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Math Help - rates of change - voltage over time

  1. #1
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    rates of change - voltage over time

    this thread has been moved for the pre calculus forum. - sorry

    I am having trouble getting my head around this question. please tell me if i am on the right track, and no idea how to go about part b.

    the voltage across a resistor in a circuit is given by V=IR
    if R= 0.010t^2 (ohms) and I= 4.12+.020t (amps) t = time in seconds

    a) find the rate of change of the voltage with respect to time when t=2.5s

    v= (4.12+.020t)(0.010t^2)

    expanded v = 0.0412t^2 + 0.0002t^3


    dv/dt = 0.0824t + 0.0006t^2
    = 0.0824(2.5) + 0.0006(2.5)^2
    = 0.20975 volts/second (210mv/sec)

    b)at what time(s) will the voltage be a minimum?

    ????????????????????????? how do i turn this into an equation of a line? do i need to
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  2. #2
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    Quote Originally Posted by craigm View Post
    this thread has been moved for the pre calculus forum. - sorry

    I am having trouble getting my head around this question. please tell me if i am on the right track, and no idea how to go about part b.

    the voltage across a resistor in a circuit is given by V=IR
    if R= 0.010t^2 (ohms) and I= 4.12+.020t (amps) t = time in seconds

    a) find the rate of change of the voltage with respect to time when t=2.5s

    v= (4.12+.020t)(0.010t^2)

    expanded v = 0.0412t^2 + 0.0002t^3


    dv/dt = 0.0824t + 0.0006t^2
    = 0.0824(2.5) + 0.0006(2.5)^2
    = 0.20975 volts/second (210mv/sec)

    b)at what time(s) will the voltage be a minimum?

    ????????????????????????? how do i turn this into an equation of a line? do i need to
    note that dv/dt > 0 for all t > 0 ... voltage is increasing, therefore the minimum voltage will be at t = 0.
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