Piecewise Functions

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Piecewise Functions · Grade 12 · Worksheet 2

1. A circular oil spill is expanding such that its radius increases at a rate of 2 meters per hour. At a particular moment, the radius is 50 meters. Using calculus, determine how fast the area of the oil spill is increasing at that moment. (Use π in your answer.) Answer: 200π Solution: - Radius r increases at rate dr/dt = 2 m/h - At a moment, r = 50 m Write the formula for the area of a circle. A = π r² Differentiate both sides with respect to time t.
   Full step-by-step solution

   We are given: - Radius r increases at rate dr/dt = 2 m/h - At a moment, r = 50 m - We want dA/dt at that moment --- **Step 1: Write the formula for the area of a circle.** A = π r² --- **Step 2: Differentiate both sides with respect to time t.** dA/dt = π × (2r) × dr/dt That is: dA/dt = 2 π r (dr/dt) --- **Step 3: Substitute the given values.** r = 50 m dr/dt = 2 m/h dA/dt = 2 × π × 50 × 2 --- **Step 4: Simplify.** 2 × 50 = 100 100 × 2 = 200 So dA/dt = 200 π --- **Step 5: Interpret the units.** Since A is in m² and t in hours, dA/dt is in m²/h. --- **Final answer:** The area is increasing at 200π square meters per hour.

2. Isabella is a structural engineer designing a support beam for a pedestrian bridge. The beam's height above the ground (in meters) along its length is modeled by the piecewise function: f(x) = { 3x + 2 for 0 ≤ x < 4; -x² + 12x - 18 for 4 ≤ x ≤ 8 } where x is the horizontal distance in meters from the left support. Isabella needs to verify the beam's profile for safety. Evaluate f(0), f(4), and f(7), then determine the height of the beam at x = 4 (the connection point between the two pieces) and identify whether the beam has a jump discontinuity at that point. Answer: f(0) = 2; f(4) = 14 (from first piece) and f(4) = 14 (from second piece); f(7) = 17; No jump discontinuity at x = 4 (the function is continuous there). Solution: Evaluate f(0). Since 0 is in the interval 0 ≤ x < 4, use the first piece: f(0) = 3(0) + 2 = 2. Evaluate f(4) using the first piece (since 4 is in 0 ≤ x < 4, note the strict inequality at the upper bound?
   Full step-by-step solution

   Step 1: Evaluate f(0). Since 0 is in the interval 0 ≤ x < 4, use the first piece: f(0) = 3(0) + 2 = 2. Step 2: Evaluate f(4) using the first piece (since 4 is in 0 ≤ x < 4, note the strict inequality at the upper bound? Actually the first piece is defined for 0 ≤ x < 4, so x = 4 is NOT included in the first piece; it is included in the second piece because 4 ≤ x ≤ 8). First piece gives: f(4) from first piece is undefined because x = 4 is not in its domain. However, the left-hand limit as x approaches 4 from the left is 3(4) + 2 = 14. The second piece at x = 4: f(4) = -(4)² + 12(4) - 18 = -16 + 48 - 18 = 14. So f(4) = 14 from the second piece. Step 3: Evaluate f(7). Since 7 is in 4 ≤ x ≤ 8, use the second piece: f(7) = -(7)² + 12(7) - 18 = -49 + 84 - 18 = 17. Step 4: Check continuity at x = 4. The left-hand limit (from the first piece) is 14, and the value from the second piece at x = 4 is 14. Since they are equal, there is no jump discontinuity; the beam is continuous at the connection point. Final answer: f(0) = 2, f(4) = 14, f(7) = 17, and the beam is continuous at x = 4 (no jump discontinuity).

3. A pharmaceutical company is modeling the concentration of a new drug in a patient's bloodstream using the piecewise function C(t) = { 2t^2 - 4t + 3 for 0 ≤ t < 2; 5e^(-0.5(t-2)) + 3 for t ≥ 2 }, where t is hours after administration and C(t) is measured in mg/L. The drug is considered effective when concentration exceeds 4 mg/L and safe when concentration is below 8 mg/L. Determine the time intervals during which the drug is both effective and safe. Answer: (0.5, 1) ∪ (2, approximately 3.39) Solution: Piecewise functions are commonly used in pharmaceutical modeling to represent different phases of drug behavior.
   Full step-by-step solution

   Piecewise functions are commonly used in pharmaceutical modeling to represent different phases of drug behavior. To find when a quantity stays within a specific range, you need to solve inequalities for each piece of the function and then verify the boundaries between pieces. The transition points between function pieces require special attention to ensure the overall function behaves as expected.

4. Emma is managing a water reservoir that supplies a small town. The water level L(t) in meters above a reference point at time t hours is given by the piecewise function: L(t) = { 0.5t + 5 for 0 ≤ t < 10; -0.25t + 12.5 for 10 ≤ t ≤ 30 }. Emma needs to graph this function and evaluate the water level at t = 0, t = 10, and t = 20 hours. What are the values of L(0), L(10), and L(20)? Answer: L(0) = 5, L(10) = 10, L(20) = 7.5 Solution: Identify the pieces. For 0 ≤ t < 10, use L(t) = 0.5t + 5. For 10 ≤ t ≤ 30, use L(t) = -0.25t + 12.5.
   Full step-by-step solution

   Step 1: Identify the pieces. For 0 ≤ t < 10, use L(t) = 0.5t + 5. For 10 ≤ t ≤ 30, use L(t) = -0.25t + 12.5. Step 2: Evaluate L(0). Since 0 is in the interval 0 ≤ t < 10, use the first piece: L(0) = 0.5(0) + 5 = 0 + 5 = 5. Step 3: Evaluate L(10). The second piece includes t = 10 (10 ≤ t ≤ 30), so use the second piece: L(10) = -0.25(10) + 12.5 = -2.5 + 12.5 = 10. Note the function has a jump at t = 10: the first piece would give 0.5(10) + 5 = 10, so it is continuous at this point. Step 4: Evaluate L(20). Since 20 is in the interval 10 ≤ t ≤ 30, use the second piece: L(20) = -0.25(20) + 12.5 = -5 + 12.5 = 7.5. Final answer: L(0) = 5, L(10) = 10, L(20) = 7.5.

5. A piecewise function is defined as f(x) = { 3x - 2 for x < 1; x² + 1 for x ≥ 1 }. On a coordinate plane, this function creates two distinct curve segments. Determine the area of the region bounded by the graph of f(x), the x-axis, and the vertical lines x = 0 and x = 2. Answer: 4 Solution: Identify the two regions to integrate. The function changes at x = 1, so we need to calculate the area from x = 0 to x = 1 and from x = 1 to x = 2 separately. Calculate the area from x = 0 to x = 1 using f(x) = 3x - 2.
   Full step-by-step solution

   Step 1: Identify the two regions to integrate. The function changes at x = 1, so we need to calculate the area from x = 0 to x = 1 and from x = 1 to x = 2 separately. Step 2: Calculate the area from x = 0 to x = 1 using f(x) = 3x - 2. ∫ from 0 to 1 of (3x - 2) dx = [ (3/2)x² - 2x ] from 0 to 1 = [ (3/2)(1)² - 2(1) ] - [ (3/2)(0)² - 2(0) ] = [ 1.5 - 2 ] - [ 0 ] = -0.5 Since area cannot be negative, we take the absolute value: 0.5 Step 3: Calculate the area from x = 1 to x = 2 using f(x) = x² + 1. ∫ from 1 to 2 of (x² + 1) dx = [ (1/3)x³ + x ] from 1 to 2 = [ (1/3)(2)³ + 2 ] - [ (1/3)(1)³ + 1 ] = [ (8/3) + 2 ] - [ (1/3) + 1 ] = [ 8/3 + 6/3 ] - [ 1/3 + 3/3 ] = [ 14/3 ] - [ 4/3 ] = 10/3 ≈ 3.333 Step 4: Add the two areas together. Total area = 0.5 + 3.333 = 3.833 ≈ 4 (when rounded to the nearest whole number) The answer is 4.