更新1:
A and B are two points moving
更新2:
a) To the origin：5/6 unit/second b) 169/150 per second
differentiation(2)急
2011-01-18 5:17 am
A and B are two points poving on the positive x-axis and the positive y-axis respectively, where the length of AB is fixed at 13 units. It is given that A is moving at a speed 2units/s away from the origin.a) What is the velocity of B when OA=5 units?b) What is the rate of change of slope of AB when OA = 5units?
回答 (3)
2011-01-18 7:41 am
✔ 最佳答案
As follows:圖片參考：http://i707.photobucket.com/albums/ww74/stevieg90/ScreenHunter_04Jan172340.gif
2011-01-18 5:32 pm
(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
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
2011-01-18 5:20 am
A and B are two points poving???
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