Asked by nandni
A circular ink blot grows at the rate of 2 cm per sec. Find the rate at which the radius is increasing after 28/11 sec.
Answers
Answered by
Reiny
I will assume you meant to say that the area of the ink blot is growing at the rate of 2 cm^2 /s
A = πr^2
dA/dt = 2πr dr/dt
2 = 2π (28/11) dr/dt
dr/dt = 11/(28π) cm/sec
= appr .125 cm/s
A = πr^2
dA/dt = 2πr dr/dt
2 = 2π (28/11) dr/dt
dr/dt = 11/(28π) cm/sec
= appr .125 cm/s
Answered by
RAJU BANERJEE
πr^2=2×28/11
r=14/11
let,a=Ï€r^2
da/dt=2Ï€r dr/dt
2 =2π×14/11×dr/dt
dr/dt=11/14Ï€ =1/4
=.25
r=14/11
let,a=Ï€r^2
da/dt=2Ï€r dr/dt
2 =2π×14/11×dr/dt
dr/dt=11/14Ï€ =1/4
=.25
Answered by
RAJU BANERJEE
πr^2=2×28/11=56/11,r=14/11
Now a=πr^2
And derivative w.r.t t,we get
da/dt=2πr dr/dt
2=2π×14/11×dr/dt
dr/dt=11/14π=1/4=.25
Now a=πr^2
And derivative w.r.t t,we get
da/dt=2πr dr/dt
2=2π×14/11×dr/dt
dr/dt=11/14π=1/4=.25
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