The Number e and Differentiating Exponentials — Solutions

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1. Estimate the limit numerically. Fluency

   1. (a) f(h) = (3^h − 1)/h values. f(0.1) = (3^0.1 − 1)/0.1 = (1.1161 − 1)/0.1 = 1.1610
      f(0.01) = (3^0.01 − 1)/0.01 = (1.01105 − 1)/0.01 = 1.1047
      f(0.001) = (3^0.001 − 1)/0.001 ≈ 1.0992
      f(0.0001) ≈ 1.0987
   2. (b) What value does the limit approach? The values are decreasing towards approximately 1.0986. This equals ln(3) ≈ 1.0986.
   3. (c) Why is this not 1? The limit lim_h→0(a^h−1)/h = ln(a). For base 3, ln(3) ≈ 1.099 ≠ 1.
      Only base e gives ln(e) = 1, which is what makes e special. With base 3, d/dx(3^x) = ln(3) × 3^x — there is an extra factor of ln(3), making it less “clean” than e^x.

2. Differentiate each function. Fluency

   1. (a) y = e^4x dy/dx = 4e^4x
   2. (b) y = e^−2x dy/dx = −2e^−2x
   3. (c) y = 5e^3x dy/dx = 5 × 3 × e^3x = 15e^3x
   4. (d) y = −2e^x dy/dx = −2e^x
   5. (e) y = e^x/2 k = 1/2, so dy/dx = (1/2)e^x/2
   6. (f) y = 4e^−x/3 k = −1/3, so dy/dx = 4 × (−1/3) × e^−x/3 = −(4/3)e^−x/3

3. Key features. Fluency

   1. (a) y = e^x + 2 y-intercept: x=0, y = e⁰ + 2 = 1 + 2 = 3, so (0, 3).
      Asymptote: y = 2 (shifted up 2 from y = 0).
      Range: y > 2.
   2. (b) y = e^−x: increasing or decreasing? dy/dx = −e^−x < 0 for all x, so the function is strictly decreasing.
      Asymptote: y = 0 as x → +∞ (now the right side approaches the asymptote).
   3. (c) y = 3e^x y-intercept: x=0, y = 3e⁰ = 3 × 1 = 3, so (0, 3).
      Range: y > 0 (vertical stretch by 3 does not affect the sign).

4. Differentiate using sum and constant-multiple rules. Fluency

   1. (a) y = 2e^x + 3x² dy/dx = 2e^x + 6x
   2. (b) y = e^2x − e^−2x dy/dx = 2e^2x − (−2)e^−2x = 2e^2x + 2e^−2x
   3. (c) y = (e^x + 1)/4 = (1/4)e^x + 1/4 dy/dx = (1/4)e^x + 0 = (1/4)e^x
   4. (d) y = 6 − e^5x dy/dx = 0 − 5e^5x = −5e^5x

5. Tangent line to an exponential curve. Understanding

   1. (a) Find dy/dx for y = 2e^3x. dy/dx = 2 × 3 × e^3x = 6e^3x
   2. (b) Gradient at x = 0. dy/dx|_x=0 = 6e⁰ = 6 × 1 = 6
   3. (c) Equation of tangent at (0, 2). m = 6, point (0, 2). Using y − y₁ = m(x − x₁):
      y − 2 = 6(x − 0)
      y = 6x + 2
   4. (d) When is dy/dx = 6e³? Set 6e^3x = 6e³
      e^3x = e³
      3x = 3, so x = 1

6. Population growth model. Understanding

   1. (a) Initial population. P(0) = 500e⁰ = 500 × 1 = 500 bacteria
   2. (b) Find P′(t) and interpret. P′(t) = 500 × 0.4 × e^0.4t = 200e^0.4t
      This gives the rate of increase of the bacterial population in bacteria per hour at time t.
   3. (c) Rate at t = 2 hours. P′(2) = 200e^0.8 = 200 × 2.2255 ≈ 445 bacteria per hour
   4. (d) Show P′(t) = 0.4P(t). P′(t) = 200e^0.4t
      0.4 × P(t) = 0.4 × 500e^0.4t = 200e^0.4t ✓
      Significance: the rate of growth is always 40% of the current population. This is the hallmark of exponential (proportional) growth — the more bacteria there are, the faster they reproduce.

7. Comparing exponential derivatives. Understanding

   1. (a) Derivatives of y₁ and y₂. dy₁/dx = 2e^2x
      dy₂/dx = e^x
   2. (b) Which is steeper at x = 0? At x = 0: dy₁/dx = 2e⁰ = 2; dy₂/dx = e⁰ = 1.
      y₁ = e^2x has the steeper gradient (gradient = 2 vs 1).
   3. (c) When is dy₁/dx = 10? 2e^2x = 10
      e^2x = 5
      2x = ln(5)
      x = ln(5)/2

8. Analysing a transformed exponential. Understanding

   1. (a) Find A using (0, 3). f(0) = Ae⁰ = A = 3, so A = 3.
   2. (b) Find k using (1, 6). f(1) = 3e^k = 6
      e^k = 2
      k = ln(2)
   3. (c) Write f(x). f(x) = 3e^ln(2)x = 3 × 2^x (since e^ln(2)x = (e^ln2)^x = 2^x)
   4. (d) Find f′(x) and f′(2). f(x) = 3e^ln(2)x, so f′(x) = 3ln(2)e^ln(2)x = 3ln(2) × 2^x
      f′(2) = 3ln(2) × 2² = 12ln(2) ≈ 12 × 0.6931 ≈ 8.32

9. Drug concentration model. Problem Solving

   1. (a) C(0) and its meaning. C(0) = 8e⁰ = 8 mg/L. This is the initial concentration of the drug in the bloodstream immediately after administration.
   2. (b) C′(t) and rates at t = 1 and t = 3. C′(t) = 8 × (−0.5) × e^−0.5t = −4e^−0.5t
      C′(1) = −4e^−0.5 ≈ −4 × 0.6065 ≈ −2.43 mg/L per hour
      C′(3) = −4e^−1.5 ≈ −4 × 0.2231 ≈ −0.89 mg/L per hour
   3. (c) Why C′(t) is always negative. C′(t) = −4e^−0.5t. Since e^−0.5t > 0 always, C′(t) = −4 × (positive) < 0 always.
      In context: the drug concentration is continuously decreasing over time, which is expected as the body metabolises and eliminates the drug.
   4. (d) When does C(t) first fall below 1 mg/L? Solve 8e^−0.5t = 1
      e^−0.5t = 1/8
      −0.5t = ln(1/8) = −ln(8)
      t = ln(8)/0.5 = 2ln(8) = 6ln(2) (exact)
      t = 2 × ln(8) ≈ 2 × 2.0794 ≈ 4.16 hours (2 d.p.)

10. Exploring the limit definition of e. Problem Solving

    1. (a) Calculate (1 + 1/n)ⁿ. n = 10: (1.1)¹⁰ ≈ 2.5937
       n = 100: (1.01)¹⁰⁰ ≈ 2.7048
       n = 1000: (1.001)¹⁰⁰⁰ ≈ 2.7169
       These approach e ≈ 2.71828 from below as n increases.
    2. (b) Limiting amount for continuously compounded interest. As n → ∞, (1 + 1/n)ⁿ → e ≈ $2.72.
       So $1 invested at 100% p.a. continuously compounded grows to $e in 1 year.
    3. (c) A = Pe^rt with P = 5000, r = 0.06, t = 10. A = 5000e^{0.06 × 10} = 5000e^0.6 ≈ 5000 × 1.8221 ≈ $9110
       dA/dt = Pe^rt × r = 5000 × 0.06 × e^0.6 ≈ 300 × 1.8221 ≈ $547 per year at t = 10.