How do I start this problem?

• Oct 16th 2010, 09:58 PM
ZeroVector
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.
• Oct 16th 2010, 10:14 PM
CaptainBlack
Quote:

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
• Oct 16th 2010, 10:21 PM
mr fantastic
Quote:

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?
• Oct 16th 2010, 10:41 PM
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?
• Oct 16th 2010, 11:15 PM
mr fantastic
Quote:

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:
Quote:

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.
• Oct 16th 2010, 11:21 PM
CaptainBlack
Quote:

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
• Oct 16th 2010, 11:30 PM
mr fantastic
Quote:

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 ....
• Oct 17th 2010, 12:45 AM
ZeroVector
Just to make sure, is the velocity vector:
v = < SQRT(400/17), -0.25 * SQRT(400/17) >
• Oct 17th 2010, 01:08 AM
CaptainBlack
Quote:

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