Rate of Change in a coordinate plane

[NYmike]

Hi. I am drawing a complete blank on this calc problem.

Point a moves along the x-axis at the constant rate of 'a' ft/sec
while point b moves along the y-axis at the constant rate of 'b'
ft/sec. Find how fast the distance between them is changing when A is
at the point (x,0)and B is at the point (0, y).

I know that dx / dt = a, and dy / dt = b, however I am stuck on where to go afterwards. I drew a graph and thought that the slope would lead
me to an answer, but I can't quite figure it out. Since I am looking for the change in distance between the two points I figured that I may have to incorporate the distance formula rather than some formula with slope, however I can't quite figure out how to go about it.

Thanks in advance for any help.

 

[hotvette]

The wording of the problem provides the clue as to how to proceed.

Find how fast the distance between them is changing

This tells you two things:

1. You need to develop an equation for the distance between the points

2. differentiate (1) and evaluate at the desired x & y

 

[NYmike]

d = \sqrt{\Delta x^2 + \Delta y^2}

\frac{dx}{dt} = a

\frac{dy}{dt} = b

That's what I've come up with so far.

Now, since I am trying to find the rate of change of the distance, then I believe I am looking for \frac{dd}{dt}

Taking the derative of the distance formula directly would be pretty sloppy, so I think I should plug in for the values of x and y and then evaluate the deriv. I believe that the two points in question are: (x+a, 0) and (0, y+b). Is this correct? This is where the points A and B are after 1 second.

If so, then \Delta x = (x+a) - (0) = (x+a) and \Delta y = (0) - (y+b) = -y-b

 

[NYmike]

Taking the derative of the distance formula directly would be pretty sloppy, so I think I should plug in for the values of x and y and then evaluate the deriv.

Even after plugging in the deriv is ridiculous.

There must be a way to plug in something else. Am I using the wrong coordinates/formula?

 

[Dorothy Weglend]

Even after plugging in the deriv is ridiculous.

There must be a way to plug in something else. Am I using the wrong coordinates/formula?

I don't see a "t" anywhere in there. Maybe I am wrong, but it seems that somehow your distance equation needs to reflect that X and Y are changing as a function of t.

 

[CarlB]

Try replacing d = \sqrt{\Delta x^2 + \Delta y^2} with d = \sqrt{x^2 + y^2}. Maybe you've been studying partial derivatives recently.

Carl

 

[NYmike]

I don't see a "t" anywhere in there. Maybe I am wrong, but it seems that somehow your distance equation needs to reflect that X and Y are changing as a function of t.

I was taught that \frac{d(some var)}{dt} = the Rate of Change of some var. with respect to time. The rate of change of x with respect to time is 'a', and of y with respect to time is 'b'. I am trying to find the rate of change of distance (var. d) with respect to time. I am pretty sure that this part is correct?

 

[NYmike]

Try replacing d = \sqrt{\Delta x^2 + \Delta y^2} with d = \sqrt{x^2 + y^2}. Maybe you've been studying partial derivatives recently.

Carl

Sorry, but I am not sure where you are going with this.

When d = \sqrt{x^2 + y^2} :

\frac{dd}{dt} = \frac{1}{2} (x^2 + y^2)^\frac{-1}{2} (2x\frac{dx}{dt} + 2y\frac{dy}{dt})

\frac{dd}{dt} = \frac{1}{2} * \frac{1}{\sqrt{x^2+y^2}} * 2xa + 2yb

\frac{dd}{dt} = \frac{xa + yb}{\sqrt{x^2+y^2}}

I have 2 points, so which would i plug in for x and y??

 

[CarlB]

I have 2 points, so which would i plug in for x and y??

x and y were given as data to the problem. You don't plug anything in for them. You're done and I think you got the right answer.

Carl