ∫(dS/dt)dt S(∞) = 0 + ∫0^∞ 0 dt S(∞) = 0 Therefore, the limiting amount of salt in the tank as t approaches infinity is 0 g. Answer 3 Explanation: Let t be the time in minutes. Let V(t) be the volume of water in the tank at time t in liters. Let S(t) be the amount of salt in the tank at time t in grams. Given that the tank initially contains 120 L of pure water, we can express V(t) as follows: V(t) = 120 + (2L/min * t) The concentration of salt in the mixture entering the tank is y g/L. Since the mixture enters and leaves the tank at the same rate, the concentration of salt in the tank remains constant over time. Therefore, the amount of salt in the tank at any time t is given by: S(t) = γ * V(t) Substituting the expression for V(t), we have: S(t) = γ * (120 + (2L/min) * t) Simplifying this expression, we get: S(t) = 120γ + 2γt Thus, the amount of salt in the tank at any time t is given by the expression 120γ + 2γt. Now, let's determine the limiting amount of salt in the tank as t→ ∞. In the long run, the tank will continuously receive the mixture with concentration y, while an equal volume of mixture leaves the tank. Therefore, the concentration of salt in the tank will stabilize at y g/L. As t approaches infinity, the amount of salt in the tank can be calculated as: lim t→∞ S(t) = lim t→∞ (120γ + 2γt) Since the term 2γt becomes infinitely large as t approaches infinity, the limiting amount of salt in the tank as t→ ∞ is infinity. Answer 4 To solve the given problem, we'll first find an expression for the amount of salt in the tank at any time t, and then determine the limiting amount of salt as t approaches infinity. Let's denote the amount of salt in the tank at time t as Q(t). Initially, the tank contains 120 L of pure water, so there is no salt present: Q(0) = 0. The rate at which the mixture enters the tank is 2 L/min, and the concentration of salt in the entering mixture is y g/L. This means that the amount of salt entering the tank per minute is 2y g/min . The mixture is well-stirred, so the concentration of salt remains uniform throughout the tank. Since the mixture is leaving the tank at the same rate of 2 L/min, the amount of salt leaving the tank per minute is also 2y g/min . Therefore, the rate of change of the amount of salt in the tank can be expressed as: dQ/dt = {rate in} - {rate out} = 2γ - 2γ = 0 This implies that the amount of salt in the tank remains constant over time, as there is no net change in the salt content. Hence, the expression for the amount of salt in the tank at any time t is: Q(t) = Q(0) = 0 The limiting amount of salt in the tank as t→ ∞ can be found by taking the limit of Q(t) as t approaches infinity: lim t→∞ Q(t) = lim t→∞ 0 = 0 Therefore, the limiting amount of salt in the tank as t→ ∞ is 0 g. Question 2 Let aₙ → 0, and use the Algebraic Limit Theorem to compute each of the following limits (assuming the fractions are always defined): lim n→∞ (1 + 2aₙ) / (1 + 3aₙ - 4(aₙ)²) Answer To compute, lim n→∞ (1 + 2aₙ) / (1 + 3aₙ - 4(aₙ)²) Given that, lim n→∞ aₙ = 0 Algebraic limit theorem: Let aₙ and bₙ be sequences of real numbers such that lim n→∞ aₙ = a and lim n→∞ bₙ = b, Then following statements hold. a) lim n→∞ c * aₙ = c * a b) lim n→∞ (aₙ + bₙ) = a + b c) lim n→∞ (aₙ * bₙ) = a * b d) If bₙ ≠ 0, for all n, then lim n→∞ aₙ/bₙ = a/b Using the algebraic limit theorem, lim n→∞ (1 + 2aₙ) / (1 + 3aₙ - 4(aₙ)²) = (lim n→∞ 1 + lim n→∞ 2aₙ) / (lim n→∞ 1 + lim n→∞ 3aₙ - lim n→∞ 4(aₙ)²) = (lim n→∞ 1 + 2 * lim n→∞ aₙ) / (lim n→∞ 1 + 3 * lim n→∞ aₙ - 4 * lim n→∞ (aₙ)²) Using given lim n→∞ aₙ = 0 lim n→∞ (1 + 2aₙ) / (1 + 3aₙ - 4(aₙ)²) = (1 + 2 * (0)) / (1 + 3 * (0) - 4 * (0)²) = 1 / 1 = 1 Therefore, lim n→∞ (1 + 2aₙ) / (1 + 3aₙ - 4(aₙ)²) = 1 Question 3 Use continuity to evaluate the limit. lim x→π 8sin(x + sin x) Don't really understand this. My trig func, knowledge is low but an explanation to look back to always helps me move forward. Answer 1 "Using continuity" means use the fact that if f is continuous, then f(a) = lim x→a f(x). In your case f(x) = 8sin(x + sin(x)) is continuous, so lim x→π f(x) = f(π) = 8sin(π + sin(π)) = 8sin(π) = 0 Answer 2 The phrase "f is continuous at a" really means lim x→a f(junk) = f(lim x→a junk) Well, note quite. This is what "f is continuous at whatever lim x→a junk is." So the start to your exercise is to write lim x→π 8sin(x + sin x) = 8sin(lim x→π (x + sin x)) Question 4 Evaluate the limit lim h→0 ((x - h)³ - x³ )/ h Answer 1 lim h→0 ((x - h)³ - x³ )/ h L = lim h→0 (x³ - 3x²h + 3xh² – h³ – x³)/ h = lim h→0 (-3x²h + 3xh² – h³)/ h = lim h→0 -3x² + 3xh – h² = lim h→0 -3x² + 0 + 0 = lim h→0 -3x² Answer 2 lim h→0 ((x - h)³ - x³ )/ h Simplify = lim h→0 (-3x²h + 3xh² - h²) / h = lim h→0 -3x² + 3x * 0 - 0² = lim h→0 -3x² Question 5 Check if the following sequences converge to the suggested limit using the notion of a sequence's convergence. a) lim n→∞ (2n + 1) / (5n + 4) = 2/5 b) lim n→∞ (2n²)/ (n³ + 3) = 0 c) lim n→∞ sin(n²)/n = 0 Answer A sequence (xₙ) converges to L if for every ε > 0, there exists N ∈ N such that |xₙ - L| < ε for all n > N. a) lim n→∞ (2n + 1) / (5n + 4) = 2/5 Solution: Let ε > 0, we have |(2n + 1) / (5n + 4) - 2/5| = |(5(2n+1) - 2(5n+4)) / 5(5n+4)| = |(10n+5 - 10n-8) / (25n+20)| = 3 / (25n+20) We wish to show that there exists N ∈ N such that for all n > N, |(2n + 1) / (5n + 4) - 2/5| < ε Choose N such that 3/(25n+20) < ε 3 / (25N+20) < ε N > (3/ε - 20) / 25 So for all n > N When we get |(2n+1)/(5n+4) - 2/5| < ε Which shows the required result. b) lim n→∞ (2n²)/ (n³ + 3) = 0 We wish to show that for all ε > 0, there exists N ∈ N such that |2n²/ (n³ + 3) - 0| < ε for all n > N We have |2n² / (n³ + 3)| = 2n² / (n³ + 3) Choose N such that 2n² / (n³ + 3) < ε for all n > N, then |2n² / (n³ + 3) - 0| < ε c) lim n→∞ sin(n²)/n = 0 We wish to show that for all ε > 0, there exists N ∈ N such that |sin(n²)/n - 0| < ε for all n > N. We have |sin(n²)/n| = |sin(n²)| / |n| Choose N > (1 / ε), then for all n > N, we have |sin(n²)/n - 0| < ε which is what we wished to show.