Results 1 to 2 of 2

Math Help - Isolate "t" in a cubic function (and sketching equation)

  1. #1
    Junior Member
    Joined
    Nov 2010
    Posts
    58

    Isolate "t" in a cubic function (and sketching equation)

    The current through a resistor, in amperes, is I = 4.85 - 0.001t^2
    The resistance, in ohms, is R = 15 +0.11t ( t is in seconds).
    The voltage, in volts, is the product of the current and the resistance.

    How long will it take for voltage to reach 77V, to the nearest second?

    I found V(t) to be V(t) = IR
    Therefore, I just multiplied and brought everything out of the bracket to get:
    v(t) = -0.00011t^3 - 0.015t^2 + 0.5335t + 72.75

    which is what is says in the Answers.

    I subbed V(t) = 77, but now I don't know how to isolate for t. We aren't too familiar with factoring cubic equations, so I'd like some help.
    Thanks. : )

    Also, how would you sketch this function?
    For example, if this was a quadratic equation, I could complete the square to find where the vertex and zeroes are and etc. to sketch it.

    Is there a method similar to completing the square for cubic equations, though?
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Master Of Puppets
    pickslides's Avatar
    Joined
    Sep 2008
    From
    Melbourne
    Posts
    5,236
    Thanks
    28
    Quote Originally Posted by TN17 View Post
    v(t) = -0.00011t^3 - 0.015t^2 + 0.5335t + 72.75
    You can google 'the cubic equation' this will isolate t but I don't recommended it.

    You can graph this by hand using a table of values and the general shape of a cubic.

    If you need to find the zeros of this equation I suggest using technology as it is quite nasty. There is no real equivavlent to completeing the square here that will be nice and clean.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Replies: 1
    Last Post: September 16th 2011, 01:08 AM
  2. Replies: 2
    Last Post: April 24th 2011, 07:01 AM
  3. How to isolate "R" (from a log equation)
    Posted in the Pre-Calculus Forum
    Replies: 4
    Last Post: December 8th 2010, 06:34 PM
  4. Replies: 1
    Last Post: October 25th 2010, 04:45 AM
  5. "Primary", "Secondary" equation
    Posted in the Calculus Forum
    Replies: 1
    Last Post: November 21st 2009, 10:16 PM

Search Tags


/mathhelpforum @mathhelpforum