Answer:
-50 cm^3/min
Explanation:
The volume of the lollipop is given by
[tex]V=\frac{4}{3}\pi r^3[/tex]
where r is the radius in centimeters.
To find how fast the volume changes, we need to find the rate of change of the volume, which is given by its derivative with respect to time:
[tex]\frac{dV}{dt}=4\pi r^2 \frac{dr}{dt}[/tex] (1)
where
dr/dt is the rate of change of the radius
We know that the circumference is
[tex]C=2\pi r[/tex]
and it decreases by 1 cm per minute, so
[tex]\frac{dC}{dt}=2\pi \frac{dr}{dt}=-1[/tex] (cm/min)
from which we find dr/dt:
[tex]\frac{dr}{dt}=-\frac{1}{2\pi}[/tex]
Substituting into (1),
[tex]\frac{dV}{dt}=4\pi r^2 \frac{1}{2\pi}=-2r^2[/tex] (cm^3/min)
And substituting r = 5, we find the rate of change of the volume when the radius is 5 cm:
[tex]\frac{dV}{dt}=-2(5)^2=-50[/tex] (cm^3/min)
The volume of her lollipop changes at [tex]-50 \dfrac{cm^{-3} }{min}[/tex]
The volume of the lollipop is given by
[tex]V=\dfrac{4}{3} \pi r^{3}[/tex]
where r is the radius in centimeters.
Since volume will reduce with the time and the volume is the function of time so
[tex]\dfrac{dv}{dt} = 4\pi r^{2} \dfrac{dr}{dt}[/tex] .............................(1)
where,
[tex]\dfrac{dr}{dt}[/tex] is the rate of change of the radius
We know that the circumference is
[tex]C=2\pi r[/tex]
and it is decreasing by 1 cm per minute, so
[tex]\dfrac{dc}{dt}= 2\pi \dfrac{dr}{dt}=-1[/tex]
we will find the rate of change of radius with time
[tex]\dfrac{dr}{dt}=\dfrac{1}{2\pi}[/tex]
Now put this value in equation (1)
[tex]\dfrac{dv}{dt}=-4\pi r^{2} \dfrac{1}{2\pi}=-2r^{2} \dfrac{cm^{-3} }{min}[/tex]
Now put the value of r=5cm
[tex]\dfrac{dv}{dt}=-2\times 5^{2} =-50 \dfrac{cm^{3} }{min}[/tex]
Thus the volume of her lollipop changes at [tex]-50 \dfrac{cm^{-3} }{min}[/tex]
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