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ElijahRockers
Gold Member
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EDIT:
I just found this thread, which handily has the same exact problem. The OP says that dh is = 0 though, and I don't quite understand why he does that.
Use differentials to estimate the amount of tin in a closed tin can with diameter 8 cm and height 12 cm if the tin is 0.04 cm thick. (Round the answer to two decimal places.)
I wasn't really sure how to start this one, but I took a stab at it.
[itex]V=\pi*r^{2}*h[/itex] where r = D/2. I then found the differential of this.
[itex]dV=\frac{\pi}{4}D^{2}*dh + \frac{\pi}{2}hD*dD[/itex]
I put in 8 for D, 12 for h, and .04 for dH and dD. I got 8.04cm3, but that is apparently wrong, according to the homework.
So I tried using the area instead.
[itex]A=hD\pi + \frac{\pi}{2}D^{2}*dD[/itex]
Differentiating, I get
[itex]dA=\pi*Ddh + \pi(h+D)*dD[/itex]
I put in 8 for D, 12 for h, and .04 for dH and dD. I got 3.52, but this answer is also incorrect.
Under normal circumstances I would calculate the area of the can and multiply by .04, but my gut feeling tells me this is wrong as well. Am I supposed to multiply dA by dV?
I only have one chance left to get it right, so I was hoping to figure it out here first. Thanks.
I just found this thread, which handily has the same exact problem. The OP says that dh is = 0 though, and I don't quite understand why he does that.
Homework Statement
Use differentials to estimate the amount of tin in a closed tin can with diameter 8 cm and height 12 cm if the tin is 0.04 cm thick. (Round the answer to two decimal places.)
The Attempt at a Solution
I wasn't really sure how to start this one, but I took a stab at it.
[itex]V=\pi*r^{2}*h[/itex] where r = D/2. I then found the differential of this.
[itex]dV=\frac{\pi}{4}D^{2}*dh + \frac{\pi}{2}hD*dD[/itex]
I put in 8 for D, 12 for h, and .04 for dH and dD. I got 8.04cm3, but that is apparently wrong, according to the homework.
So I tried using the area instead.
[itex]A=hD\pi + \frac{\pi}{2}D^{2}*dD[/itex]
Differentiating, I get
[itex]dA=\pi*Ddh + \pi(h+D)*dD[/itex]
I put in 8 for D, 12 for h, and .04 for dH and dD. I got 3.52, but this answer is also incorrect.
Under normal circumstances I would calculate the area of the can and multiply by .04, but my gut feeling tells me this is wrong as well. Am I supposed to multiply dA by dV?
I only have one chance left to get it right, so I was hoping to figure it out here first. Thanks.
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