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  • 2 years ago

The radius, r, of the base of a circular cylinder increases by 2 feet per second while the height, h, decreases by 1 foot per second. How fast is the surface area of the cylinder changing when the height of the cylinder is 50 feet and the radius of the base is 40 feet? I got \(440\pi\) square feet per second, am I right?

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  1. geerky42
    • 2 years ago
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    My work: __________________________________________________________________ Since it is known that surface area of this cylinder is \(SA = 2\pi r^2 + 2\pi rh\)\[\dfrac{dSA}{dt} = 4\pi r\dfrac{dr}{dt} + 2\pi \dfrac{dr}{dt} +2\pi r \dfrac{dh}{dt}\]We are given that r = 40, h = 50, \(\dfrac{dr}{dt} = 2\), \(\dfrac{dh}{dt} = -1\) I plugged in these values and I got \(\boxed{440\pi}\) __________________________________________________________________ But there is no option for \(440\pi\), I'm not sure if there is typo in options or I made mistake somewhere in my work...

  2. klimenkov
    • 2 years ago
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    Your solution seems to be okay, (but you forgot \(h\) in the second addend). Maybe there is a typo or the surface area is computed without bottom and top of the cylinder.. Better to ask the person that has gived you this task.

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