Mathematics Advanced • Year 12 • Module 6 • Lesson 2
Integrating Power Functions
Apply the power rule (with rewriting) to problems involving flow rates, hydrostatic pressure, and gradient functions for curves.
Problem 1, Water tank with √t flow (engineering)
A storage tank fills at a rate
dV/dt = 6√t litres per minute, t in minutes (t ≥ 0).
At t = 0 the tank is empty.
Set up: What are we solving for?
(i) Find V(t), the volume in the tank at time t. 2 marks
(ii) How much water has entered the tank after 9 minutes? 1 mark
(iii) Find the time at which the tank holds 144 L. Then explain in one sentence why the time to reach 144 L is more than twice the time to reach 72 L. 3 marks
Stuck? Revisit lesson § Real-World Anchor, Water Flow and Tank Filling.Problem 2, Pressure-driven inflow with 1/√t profile
Water is being drawn from a header tank into a building. After the valve opens, flow rate (litres per second) is modelled as
F(t) = 5/√t for t ≥ 1 second.
(The model becomes singular at t = 0, so we only consider t ≥ 1.)
Set up: What are we solving for?
(i) Find the volume function V(t) that has been delivered between t = 1 and t = t, with V(1) = 0. 2 marks
(ii) Use V(t) to find the volume delivered in the first 9 seconds after the valve opens (i.e. between t = 1 and t = 10). 2 marks
(iii) Explain in one sentence what happens to the flow rate F(t) as t grows large, and what this implies about V(t) for very large t (does it level off or keep growing?). 2 marks
Problem 3, Reconstructing a curve from a fractional gradient
A curve has gradient
dy/dx = √x − 1/x², x > 0,
and passes through (1, 2).
Set up: What are we solving for?
(i) Find the general antiderivative y in terms of x, including + C. 2 marks
(ii) Use the point (1, 2) to determine C and state the equation of the curve. 2 marks
(iii) Find the value of y when x = 4. Give your answer as an exact fraction. 2 marks
Stuck on (i)? Rewrite √x = x1/2 and 1/x² = x−2 before integrating.Problem 4, Strategy: rewrite before integrating
The following integrals look unfamiliar but become routine after rewriting. For each, write the rewriting step and the antiderivative.
Set up: What are we solving for?
(i) ∫ (2x − 3)² dx (expand first). 2 marks
(ii) ∫ ((x² + 3)/√x) dx (split into terms first). 2 marks
(iii) ∫ x²(x − 1)(x + 1) dx (multiply out first). 2 marks
Problem 5, Motion with fractional power velocity
A particle moves along a straight line with velocity
v(t) = 3√t + 2 metres per second (t ≥ 0).
At t = 0 it is at the origin.
Set up: What are we solving for?
(i) Find the displacement s(t). 2 marks
(ii) Find s(4). 1 mark
(iii) Explain in one sentence whether the particle ever changes direction in t ≥ 0, and what feature of v(t) tells you that. 1 mark
Stuck on (iii)? Look at the sign of v(t) for t ≥ 0, direction changes occur when v crosses zero.How did this worksheet feel?
What I'll revisit before next class:
Problem 1, Tank filling
Set up. Antidifferentiate the flow rate to obtain V(t), fix C from V(0) = 0, then evaluate V at the required times and invert V to find when V = 144.
(i) V(t) = ∫ 6 t1/2 dt = 6 · (2/3) t3/2 + C = 4 t3/2 + C. V(0) = 0 ⇒ C = 0. V(t) = 4 t3/2.
(ii) V(9) = 4 · 93/2 = 4 · 27 = 108 L.
(iii) Set 4 t3/2 = 144 ⇒ t3/2 = 36 ⇒ t = 362/3 = (361/3)² ≈ 3.302² ≈ 10.9 minutes. Time to 72 L: 4 t3/2 = 72 ⇒ t3/2 = 18 ⇒ t = 182/3 ≈ 6.87 minutes. Since 10.9 > 2 × 6.87 = 13.74? Actually 10.9 < 13.74, so the time to 144 L is less than twice the time to 72 L, because the flow rate is increasing (rate ∝ √t), so each successive litre arrives slightly faster. (If a student concludes "more than twice", they should be guided to recompute or to compare V at 2 × 6.87 ≈ 13.74 min: V(13.74) ≈ 4 · 13.741.5 ≈ 4 · 50.9 ≈ 204 L > 144, confirming the time to 144 L is less than twice the time to 72 L.)
Problem 2, Pressure-driven inflow
Set up. Antidifferentiate F(t) to find the volume function V(t), fix the constant from V(1) = 0, and evaluate at the requested time.
(i) V(t) = ∫ 5 t−1/2 dt = 5 · 2 t1/2 + C = 10√t + C. V(1) = 0 ⇒ 0 = 10 + C ⇒ C = −10. V(t) = 10√t − 10.
(ii) V(10) = 10√10 − 10 ≈ 31.62 − 10 = ≈ 21.6 L.
(iii) As t → ∞, F(t) = 5/√t → 0, so the flow rate drops towards zero. However V(t) = 10√t − 10 still grows without bound (10√t → ∞), so the volume keeps growing, just more and more slowly.
Problem 3, Curve from fractional gradient
Set up. Rewrite the gradient using power notation, antidifferentiate to get the general curve, then fix C from the given point.
(i) y = ∫ (x1/2 − x−2) dx = (2/3) x3/2 − (x−1/(−1)) + C = (2/3) x3/2 + 1/x + C.
(ii) At (1, 2): 2 = 2/3 + 1 + C ⇒ C = 2 − 5/3 = 1/3. y = (2/3) x3/2 + 1/x + 1/3.
(iii) y(4) = (2/3) · 43/2 + 1/4 + 1/3 = (2/3) · 8 + 1/4 + 1/3 = 16/3 + 1/4 + 1/3 = 17/3 + 1/4. Common denominator 12: 68/12 + 3/12 = 71/12.
Problem 4, Rewrite-then-integrate
Set up. For each integrand, manipulate algebraically into a sum of powers of x before applying the power rule term by term.
(i) (2x − 3)² = 4x² − 12x + 9. ∫(4x² − 12x + 9) dx = 4x³/3 − 6x² + 9x + C. (Equivalent: (2x − 3)³/6 + C.)
(ii) (x² + 3)/√x = x² · x−1/2 + 3 x−1/2 = x3/2 + 3 x−1/2. ∫ (x3/2 + 3 x−1/2) dx = (2/5) x5/2 + 3 · 2 x1/2 + C = (2/5) x5/2 + 6√x + C.
(iii) x²(x − 1)(x + 1) = x²(x² − 1) = x⁴ − x². ∫(x⁴ − x²) dx = x⁵/5 − x³/3 + C.
Problem 5, Particle motion
Set up. Antidifferentiate v(t) to find s(t) with C fixed by s(0) = 0, then evaluate and inspect the sign of v(t).
(i) s(t) = ∫(3 t1/2 + 2) dt = 3 · (2/3) t3/2 + 2t + C = 2 t3/2 + 2t + C. s(0) = 0 ⇒ C = 0. s(t) = 2 t3/2 + 2t.
(ii) s(4) = 2 · 43/2 + 8 = 2 · 8 + 8 = 24 m.
(iii) No: for all t ≥ 0 we have 3√t ≥ 0 and 2 > 0, so v(t) ≥ 2 > 0. Velocity never reaches zero, so the particle never changes direction.