If a snowball melts so its surface area decreases at a rate of 1 cm^2 /min , find the rate at which the diameter decreases when the diameter is 6 cm?
2 Answers
Explanation:
Area _((snowball))=4πr^2 A(t)=4πr^2(t) A'(t)=2*4πr(t)r'(t) = 4π*2r(t)r'(t)
For
,
,
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A'(t_0)=4π*2r(t_0)r'(t_0) <=>
The diameter is decreasing at a rate of
Explanation:
Let us setup the following variables:
{ (D, "Diameter of snowball at time t","(cm)"), (S, "Surface area of snowball at time t", "(cm"^3")"), (t, "time", "(min)") :}
The standard formula for Surface Area of a sphere with radius
S = 4pir^2
\ \ = 4pi(D/2)^2
\ \ = piD^2
Differentiating wrt
(dS)/(dD) = 2piD
We are given that the constant rate at which surface area decreases is 1, Thus:
(dS)/(dt) = -1
And, we seek the value of
(dD)/(dt) = (dD)/(dS) * (dS)/(dt)
\ \ \ \ \ \ = 1/(2piD) * (-1)
\ \ \ \ \ \ = -1/(2piD)
So then:
[ (dD)/(dt) ]_(D=5) = -1/(2pi * 6) = -1/(12pi)
Thus the diameter is decreasing at a rate of