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Word problem with exponential and quadratic models

  1. Jul 14, 2014 #1
    1. The problem statement, all variables and given/known data

    a6nQNvO.png

    2. Relevant equations



    3. The attempt at a solution

    Confused with (d) a little.

    Rocket explodes at ##h=3.85262 ## miles

    ## -16t^2+1400\sin(\alpha)t=3.852624*5280##

    ## \alpha=\arcsin\left(\dfrac{3.852624*5280+16t^2}{1400t}\right) ##


    But what do I do from here? Looks a bit complicated.
     
  2. jcsd
  3. Jul 15, 2014 #2
    What formulas do you know for working with parabolas?

    You'll need one of these formulas to help you. The vertex formula, possibly? The quadratic formula is another possibility. The derivatives, but you probably aren't doing that yet.
     
  4. Jul 15, 2014 #3
    Would you be more specific? I don't exactly know how to implement vertex formula here. The graph of this rational function is a kind of hyperbola but with a vertex, kind of like a mix of a hyperbola and parabola. But I don't have tools to finding such a vertex.
     
  5. Jul 15, 2014 #4
    At what altitude h is the pressure p = 10 psi? This is the altitude at which the rocket will explode.

    Chet
     
  6. Jul 15, 2014 #5
    bump
     
  7. Jul 15, 2014 #6

    Char. Limit

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    Gold Member

    You already have the altitude that the rocket will explode at, that's good. Though you will obviously want to get that in feet, given that the rest of the problem works in feet. You also have a formula for the altitude the rocket will take after t seconds as a function of t and α, so also good. What you basically need to do is find what the maximum of this function can be. The maximum will be a function of sin(α), which you can then find the angle from.

    Before I continue, though, do you know any particular methods to finding the maximum* of a quadratic function?

    *: Or minimum, if the function has a positive t^2 coefficient. But this one's negative, so it's a maximum.
     
  8. Jul 16, 2014 #7
    Well, the function ##y(t)## is a parabola. I think it helps to draw it and figure out what you need. And ways of getting that information.

    I'm not certain what you are graphing to get the hyperbola. It sounds like you already have all the information you need already. You just need to put it together to describe the parabola -- and then you will have alpha.
     
  9. Jul 16, 2014 #8
    From what you've learned in math class, were you aware that, if y = ax2+bx+c represents the equation for a parabola,
    1. The roots of the equation ax2+bx+c =0 is where the parabola crosses the x axis (in two places)
    2. The maximum point of the parabola is at the average of the two roots.
    3. The sum of the two roots is -b/a

    Chet
     
    Last edited: Jul 16, 2014
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