Using quadratic equation to find diameter of base of cylinder

[Simonsky]

The question:

The height of a closed cylinder is 5cm and its surface area is 100cm^2

Find the diameter of the base of the cylinder.

What I've tried so far: Take the formula for surface area of cylinder: 2Pir^2 +5(2Pir) =100. I then rearranged it to: 2Pir^2 +10Pir -100 =0 I then thougjht I could substitute 'x' for Pir ; 2x^2 +10x -100 = 0.

But then I though: Is this legitimate as Pi x r^2 isn't the same as (Pir)^2.

Might need a pointer here -thanks.

 

[Dr.Peterson]

The question:

The height of a closed cylinder is 5cm and its surface area is 100cm^2

Find the diameter of the base of the cylinder.

What I've tried so far: Take the formula for surface area of cylinder: 2Pir^2 +5(2Pir) =100. I then rearranged it to: 2Pir^2 +10Pir -100 =0 I then thougjht I could substitute 'x' for Pir ; 2x^2 +10x -100 = 0.

But then I though: Is this legitimate as Pi x r^2 isn't the same as (Pir)^2.

Might need a pointer here -thanks.

You're right -- that isn't valid. That is, replacing pi r with x will not eliminate all pi's from the equation. So I would leave it as it is, using the quadratic formula with a=2pi, b=10pi, and c=-100. The answer won't be pretty.

Alternatively, you could divide all terms by 2pi; but that might make it a little uglier. You're being asked for a decimal answer anyway, right?

 

[sinx]

The question:

The height of a closed cylinder is 5cm and its surface area is 100cm^2

Find the diameter of the base of the cylinder.

What I've tried so far: Take the formula for surface area of cylinder: 2Pir^2 +5(2Pir) =100. I then rearranged it to: 2Pir^2 +10Pir -100 =0 I then thougjht I could substitute 'x' for Pir ; 2x^2 +10x -100 = 0.

But then I though: Is this legitimate as Pi x r^2 isn't the same as (Pir)^2.

Might need a pointer here -thanks.

if i read the problem correctly, the surface area of a cylinder should be circumference x ht; i.e.
surface area = 2(pi)R x ht.
or/ surface area =Dia.(pi) x ht=100
[(pi)r² x ht = volume]

 

[Dr.Peterson]

if i read the problem correctly, the surface area of a cylinder should be circumference x ht; i.e.
surface area = 2(pi)R x ht.
or/ surface area =Dia.(pi) x ht=100
[(pi)r² x ht = volume]

That would be the lateral surface area. Simonsky is using the total surface area of a closed cylinder (that is, including both ends).

 

[Simonsky]

You're right -- that isn't valid. That is, replacing pi r with x will not eliminate all pi's from the equation. So I would leave it as it is, using the quadratic formula with a=2pi, b=10pi, and c=-100. The answer won't be pretty.

Alternatively, you could divide all terms by 2pi; but that might make it a little uglier. You're being asked for a decimal answer anyway, right?

Thanks. For some reason I didn't think of the Coefficient as the integer x Pi. Yet again an example of how I notice my thinking is very rigid and not flexible enough, as yet, to interpret the question. because I'm used to seeing, generally, an integer as a coefficient, I dissociated from the Pi.

I got the right answer but as you said, the numbers looked cumbersome.

Thanks.

 

[Dr.Peterson]

Thanks. For some reason I didn't think of the Coefficient as the integer x Pi. Yet again an example of how I notice my thinking is very rigid and not flexible enough, as yet, to interpret the question. because I'm used to seeing, generally, an integer as a coefficient, I dissociated from the Pi.

I got the right answer but as you said, the numbers looked cumbersome.

We tend to give examples that are simpler than the definitions allow, which leads to that kind of issue!

A coefficient, in the most general sense, is whatever is multiplying something. In a polynomial, it's everything in a term other than the power of the variable. But textbooks often use "coefficient" to mean "numerical coefficient" (as if the coefficient has to be a number), and give only examples where it's an "integer coefficient". You could have a term like πb√(3)x⁴, where the coefficient of the variable is all of πb√(3).

We also tend to give examples where the answers are neat simple numbers or expressions, whereas in real life most numbers look random.

I guess we need to make things hard more quickly than we do. Tough love ...