Rate of distance between two objects

[holezch]

Homework Statement

An object A moves along the positive horizontal axis, and object B along the graph of f(x) =
-sqrt(3)x for x <= 0. At a certain time, A is at the point (5,0) and moving with the speed 3 units/sec and B is at a distance of 3 units from the origin and moving with speed 4 units/ sec. At what rate is the distance between A and B changing?

Homework Equations

line equation

The Attempt at a Solution

don't I only have to draw a line from A to B at that point and take the slope? Probably not, since I;'m given so much information... anybody want to help me out? thank you

 

[clamtrox]

What are
\frac{d A_x}{dt}, \frac{d A_y}{dt}, \frac{d B_x}{dt} \mathrm{and} \frac{d B_y}{dt}?
What is the rate at which the distance changes in terms of these?

 

[holezch]

thank you for the reply, I realized that I will have to have a distance function dependent on A and B... I don't know how to get a distance function.. thank you

 

[holezch]

What are
\frac{d A_x}{dt}, \frac{d A_y}{dt}, \frac{d B_x}{dt} \mathrm{and} \frac{d B_y}{dt}?
What is the rate at which the distance changes in terms of these?

well Ay/dt and By/dt are given , 3 and 4 units per second right

 

[holezch]

please help, I'm stuck here

 

[clamtrox]

What you are given is the velocity, which is \left| \frac{d \mathbf{A}}{dt} \right|.

 

[holezch]

okay, I have the following:
d(t) = distance with respect to time
d(t)^2 = (xA - xB)^2 + (yA-yB)^2

d(t)^2/dt ...

2d(t)d'(t) = 2 (5 - 3/2)(3-dxB/dt) + 2(0-(-sqrt(3)(3/2)))(0 - dyB/dt)
so now I solve for dxB/dt and dyB/dt?

thank you

 

[holezch]

not sure how to find dxB/dt and dyB/dt.. anybody want to help? thank you

 

[clamtrox]

Use Pythagoras. You know what shape curve you are doing so just start by

\sqrt{(dx/dt)^2 + f&#039;(x)^2} = 4 and from there, solve for dx/dt.