A particle is on the xaxis and is moving to the rt. at 2 units/sec. A sec. parti

Question

A particle is on the xaxis and is moving to the rt. at 2 units/sec. A sec. particle is moving up the yaxis at the rate of 3 units/sec. At a certain instant the first particle is at the pt. (5,0) and the 2nd is @ the pt. (0,7). A.) how rapidly is the distance between the particle changing at that instant? B.) Are the particles moving towards or away from each other at that instant? Please show all steps. Thanks

Explanation / Answer

So the particle is always on the x-axis, so use the distance formula to write down an expression (function really) for the distance, then find the derivative of that expression with respect to time: Using Pythagorean theorem: d² = (x0 - x)² + y0² (since it's always on the x-axis, y is always zero --> (y0 - 0)² = y0²) Now, you probably don't want to deal with square roots, right? So use implicit differentiation --> differentiate with respect to t (remember x0 and y0 are constants, therefore dx0/dt = dy0/dt = 0) 2d * dd/dt = 2(x0 - x)dx/dt + 0 --> 2's cancel dd/dt = (x0 - x) / d * dx/dt So find x, x0, d and dx/dt You are given x, x0, and dx/dt (x0 = 0 because the "reference point" is (0, 9) --> x0 = 0, y0 = 9) x = 5, dx/dt = 2 Now you just need to calculate d at that point: d² = x² + y0² --> d² = 5² + 9² = 25 + 81 = 106 --> d = v106 dd/dt = 5/v106 * 2 = 10 / v106 there maybe should have been a negative, but it should be clear that if it's moving to the right, the distance is getting bigger so we know that the rate should be positive...this stems from the fact that you could just as easily do: d² = (x0 - x)² + y0² or d² = (x - x0)² + y0²

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