Blackboard Home Courses Help MAT 265: Calculus for Engineers 1 (2018 Spring-A) 2

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Blackboard Home Courses Help MAT 265: Calculus for Engineers 1 (2018 Spring-A) 2018SpringA-X-MAT265-15982a WeBWork Entered Answer Preview Problem 7. PREVIEW ONLY - ANSWERS NOT RECORDED (1 point) Hellum is pumped into a spherical balloon at a rate of 2 cubic feet per second. How fast is the radius increasing after 3 minutes? Note: The volume of a sphere is given by V (4/3)xr Rate of change of radius (in feet per second) Entered Answer Preview Problem 8. -PREV EVON yHANSWERSNOTRECORDED (1 point) The radius of a spherical bailoon is increasing at a rate of 2 centimeters per minute. How fast is the volume changing when the radius is 14 centimeters? Note: The volume of a sphere is given by V (4/3) Rate of change of volume - Entered Answer Preview

Explanation / Answer

1)V=(4/3) r³

Differentiating the formula for the volume of the sphere gives:

dV = 4r^2(dr/dt).

Solving this for dr/dt gives:

dr/dt = dV/(4r^2).

Assuming that the balloon was empty when we started, the volume of the balloon after 3 minutes is 3*120 =360 cubic units.

360 = (4/3)r^3
==> 270 = r^3
==> r = (270/)^(1/3).

Thus, with dV = 2 and r = (270/)^(1/3), we have:

dr/dt = dV/(4r^2) = 2/[4(270/)^(2/3)] = 1/[2(270/)^(2/3)] 0.00187 units/sec.

2) V=(4/3) r³

Differentiate wrt t (time)

dV/dt = 4r² * dr/dt
given
dr/dt =2cm/min
r=14cm

dV/dt=4*14*14*2=1568 cc/minute

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