If I think I’m understanding it, the new members out of the 200 original would be adding 28% to the mix
Answer:
the rate of change in volume is dV/dt = 4π mm³/s = 12.56 mm³/s
Step-by-step explanation:
since the volume V of a cylinder is related with the height H and the radius R through:
V = πR²*H
then the change in time is given by the derivative with respect to time t
dV/dt = (∂V/∂R)*(dR/dt) + (∂V/∂H)*(dH/dt)
the change in volume with radius at constant height is
(∂V/∂R) = 2*πR*H
the change in volume with height at constant radius is
(∂V/∂H) = πR²
then
dV/dt = 2π*R*H *(dR/dt) + πR²*(dH/dt)
replacing values
dV/dt = 2π* 2 mm * 20 mm * (-0.1 mm/s) + π (2 mm) ²* 3 mm/s = 4π mm³/s
dV/dt = 4π mm³/s = 12.56 mm³/s
Given:
To find:
The rate of change in volume at 
Solution:
We know that, volume of a cone is
Differentiate with respect to t.
![\dfrac{dV}{dt}=\dfrac{1}{3}\pi\times \left[(r^2\dfrac{dh}{dt}) + h(2r\dfrac{dr}{dt})\right]](https://tex.z-dn.net/?f=%5Cdfrac%7BdV%7D%7Bdt%7D%3D%5Cdfrac%7B1%7D%7B3%7D%5Cpi%5Ctimes%20%5Cleft%5B%28r%5E2%5Cdfrac%7Bdh%7D%7Bdt%7D%29%20%2B%20h%282r%5Cdfrac%7Bdr%7D%7Bdt%7D%29%5Cright%5D)
Substitute the given values.
![\dfrac{dV}{dt}=\dfrac{1}{3}\times \dfrac{22}{7}\times \left[(120)^2(-2.1) +175(2)(120)(1.4)\right]](https://tex.z-dn.net/?f=%5Cdfrac%7BdV%7D%7Bdt%7D%3D%5Cdfrac%7B1%7D%7B3%7D%5Ctimes%20%5Cdfrac%7B22%7D%7B7%7D%5Ctimes%20%5Cleft%5B%28120%29%5E2%28-2.1%29%20%2B175%282%29%28120%29%281.4%29%5Cright%5D)
![\dfrac{dV}{dt}=\dfrac{22}{21}\times \left[-30240+58800\right]](https://tex.z-dn.net/?f=%5Cdfrac%7BdV%7D%7Bdt%7D%3D%5Cdfrac%7B22%7D%7B21%7D%5Ctimes%20%5Cleft%5B-30240%2B58800%5Cright%5D)
Therefore, the volume of decreased by 29920 cubic inches per second.
Answer: 1232
Step-by-step explanation: ( 8x10 -3) x(2x10 -4)
( 8x10 -3) = 77
(2x10 -4) = 16
77 x 16 = 1232
Answer = 1232
