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Homework Help: Definite integration problem

  1. Nov 29, 2005 #1
    This is a question that is bugging me because trying to solve it the spirit of the topic I am on (backed up by one example under simple circumstances and a frustratingly complicated explaination) is proving really difficult. Trying to tackle it on my own terms yields me a wrong answer :frown:

    A curve is given parametrically: x= t^3, y= t^2
    a tangent to the curve is drawn at the point where t=2
    find a) the equation of the tangent
    b) the area bounded between the curve, tangent and y-axis

    (a) isn't a problem..the answer is 3y = x+4
    (b) is a problem though. My first inclination is to express y in terms of x, if x = t^3 then x^(2/3) = t^2 and so y=x^(2/3) or 3y =3(x^(2/3))
    now I have two equations with the same variables that I should be able to integrate without a problem subtracting 3(x^(2/3)) from x+4 I get:
    x+4 - 3(x^(2/3))...
    integrating this w.r.t.x I get [1/2(x^2) +4x - 9/5(x^(5/3))]
    as x = t^3, if t = 2 then x = 8 and so my limits should be 0 and 8
    plugging x=8 into the above however yields an answer of 6.4, the books answer is 2.13 (and by plotting the damned thing and measuring trapeziums I find that the book answer is correct)

    solving it in terms of t though is mega uncomfortable...the book example isn't very helpful and I just don't know what I'm doing.
    am I right that the tangent can be expressed as 3t^2=t^3 +4?
    Even if this is true how do I work with the curve?
     
  2. jcsd
  3. Nov 29, 2005 #2

    HallsofIvy

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    Science Advisor

    Sure. Since the tangent line is always above the curve, the area is
    [tex]\int_0^8 (y_1- y_2)dx[/tex]
    where y1 is the line 3y= x+4 and y2 is the parametric curve y= t2. Since x is measured by x= t3, 3y= x+ 4 becomes 3y= t2+ 4 and dx= 3t2dt. Of course, as x goes from 0 to 8, t goes from 0 to 2. The area is:
    [tex]\int_0^2(\frac{1}{3}t^3+ \frac{4}{3}- t^2)(3t^2)dt= \int_0^2(\t^5+ 4t^2- 3t^4)dt[/tex]
    That is
    [tex] \frac{1}{6}t^6+ \frac{4}{3}t^3- \frac{3}{5}t^5[/tex]
    evaluated between 0 and 2:
    [tex]\frac{64}{6}+ \frac{32}{3}- \frac{96}{5}[/tex]
    [tex]= \frac{32}{3}+ \frac{32}{3}- \frac{96}{5}[/tex]
    [tex]= \frac{64}{3}- \frac{96}{5}[/tex]
    [tex]= \frac{320- 288}{15}= \frac{32}{15}= 2 \frac{2}{15}[/tex]
     
  4. Nov 29, 2005 #3
    lol...just my luck!...Thanks for replying HallsofIvy! but this computer won't display the laTeX graphics...and I get booted off in 5 mins, gonna need to travel a bit before I can find another computer and read the full reply. :frown:
     
  5. Nov 29, 2005 #4
    HallsofIvy...Thankyou very much for working through it :smile: (though I had to hit the quote button so that I could read the TeX :biggrin: )
    I realise also, the reason why my first attempt was wrong...I should not have worked in terms of 3y instead should have worked in terms of just y...one of those things I just couldn't see at the time.
    Thanks again!
     
    Last edited: Nov 29, 2005
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