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myria36
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myria36
myria36 is Online:
Posts: 1
calculus - derivatives
1. Homework Statement
the total surface area of a right circular cylinder is given by the formula: (A = 2Pir(r + h) ).
where r is the radius and h is the height.
sub part a) find the rate of change of A with respect to h is r remains constant
**i know how to take derivatives. the only thing is that in this case, I am not sure how to take the derivative of h since it is only present in one term.
2. Homework Equations
the derivative equation
dA / dr
3. The Attempt at a Solution
i first distributed the 2pir, to yield
2pir^2 + 2pirh
2pir^2(h/h) + 2pirh -- i added the (h/h), which is like multiplying by 1, to add h to the first term- I am not sure if this is correct, but i was just guessing.
h (2pir^2 h^-1 + 2pi r)
now i am stuck here. i can't take the derivative of all the h's in my problem, because one h is still present in the equation.
**below is my attempt to still work with it.
dA/dh = 1 times [-1(2pir^2h^-2) + 2pir
final answer: (-2pir^2h^-2) + 2pir
please can someone guide me on the technique i should use for getting the area to be in terms of h. any and all replies are welcome and appreciated
myria36
myria36 is Online:
Posts: 1
calculus - derivatives
1. Homework Statement
the total surface area of a right circular cylinder is given by the formula: (A = 2Pir(r + h) ).
where r is the radius and h is the height.
sub part a) find the rate of change of A with respect to h is r remains constant
**i know how to take derivatives. the only thing is that in this case, I am not sure how to take the derivative of h since it is only present in one term.
2. Homework Equations
the derivative equation
dA / dr
3. The Attempt at a Solution
i first distributed the 2pir, to yield
2pir^2 + 2pirh
2pir^2(h/h) + 2pirh -- i added the (h/h), which is like multiplying by 1, to add h to the first term- I am not sure if this is correct, but i was just guessing.
h (2pir^2 h^-1 + 2pi r)
now i am stuck here. i can't take the derivative of all the h's in my problem, because one h is still present in the equation.
**below is my attempt to still work with it.
dA/dh = 1 times [-1(2pir^2h^-2) + 2pir
final answer: (-2pir^2h^-2) + 2pir
please can someone guide me on the technique i should use for getting the area to be in terms of h. any and all replies are welcome and appreciated