# Math Help - How do I start this problem?

1. ## How do I start this problem?

What is the velocity vector of a particle traveling to the right
along the hyperbola y = 1/x with constant speed 5cm/s when the
particle's location is (2, 1)?

I'm not sure where to start or what formula(s) I should use.

2. Originally Posted by ZeroVector
What is the velocity vector of a particle traveling to the right
along the hyperbola y = 1/x with constant speed 5cm/s when the
particle's location is (2, 1)?

I'm not sure where to start or what formula(s) I should use.
Start with the speed is 5 cm/s so the square of the speed:

$v^2=\left(\dfrac{dx}{dt}\right)^2+\left(\dfrac{dy} {dt}\right)^2=25$

Now you know that $y=1/x$ so:

$\dfrac{dy}{dt}=-\;\dfrac{1}{x^2}\dfrac{dx}{dt}$

etc..

CB

3. Originally Posted by ZeroVector
What is the velocity vector of a particle traveling to the right
along the hyperbola y = 1/x with constant speed 5cm/s when the
particle's location is (2, 1)?

I'm not sure where to start or what formula(s) I should use.
(2, 1) does not lie on y = 1/x.

And what coordinate system is to be used? Intrinsic coordinates?

4. Sorry, I was mindlessly altering it so that I could (try and) reproduce it by myself. The coordinates my book gives is (2, 1/2).

Edit: I'm still sort of confused. What do I do with dy/t = -1/x^2 dx/dt? And (dx/dt)^2 + (dy/dt)^2 = 25?

5. Originally Posted by ZeroVector
Sorry, I was mindlessly altering it so that I could (try and) reproduce it by myself. The coordinates my book gives is (2, 1/2).

Edit: I'm still sort of confused. What do I do with dy/t = -1/x^2 dx/dt? And (dx/dt)^2 + (dy/dt)^2 = 25?
Refer to my edit of CB's reply:
Originally Posted by CaptainBlack
[snip]
$v^2=\left(\dfrac{dx}{dt}\right)^2+\left(\dfrac{dy} {dt}\right)^2=25$ .... (1)

Now you know that $y=1/x$ so:

$\dfrac{dy}{dt}=-\;\dfrac{1}{x^2}\dfrac{dx}{dt}$ .... (2)
Substitute x = 2 into (2). Then substitute the result into (1) and solve for dx/dt. Substitute your value of dx/dt into (2) and solve for dy/dt. Use the values to get the velocity vector.

6. Originally Posted by mr fantastic
Refer to my edit of CB's reply:

Substitute x = 2 into (2). Then substitute the result into (1) and solve for dx/dt. Substitute your value of dx/dt into (2) and solve for dy/dt. Use the values to get the velocity vector.
Doesn't the Bible say something about not numbering the equations of your posts!?

CB

7. Originally Posted by CaptainBlack
Doesn't the Bible say something about not numbering the equations of your posts!?

CB
I think the Book of Numbers says somewhere that it's OK ....

8. Just to make sure, is the velocity vector:
v = < SQRT(400/17), -0.25 * SQRT(400/17) >

9. Originally Posted by ZeroVector
Just to make sure, is the velocity vector:
v = < SQRT(400/17), -0.25 * SQRT(400/17) >
since $\sqrt{400}=20$ might I suggest you try simplifying the above (otherwise it looks OK)

CB