Question involving mass/weight and upper/lower bounds

[Simonsky]

Here is the question it is no. 10 below:



The first issue that bugged me was the fact that it gives the weight of the cylinder and not the mass, so if you want to work out the volume (which I think you need to do to find the height using Pi r^2 h, you need to use Density = M/V. I can't believe the question really wants you to convert weight to mass which means using a gravitational field strength of 9.8N/kg!!!

So I started by trying to find the volume: Density = 19.3 g/cm^3 = 1200/V . 1200/19.3 = 62.2 cm^3(1 dp)

Now the first bit asks to find the the upper bound on the height of the cylinder. So the upper bound of the measurements that have been rounded to 1dp will be:

Weight= 1250 g. Density= 19.35 g/cm^3. So volume will be : 1250/19.35 = 64.6(1dp) so using the volume formula we get:

Pi x 2.5(radius upper bound) x h = 64.6 = 64.6/(Pi x 2.5) =8.28. BUT the answer given is 5.4 - I've obviously made a blunder I cannot see!

 

[Dr.Peterson]

Here is the question it is no. 10 below:
The first issue that bugged me was the fact that it gives the weight of the cylinder and not the mass, so if you want to work out the volume (which I think you need to do to find the height using Pi r^2 h, you need to use Density = M/V. I can't believe the question really wants you to convert weight to mass which means using a gravitational field strength of 9.8N/kg!!!

So I started by trying to find the volume: Density = 19.3 g/cm^3 = 1200/V . 1200/19.3 = 62.2 cm^3(1 dp)

Now the first bit asks to find the the upper bound on the height of the cylinder. So the upper bound of the measurements that have been rounded to 1dp will be:

Weight= 1250 g. Density= 19.35 g/cm^3. So volume will be : 1250/19.35 = 64.6(1dp) so using the volume formula we get:

Pi x 2.5(radius upper bound) x h = 64.6 = 64.6/(Pi x 2.5) =8.28. BUT the answer given is 5.4 - I've obviously made a blunder I cannot see!

I don't get 5.4 either. I get 10.6.

But to get the upper bound, you want to use the upper bound of values in the numerator, and the LOWER bound of numbers in the denominator. Do you see why?

Also, the radius is given as 2.0, so the upper bound would be 2.05, wouldn't it? Why did you use 2.5?

 

[Simonsky]

I don't get 5.4 either. I get 10.6.

But to get the upper bound, you want to use the upper bound of values in the numerator, and the LOWER bound of numbers in the denominator. Do you see why?

Also, the radius is given as 2.0, so the upper bound would be 2.05, wouldn't it? Why did you use 2.5?

Thanks for answering, Dr. Peterson. It might well be that the text book is wrong (again!).

To be honest, I can't immediately grasp why you use the upper bound in the numerator and the lower in the denominator-definite lack of intuition there! If you can give me some pointers I'd be grateful.

I tend to get a bit flakey around the edges with this upper and lower bound decimal place thing. I thought of 2.0 as being rounded to 2 rather than 2.0 I guess, so was my mistake to not count the .O bit? Oh well, more of a reminder that Maths is still a bit of a foreign language to me!

Also-what about the mass/weight issue I mentioned at the beginning-is the book making an error conflating mass with weight?

Thanks for you time and help as always.

 

[JeffM]

Thanks for answering, Dr. Peterson. It might well be that the text book is wrong (again!).

To be honest, I can't immediately grasp why you use the upper bound in the numerator and the lower in the denominator-definite lack of intuition there! If you can give me some pointers I'd be grateful.

I tend to get a bit flakey around the edges with this upper and lower bound decimal place thing. I thought of 2.0 as being rounded to 2 rather than 2.0 I guess, so was my mistake to not count the .O bit? Oh well, more of a reminder that Maths is still a bit of a foreign language to me!

Also-what about the mass/weight issue I mentioned at the beginning-is the book making an error conflating mass with weight?

Thanks for you time and help as always.

I am not going to try to answer any questions about physics: my physics instructor in high school induced a loathing of the subject in me. Let's work with a numeric example.

I want to calculate

\(\displaystyle \dfrac{a}{b} \text { given } 29.95 \le a \le 30.05 \text { and } 1.95\le b \le 2.05\)

Now let's calculate five quotients.

\(\displaystyle \dfrac{30.0}{2.0} = 15.0\) seems intuitive, but that .0 is deceptive.

\(\displaystyle 15.0 < \dfrac{29.95}{1.95} < 15.359.\)

\(\displaystyle 14.6585 < \dfrac{30.05}{2.05} < 15.0.\)

\(\displaystyle 15.0< \dfrac{30.05}{1.95} < 15.411.\)

\(\displaystyle 14.609 < \dfrac{29.95}{2.05} < 15.0\)

\(\displaystyle \therefore 14.609 < \dfrac{a}{b} < 15.411.\)

When I reduce (increase) the absolute value of the denominator, I increase (reduce) the absolute value of the quotient. When I reduce (increase) the absolute value of the numerator, I reduce (increase) the absolute value of the quotient. Therefore, when I want determine the greatest possible change in the absolute value of the quotient, I have to use opposite changes in the numerator and denominator so they both affect the quotient in the same direction instead of partially offsetting.

 

[Dr.Peterson]

Thanks for answering, Dr. Peterson. It might well be that the text book is wrong (again!).

To be honest, I can't immediately grasp why you use the upper bound in the numerator and the lower in the denominator-definite lack of intuition there! If you can give me some pointers I'd be grateful.

I tend to get a bit flakey around the edges with this upper and lower bound decimal place thing. I thought of 2.0 as being rounded to 2 rather than 2.0 I guess, so was my mistake to not count the .O bit? Oh well, more of a reminder that Maths is still a bit of a foreign language to me!

Also-what about the mass/weight issue I mentioned at the beginning-is the book making an error conflating mass with weight?

Thanks for you time and help as always.

When they say it "weighs 1.7 kg", they just mean that weighs as much (under the same conditions) as a mass of 1.7 kg; in other words, its mass is 1.7 kg. They can't be talking about weight in the technical sense, because that is not measured in kg. So ignore this issue.

Suppose I had the fraction 5/7. If I increase the 5, or decrease the 7, I will increase the quotient. So that is what I want to do to find the largest possible result.

 

[Simonsky]

When they say it "weighs 1.7 kg", they just mean that weighs as much (under the same conditions) as a mass of 1.7 kg; in other words, its mass is 1.7 kg. They can't be talking about weight in the technical sense, because that is not measured in kg. So ignore this issue.

Suppose I had the fraction 5/7. If I increase the 5, or decrease the 7, I will increase the quotient. So that is what I want to do to find the largest possible result.

Yes, I get it! Thanks so much to both of you (Jeff+ Dr.P) for the help. It's funny how I just applied the upper bound number to the denominator mechanically-it's as if one's thinking blindly does something which is not real thinking at all!

 

[Dr.Peterson]

Yes, I get it! Thanks so much to both of you (Jeff+ Dr.P) for the help. It's funny how I just applied the upper bound number to the denominator mechanically-it's as if one's thinking blindly does something which is not real thinking at all!

Yes, thinking without thinking is not thinking. A lot of students need time to learn how to distinguish thinking from its imitators; I sometimes think that this is one of the main purposes of teaching math to many people. Unfortunately, too many math classes only teach mechanical "thinking".