Chapter 36 Assorted problems
Solutions to problems which have yet to be categorised.
36.0.1 Cones problem
Problem: Two similar cones, and , have surface areas and respectively.
The volume of cone is .
Show that the volume of cone Y is .
We have two cones, let’s let be this one, with radius and height
Then we can draw (which is smaller than , but not drawn to scale here). As they are similar cones, the radius of is equal to (where is the scale factor between the two cones) and the height of is equal to
First, the volume of a cone is and the surface area of a cone is . These can be derived using integration (but not in Single Maths A Level at least).
Using this, we can write this equation relating the radius and height of to the value given in the question
We can also do the same for
This can be simplified a bit by factoring the from the square rooted part of the equation, giving
and thus that
We can solve this for by dividing the two equations, giving that
and thus, after cancelling, that
which means that
Given this, we can then turn to the volume formula.
We know that the volume of is
and the formula for the volume of is equal to
Thus, by multiplying Equation 36.8 through by , we have
The left-hand side is just the volume of , and the right-hand side is just
Which is what we needed to show.
36.0.2 Divisibility of
When we represent 789 in decimal, we mean 7 hundreds plus 8 tens plus 9 units. If is taken to represent a decimal number (not times times as in algebra), we mean the value . Show that if is divisible by 9, then is divisible by 9.
Solution: Anything that is divisible by 9 can be written as . As this is algebra, we can use a letter (e.g. ) to represent \saysomething. As we are assuming that is divisible by 9, we can write that
Where is some number (which depends on the value of , and ). Then we can consider . First, we can write this (as given in the question) as
To show that this is divisible by we need to write it as . We need to apply the fact that here (because we are trying to show that, given this assumption, is divisible by ). We want to somehow extract an , here. Currently, we have a , which is great (because it’s what we’re after), and all we need is a and a , which we can get by breaking into and into . When we apply this to the expression as a whole, our desired result pops out fairly quickly.
Therefore, if (i.e. is divisible by ) is also divisible by .
Show that , and hence rationalise the denominator of
The first part is just algebraic manipulation, i.e.
For the second part, we want to apply this identity to the expression to rationalise. We have an expression in the form
And using the identity (Equation 36.20), we know that
Which implies (using the properties of reciprocals and fractions11 1 See Section 4.1.1 if you’re unsure about this) that
Given that our and , this is exactly what we want! As (which is rational) and which is rational, is also rational (specifically ). Therefore, we have that