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Differentiation of surface area of a right cylinder

  • Thread starter tmlrlz
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  • #1
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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.
a) Find the rate of change of A with respect to h if r remains constant.
b) Find the rate of change of A with respect to r if h remains constant.
c) Find the rate of change of h with respect to r if A remains constant.


Homework Equations


A = 2pir(r + h)


The Attempt at a Solution


I thought i had this all right but then realized that what i had done for part a was wrong, i think that part b and c are right but i will just put my answers up to check.
a) A = 2pir(r + h)
r = c (constant)
A = 2pic(c + h)
A = 2pi(c2 + ch)
A/2pi = c2 + ch
however, i do not know at this point, how i would isolate c!
What i did was bring the A/2pi over and get c2 + ch - A/2pi = 0 and use the quadratic formula. From that i got that c = (h +- h)/4pi and reason that c cannot be zero or else the surface area would be zero so c must be h/2pi. I think that this process of my work is wrong because i dont think you can use the quadratic formula and solve for c because c is a constant.
We know that
A = 2pic(c + h)
A = 2pic2 + 2pich
dA/dh = 2pic = 2pi(h/2pi) = h
As you can see i dont think that what i did was correct, please help me with this one, a) is what i need most help with

b) h = c (constant)
A = 2pir(r + c)
A/2pir = r + c
c = (A/2pir) - r
dA/dr = 2pi(r + c) + 2pir = 4pir + 2pic = 4pir + 2pi((A/2pir) - r) = 4pir + A/r - r

c) A = c (constant)
c = 2pir(r + h)
c/2pir - r = h
dh/dr = (-c/2pir2) - 1
= (-(2pir(r + h))/2pir2) - 1
= ((- r - h) / r) - 1
= (-2r - h) / r
 

Answers and Replies

  • #2
33,270
4,966

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.
a) Find the rate of change of A with respect to h if r remains constant.
b) Find the rate of change of A with respect to r if h remains constant.
c) Find the rate of change of h with respect to r if A remains constant.


Homework Equations


A = 2pir(r + h)


The Attempt at a Solution


I thought i had this all right but then realized that what i had done for part a was wrong, i think that part b and c are right but i will just put my answers up to check.
a) A = 2pir(r + h)
r = c (constant)
A = 2pic(c + h)
A = 2pi(c2 + ch)
A/2pi = c2 + ch
however, i do not know at this point, how i would isolate c!
What i did was bring the A/2pi over and get c2 + ch - A/2pi = 0 and use the quadratic formula. From that i got that c = (h +- h)/4pi and reason that c cannot be zero or else the surface area would be zero so c must be h/2pi. I think that this process of my work is wrong because i dont think you can use the quadratic formula and solve for c because c is a constant.
We know that
A = 2pic(c + h)
A = 2pic2 + 2pich
dA/dh = 2pic = 2pi(h/2pi) = h
As you can see i dont think that what i did was correct, please help me with this one, a) is what i need most help with
Nope, not right, and it's incorrect from almost the very beginning. What they're looking for is dA/dh, or if you know about partial derivatives, [itex]\frac{\partial A}{\partial h}[/itex].


b) h = c (constant)
A = 2pir(r + c)
A/2pir = r + c
c = (A/2pir) - r
dA/dr = 2pi(r + c) + 2pir = 4pir + 2pic = 4pir + 2pi((A/2pir) - r) = 4pir + A/r - r

c) A = c (constant)
c = 2pir(r + h)
c/2pir - r = h
dh/dr = (-c/2pir2) - 1
= (-(2pir(r + h))/2pir2) - 1
= ((- r - h) / r) - 1
= (-2r - h) / r
b and c are incorrect, as well.

For a, suppose you had A = 2π*6(h + 6). What would dA/dh be? If it was A = 2πk(h + k), with k a constant, what would dA/dh be?

For b, suppose you had A = 2πr(r + 3). What would dA/dr be? If is was A = 2πr(r + k), what would dA/dr be?
 
  • #3
29
0
Nope, not right, and it's incorrect from almost the very beginning. What they're looking for is dA/dh, or if you know about partial derivatives, [itex]\frac{\partial A}{\partial h}[/itex].




b and c are incorrect, as well.

For a, suppose you had A = 2π*6(h + 6). What would dA/dh be? If it was A = 2πk(h + k), with k a constant, what would dA/dh be?

For b, suppose you had A = 2πr(r + 3). What would dA/dr be? If is was A = 2πr(r + k), what would dA/dr be?
A = (2pi * 6)(h+6)
= (12pi)(h + 6)
= 12pih + 72pi
dA/dh = 12pi

A = (2pi * k) (h + k)
= 2pikh + 2pik2
dA/dh = 2pi * k

A = 2pir(r + 3)
= 2pir2 + 6pir
da/dr = 4pir + 6pi

A = 2pir(r+k)
= 2pir2 + 2pirk
dA/dr = 4pir + 2pik

right?
 
  • #4
HallsofIvy
Science Advisor
Homework Helper
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A = (2pi * 6)(h+6)
= (12pi)(h + 6)
= 12pih + 72pi
dA/dh = 12pi
Where in the world did the "6" come from?

A = (2pi * k) (h + k)
= 2pikh + 2pik2
dA/dh = 2pi * k
and now where did "k" come from?

A = 2pir(r + 3)
= 2pir2 + 6pir
da/dr = 4pir + 6pi
where did this "3" come from?

A = 2pir(r+k)
= 2pir2 + 2pirk
dA/dr = 4pir + 2pik
where did "k" come from?

right?
Why not just use the original variables "r" and "h" rather than introducing numbers or other symbols?
 
  • #5
29
0
Where in the world did the "6" come from?


and now where did "k" come from?


where did this "3" come from?


where did "k" come from?


Why not just use the original variables "r" and "h" rather than introducing numbers or other symbols?
but isn't that what i did in my originial answer, the only difference was that i was trying to isolate r or h so that when i got the derivative i could plug their original values back in to the derivative to simplify the derivative.
 
  • #6
33,270
4,966
Where in the world did the "6" come from?

and now where did "k" come from?
where did this "3" come from?
where did "k" come from?

Why not just use the original variables "r" and "h" rather than introducing numbers or other symbols?
They came from the revised problems that I suggested in post 2.

but isn't that what i did in my originial answer, the only difference was that i was trying to isolate r or h so that when i got the derivative i could plug their original values back in to the derivative to simplify the derivative.
 
  • #7
33,270
4,966
A = (2pi * 6)(h+6)
= (12pi)(h + 6)
= 12pih + 72pi
dA/dh = 12pi
Right. So in the original problem, with r considered to be constant, you have A = 2πr(h + r), so dA/dh = 2πr.

A = (2pi * k) (h + k)
= 2pikh + 2pik2
dA/dh = 2pi * k
Right.

A = 2pir(r + 3)
= 2pir2 + 6pir
da/dr = 4pir + 6pi
Should be dA/dr, not da/dr, but otherwise OK.
A = 2pir(r+k)
= 2pir2 + 2pirk
dA/dr = 4pir + 2pik
Both fine. Now if A = 2πr(r + h), and h is considered to be a constant, what is dA/dr?
 

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