Lesson G6

G6 Constructing differential equations Quiz: AQA Maths, Unit 7

20 questions

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Lesson G6, Constructing differential equations: 20 multiple choice questions for the AQA Maths (7357), Unit 7: Differentiation, written with Revision Ninja.

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The 20 questions

  1. A population P grows at a rate proportional to its current size, with constant of proportionality k. Which differential equation models this?

    • dP/dt = k/P
    • dP/dt = kt
    • d^2P/dt^2 = -kP
    • dP/dt = kP
  2. Acceleration is the rate of change of velocity v with respect to time t. Which expression gives the acceleration a?

    • a = dv/dt
    • a = dt/dv
    • a = v/t
    • a = dv/ds
  3. Velocity is the rate of change of displacement s with respect to time t. Which expression gives v?

    • v = dt/ds
    • v = ds/dt
    • v = s/t
    • v = d^2s/dt^2
  4. A particle's velocity satisfies dv/dt = -kv for a constant k > 0. Which description fits this model?

    • Velocity increasing at a constant rate
    • Constant acceleration equal to k
    • Velocity directly proportional to time
    • Deceleration proportional to velocity, as with a resistance force proportional to speed
  5. Demand Q falls at a rate proportional to Q as the price p rises, with constant of proportionality 0.4. Which differential equation models this?

    • dp/dQ = 0.4
    • dQ/dp = 0.4Q
    • dQ/dp = -0.4p
    • dQ/dp = -0.4Q
  6. Which first-order differential equation is satisfied by y = Ax^2 for every constant A?

    • dy/dx = 2x
    • x dy/dx = y
    • x dy/dx = y^2
    • x dy/dx = 2y
  7. Eliminate the constant C from y = Cx^3 to form a differential equation.

    • x dy/dx = y
    • x dy/dx = 3y
    • dy/dx = 3x^2
    • dy/dx = 3y
  8. A population satisfies dP/dt = 0.02P with P = 500 when t = 0. Which expression gives P in terms of t?

    • P = 0.02e^(500t)
    • P = 500e^(0.02t)
    • P = 500 + 0.02t
    • P = 500t^0.02
  9. Which of these is a solution of dQ/dp = -2Q/p?

    • Q = Ap^2
    • Q = Ap^(-1/2)
    • Q = Ap^(-2)
    • Q = Ae^(-2p)
  10. A particle moves in a straight line with displacement s = 4t - t^2 metres. What is its constant acceleration?

    • 2 m/s^2
    • -1 m/s^2
    • 4 m/s^2
    • -2 m/s^2
  11. For a falling object, dv/dt = 9.8 - 0.2v in SI units. What is the terminal velocity?

    • 0.2 m/s
    • 9.8 m/s
    • 1.96 m/s
    • 49 m/s
  12. Which second-order differential equation is satisfied by y = Ae^(2x) + Be^(-x) for constants A and B?

    • d^2y/dx^2 + dy/dx - 2y = 0
    • d^2y/dx^2 - dy/dx - 2y = 0
    • d^2y/dx^2 + dy/dx + 2y = 0
    • d^2y/dx^2 - dy/dx + 2y = 0
  13. A curve has gradient at each point (x, y) equal to 2y/x. Which differential equation describes the curve?

    • dy/dx = 2x/y
    • dy/dx = 2y/x
    • dy/dx = y/(2x)
    • dy/dx = 2y + x
  14. Eliminate C from y = Cx^2 + 1 to form a differential equation.

    • x dy/dx = 2y
    • x dy/dx = y - 1
    • x dy/dx = 2(y - 1)
    • dy/dx = 2(y + 1)/x
  15. By Newton's second law, a particle of mass m under resultant force F satisfies which differential equation for its velocity?

    • m dv/dt = F/m
    • m v = F t
    • m dv/dt = F
    • dv/dt = m/F
  16. A population grows according to dP/dt = kP and doubles every 10 years. What is k to 4 significant figures?

    • 2/10, about 0.2 per year
    • ln 2 / 10, about 0.06931 per year
    • 10 ln 2, about 6.931 per year
    • ln 10 / 2, about 1.151 per year
  17. A particle has velocity v = 3t^2 + 2 m/s. What is its acceleration at t = 2 s?

    • 16 m/s^2
    • 6 m/s^2
    • 14 m/s^2
    • 12 m/s^2
  18. A population satisfies dP/dt = 0.05P with P = 1000 at t = 0. Estimate P after 10 years to the nearest whole number.

    • 1500
    • 1649
    • 1051
    • 2718
  19. Velocity satisfies dv/dt = 9.8 - 0.2v with v = 0 when t = 0. Which expression gives v in terms of t?

    • v = 49e^(-0.2t)
    • v = 9.8t - 0.1t^2
    • v = 49(1 + e^(-0.2t))
    • v = 49(1 - e^(-0.2t))
  20. A curve passes through (1, 2) and satisfies dy/dx = 2xy. Which equation describes the curve?

    • y = 2e^(2x - 1)
    • y = 2e^(x^2 - 1)
    • y = 2e^(x^2)
    • y = x^2 + 1

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