Math Help - orthogonal trajectories

1. orthogonal trajectories

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?

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

3. 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?

4. 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

5. 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 $\sqrt[3]{3}$

6. 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

7. 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?

8. 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}
$

9. 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

10. You're welcome