Why Is Calculating the Area in a Parametric Integral Problem Challenging?

[GregA]

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 explanation) is proving really difficult. Trying to tackle it on my own terms yields me a wrong answer

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?

 

[HallsofIvy]

Sure. Since the tangent line is always above the curve, the area is
\int_0^8 (y_1- y_2)dx
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:
\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
That is
\frac{1}{6}t^6+ \frac{4}{3}t^3- \frac{3}{5}t^5
evaluated between 0 and 2:
\frac{64}{6}+ \frac{32}{3}- \frac{96}{5}
= \frac{32}{3}+ \frac{32}{3}- \frac{96}{5}
= \frac{64}{3}- \frac{96}{5}
= \frac{320- 288}{15}= \frac{32}{15}= 2 \frac{2}{15}

 

[GregA]

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, going to need to travel a bit before I can find another computer and read the full reply.

 

[GregA]

HallsofIvy...Thankyou very much for working through it (though I had to hit the quote button so that I could read the TeX )
I realize 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!