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Sagot :
The rate at which the diameter decreases is 0.0159cm/min.
Given that a snowball melts and its surface area decreases at a rate of 1cm²/min.
The surface area of a solid object may be a measure of the entire area that the surface of the thing occupies.
The surface area of the snowball is A=4πr² because the snowball is spherical.
Here, r is the radius of the snowball.
Let D be the diameter of the snowball.
As the radius is half of the diameter
The surface area of a snowball can also be rewritten as
A=4π(D÷2)²
A=4π(D²÷4)
A=πD²
Given that the surface area decreases by 1cm²/min, so,
dA÷dt=-1cm²/min.
Differentiating both sides in A=πD² and get
dA÷dt=2πD(dD÷dt)
Given that the diameter is 12cm.
Substitute these values in the formula and find dD÷dt
-1=2π×12×(dD÷dt)
-1=24π×(dD÷dt)
-1÷24π=dD÷dt
-0.0159≈dD÷dt
Hence, the diameter decreases at the rate of 0.0159 cm/min when the diameter is 12 cm.
Learn about the surface area from here brainly.com/question/15169645
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