If $A = \frac{9}{16} (4 r - sin (4 r))$ $A = 9/16(4r-sin(4r))$ and $\frac{d r}{d t} = 0.7$ $(dr)/dt=0.7$ when $r = \frac{π}{4}$ $r=pi/4$ then evaluate $\frac{d A}{d t}$ $(dA)/dt$ when $r = \frac{π}{4}$ $r=pi/4$ ?

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Answer 1 · 2017-11-20T22:47:30

${[\frac{d A}{d t}]}_{r = \frac{π}{4}} = 3.15$ $[ (dA)/dt ]_(r=pi/4) = 3.15$

We have:

$A = (\frac{9}{16}) (4 r - sin (4 r))$ $A = (9/16)(4r-sin(4r))$

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

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

If A = 9/16(4r-sin(4r)) A=916(4r−sin(4r)) A = 9/16(4r-sin(4r)) and (dr)/dt=0.7drdt=0.7(dr)/dt=0.7 when r=pi/4r=π4r=pi/4 then evaluate (dA)/dt dAdt (dA)/dt when r=pi/4r=π4r=pi/4?