Chapter 36 Assorted problems

Solutions to problems which have yet to be categorised.

36.0.1 Cones problem

Problem: Two similar cones, $X$ and $Y$ , have surface areas $270\text{cm}^{2}$ and $120\text{cm}^{2}$ respectively.

The volume of cone $X$ is $1215\text{cm}^{3}$ .

Show that the volume of cone Y is $360\text{cm}^{3}$ .

Solution:

We have two cones, let’s let $X$ be this one, with radius $r$ and height $h$

Then we can draw $Y$ (which is smaller than $X$ , but not drawn to scale here). As they are similar cones, the radius of $Y$ is equal to $ar$ (where $a$ is the scale factor between the two cones) and the height of $Y$ is equal to $ah$

First, the volume of a cone is $\frac{\pi r^{2}h}{3}$ and the surface area of a cone is $\pi r(r+\sqrt{h^{2}+r^{2}})$ . 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 $X$ to the value given in the question

\pi r(r+\sqrt{h^{2}+r^{2}})=270

(36.1)

We can also do the same for $Y$

\pi ar(ar+\sqrt[]{(ah)^{2}+(ar)^{2}})=120

(36.2)

This can be simplified a bit by factoring the $a^{2}$ from the square rooted part of the equation, giving

\pi ar(ar+\sqrt{a^{2}}\sqrt[]{(h)^{2}+(r)^{2}})=120

(36.3)

and thus that

\pi a^{2}r(r+\sqrt{(h)^{2}+(r)^{2}})=120

(36.4)

We can solve this for $a$ by dividing the two equations, giving that

\frac{\pi a^{2}r(r+\sqrt{(h)^{2}+(r)^{2}})}{\pi r(r+\sqrt{h^{2}+r^{2}})}=\frac% {120}{270}

(36.5)

and thus, after cancelling, that

a^{2}=\frac{9}{4}

(36.6)

which means that

a=\frac{3}{2}

(36.7)

Given this, we can then turn to the volume formula.

We know that the volume of $X$ is

\pi r^{2}h=1215

(36.8)

and the formula for the volume of $Y$ is equal to

\pi(ar)^{2}(ah)=a^{3}\pi r^{2}h

(36.9)

Thus, by multiplying Equation 36.8 through by $a^{3}$ , we have

a^{3}\pi r^{2}h=a^{3}1215

(36.10)

The left-hand side is just the volume of $Y$ , and the right-hand side is just

	$\displaystyle a^{3}1215$	$\displaystyle=\left(\frac{2}{3}\right)^{3}\cdot 1215$		(36.11)
		$\displaystyle=360$		(36.12)

Which is what we needed to show.

36.0.2 Divisibility of $abc_{10}$

Example 36.0.1

\say

When we represent 789 in decimal, we mean 7 hundreds plus 8 tens plus 9 units. If $abc_{10}$ is taken to represent a decimal number (not $a$ times $b$ times $c$ as in algebra), we mean the value $100a+10b+c$ . Show that if $a+b+c$ is divisible by 9, then $abc_{10}$ is divisible by 9.

Solution: Anything that is divisible by 9 can be written as $9\cdot\text{something}$ . As this is algebra, we can use a letter (e.g. $m$ ) to represent \saysomething. As we are assuming that $a+b+c$ is divisible by 9, we can write that

a+b+c=9m

(36.13)

Where $m$ is some number (which depends on the value of $a$ , $b$ and $c$ ). Then we can consider $abc_{10}$ . First, we can write this (as given in the question) as

abc_{10}=100a+10b+c

(36.14)

To show that this is divisible by $9$ we need to write it as $9\cdot\text{something}$ . We need to apply the fact that $a+b+c=9m$ here (because we are trying to show that, given this assumption, $abc_{10}$ is divisible by $9$ ). We want to somehow extract an $a+b+c$ , here. Currently, we have a $c$ , which is great (because it’s what we’re after), and all we need is a $b$ and a $c$ , which we can get by breaking $100a$ into $99a+a$ and $10b$ into $9b+b$ . When we apply this to the expression as a whole, our desired result pops out fairly quickly.

$\displaystyle abc_{10}$	$\displaystyle=100a+10b+c$	(36.15)
	$\displaystyle=99a+a+9b+b+c$	(36.16)
	$\displaystyle=9(11a+b)+a+b+c$	(36.17)
	$\displaystyle=9(11a+b)+9m$	(36.18)
	$\displaystyle=9(11a+b+m)$	(36.19)

Therefore, if $a+b+c=9m$ (i.e. $a+b+c$ is divisible by $9$ ) $abc_{10}$ is also divisible by $9$ .

36.0.3 Rationalising

Example 36.0.2

Show that $a^{3}-b^{3}=(a-b)(a^{2}+ab+b^{2})$ , and hence rationalise the denominator of

\frac{1}{\sqrt[3]{3}-\sqrt[3]{2}}

(36.20)

Solution

The first part is just algebraic manipulation, i.e.

$\displaystyle(a-b)(a^{2}+ab+b^{2})$	$\displaystyle=a^{2}+a^{2}b+ab^{2}-ba^{2}-ab^{2}-b^{3}$	(36.21)
	$\displaystyle=a^{2}+(a^{2}b-ab^{2})+(ab^{2}-ab^{2})-b^{3}$	(36.22)
	$\displaystyle=a^{3}-b^{3}$	(36.23)

For the second part, we want to apply this identity to the expression to rationalise. We have an expression in the form

\frac{1}{a-b}

(36.24)

And using the identity (Equation 36.20), we know that

a-b=\frac{a^{3}-b^{3}}{a^{2}+ab+b^{2}}

(36.25)

Which implies (using the properties of reciprocals and fractions¹¹ 1 See Section 4.1.1 if you’re unsure about this) that

\displaystyle\frac{1}{a-b}

\displaystyle=\frac{a^{2}+ab+b^{2}}{a^{3}-b^{3}}

(36.26)

Given that our $a=\sqrt[3]{3}$ and $b=\sqrt[3]{2}$ , this is exactly what we want! As $a^{3}=3$ (which is rational) and $b^{3}=2$ which is rational, $a^{3}-b^{3}$ is also rational (specifically $a-b=1$ ). Therefore, we have that

\frac{1}{\sqrt[3]{3}-\sqrt[3]{2}}=\frac{(\sqrt[3]{3})^{2}+\sqrt[3]{3}\sqrt[3]{% 2}+(\sqrt[3]{2})^{2}}{1}