differentiation(2)急

As follows:


圖片參考：http://i707.photobucket.com/albums/ww74/stevieg90/ScreenHunter_04Jan172340.gif

(a)
OA^2 + OB^2 = 13^2
x^2 + y^2 = 169 ... (*)

Differentiating (*) with respect to t, we obtain
2x dx/dt + 2y dy/dt = 0
x dx/dt + y dy/dt = 0

When OA = x = 5 units, y = 12 units.
Also, dx/dt = 2 (positive when A is moving aways from the origin)

So, 5(2) + 12 dy/dt = 0
10 + 12 dy/dt = 0
12 dy/dt = -10
dy/dt = -10/12
= -5/6

B is moving towards the origin at a rate of 5/6 units per second.

(b)
x^2 + y^2 = 169
1 + (y/x)^2 = 169/x^2
1 + m^2 = 169/x^2, where m is the slope of AB ... (**)


Differentiating (**) with respect to t, we obtain
2m dm/dt = 169 * -(2x dx/dt)/x^4
m dm/dt = -(169x dx/dt)/x^4

When OA = x = 5, y = 12, m = -12/5.
-12/5 dm/dt = -169(5) * 2 / 5^4
= -338/125
dm/dt = 169/150

The slope is increasing at a rate of 169/150 units per second.

2011-01-18 17:54:39 補充：
睇番我同STEVIE-G™ 既答案...

If x, y is moving away from the origin,
x, y increase with time.
So, dx/dt, dy/dt > 0

If x, y is moving towards the origin,
x, y decrease with time, but x, y >= 0
So, dx/dt, dy/dt < 0

2011-01-18 17:55:20 補充：
ALso, if x, y stop momentarily at time t,
dx/dt and dy/dt = 0

A and B are two points poving???