Apply of differentiation(1Q)!!
2011-08-24 2:28 am
from position A, Peter starts to run due east and David starts to run due south at the same time.When Peter is 3km away from A, David is 4 km away from A. The velocities of Peter and David at moment are 6 km/h and 7km/h respectively. Find the rate of change of the distance between Peter and David w.r.t time at that moment.
回答 (1)
2011-08-24 4:33 am
✔ 最佳答案
Let x km be distance of Peter from A, and y be the distance of David from A,and let distance between Peter and David be z km,
z = √(x^2 + y^2)
At that time, x = 3, y = 4, dx/dt = 6, dy/dt = 7
dz/dt
= (1/2) [1/√(x^2+y^2)] * (2x*dx/dt+2y*dy/dt)
= (1/2) [1/√(3^2 + 4^2)] * (2*3*6+2*4*7)
= (1/2) (1/5) (36+56)
= (1/10)(92)
= 9.2 km/h
參考: Hope the solution can help you^^”
收錄日期: 2021-04-13 18:11:34
原文連結 [永久失效]:
https://hk.answers.yahoo.com/question/index?qid=20110823000051KK00963