ce amath---rate of change

3.a)
Let the coordinates of A and B be (x, 0) and (0, y) respectively.
Then √[(x - 0)² + (0 - y)²] = 13,
i.e. x² + y² = 169
Differentiating both sides with respect to t,
2x(dx/dt) + 2y(dy/dt) = 0
When OA = 5 units, i.e. x = 5, 5² + y² = 169, y = 13
From the given, dx/dt = 2
We have 2(5)(2) + 2(13)(dy/dt) = 0, dy/dt = -10/13
Therefore, B is moving at a rate of 10/13 unit/s towards the origin when OA = 5 units.

b)
Slope of AB = m = -y/x
dm/dt = -[x(dy/dt) - y]/x² = [y - x(dy/dt)]/x²
When OA = 5 units, from the result of (a), dy/dt = -10/13
We have dm/dt = [13 - (5)(-10/13)]/(5²) = 219/325
Therefore, rate of change of slope of AB = 219/325 (= 0.674) per second

d答案咁奇怪...希望冇計錯@@

2007-09-16 19:35:04 補充：
係。y = 12From the given, dx/dt = 2We have 2(5)(2) 2(12)(dy/dt) = 0, dy/dt = -5/6Therefore, B is moving at a rate of 5/6 unit/s towards the origin when OA = 5 units.

2007-09-16 19:35:52 補充：
上面漏了 號b)Slope of AB = m = -y/xdm/dt = -[x(dy/dt) - y]/x² = [y - x(dy/dt)]/x²When OA = 5 units, from the result of (a), dy/dt = -5/6We have dm/dt = [12 - (5)(-5/6)]/(5²) = 97/150Therefore, rate of change of slope of AB = 97/150 (= 0.647) per second

佢個a part有啲問題....
之後個y 唔係等於13.....
係咁先啱
x^2+y^2=13^2
5^2+y^2=13^2
y=12

跟住你改返下面就計到架啦~~
努力啦~~