If # A = 9/16(4r-sin(4r)) # and #(dr)/dt=0.7# when #r=pi/4# then evaluate # (dA)/dt # when #r=pi/4#?

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Answer 1 · 2017-11-20T22:47:30

# [ (dA)/dt ]_(r=pi/4) = 3.15 #

We have:

# A = (9/16)(4r-sin(4r)) #

If we denote time in minutes by #t#, Then differentiating implicitly wrt #t# we have:

# \ \ \ (d)/dtA = (9/16)d/dt(4r-sin(4r)) #

# :. (dA)/dt = (9/16)(dr)/dt d/(dr)(4r-sin(4r)) #

# :. (dA)/dt = (9/16)(dr)/dt (4-4cos(4r)) #

And we are given that #(dr)/dt=0.7#

# \ \ \ \ (dA)/dt = (9/4)(0.7) (1-cos(4r)) #
# :. (dA)/dt = (1.575) (1-cos(4r)) #

So, when #r=pi/4#, we have:

# [ (dA)/dt ]_(r=pi/4) = (1.575) (1-cos(pi)) #
# " " = (1.575) (1-(-1)) #
# " " = 3.15 #