Answer:
The diameter decreases at the rate of 0.0796 cm/min when the diameter is 10 cm.
Step-by-step explanation:
A snowball is spherical, so it's area is given by the following formula:
[tex]A = 4\pi r^{2}[/tex]
The radius is half the diameter so:
[tex]A = 4\pi (\frac{d}{2})^{2}[/tex]
[tex]A = 4\pi \frac{d^{2}}{4}[/tex]
[tex]A = \pi d^{2}[/tex]
If a snowball melts so that its surface area decreases at a rate of 5 cm2/min, find the rate at which the diameter decreases when the diameter is 10 cm.
We have to find [tex]\frac{dd}{dt}[/tex] when [tex]\frac{dA}{dt} = -5, d = 10[/tex]
[tex]A = \pi d^{2}[/tex]
Applying implicit differentiation:
We have to variables(A and d), so:
[tex]\frac{dA}{dt} = 2\pi d \frac{dd}{dt}[/tex]
[tex]-5 = 2\pi (10) \frac{dd}{dt}[/tex]
[tex]\frac{dd}{dt} = -\frac{5}{20\pi}[/tex]
[tex]\frac{dd}{dt} = -0.0796[/tex]
The diameter decreases at the rate of 0.0796 cm/min when the diameter is 10 cm.