Answer:
The diameter decreases at a rate of 0.143cm/min when it is of 10 cm.
Step-by-step explanation:
Surface area of an snowball:
An snowball has spherical format. The surface area of an sphere is given by:
[tex]S = d^2\pi[/tex]
In which d is the diameter of the sphere.
In this question:
We need to differentiate S implicitly in function of time. So
[tex]\frac{dS}{dt} = 2d\pi\frac{dd}{dt}[/tex]
Surface area decreases at a rate of 9 cm2/min
This means that [tex]\frac{dS}{dt} = -9[/tex]
At which the diameter decreases when the diameter is 10 cm?
This is [tex]\frac{dd}{dt}[/tex] when [tex]d = 10[/tex]. So
[tex]\frac{dS}{dt} = 2d\pi\frac{dd}{dt}[/tex]
[tex]-9 = 2(10)\pi\frac{dd}{dt}[/tex]
[tex]\frac{dd}{dt} = -\frac{9}{20\pi}[/tex]
[tex]\frac{dd}{dt} = -0.143[/tex]
Area in cm², so diameter in cm.
The diameter decreases at a rate of 0.143cm/min when it is of 10 cm.
