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5.9 Differentiation rules and related ratesIB Maths: Applications and Interpretation HL: Subtopic test

10 questions, 27 marks

IB Maths: Applications and Interpretation HL

5.9 Differentiation rules and related rates

Total 27 marks

Name

Class

Date

  1. 1
    The depth of water, hh metres, in a harbour tt hours after midnight is modelled by h(t)=6+2sin⁡(0.5t)h(t)=6+2\sin(0.5t) for 0≤t≤120\le t\le 12, where the angle is in radians.
    (a)
    Find dhdt\dfrac{dh}{dt}.
    [1 mark]
    • A2cos⁡(0.5t)2\cos(0.5t)
    • B−cos⁡(0.5t)-\cos(0.5t)
    • Ccos⁡(0.5t)\cos(0.5t)
    • D0.5cos⁡(0.5t)0.5\cos(0.5t)
    (b)
    Use your GDC, in radian mode, to find the rate of change of the depth at 04:00.
    [1 mark]
    • A−0.416-0.416 m h−1^{-1}
    • B0.4160.416 m h−1^{-1}
    • C0.9990.999 m h−1^{-1}
    • D−0.909-0.909 m h−1^{-1}
    (c)
    Find the first time after midnight at which the depth is greatest, and state the greatest depth.
    [2 marks]

    Total for question 1: 4 marks

  2. 2
    The concentration of a drug in a patient's blood, CC mg l−1^{-1}, is modelled by C(t)=20tt2+4C(t)=\dfrac{20t}{t^2+4} for t≥0t\ge0, where tt is the time in hours after the drug is given.
    (a)
    Find dCdt\dfrac{dC}{dt}.
    [1 mark]
    • A20(t2−4)(t2+4)2\dfrac{20(t^2-4)}{(t^2+4)^2}
    • B202t\dfrac{20}{2t}
    • C20(t2+4)+40t2(t2+4)2\dfrac{20(t^2+4)+40t^2}{(t^2+4)^2}
    • D20(4−t2)(t2+4)2\dfrac{20(4-t^2)}{(t^2+4)^2}
    (b)
    Find the time at which the concentration is greatest.
    [1 mark]
    • A44 hours
    • B22 hours
    • C55 hours
    • D00 hours
    (c)
    Find the rate of change of the concentration at t=1t=1 and state whether the concentration is increasing or decreasing.
    [2 marks]

    Total for question 2: 4 marks

  3. 3
    The concentration of a chemical in a reaction vessel, gg mol dm−3^{-3}, is modelled by g(t)=t e−0.5tg(t)=\sqrt{t}\,e^{-0.5t} for t>0t>0, where tt is the time in minutes after mixing.
    (a)
    Find an expression for dgdt\dfrac{dg}{dt}.
    [3 marks]
    (b)
    Find the greatest concentration, justifying that it is a maximum.
    [4 marks]

    Total for question 3: 7 marks

  4. 4
    A spherical balloon is inflated so that its volume, VV cm3^3, increases at a constant rate of 5050 cm3^3 s−1^{-1}. The radius of the balloon is rr cm. The volume of a sphere is V=43πr3V=\frac43\pi r^3 and its surface area is A=4πr2A=4\pi r^2.
    (a)
    (i) Find the rate of change of the radius when r=10r=10.
    (ii) Find the rate of change of the surface area when
    r=10r=10.
    [6 marks]
    (b)
    The balloon is empty when t=0t=0, where tt is in seconds.
    (i) Find the radius when
    t=20t=20.
    (ii) Show that
    r=(150t4π)1/3r=\left(\dfrac{150t}{4\pi}\right)^{1/3}, and hence find drdt\dfrac{dr}{dt} when t=20t=20.
    [6 marks]

    Total for question 4: 12 marks

End of questions

Written by the Exaim team, led by Shaun Daswani (Head of Upper Secondary, Improve ME Institute; MSc Financial Mathematics, Imperial College London; BSc, UCL) and Jason Daswani (operational lead, Improve ME Institute; LSE).