Curve not symmetrical about y = x?

Hello, I am stuck on 8ii) on why the curve has to have a gradient of -1 at point P to be symmetrical about y = x. I'm sorry if the answer is really obvious but I cant seem to understand why it is. Thanks any help would be appreciated.

Reply 1

ghostwalker

17

Original post by znx

Hello, I am stuck on 8ii) on why the curve has to have a gradient of -1 at point P to be symmetrical about y = x. I'm sorry if the answer is really obvious but I cant seem to understand why it is. Thanks any help would be appreciated.

For symmetry, at the point where the curve crosses the alleged line of symmetry, the curve must be perpendicular to said line, otherwise it won't reflect onto itself.

Gradient of line is 1, so perpendicular gradient is -1.

Reply 2

Deranging

9

Original post by znx

...

This is an interesting question :smile:

.
In order to be symmetrical about the line

y = x

, what must the function look like very close to any point where it intersects the line

y = x

in order to be symmetrical?

Reply 3

znx

OP

10

Original post by ghostwalker

For symmetry, at the point where the curve crosses the alleged line of symmetry, the curve must be perpendicular to said line, otherwise it won't reflect onto itself.

Gradient of line is 1, so perpendicular gradient is -1.

Ah yes, thank you very much

Reply 4

znx

OP

10

Original post by Deranging

This is an interesting question :smile:

.
In order to be symmetrical about the line

y = x

, what must the function look like very close to any point where it intersects the line

y = x

in order to be symmetrical?

Either side of the line y=x has to be a mirror image of the other side?

Reply 5

ghostwalker

17

Original post by znx

Ah yes, thank you very much

I would add one minor caveat. This is true if the curve is differentiable at that point.

If it's not, you'll have to work harder. E.g. the modulus function, |x| is symmetrial about x=0, but it's not differentiable there.

Reply 6

Deranging

9

Original post by znx

Either side of the line y=x has to be a mirror image of the other side?

Yeah, now think about what that tells you about the gradient of curve very close to an intersect with y = x. By very close I mean such that you can approximate the curve by its tangent.

Reply 7

znx

OP

10

Original post by Deranging

Yeah, now think about what that tells you about the gradient of curve very close to an intersect with y = x. By very close I mean such that you can approximate the curve by its tangent.

The curve either side would be perpendicular to each other, so the gradients of either side would have to be negative reciprocals of eachother? So in the example above since the gradient of the curve at p is not the negative reciprocal of y = x, its not symmetrical?

(edited 6 years ago)

Reply 8

Deranging

9

Original post by znx

The curve either side would be perpendicular to each other, so the gradients of either side would have to be negative reciprocals of eachother? So in the example above since the gradient of the curve at p is not the negative reciprocal of y = x, its not symmetrical?

I think you have the broad strokes of the idea :smile:

.
Note that it's the tangent of the curve which is perpendicular to line of symmetry. So the gradient of the curve and the gradient of the line of symmetry have to be negative reciprocals.

(As with what ghostwalker said, this relies on us being able to locally approximate the curve by it's tangent - so this works as long as your function is sufficiently 'friendly' and doesn't do funky stuff at the intersect.)

(edited 6 years ago)

Original post by znx

Hello, I am stuck on 8ii) on why the curve has to have a gradient of -1 at point P to be symmetrical about y = x. I'm sorry if the answer is really obvious but I cant seem to understand why it is. Thanks any help would be appreciated.

Just to slot an alternative way to think about it:

We know that the gradient at

P(3,3)

is

\dfrac{dy}{dx}=-\frac{1}{2}

evaluated at this point.
Reflecting everything in the line

y=x

means our gradient at that point is now

\dfrac{dx}{dy} = \dfrac{2(x-2)^{3/2}}{x-4}

and it should give

-\frac{1}{2}

IF

y=x

is a line of symmetry (because P is on the line of symmetry, so it shouldn't change, hence the gradient should remain the same). Clearly, we get -2 instead, hence