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Derivatives: Find Q and its tangent

  1. Oct 18, 2015 #1
    1. The problem statement, all variables and given/known data
    Disclaimer: English is not my first language, so i apologize for any wrong math-terms.

    We look at the function f(x) = x^3. On the graph for f we have a point, P(a,a^3), where a =/= 0. The tangent to f through P cuts through f in another point, Q. Find Q and show, that the tangent to f through Q has a slope which is 4 times as steep as the tangent to f through P.

    2. Relevant equations


    3. The attempt at a solution

    Tangent through P cuts the function f in Q, which means in Q tangentP=x^3.

    3a^2(x-a)+a^3=x^3 <=>
    3a^2=(x^3-a^3)/(x-a) <=>
    3a^2=x^2-a^2 <=>
    4a^2=x^2
    2a=x

    Which means Q is located in (2a,(2a)^3)<=> Q(2a,8a^3)

    But this means that the tangent through Q has a slope which is twice as steep as it should be, since:

    tangentQ=8a^3+24a^2(x-a)

    I am not sure how i should solve this problem, and i hope you are able to understand my sloppy english/and math.
     
  2. jcsd
  3. Oct 18, 2015 #2

    vela

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    Looks fine up to here. However, you apparently think ##\frac{x^3-a^3}{x-a} = x^2-a^2##. Try it with x=2 and a=1, for instance. You get 7 = 3. That doesn't work, does it?

     
  4. Oct 18, 2015 #3

    BvU

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    Hello Zebo, welcome to PF :smile: !

    Your english is excellent, no problem reading and understanding at all. Now for the math :
    If I make a sketch of f(x) = x^3 and draw a tangent at x=1 (so your a = 1), then I see that that tangent intersects the curve of f(x) at a negative value of x, and not at x= 2. Conclusion: there must be something wrong with your solving ##3a^2 (x-a)+a^3 =x^3##. Do you agree ?

    [edit]vela was faster but (fortunately :rolleyes:)the replies match
     
  5. Oct 18, 2015 #4
    Thank you for the fast response. Oh yeah i see, im sorry about that i am snotty and heavy-headed.

    My biggest problem right now is i have no clue which direction i need to take to solve this. I want to find Q, without using any calculators, and the chapter which this problem is a part of does not explain anything resembling this. Am i on the right path or is there a much simpler way?

    Btw thanks again for the fast responses and thank you for welcoming me to the forum :)
     
  6. Oct 18, 2015 #5

    BvU

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    Well, that is nice of you. So I'll throw in a hint: there is some information in the problem statement that you haven't used yet and that will give you a clue what the solution of this cubic equation might be .. :wink:
     
  7. Oct 18, 2015 #6

    vela

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    You've taken a reasonable approach to the problem. You just need to get the algebra right.

    As far as calculators go, we're not saying you need to use a calculator to solve the problem, but when you make mistakes, using a calculator to check and test your work is a good way to figure out where you went wrong. Also, looking at a specific example like a=1, can furnish you with insights on how to solve the general case. Solving a simpler but related problem can help you eventually solve the more general problem.
     
  8. Oct 18, 2015 #7
    I've tried again:

    3a^2=(x^3-a^3)/(x-a) <=> 3a^2 = x^2+a^2+ax <=> x^2-ax-2a^2 = 0

    d = b^2 -4ac <=> d=9a^2

    x = (a+3a)/2a v (a-3a)/2a <=> x = 2 v x = -1

    So one of the x is where P is and the other is where Q is ? or have i gone down a wrong path?

    I apologize if i am fumbling about with this, it is pretty late where i live

    edit: If Q is to be found in (2,f(2)) the tangent would have a slope 4 times as steep as the tangent in P, and then P would be found in (-1,f(-1))?
     
    Last edited: Oct 18, 2015
  9. Oct 18, 2015 #8

    vela

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    You made a sign error somewhere. You should end up with ##x = \frac{-a \pm |3a|}{2}.## The ##a## that appears in the bottom of the quadratic formula is not the same as the ##a## you're using.

    Remember how you defined your variables. You have ##x## as the x-coordinate of Q while ##a## is the x-coordinate of P.
     
  10. Oct 19, 2015 #9
    Thanks!

    So as of now i have the following solution:

    f(x) = x^3, P(a,a^3), a=/=0, f'(x)=3x^2

    TangentP = a^3+3a^2(x-a)

    In the point Q, TangentP = f(x)

    a^3+3a^2(x-a) = x^3 <=> x^2+ax-2a^2 = 0

    x = a v x = -2a

    The tangent equals f(x) in P and Q, and P is located at (a,a^3). This means Q must be located at x = -2a

    Q is located at (-2a,(-2a)^3) = Q(-2a,-8a^3)

    The tangent in Q is the following:

    TangentQ = -8a^3 + 3(-2a)^2(x-a) = -8a^3 + 12a^2(x-a)

    The slope of tangentP was 3 and the slope of tangentQ is 12.

    I conclude that Q is located at (-2a,-8a^3). The tangent to the point Q on f(x) has a slope that is 12/3 = 4 times as steep as the slope of tangentP.
     
  11. Oct 19, 2015 #10

    BvU

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    Yes. The 4 is mentioned in the problem statement, so in fact you can get by just solving 3x^2 = 4 * 3 a^2 .

    Old dutch saying: laziness makes inventive....
     
  12. Oct 19, 2015 #11
    Im not sure my math professor would approve, i did something similar before and he did not like it.
     
  13. Oct 19, 2015 #12

    BvU

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    Sure, but once you know the answer is x = -2a you can check that it also satisfies the cubic equation :smile: (the other two solutions are x = a) and you don't have to tell explicitly how you found these factors...

    from the picture it's evident
     
  14. Oct 19, 2015 #13
    Thank you BvU and vela for your help :)
     
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