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.