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Math Help - related rates question

  1. #1
    Junior Member
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    related rates question

    ok so i have this rate of change problem, driving me mad . its probably because of my algebra, but lets see
    it is the altitude of a triangle decreases at a rate of 2cm/s while the base increases at a rate of 4 cm/s. how fast is the area changing when the altitude is 25 cm and the base is 10 cm.
    what i have is dh/dt is 2cm/s and db/dt is 4
    da/dt | b=10, h=25

    a= 1/2 bh
    a=bh/2
    da/dt= ((db/dt*h + dh/dt*b)2 - (0)bh )/2
    after canceling the 2's and subing in i get da/dt=(4)25+2(10)
    da/dt = 120 cm^2/s
    the answer says its 40 cm ^2/s not sure what im doin wrong here
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  2. #2
    MHF Contributor
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    ive never seen a question like this before, so sorry if i am wrong.

     \frac{da}{dt} = \frac {d(0.5bh)}{dt} = 0.5 \frac{d(bh)}{dt}

     = 0.5 \left( h \frac{db}{dt} + b \frac{dh}{dt} \right) (product rule)

     = 0.5(25*4 + 10*-2) = 40
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