Differentiation

用戶225241

2008-10-13 7:00 am

A amn 2 meters tall is approaching the foot of a tower over lever ground at the
constant rate of 2m/sec. If the tower is 7 meters high, at what rate is the
distance between the top of the tower and his head changing at the instant that
he is 12 meters from the foot of the tower??

回答 (1)

2008-10-13 7:30 am

✔ 最佳答案

H=7m = height of tower
h=2m = height of man
H-h=Vertical distance between the top of tower and head of man
x=horizontal distance from the tower
S(x)=sloping distance from the tower using Pythagoras theorem
=sqrt((H-h)^2+x^2)
dx/dt=2 horizontal speed of man

V(x)
=dS(x)/dt
=dX/dx.dx/dt
=d(sqrt((H-h)^2+x^2))/dx. dx/dt
=x/sqrt(5^2+x^2) . 2
=2x/sqrt(25+x^2)

V(12)
=24/(sqrt(5^2+12^2)
=24/13
=1.846 m/s

2008-10-12 23:45:31 補充：
The above answer was inspired by the title differentiation.

The same answer can be obtained simply by trigonometry:
Horizontal speed = 2 m/s
Rate of change of distance on hypothenuse
= 2 m/s * cos(theta)
= 2 m/s * 12/sqrt(12^2+5^2)
= 2 m/s * 12/13
= 24/13
= 1.846 m/s

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