- #1
ktpr2
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Given, "if a snowball melts so that its surface area decreases at a rate of 1 cm^2/min, find the rate at which the diameter decreases when the diameter is 10 cm."
[tex]\frac{dD}{dt} = \frac{dD}{dA} \frac{dA}{dt}[/tex] were D is my diameter, A is (surface) area and t is time.
I relate D to A by, [tex]A = 4pi(\frac{D}{2})^2 = piD^2; A' = 2piD[/tex] and then solve,
[tex]\frac{dD}{dt} = 2piD * -1 cm^2/min =[/tex]
[tex]2pi(10cm) * -1cm^2/min = -20 cm^3/min[/tex]
I know I am wrong because I got volume, not length. When I solve for variables, like A, and differentiate, do i always place result where ever A shows up in the differentiation equation? In this case, dD/dA, so I should write 1/(2piD)?
The book give the answer as [tex]\frac{1}{20 cm/min}[/tex]
[tex]\frac{dD}{dt} = \frac{dD}{dA} \frac{dA}{dt}[/tex] were D is my diameter, A is (surface) area and t is time.
I relate D to A by, [tex]A = 4pi(\frac{D}{2})^2 = piD^2; A' = 2piD[/tex] and then solve,
[tex]\frac{dD}{dt} = 2piD * -1 cm^2/min =[/tex]
[tex]2pi(10cm) * -1cm^2/min = -20 cm^3/min[/tex]
I know I am wrong because I got volume, not length. When I solve for variables, like A, and differentiate, do i always place result where ever A shows up in the differentiation equation? In this case, dD/dA, so I should write 1/(2piD)?
The book give the answer as [tex]\frac{1}{20 cm/min}[/tex]