Intereting Posts

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Translating a graph in the $xy$ plane given by,

$$f(x,y)=c$$

With $c \in \mathbb{R}$, $k$ units right and $h$ units up one gets,

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$$f(x-k,y-h)=c$$

The best explanation as to why this is the one I got in Algebra, to see why this is we can consider, the origin. Suppose $f(0,0)=c$. Translating the origin $k$ units right and $h$ units up, gives $(h,k)$. This means the new coordinate is $x=h$ and $y=k$. So $f(k-k,h-h)=c$ works.

Do you have a more general, better, explanation? In the above explanation it doesn’t really explain how we get the new equation from scratch , but it shows that once we get it, it works.

Also what happens to the equation,

$$f(x,y)=c$$

When we rotate the graph of this curve by $\theta$ counterclockwise, and why.

I know to rotate we need $(x,y) \to (x\cos(\theta)-y\sin(\theta), x\sin (\theta))+y\cos (\theta))$. But what becomes the new equation of the new curve? My guess is it has something to do with the “inverse rotation=clockwise” as I see in the case of translating by $T$ the graph of $f(x,y)=c$ we get a new graph whose equation is given by $f(T^{-1}(x,y))$. This is what is intuitive to me, but I can’t grasp how to explain my intuition

mathematically.

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Let $(x,y)$ be a point of the plane. If you want to rotate it around the origin, then translate it, the equations are

$$x’=x\cos\theta-y\sin\theta+x_0,\\y’=x\sin\theta+y\cos\theta+y_0, $$ and by inversion

$$x=\ \ (x’-x_0)\cos\theta+(y’-y_0)\sin\theta,\\y=-(x’-x_0)\sin\theta+(y’-y_0)\cos\theta.$$

(This is simply the inverse translation followed by the inverse rotation.)

As the original curve is described by $$f(x,y)=c,$$ by substitution the transformed curve has the equation

$$f((x’-x_0)\cos\theta+(y’-y_0)\sin\theta,-(x’-x_0)\sin\theta+(y’-y_0)\cos\theta)=c.$$

Here’s my go on the first part. The second part would I imagine be similar.

Suppose $y=f(x)$ is a function. For each $x_1$ in the domain of $f$ there exists a $y_1$ such that $f(x_1)=y_1$ giving the ordered pair $(x_1,y_1)$ on the graph of $f$.

Now consider $f_1(x)=f(x-h)+k$. For every $x_1+h$ in the domain of $f$:

$f_1(x_1+h)=f(x_1+h-h)+k=f(x_1)+k=y_1+k$

This means the point $(x_1+h,y_1+k)$ is on $f_1$.

We can say the point $(x_1,y_1)$ *shifted* to $(x_1+h,y_1+k)$ and that to obtain $f_1$ from $f$ we shift $f$ to the right $h$ units and up $k$ units.

Consider,

$$y=f(x)$$

Suppose we want to shift all points to the right by $h$ to get a “new” point from every point . Given a point $(x,y)$ on the graph of this function, this will become $(x_{\text{new}},y_{\text{new}})=(x+h,y)$. We have,

$$x_{n}=x+h$$

$$y_{n}=y$$

So,

$$x_{n}-h=x$$

$$y_{n}=y$$

But $f(x)=y$ means $f(x_{n}-h)=y_{n}$. Redefining the new coordinates we have $f(x-h)=y$.

Now we use the same idea for $f(x,y)=c$:

Consider the graph $G$ that consists of points $(x,y)$ such that $f(x,y)=c$. Suppose we translate/rotate every point in $G$ using a “linear function” $L$ on $(x,y)$. Then the result is the set of points $X=L(x,y)$ with $f(x,y)=c$.

Clearly $L^{-1}L((x,y))=(x,y)$. So that we have $f(L^{-1}L((x,y)))=f(x,y)=c$. This means $f(L^{-1}(X))=c$. Now redefining $X$ to be $(x,y)$ after rotation/transformation then we have,

$$f(L^{-1}(x,y))=c$$

This shows that if you want to rotate a curve given by $f(x,y)=c$ counterclockwise about the origin using $L : (x,y) \to (x \cos \theta-y\sin \theta, x\sin \theta+y\cos \theta)$. The resulting equation of the new curve is,

$$f(x \cos \theta+y \sin \theta, y \cos \theta-x \sin \theta)=c$$

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