# [SOLVED] Curve Sketching

• Sep 16th 2008, 12:56 AM
ose90
[SOLVED] Curve Sketching
a)Find the equation of the tangent at the point (t,$\displaystyle t^3$) on the curve $\displaystyle y=x^3$.Find also the coordinates of the point where this tangent meets the curve again.

b) For the curve, $\displaystyle y=(4/x^2)+x$ , show the following ,
i) the entire curve lies above the line y=x

Red colour fonts are parts which I need help, thanks in advance!
• Sep 16th 2008, 01:13 AM
Jameson
a) $\displaystyle y' = 3x^2$, so the slope at point x=t is $\displaystyle 3t^2$. Now we have a point and a slope, thus we're able to make a line.

$\displaystyle y-t^3=3t^2(x-t)$, plugging in the info into the point-slope equation for a line. Simplify that out and you'll get your tangent equation.

To see where these two lines meet again, set them equal to each other and solve.
• Sep 16th 2008, 01:21 AM
ose90
Quote:

Originally Posted by Jameson
a) $\displaystyle y' = 3x^2$, so the slope at point x=t is $\displaystyle 3t^2$. Now we have a point and a slope, thus we're able to make a line.

$\displaystyle y-t^3=3t^2(x-t)$, plugging in the info into the point-slope equation for a line. Simplify that out and you'll get your tangent equation.

To see where these two lines meet again, set them equal to each other and solve.

Set these two equations together?
$\displaystyle y=x^3$ and $\displaystyle y=3t^2(x-t)+t^3$
• Sep 16th 2008, 01:26 AM
Jameson
Yessir.
• Sep 16th 2008, 01:31 AM
Chop Suey
$\displaystyle y=(4/x^2)+x$

b) Assume that the curve $\displaystyle y=(4/x^2)+x$ and the line y=x intersects. Therefore:
$\displaystyle x = (4/x^2)+x$

Canceling x, you would get the ridiculous statement that:
$\displaystyle 0 = \frac{4}{x^2}$

$\displaystyle 0 = 4$

There is a contradiction, therefore our original assumption is not true.

EDIT: Continue with the next post below.
• Sep 16th 2008, 01:34 AM
ose90
Quote:

Originally Posted by Jameson
Yessir.

I'm sorry to trouble you again, but I cannot solve the equation for x. (Crying)
• Sep 16th 2008, 01:37 AM
ose90
Quote:

Originally Posted by Chop Suey
$\displaystyle y=(4/x^2)+x$

b) Assume that the curve $\displaystyle y=(4/x^2)+x$ and the line y=x intersects. Therefore:
$\displaystyle x = (4/x^2)+x$

Canceling x, you would get the ridiculous statement that:
$\displaystyle 0 = \frac{4}{x^2}$

$\displaystyle 0 = 4$

There is a contradiction, therefore our original assumption is not true.

Isn't that an inclined asymptote for the curve?
I can't explain why it is above the line y=x.
• Sep 16th 2008, 01:53 AM
Jameson
No problem :)

So $\displaystyle x^3=3t^2(x)-3t^3+t^3$

Simply $\displaystyle x^3=3t^2(x)-2t^3$

Group and factor $\displaystyle x(x^2-3t^2)=-2t^3$

Made a mistake... working on it
• Sep 16th 2008, 01:59 AM
Chop Suey
Quote:

Originally Posted by ose90
Isn't that an inclined asymptote for the curve?
I can't explain why it is above the line y=x.

$\displaystyle f(x) = \frac{4}{x^2} + x$

$\displaystyle g(x) = x$

Let's assume that f(x) < g(x). So that:
$\displaystyle \frac{4}{x^2} + x < x$

But if you cancel the x, you see that:
$\displaystyle \frac{4}{x^2}<0$

$\displaystyle 4<0$

$\displaystyle \therefore f(x) > g(x)$

The reason why I didn't put the equality sign is because I showed that they do not intersect.
• Sep 16th 2008, 02:08 AM
ose90
Quote:

Originally Posted by Chop Suey
$\displaystyle f(x) = \frac{4}{x^2} + x$

$\displaystyle g(x) = x$

Let's assume that f(x) < g(x). So that:
$\displaystyle \frac{4}{x^2} + x < x$

But if you cancel the x, you see that:
$\displaystyle \frac{4}{x^2}<0$

$\displaystyle 4<0$

$\displaystyle \therefore f(x) > g(x)$

The reason why I didn't put the equality sign is because I showed that they do not intersect.

Thanks a lot (Nod)
• Sep 16th 2008, 02:11 AM
ose90
Quote:

Originally Posted by Jameson
No problem :)

So $\displaystyle x^3=3t^2(x)-3t^3+t^3$

Simply $\displaystyle x^3=3t^2(x)-2t^3$

Group and factor $\displaystyle x(x^2-3t^2)=-2t^3$

Made a mistake... working on it

Thanks for helping me. I have tried to solve it for many times but still can't get the value of x. Maybe there is no solution.
• Sep 16th 2008, 02:19 AM
Jameson
You're welcome. I think there's a trick I'm not seeing. Cubics that have no x^2 term have ways of being solved. Either someone else will see it or I'll solve it soon.
• Sep 16th 2008, 02:27 AM
Chop Suey
Notice that x-t is a factor, and by polynomial division, you can reduce it to:
$\displaystyle (x-t)(x^2+tx-2t^2)=0$

And it can be factored even further to:
$\displaystyle (x-t)(x-t)(x+2t)=0$
• Sep 16th 2008, 03:43 AM
ose90