anonymous
  • anonymous
Can someone help me find the value of r in this equation: 2 x pi x r(.5) + 4pi r^2= pi
Precalculus
  • Stacey Warren - Expert brainly.com
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SOLVED
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chestercat
  • chestercat
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anonymous
  • anonymous
Is this what you equation looks like? \(2\pi r(.5) + 4\pi r^2 = \pi\)
anonymous
  • anonymous
yess
anonymous
  • anonymous
OK. In the first term, the constants can be combined.

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anonymous
  • anonymous
Do you know how to do that?
anonymous
  • anonymous
it would be πr+ 4πr^2 right?
anonymous
  • anonymous
sorry i'm new to this so i'm not sure how to add the pi sign and such
anonymous
  • anonymous
You can use the equation editor or learn \(\LaTeX\). The minimal amount of \(\LaTeX\) needed here is easy to learn. What is \(2 \times 0.5\)?
anonymous
  • anonymous
Nevermind, I'm an idiot and you are correct.
anonymous
  • anonymous
So you have \(\pi + 4\pi r^2=\pi\) correct?
anonymous
  • anonymous
pi r
anonymous
  • anonymous
http://openstudy.com/study#/groups/LaTeX%20Practicing!%20%3A) For help learning \(\LaTeX\) if you so desire.
anonymous
  • anonymous
alright thanks
anonymous
  • anonymous
If you divide the entire equation by \(\pi\), what do you get?
anonymous
  • anonymous
r+4r^2=1
anonymous
  • anonymous
At this point, I would move everything to the left side and factor (or use the quadratic formula).
anonymous
  • anonymous
Unless I've made a stupid mistake, this answer is going to have a radical in it.
anonymous
  • anonymous
uh oh, this is part of a larger question and r means radius... but the work makes sense so thanks for your help!
anonymous
  • anonymous
OK. Feel free to ask another question if you can't work it out. :-)
anonymous
  • anonymous
do you mind if i explain the bigger question? maybe i set this up wrong
anonymous
  • anonymous
Sounds like a good idea.
anonymous
  • anonymous
overview: we're working with a vitamin c tablet thats the shape of a right circular cyinder w a hemisphere attached at each end
anonymous
  • anonymous
one brand of tablets has a total length of 2 cm and a diameter of .5 cm. i found the surface area for it, which i think is \[\pi\]
anonymous
  • anonymous
now i have a tablet in which the cylinder has an altitude of .5 cm. find the diameter of the hemispheres of the new tablet if the surface area is the same as the first one
anonymous
  • anonymous
By total length, do you mean height? And what do you mean by altitude? If you could draw a picture using the draw button below, it may make more sense to me.
anonymous
  • anonymous
i found all that confusing too. total length of the cylinder and hemispheres. altitude means height of the cylinder i think|dw:1378090929678:dw|
anonymous
  • anonymous
|dw:1378091069129:dw|
anonymous
  • anonymous
Pill-1: diameter = 0.5; height = 2; Surface Area = circumference of the circle times height plus surface area of the circle; Pill-2: height = 0.5; Surface Area = Surface Area of pill-1; How did you find the surface area of pill-1?
anonymous
  • anonymous
i used the formula: 2 \[\pi \] r h + 4\[\pi \]r^2 where r is (1/4) and h is (1.5) because total means the cylinder and the two hemispheres so i had to take out the height of the hemis from the whole (did that make any sense?)
anonymous
  • anonymous
I'm thinking you have a cylinder and a sphere (two hemispheres make a whole sphere) so... \(SA_{Pill-1}=4\pi r^2+2\pi r^2+2\pi r h\) That is adding the formulas for surface area of a sphere and cylinder. Does that make sense?
anonymous
  • anonymous
Your right, take out the surface area of the circles....
anonymous
  • anonymous
i originally had the same equation as you but realized i didn't need the circles in it
anonymous
  • anonymous
\(SA_{Pill-1}=4\pi r^2+2\pi r h\) Substitute the correct values: diameter = 0.5 so r=0.25. \(SA_{Pill-1}=4\pi (0.25)^2+2\pi (0.25)(2)\) \(SA_{Pill-1}=0.25\pi + \pi\) \(SA_{Pill-1}=1.25\pi\)
anonymous
  • anonymous
Pill-1: diameter = 0.5; height = 2; Surface Area = \(1.25\pi\) Pill-2: height = 0.5; Surface Area = \(1.25\pi\); Do we both agree?
anonymous
  • anonymous
so "total length" should be used as the height? i don't need to take out the sphere to make it the height of the cylinder?
anonymous
  • anonymous
You are talking about a capsule and not a flat pill right? So, the surface area will be the surface area of the cylinder minus the area of the 2 circles on either end, plus the surface area of the sphere (one half on either end). And, you are correct about subtracting out the diameter of the circle. Total length would include the sphere. Back to the drawing board....
anonymous
  • anonymous
|dw:1378092434763:dw| Pill-1: diameter = 0.5; height = 1; Surface Area = 0.75π Pill-2: height = 0.5; Surface Area = 0.75π;
anonymous
  • anonymous
oh you subtract the diameter not the radius from the total length?
anonymous
  • anonymous
Right, because you have one half of the sphere on each end so the total length will include the full diameter.
anonymous
  • anonymous
\(SA_{Pill-2}=4\pi r^2+2\pi r h\) Plug in what we know for pill-2: \(0.75\pi=4\pi r^2 + 2\pi r\)
anonymous
  • anonymous
I didn't get a radical in this answer, so I'm more confident with it. :-)
anonymous
  • anonymous
okay thank you so much for all your help
anonymous
  • anonymous
Let me know what you end up with. I'm going over it one more time, just to be sure. I love math questions that make me think.... :-)
anonymous
  • anonymous
Re-check all the calculations - I think I've found a mistake...
anonymous
  • anonymous
i've been going over it with my dad. you would subtract the radius twice not diameter from the total length. i'm starting to think that we need to add the whole surface area equation of the cylinder because things aren't adding up
anonymous
  • anonymous
The radius twice is equal to the diameter, but that give a height of 1.5 instead of two - that ends up with a radical for an answer. Can you post a picture of the question? Maybe there is some small thing that has been missed?
anonymous
  • anonymous
yeah that's the issue. let me try to upload the assignment. again thank you for your help
anonymous
  • anonymous
1 Attachment
anonymous
  • anonymous
#5 has a typo. the sentence ends with "equal to the surface area of number 4"
anonymous
  • anonymous
OK. I'm not seeing anything obvious. I'll keep looking though.
anonymous
  • anonymous
thanks
anonymous
  • anonymous
It really depends on the definition of altitude, I think. That is not a word normally associated with an object.
anonymous
  • anonymous
yeah that threw me off too
anonymous
  • anonymous
Let me just clarify...a right circular cylinder is like a soup can right? the side is perpendicular to the end?
anonymous
  • anonymous
correct. and right i think means that the radius and height are perpendicular
anonymous
  • anonymous
|dw:1378097168919:dw|
anonymous
  • anonymous
They are talking about a tablet, but clearly describing a capsule. Those are two different shapes. Assuming a capsule is what is meant, there is no way the area of the circle is included in the surface area of the capsule since it is covered by the hemispheres.
anonymous
  • anonymous
so we set it up correctly
anonymous
  • anonymous
I believe we did. @DebbieG Can you look over our work here? Something isn't quite right...
anonymous
  • anonymous
Now I'm nitpicking because I can't find anything else to check... How do you know that #5 is supposed to end "equal to the surface area of number 4"?
anonymous
  • anonymous
my teacher posted it on the website that she forgot and she announced it in class.. let me double check that its surface area and not volume
anonymous
  • anonymous
yeah it's surface area
anonymous
  • anonymous
And it says #4 not some portion of #4?
anonymous
  • anonymous
OK. So, Pill-1 has a diameter of 0.5; radius of 0.25; height of 1.5(because 2-diameter ).5 = 1.5; \(SA=4\pi r^2+2\pi r h\) Substituting what we know: \(SA=4\pi (0.25)^2 + 2\pi (0.25)(1.5)\) \(SA=0.25\pi+0.75\pi\) \(SA=\pi\) Do we agree on this surface area?
anonymous
  • anonymous
yeah, that's what i did
anonymous
  • anonymous
Now for Pill-2: height = 0.5; SA = \(\pi\); \(SA=4\pi r^2+2\pi r h\) Substituting what we know: \(\pi=4\pi r^2+2\pi r(0.5)\) \(\pi=4\pi r^2 + \pi r\) divide through by \(\pi\) \(1=4r^2+r\) or \(4r^2+r-1=0\) Are we still in agreement?
anonymous
  • anonymous
correct
anonymous
  • anonymous
Using the quadratic formula: \(\dfrac{-b\pm \sqrt{b^2-4ac}}{2a}\) \(\dfrac{-1 \pm \sqrt{1-4(4)(-1)}}{2(4)}\) \(\dfrac{-1\pm\sqrt{1+16}}{8}\) \(\dfrac{-1\pm\sqrt{17}}{8}\) I can't find a way to avoid the radical - The negative answer doesn't make sense, so: \(\dfrac{-1+4.1231056}{8}\) \(\dfrac{3.1231056}{8}\) r = 0.3903882 This answer could make sense because it is larger than the radius of pill-1, but not by much and I'm having a hard time believing it.
anonymous
  • anonymous
that's the problem, the surface areas are supposed to be equal in some way. while i did the work this way (at least five times) i was thinking if there's a different way to solve it.
anonymous
  • anonymous
possibly ratios? or somehow using volume?
anonymous
  • anonymous
You are asked to find the volume of both pills. It is the surface areas that are to be equal...
anonymous
  • anonymous
correct
anonymous
  • anonymous
I just calculated the surface area of a pill with a radius of 0.5 and a height of 0.5 and got \(1.5\pi\). Now a radius of 0.5 is a bit larger than the radius we calculated for pill-2 and a larger surface area. Which means that even though the answer does not come out close to even as expected, it does make sense.
anonymous
  • anonymous
That answer has to be correct.
anonymous
  • anonymous
your reasonin does make sense. i feel like there's something wrong with the assignment because we've been working on this for too long
anonymous
  • anonymous
We've been working on it for so long because we didn't believe the answer. There are problems with the assignment though, for example the pill described is a capsule and not a tablet. My only major concern is that there is no instruction to round to a certain place value or mention of significant digits. These questions usually come with some instructions on the format of the answer.
anonymous
  • anonymous
I would definitely talk to the teacher about this. Maybe there is a standard for the class, that was not mentioned in this assignment, like rounding to the nearest hundredth or something like that.
anonymous
  • anonymous
i wish i could talk to my teacher. it's a precalc class and we're given assignments worth 20 points that are preparing us for calc. it's due tuesday so i can't ask her
anonymous
  • anonymous
Maybe ask when you turn in your assignment so you will know for next time. It's hard to read the teacher's mind sometimes...
anonymous
  • anonymous
you're right. thank you for all your help!
anonymous
  • anonymous
You're welcome.
DebbieG
  • DebbieG
@gypsy1274, I got the same answer, \(\Large r=\dfrac{-1+\sqrt{17}}{8}\approx0.39039\). I guess I'm not clear on why you guys are 2nd-guessing that? :) All the work appears to be correct to me. There is no reason that the answer can't have a radical in it, and it makes the radius of the 2nd pill larger than that of the first by more than 50%. So the diameter of the 2nd tablet would be \(\Large d=2\cdot\dfrac{-1+\sqrt{17}}{8}=\dfrac{-1+\sqrt{17}}{4}\) You could give the approximate answer too, but no need to. This answer is exact. :)
DebbieG
  • DebbieG
I will add though, it isn't hard (and what a sinking feeling!) to write a problem, and not realize until you have asked students to do it, that it doesn't work out "correctly" or the way you intended it to. I'll never quite get over having asked them to prove an identity on a trig exam, that turned out not to be true..... OOPS! lol
anonymous
  • anonymous
Thanks Debbie. My reason for second guessing the answer was that there were no instructions for rounding. In my classes we were always given instructions on the format of the answer. (exact or estimated) I've made similar mistakes in tutoring and pointed them out a "This is why you always check your work!" It usually makes my students smile.
DebbieG
  • DebbieG
Haha.... indeed, ALWAYS check your work! :) I like to stress that, unless a decimal approximation is clearly needed or asked for, exact answers should be the "default" because, well.... they are EXACT. Unfortunately, some students think that "exact answer" means "write down alllllllllllll the digits that are after the decimal point on my calculator display." ;)
anonymous
  • anonymous
thank you for all your help. we did it all correctly and i got a good grade on the assignment
anonymous
  • anonymous
Great! Thanks for the feedback. It's much appreciated.

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