1. ## Rates

Hi everyone, I can't seem to figure htis problem out.

A dog heading due north at a constant speed of 2 meters per second trots past a fire hydrant at t = 0 sec. Another dog heading due east at a constant speed of 3 meters per second trots by the hydrant at t = 1 sec. At t = 9 sec, the rate of change of the distance between the two dogs is?

2. You're given a velocity. Integrate it to get position and use the extra information for your constant of integration.

After that, you don't need calculus. Just take y2-y1/x2-x1 of the two dogs' positions.

Edit: An added hint, their velocities are constant. So you'll be integrating constants. Oh, and remember to account for their direction (north and east).

3. So my two intergrals are 2x + C and 3x + C, right? How do I incorporate the t = 9 into those and finally come out with an answer? I'm sorry I just don't think I understand what you're saying

4. Let the hydrant be our origin.

$\int v(t)dt = s(t) = 2dt = 2t + c$

To solve for c, use what they provided: 2 meters at t = 0

2 = 2(0) + c

c = 2

s(t) = 2x + 2

This dog is running north. So s(t) is returning a y-value.

The other equation is s(t) = 3x + c, 3 = 3(1) + c, c = 0. So s(t) = 3x.

This dog is running east, so you're getting an x-value.

Now, at t = 9, dog1 is 2(9) + 2 away from the origin. dog2 is 3(9) from the origin.

Now use y2-y1/x2-x1

20 - 0 / 0 - 27 = -20/27