dv/dr=.06r => v(r)=C-.03r^{2}

v(13)=0=C-.03(169) => C= 5.07

therefore,

v(r)=5.07-.03r^{2}

Erik K.

asked • 01/15/19As water flows through a tube of radius R = 13 cm, the velocity v of an individual water particle depends only on its distance r from the center of the tube. The particles at the walls of the tube have zero velocity and dv/dr = −0.06r, where v is measured in cm/s. Determine v(r).

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dv/dr=.06r => v(r)=C-.03r^{2}

v(13)=0=C-.03(169) => C= 5.07

therefore,

v(r)=5.07-.03r^{2}

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