Submitted by Edogowa Conan on Mon, 11/07/2016 - 10:48
Forums:
There is an article in the NRICH website called the why and how of substitution. In it they give a few examples of how to simplify difficult expressions. But I don't understand how they simplified example 5, which is a polynomial to the power n, and example 6 which is the formula of conic.
Please help!
Q5
They start by making the substitution $x=t+k$ where $k$ is a constant. This gives:
\[
p(t+k) = a_n(t+k)^n+a_{n-1}(t+k)^{n-1}+ \cdots + a_1(t+k)+a_0
\]
It we imagined expanding all these brackets, and remembering that $k$ is a constant, we would get a polynomial in $t$. Let's call this polynomial $q(t)$.
If we expanded the brackets, coefficients of $t^n$ can only come from $a_n(t+k)^n$, and coefficients of $t^{n-1}$ can come from $a_n(t+k)^n$ and $a_{n-1}(t+k)^{n-1}$. Considering these expansions gives:
\[
q(t) = a_nt^n + (na_nk + a_{n-1})t^{n-1} + \cdots
\]
This means that we can pick $k$ so that there is no $t^{n-1}$ term.
Does this help any?
Example 6
Using the given substitutions for $x$ and $y$ we have:
\[
\frac 1 2 (u-v)^2+2b \times \frac 1 2 (u-v)(u+v) + \frac 1 2 (u+v)^2 = 1
\]
Which becomes
\[
\frac 1 2 (u^2 - 2uv +v^2 + 2bu^2 - 2bv^2 + u^2 + 2uv + v^2) = 1
\]
or
\[
u^2 + bu^2 + v^2 - bv^2=1
\]
Try using https://www.desmos.com/ to plot $x^2(1+b)+y^2(1-b)=1$ and use a "slider" to see what happens as $b$ varies.
Let me know if you want me to attempt any further explanation!
Thanks for the reply,
Thanks for the reply,
Why is the equation an ellipse if |b|<1, a hyperbola if |b|>1, and a pair of lines if |b|=1.
Also, why is it a pair of lines, why not just a single line.
Thanks for your help
Conic sections
If $|b|=1$ we have two cases:
$b=1$ gives the equation $2x^2=1$ so $x=\pm \frac 1 {\sqrt {2}}$.
$b=-1$ gives the equation $2y^2=1$ so $y=\pm \frac 1 {\sqrt {2}}$.
If $|b| \lt 1$ then the equation has the form $Ax^2+By^2=1$ where both $A$ and $B$ are positive. Also, as $x^2$ increases, $y^2$ decreases and vice versa.
This is an ellipse equation - it crosses the axes at $(0, \pm \frac 1 {\sqrt{B}})$ and $(\pm \frac 1 {\sqrt{A}}, 0)$.
If $b >1$ then the equation has the form $Ax^2 - By^2=1$ where $A$ and $B$ are positive. This is a hyperbola. It crosses the $x$ axis at $(\pm \frac 1 {\sqrt{A}}, 0)$ and does not cross the $y$ axis (as then we would have $y^2 = - \frac 1 B$). We can rewrite the equation as $Ax^2=1+By^2$ and so as $x^2$ increases, $y^2$ decreases. As $x \to \infty$ we have $Ax^2 \approx By^2$ so there are asymptotes at $y=\pm \frac A B x$. The situation is very similar when $b \lt -1$.
Much appreciated.
Much appreciated.
What do they mean when they say 'rotation of the plane' in example 6. I've never come across this before. And how will that help us graphically.
Also even if we rotate the plane and solve the equation, how will we write the answer in terms of the original variables.
Thanks
I'll get back to you
Might not have time to answer that today - I'll have a look tomorrow or Thursday.
Rotations
If you use https://www.desmos.com/calculator to plot $x^2+2bxy+y^2=1$ and use a slider to see what happens for different values of $b$ you will see the curve changing from a hyperbola to an ellipse and back again, but on a slant rather than as you would expect them to look. This implies that is we rotate everything around we might get to the standard equation of an ellipse ($a^2x^2+b^2y^2=1$) or a hyperbola ($a^2x^2-b^2y^2=1$ or $=-1$).
If you imagine a point with coordinates $(u,v)$ and then rotate this by angle $\theta$ clockwise about the origin then you end up at the point $(u \cos \theta - v \sin \theta, u \sin \theta + v \cos \theta)$. If you know anything about matrix transformations you will be able to see why this is true, otherwise you can draw a diagram and convince yourself.
Substituting these in for $x$ and $y$ shows what the equation will look like after this rotation, and then we can pick the value of $\theta$ for which the cross term disappears. The point they are making by talking about the rotation is trying to show you where the first substitutions came from - if we let $\theta = \pi/4$ then $x=u \cos \theta - v \sin \theta$ becomes $x=\frac 1 {\sqrt{2}}u - \frac 1 {\sqrt{2} }v$ and similar for $y$.