Answer:
The diameter decreases at a rate of [tex]\frac{-5}{22\pi } \:\frac{cm}{min}[/tex].
Step-by-step explanation:
The surface area of a sphere is given by
[tex]A=4\pi r^2[/tex]
and its diameter is [tex]d=2r[/tex].
[tex]A=4\pi (\frac{d}{2} )^2=\pi d^2[/tex]
We assume that the snowball is a sphere.
If we take the derivative with respect with time using the chain rule, we get:
[tex]\frac{dA}{dt}= 2\pi d\frac{dd}{dt}[/tex]
We know that the area decreases at a rate of [tex]5 \:\frac{cm^2}{min}[/tex], and we want to find the rate at which the diameter decreases when the diameter is 11 cm [tex]\frac{dd}{dt} \Bigr\rvert_{d = 11}[/tex]
So,
[tex]-5= 2\pi (11)\frac{dd}{dt}\\\\\frac{dd}{dt}=\frac{-5}{22\pi } \:\frac{cm}{min}[/tex]
The diameter decreases at a rate of [tex]\frac{-5}{22\pi } \:\frac{cm}{min}[/tex].
