Related rates ~

Peng

2008-10-19 11:31 am

A particle is moving along the curve y=x^1/2 .As the particle passes through the point(4,2), its x-coordinate increases at a rate of 3cm/s. how fast is the distance from the particle to the origin changing at this instant.

回答 (1)

Audrey Hepburn

2008-10-19 8:53 pm

✔ 最佳答案

y = x1/2

Differentiate both sides with respect to time t

dy/dt = 1 / 2x1/2 dx/dt

When the coordinates is (4 , 2)

Given dx/dt = 3

So, dy/dt = 1 / 2(4)1/2 (3) = 3/4

Let s be the distance of the particle from the origin.

s = (x2 + y2)1/2

ds/dt = 1 / 2(x2 + y2)1/2 (2x dx/dt + 2ydy/dt)

= (xdx/dt + ydy/dt) / (x2 + y2)1/2

Therefore, at that instant

ds/dt = [(4)(3) + (2)(3/4)] / [(4)2 + (2)2]1/2

= 27 √5 / 20

So, the rate of change of distance of the particle from the origin is 27 √5 / 20 cm per second at that instant.

參考: Myself~~~

👍 0 👎 0