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Homework Help: 1 integral (easy) + a few precalculus problems.

  1. Dec 17, 2013 #1
    1. The problem statement, all variables and given/known data

    Problem 1.

    ∫ 6x^2dx
    Problem 2.

    Solve as far as you can(simplify the problem):

    Problem 3.

    The Canadian goose came to sweden in the 1930's. Afterwhich the population of the bird increased. At the same time every year they count the ammount of goose and how much it has increased. The populations growth can be described with an exponential model.

    The chart below shows the ammount of goose "K" as a function of time in years "t", where t=0 is the year 1977.

    http://imageshack.us/a/img27/9809/6dnz.jpg [Broken]

    a) Determine an approximation to K'(30) with the help of this picture.

    b) Explain what K'(20)=800 describes.

    3. The attempt at a solution

    2 2
    ∫ 6x^2dx [2x^3] = 2*2^3 - (2*1^3) = 14 ae (area units). I realise now as I'm writing this that
    1 1

    it's solved. I was stumbled and thought I got it wrong, but I'm just tired of studying all day I guess.

    2. (x-3)(x+2)/2x-6 = x^2 -x -6/2x-6 = x^2 - x / 2x

    After these steps I'm thinking about breaking out the x, like x(x-1) / x(2) and let the x's take each other out but that would leave the answer to be x-1/2 and the answer is supposed to be x+2/2.

    3. a)
    b) I think K'(20)=800 is how much growth the gooses do per year at the year 20.

    About "a" though. I don't know how to figure this out, it's supposedly very easy but I don't know.
    1. The problem statement, all variables and given/known data

    2. Relevant equations

    3. The attempt at a solution
    Last edited by a moderator: May 6, 2017
  2. jcsd
  3. Dec 17, 2013 #2


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    Hi Fishingaxe! :smile:
    why are you making it more complicated?? :redface:

    (you're supposed to be simplifying it :rolleyes:)

    try simplifying the bottom :wink:

    (and then get some sleep! :zzz:)
    yes, except of course that's an average, it's not the amount in any particular year …

    it's approximately 16 geese a week :wink:
    the derivative is the slope of the tangent to the curve

    sooo … draw the tangent, then measure the slope by counting those little squares! :smile:
  4. Dec 17, 2013 #3
    Omg, of course, how can I miss that. That is just redicilous! Thanks man.

    About 3. a) I don't know. Am I supposed to draw the derivitive functions line? and if so, how do I get the function? I am still lost in this.
  5. Dec 17, 2013 #4


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    Hi Fishingaxe! :smile:
    No, you don't have to draw the derivative function.

    You only need to draw the tangent of the original function (at the particular pont you're interested in).

    The tan of the angle of that tangent is the derivative at that point. :wink:

    (try it eg with y = x2 on graph-paper)
  6. Dec 17, 2013 #5
    Okay, I get it, or at least some of it. The derivitive is of course the angle of that tangent. If I draw a tangent there, I still fail to see how I can solve K'(30). You said count the squares, how exactly does that work? I draw the tangent until it reaches y=0? I really need step for step here.
  7. Dec 17, 2013 #6


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    Yes, in this case you may as well extend it to y = 0, to get the maximum accuracy.

    Then count how many squares along it goes, and how many up (to the nearest one-tenth), and divide! :smile:
  8. Dec 17, 2013 #7
    The answer is supposed to be 1700. How are u supposed to get to know the number of squares going up and sideways(along)? Since depending on how you interperpt them they can go either up and sideways? I'm sorry for being slow.
  9. Dec 17, 2013 #8


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    Draw a triangle with the point in question at the top, the tangent at that point as the hypotenuse, and the vertical from that point and y = 0 as the two sides.

    Count the number of squares along the horizontal and vertical sides, and use the scale to decide what they mean. :smile:

    (and yes i get about 1700)
  10. Dec 17, 2013 #9
    I may need to take a serious nap! My problem is the squares. As I said previously, the number of squares that goes vertical and horizontal are the exact same ammount, right? I drew the triangle as you said.

    I am taking a nap now, I am too lost. Can't think straight.

    Thanks for the help tiny-tim. I really appreciate it. Hopefully I can get it when I wake up! (have a big test tomorrow)
  11. Oct 11, 2014 #10
    For your first problem, that is just a power rule.
  12. Oct 11, 2014 #11

    Ray Vickson

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    Before you take your test, make sure you force yourself to use parentheses, so you don't write things like
    [tex] \frac{(x-3)(x+2)}{2x}-6 \; \leftarrow \;\text{ yes, that IS what you wrote!}[/tex]
    instead of
    [tex] \frac{(x-3)(x+2)}{2x-6} [/tex]
    which, I guess, is what you intended. Just use parentheses, like this: (x-3)(x+2)/(2x-6). It might save you from losing marks needlessly.
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