orthogonal trajectories

**polymerase** · October 21st 2007, 06:36 AM

Could someone explain what this question really means and how to start solving it....

If m is a fixed constant such that the two families of curves $y=\frac{1}{x+c}$ and $y=m(x+k)^\frac{1}{3}$ are orthogonal trajectories of each others, what must be the value of m?

**TD!** · October 21st 2007, 06:45 AM

The functions y = g(x) and y = h(x) are orthogonal at x = a, if g'(a)h'(a) = -1.

**polymerase** · October 21st 2007, 07:01 AM

Originally Posted by TD!

The functions y = g(x) and y = h(x) are orthogonal at x = a, if g'(a)h'(a) = -1.

$y'=\frac{-1}{(x+c)^2}$ and $y'=\frac{m}{3(x+k)^\frac{2}{3}}$ therefore when i multiple and set them equal to -1 I get, $\frac{m}{3(x+c)^2(x+k)^\frac{2}{3}}=1$ but how do i solve for m?

**topsquark** · October 21st 2007, 09:14 AM

Originally Posted by polymerase

$y'=\frac{-1}{(x+c)^2}$ and $y'=\frac{m}{3(x+k)^\frac{2}{3}}$ therefore when i multiple and set them equal to -1 I get, $\frac{m}{3(x+c)^2(x+k)^\frac{2}{3}}=1$ but how do i solve for m?

Obviously for a given c and k the two functions will intersect at a set of points $\{ a_i \}$ , so we have
$\frac{m}{3(a_i + c)^2(a_i + k)^\frac{2}{3}}=1$
for each such $a_i$ .

So your solution for m depends on the c, k, and set $\{ a_i \}$ .

-Dan

**polymerase** · October 21st 2007, 10:10 AM

Originally Posted by topsquark

Obviously for a given c and k the two functions will intersect at a set of points $\{ a_i \}$ , so we have
$\frac{m}{3(a_i + c)^2(a_i + k)^\frac{2}{3}}=1$
for each such $a_i$ .

So your solution for m depends on the c, k, and set $\{ a_i \}$ .

-Dan

So are you saying there is no defined answer.....? Because it says the answer is $\sqrt[3]{3}$

**topsquark** · October 21st 2007, 10:12 AM

Originally Posted by polymerase

So are you saying there is no defined answer.....? Because it says the answer is $\sqrt[3]{3}$

I haven't actually solved it, so I couldn't tell you. But start with where the two curves will intersect and go from there.

-Dan

**polymerase** · October 21st 2007, 10:16 AM

Originally Posted by topsquark

I haven't actually solved it, so I couldn't tell you. But start with where the two curves will intersect and go from there.

-Dan

What you mean? start where the original curves will intersect?

**TD!** · October 21st 2007, 12:26 PM

Originally Posted by polymerase

$y'=\frac{-1}{(x+c)^2}$ and $y'=\frac{m}{3(x+k)^\frac{2}{3}}$ therefore when i multiple and set them equal to -1 I get, $\frac{m}{3(x+c)^2(x+k)^\frac{2}{3}}=1$ but how do i solve for m?

Ok, so far so good! You know have m/(...) = 1.
A fraction is 1 if numerator and denominator are equal.
So you can easily solve this for m. We have:

$ m = 3\left( {x + c} \right)^2 \left( {x + k} \right)^{\frac{2}{3}} $

Now, this m doesn't seem "constant", but dependent of x, c and k!
But, where do we have to look for ortogonality? At their intersection!
Let's try to find something useful, equalling both equations:

$ \frac{1}{{x + c}} = m\left( {x + k} \right)^{\frac{1}{3}} \Leftrightarrow \frac{1}{m} = \left( {x + c} \right)\left( {x + k} \right)^{\frac{1}{3}} \Rightarrow \frac{1}{{m^2 }} = \left( {x + c} \right)^2 \left( {x + k} \right)^{\frac{2}{3}} $

You recognise this? Filling in, in what we had for m earlier:

$ m = 3\left( {x + c} \right)^2 \left( {x + k} \right)^{\frac{2}{3}} \Leftrightarrow m = \frac{3}{{m^2 }} \Leftrightarrow m^3 = 3 \Leftrightarrow m = \sqrt[3]{3} $

**polymerase** · October 21st 2007, 01:42 PM

Originally Posted by TD!

Ok, so far so good! You know have m/(...) = 1.
A fraction is 1 if numerator and denominator are equal.
So you can easily solve this for m. We have:

$ m = 3\left( {x + c} \right)^2 \left( {x + k} \right)^{\frac{2}{3}} $

Now, this m doesn't seem "constant", but dependent of x, c and k!
But, where do we have to look for ortogonality? At their intersection!
Let's try to find something useful, equalling both equations:

$ \frac{1}{{x + c}} = m\left( {x + k} \right)^{\frac{1}{3}} \Leftrightarrow \frac{1}{m} = \left( {x + c} \right)\left( {x + k} \right)^{\frac{1}{3}} \Rightarrow \frac{1}{{m^2 }} = \left( {x + c} \right)^2 \left( {x + k} \right)^{\frac{2}{3}} $

You recognise this? Filling in, in what we had for m earlier:

$ m = 3\left( {x + c} \right)^2 \left( {x + k} \right)^{\frac{2}{3}} \Leftrightarrow m = \frac{3}{{m^2 }} \Leftrightarrow m^3 = 3 \Leftrightarrow m = \sqrt[3]{3} $

Thanks! I understand now

**TD!** · October 21st 2007, 11:29 PM

You're welcome

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