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Hi everyone. Is everything okay? How are you getting on? Why are we so busy and have very little time these days?
I'll need help and going over of some work I'm solving. Here goes :
1. A solid cylinder of height 10 cm and radius 4 cm is to be plated with material costing £ 11 per cm^3. Find the cost of the plating.
2. A tin of paint covers a surface area of 60 m^2 and cost £ 4.50. Find the cost of painting the outside surface of a cylinder gas holder of height 30 m and radius 18 m. The top of the gas holder is a flat circle.
For #1, I have the volume to be approximately 503 cm^3. Then, using ratio and proportion, the cost is £ 5533.
#2, I used the formula for finding Total Surface Area and my answer is approximately 5431 m^2. Again, with ratio and proportion, the cost iw £ 407.33.
Am I doing the right thing?
Confused about the next one.
3. A sphere with radius 6 cm fit exactly inside a cylinder.
a. Write
(i) the radius (ii) the height of the cylinder.
b. Work out the surface area of the sphere.
c. Work out the curved surface area of the cylinder.
d. For a sphere radius r that fits exactly inside a cylinder, find an expression for
(i) the surface area of the sphere
(ii) the curved surface area of the cylinder.
e. Explain why youranswers to (b) and (c) must be the same. You may want to use your answer to part d to help you.
For #3, I know the formulas but I'm not sure which values to compute for the solution, like finding the radius and height of the cylinder.
Thanks
Only a friend tells you your face is dirty.
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Is there a formula to use for # 1 and 2, without using the ratio and proportion method? Thanks
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Yes.
The formula for the curved surface of a cylinder (imagine it 'unfolded' to make a rectangle) is
and for the two ends, two lots of
So add up the total surface area and times by the cost per square not cube centimetre.
I get 407 for Q2.
Q3. If the sphere exactly fits inside the cylinder then they have the same diameter (and radius) and the height of the cylinder must also be the diameter of the sphere.
Children are not defined by school ...........The Fonz
You cannot teach a man anything; you can only help him find it within himself..........Galileo Galilei
Sometimes I deliberately make mistakes, just to test you! …………….Bob
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Thanks Bob for the explanations.
May I ask, your answer 407, is it an exact or approximated value?
Actually, I didn't use the exact pi constant but rather, 22/7. Could that be the reason for our different answers?
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I had rounded it off. 22/7 would make a slightly different answer, but it would come to the same when rounded. If you think about the problem in terms of accuracy, it makes sense to round things. The height and radius are given as integers. As it is impossible to measure something with absolute accuracy you should take into account what difference a small change like 18.1 rather than 18 would make. And paint coverage introduces a huge potential inaccuracy. The exact coverage will depend on how much the painter has on his brush; that figure will always be only an estimate.
In GCSE (UK exam) it says in the syllabus that candidates should round off appropriately and markers are given a range of answers to take account of this. Just because your calculator can work to (say) 10 figures, doesn't mean you have to declare them all when giving an answer.
Bob
Children are not defined by school ...........The Fonz
You cannot teach a man anything; you can only help him find it within himself..........Galileo Galilei
Sometimes I deliberately make mistakes, just to test you! …………….Bob
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Thanks Bob for that.
By the way, how do you do?
Only a friend tells you your face is dirty.
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How do you do?
Tres bien, mon ami. But you've been a member since 31st March, so isn't this a bit overdue?
The question is rarely used in England now. People usually say "How are you?" They don't usually expect an answer
Bob
Children are not defined by school ...........The Fonz
You cannot teach a man anything; you can only help him find it within himself..........Galileo Galilei
Sometimes I deliberately make mistakes, just to test you! …………….Bob
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Hi;
I was going through some questions and I can't find my way round this one. Need some help.
The radii of the base of two cylindrical tins, P and Q are r and 2r, respectively. If the water level in P is 10 cm high, what would be the height of the same quantity of water in Q?
The answer is given as 2.5 cm. I don't see how this is so.
Thanks.
Only a friend tells you your face is dirty.
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Hi;
The volume of P is
The volume of Q is
Set them equal to each other to find the new height h.
Divide both sides by
In mathematics, you don't understand things. You just get used to them.
If it ain't broke, fix it until it is.
Always satisfy the Prime Directive of getting the right answer above all else.
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Thanks. I get it now.
Only a friend tells you your face is dirty.
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It is sometimes hard to understand what the math is saying. Concepts and conclusions are very counterintuitive.
In mathematics, you don't understand things. You just get used to them.
If it ain't broke, fix it until it is.
Always satisfy the Prime Directive of getting the right answer above all else.
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