These questions mix differentiation and integration, and none tells you which to use. All numbers are original, and each answer gives the working and a check.
Work on paper, and record any wrong method choice in the mistake log and paper-error review.
Questions and answers
Question 1. For each statement, say whether you would differentiate or integrate, without solving.
- (a) Given the volume, find how fast it is changing.
- (b) Given a flow rate, find the total volume.
- (c) Find the gradient of the curve at a point.
- (d) Given dy/dx, find the equation of the curve.
Answer
(a) Differentiate: a rate from a total. (b) Integrate: a total from a rate. (c) Differentiate: a gradient. (d) Integrate: the original function from its gradient.
Question 2. The curve y = x² + px + 3 has gradient 5 at x = 2. Find p.
Answer
dy/dx = 2x + p. At x = 2: 4 + p = 5, so p = 1. Check: dy/dx = 2x + 1 = 5 at x = 2.
Question 3. The curve y = ax³ + bx has a stationary point at (1, −2). Find a and b.
Answer
Point: y = a + b = −2. Stationary: dy/dx = 3ax² + b = 3a + b = 0 at x = 1.
Subtract the first from the second: 2a = 2, so a = 1 and b = −3.
Check: y = x³ − 3x gives y(1) = −2 and dy/dx = 3 − 3 = 0.
Question 4. A curve has dy/dx = 6x − 4 and passes through (1, 5). Find its equation.
Answer
Integrate: y = 3x² − 4x + c. Substitute (1, 5): 5 = 3 − 4 + c, so c = 6.
The curve is y = 3x² − 4x + 6. Check: at x = 1, y = 3 − 4 + 6 = 5.
Question 5. The profit is P = 8x − x² for 0 ≤ x ≤ 3. Find the greatest profit.
Answer
P′ = 8 − 2x = 0 gives x = 4, which is outside the interval.
Compare the endpoints: P(0) = 0 and P(3) = 24 − 9 = 15. The greatest profit is 15 at x = 3.
Question 6. The amount of liquid is V = 80 − 5t + 0.05t² litres after t minutes, for 0 ≤ t ≤ 20. Describe how V is changing at t = 10.
Answer
dV/dt = −5 + 0.1t. At t = 10: −5 + 1 = −4.
The volume is decreasing at 4 litres per minute at t = 10.
Question 7. Water flows into a tank at R = 12 − t litres per minute for 0 ≤ t ≤ 6. Find the total volume that flows in.
Answer
Integrate: the integral of (12 − t) from 0 to 6 is [12t − t²/2] = 72 − 18 = 54 litres.
Question 8. A particle has velocity v = t² − 4t + 3 m/s. Find (a) the times when it is at rest, (b) its acceleration at t = 0, (c) its displacement from t = 0 to t = 1.
Answer
(a) v = (t − 1)(t − 3) = 0, so t = 1 and t = 3.
(b) a = dv/dt = 2t − 4, so at t = 0, a = −4 m/s².
(c) Integrate: [t³/3 − 2t² + 3t] from 0 to 1 = 1/3 − 2 + 3 = 4/3 m.
If you got these wrong
- Question 1 needs choosing between a derivative and an integral.
- Questions 2 and 3 need using a tangent condition.
- Question 5 needs comparing endpoints and stationary points.
- Question 6 needs what a negative derivative means.
Return to the mixed calculus hub for the study order. To have a teacher watch how you choose a method, see online one-to-one Additional Mathematics tuition.