On a hot summer day, a spherically shaped scoop of ice cream is left outside in a bowl. The ice cream starts?

melting. What is the rate of change with respect to time of this icecream ball's radius at the moment when the radius is 3 centimeters, if the ice creams volume is changing at a rate of -5 cubic centimeters per minute?On a hot summer day, a spherically shaped scoop of ice cream is left outside in a bowl. The ice cream starts?
The radius is changing at (-5/(4 pi r^2)) cm / min.


= .0442 cm / min





That is derived as follows:





dr/dt = dr/dV x dV/dt





We are given that dV/dt = -5 cm^3/min


We can solve for dr/dV, using the formula for the volume of a sphere and the fact that the radius is 3 cm:





V = (4/3) pi r^3


dV/dr = 4 pi r^2 (which happens to be the surface area of the sphere, which is not a coincidence, when you think about it)





If r = 3, dV/dr = 4 pi 3^3 = 36 pi (cm^3 per cm)





So dr/dt = dr/dV x dV/dt


= (1/(36 pi) cm / cm^3) x (-5 cm^3 / min)


= (-5/(36 pi)) cm / min


= .0442 cm / minOn a hot summer day, a spherically shaped scoop of ice cream is left outside in a bowl. The ice cream starts?
V = 4/3蟺.r鲁, so dV/dt = 4蟺r虏.dr/dt





But dV/dt = -5, so -5 = 4蟺r虏.dr/dt





dr/dt = -5/(4蟺r虏) - now just substitute r = 3 to get your rate of change of radius.





dr/dt = -5/(36蟺)