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

**Daredemo** · July 8th 2007, 02:05 PM

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:

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

I am asked to express the slope of L2 in terms of p.

**rualin** · July 8th 2007, 02:16 PM

Seems like the image should have been

**rualin** · July 8th 2007, 02:21 PM

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

**Jhevon** · July 8th 2007, 02:27 PM

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)

**red_dog** · July 8th 2007, 02:39 PM

Let $m$ be the slope of $L_2$ .
Then the equation of $L_2$ is: $y-2p^3=m(x-p)$ .
$L_2$ must intersect the graph in two distinct points. One point is $P$ and the other is the point where $L_2$ is tangent to the graph.
So, the system $\left\{\begin{array}{ll}y-2p^3=m(x-p)\\y=2x^3\end{array}\right.$ must have two solutions.
Replacing y from the second equation in the first equation we have
$2(x^3-p^3)=m(x-p)\Leftrightarrow (x-p)(2x^2+2px+2p^2-m)=0\Rightarrow x_1=p$ and $2x^2+2px+2p^2-m=0$ .
The last equation must have equal roots. Then
$4p^2-16p^2+8m=0\Rightarrow m=\frac{3p^2}{2}$

**rualin** · July 8th 2007, 03:10 PM

Interesting... I had never seen this done. Thanks, red_dog!

**Daredemo** · July 8th 2007, 08:43 PM

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.

**galactus** · July 9th 2007, 05:07 AM

Here's a calc way if you'd still like to see it.

We can find where L2 is tangent to C by:

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

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

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

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

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

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

Which agrees with red dogs clever method.

**rualin** · July 9th 2007, 05:52 AM

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:

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

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

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

Could we have substituted this $x$ into $m=6x^2$ to get $m=6\left(\frac{-p}{2}\right)^2=\frac{3}{2}\,p^2$ ?

**galactus** · July 9th 2007, 08:30 AM

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:

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

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