Differentiation-rate of change

1(A)
Since circumference C=πd
The derivative is
dC/dt
=d/dt (πd)
=πd d/dt
Now d d/dt=-0.4 when it is 12 cm
dC/dt=-0.4π
The rate of change of circumference is -0.4π cm
(b)
Area A=πr^2=πd^2/4
dA/dt
=d/dt(πd^2/4)
=(π/4)(2d)(d d/dt)
Substitute d=12 and d d/dt=-0.4
dA/dt
=(π/4)(24)(-0.4)
=-2.4π
The rate of change of area is -2.4π cm^2
2
(a)
A=4πr^2
dA/dt
=(4π)(2r)(dr/dt)
=(4π)(10)(0.6) [Substitute r=5 and dr/dt=0.6]
=24π
The rate of change of its surface area is 24π cm^2
V=(4/3)πr^3
dV/dt
=(4/3)π(3r^2)(dr/dt)
Substitute r=5 and dr/dt=0.6
=(4/3)π(75)(0.6)
=60π
The rate of change of its volume is 60π cm^3



2009-01-02 22:22:00 補充：
我應該將直徑用大楷D表示﹐希望不會做成誤解啦

1)a)let the circumference be C,the diameter be D
C=Dπ
dC/dt=(dD/dt)(π)
=-0.4πcm/s
so,the circumference of the spotlight is diminishing at 0.4πcm/s.

b)A=πr^2
=π(D/2)^2
=πD/4
dA/dt=2(D)(dD/dt)(π/4)
=2(12)(-0.4)(π/4)
=-2.4πcm^2/s
so,the the area of the spotlight is diminishing at 2.4πcm^2/s.

2)a)A=4πr^2
dA/dt=4(2r)(dr/dt)π
=8(5)(0.6)π
=24πcm^2/s
so,the surface area of the spherical balloon is increasing at 24πcm^2/s.

b)V=(4/3)πr^3
dV/dt=(4/3)(3r^2)(dr/dt)π
=(4)(5^2)(0.6)π
=60πcm^3/s
so,the volume of the spherical balloon is increasing at 60πcm^3/s.

希望冇錯+幫到你

2009-01-02 21:56:32 補充：
有亂碼-__-sor

π - π