# Slope of a tangent line on a curve. Too difficult for me.

• Jul 8th 2007, 02:05 PM
Daredemo
Slope of a tangent line on a curve. Too difficult for me.
So, I have this curve C: y=2x^3 and a point P(p, 2p^3), p being a positive number. With the following graphic:

http://img404.imageshack.us/my.php?i...gagoil1.th.png

L1 and L2 are tangent lines of C passing through P.

I am asked to express the slope of L2 in terms of p.
• Jul 8th 2007, 02:16 PM
rualin
Seems like the image should have been

http://img404.imageshack.us/img404/7...gagoil1.th.png
• Jul 8th 2007, 02:21 PM
rualin
hmm... I don't really understand the question since at P the tangent line is L1
• Jul 8th 2007, 02:27 PM
Jhevon
Quote:

Originally Posted by rualin
hmm... I don't really understand the question since at P the tangent line is L1

L2 is tangent to the part of the curve on the left of the y-axis. it intersects with the tangent line L1 at P, where L1 is tangent to the curve at P (it seems, that diagram is really small)
• Jul 8th 2007, 02:39 PM
red_dog
Let $\displaystyle m$ be the slope of $\displaystyle L_2$.
Then the equation of $\displaystyle L_2$ is: $\displaystyle y-2p^3=m(x-p)$.
$\displaystyle L_2$ must intersect the graph in two distinct points. One point is $\displaystyle P$ and the other is the point where $\displaystyle L_2$ is tangent to the graph.
So, the system $\displaystyle \left\{\begin{array}{ll}y-2p^3=m(x-p)\\y=2x^3\end{array}\right.$ must have two solutions.
Replacing y from the second equation in the first equation we have
$\displaystyle 2(x^3-p^3)=m(x-p)\Leftrightarrow (x-p)(2x^2+2px+2p^2-m)=0\Rightarrow x_1=p$ and $\displaystyle 2x^2+2px+2p^2-m=0$.
The last equation must have equal roots. Then
$\displaystyle 4p^2-16p^2+8m=0\Rightarrow m=\frac{3p^2}{2}$
• Jul 8th 2007, 03:10 PM
rualin
Interesting... I had never seen this done. Thanks, red_dog!
• Jul 8th 2007, 08:43 PM
Daredemo
And here was silly me thinking of derivatives and more complex stuff.

Seriously, I would have never thought of that.

Thank you very much, Red Dog.
• Jul 9th 2007, 05:07 AM
galactus
Here's a calc way if you'd still like to see it.

We can find where L2 is tangent to C by:

$\displaystyle y'=6x^{2}=m$

$\displaystyle 2x^{2}-2p^{3}=6x^{2}(x-p)$

Solve this for x and we get $\displaystyle x=\frac{-p}{2}$

Sub this back into $\displaystyle y=2x^{3}$ and we see $\displaystyle y=\frac{-p^{3}}{4}$

Now, we have two end points from which to find the slope of L2. Use the slope of a line formula.

$\displaystyle m=\frac{2p^{3}+\frac{p^{3}}{4}}{p+\frac{p}{2}}=\fr ac{3p^{2}}{2}$

Which agrees with red dogs clever method.
• Jul 9th 2007, 05:52 AM
rualin
Quote:

Originally Posted by galactus
Here's a calc way if you'd still like to see it.

We can find where L2 is tangent to C by:

$\displaystyle y'=6x^{2}=m$

$\displaystyle 2x^{2}-2p^{3}=6x^{2}(x-p)$

Solve this for x and we get $\displaystyle x=\frac{-p}{2}$

Could we have substituted this $\displaystyle x$ into $\displaystyle m=6x^2$ to get $\displaystyle m=6\left(\frac{-p}{2}\right)^2=\frac{3}{2}\,p^2$?
• Jul 9th 2007, 08:30 AM
galactus
Why yes, even shorter. I wonder why they have theta in there when it's not really needed to find the slope of L2. L1 is easy. Now that we have L2, theta can be found by:

$\displaystyle tan{\theta}=\frac{m_{2}-m_{1}}{1+m_{1}m_{2}}$, if need be.