If a snowball melts so that its surface area decreases at a rate of 5 cm2/min, then the rate at which the diameter decreases when the diameter is 12 cm will be 0.0663
Assuming the snowball is a perfect sphere, then if denotes the surface area and D the diameter then:
A = 4πr² = 4π(D/2)²= πD²
Differentiating w.r.t to D we get
[tex]\frac{dA}{dD}[/tex] = 2πD
It is also given that,
[tex]\frac{dA}{dT}[/tex] =-5
Now, [tex]\frac{dA}{dD}[/tex] = [tex]\frac{dA}{dT}[/tex]×[tex]\frac{dT}{dD}[/tex]
⇒2πD =[tex]\frac{dA/dT}{dD/dT}[/tex]
⇒[tex]\frac{dD}{dT}[/tex] = [tex]\frac{-5}{2*3.14*12}[/tex]
⇒[tex]\frac{dD}{dT}[/tex]= -0.0663
The rate at which the diameter will decrease when the area is decreasing at a rate of 5 cm²/min is 0.0663
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