Same q… again (1 Viewer)

eternallyboreduser · Dec 23, 2023

For the below q this is the supplied working out. However im still confused as to how it proves that c can have various values..

eternallyboreduser · Dec 23, 2023

Photo is hard to see so heres rhe q:
show that the curves x^2+cxy+y^2+1=0 (c cant be -2) for various values of c touch at the point (1,1)

eternallyboreduser · Dec 23, 2023

Luukas.2 · Dec 23, 2023

The working provided appears correct, but is flawed because there is only one curve (and one value of c) which actually passes through (1, 1). It is true that all curves have the same gradient because the total number of curves in the 'family' passing through the point (1, 1) is 1.

Luukas.2 · Dec 23, 2023

Note, there is no curve with $x,\ y \in \mathbb{R}$ if $c = -2$ because

$\begin{align*} x^2 + cxy + y^2 + 1 &= 0 \\ \text{Put $c = -2$:} \qquad x^2 + -2xy + y^2 + 1 &= 0 \\ (x - y)^2 + 1 &= 0 \\ (x - y)^2 &= -1 = i^2 \\ x - y &= \pm i \end{align*}$

eternallyboreduser · Dec 23, 2023

Luukas.2 said:
The working provided appears correct, but is flawed because there is only one curve (and one value of c) which actually passes through (1, 1). It is true that all curves have the same gradient because the total number of curves in the 'family' passing through the point (1, 1) is 1.

So the q isnt valid then

Luukas.2 · Dec 23, 2023

eternallyboreduser said:
So the q isnt valid then

The way it is phrased, "for various values of c," is not good as there is only one possible curve and one possible value of c that goes through the point.

If it said "show that all the curves blah that pass through (1, 1) touch", then their proof does do that, even though "all" is only one curve.

eternallyboreduser · Dec 23, 2023

Luukas.2 said:
The way it is phrased, "for various values of c," is not good as there is only one possible curve and one possible value of c that goes through the point.

If it said "show that all the curves blah that pass through (1, 1) touch", then their proof does do that, even though "all" is only one curve.

Do you mind explaining the proof im kinda confused @Luukas.2

eternallyboreduser · Dec 23, 2023

eternallyboreduser said:
Do you mind explaining the proof im kinda confused

Like the logic behind the proof doesnt make sense to me

Luukas.2 · Dec 24, 2023

eternallyboreduser said:
Like the logic behind the proof doesnt make sense to me

The proof is based on an interpretation of the word "touch"... suppose that there were multiple different curves all of which passed through the point (1, 1). Clearly, all of these curves come into contact at that point, but if they all have different gradients at that point then they cross one another, rather than "touching". By "touching", the question means that the curves are tangential to each other and do not cross, but rather all have the same tangent at that point. For example, consider the family of curves $y = x^n$ where $n \in \mathbb{Z}^+$ . All members of the family pass through the origin. All have a common tangent at the origin of the x-axis. When n is even, the curves all touch at the origin. When n is odd, the tangent is the same but the origin is a point of inflexion and so the curves cross one another, and cross all members of the family where n is even. All members of the family of curves also meet and cross at (1, 1). None touch as they all have different tangents, $y = nx - n + 1$ , and all cross - if $y = x^a$ is above $y = x^b$ for $0 < x < 1$ , with $a,\ b \in \mathbb{Z}^+$ , then $y = x^a$ must be below $y = x^b$ for $x > 1$ .

The proof given uses implicit differentiation to establish that the gradient of the tangent at (1, 1) to the curve is the same regardless of the value of $c$ , and thus concludes that all such curves touch... what it fails to do is establish that there is more than one curve through (1, 1). Indeed, it fails to establish that there is any curve at all through that point.

Take the family of curves that I mentioned above, $y = x^n$ for $n \in \mathbb{Z}^+$ . Every member of the family must have a horizontal tangent at (0, -5), provided they pass through that point. The proof is simple - show that the gradient of the tangent is zero, and thus the curve is horizontal, at the specified point. The problem is that taking the fact that the gradient is zero and using it to conclude that all members of the family touch at this point is flawed when there are no such curves that actually pass through the required point. It is like finding the slope of the unit circle at (10, 500)... I can do the calculation but that doesn't mean that the point is on the circle, and so the result is meaningless. The question posed has only one curve through the given point, so trivially all curves through that point have the same tangent. The flaw is not in the method of proof, it is in the assumption that there are multiple members of the family of curves through the given point. If there were multiple members, the conclusion would be valid.

eternallyboreduser · Dec 24, 2023

Luukas.2 said:
The proof is based on an interpretation of the word "touch"... suppose that there were multiple different curves all of which passed through the point (1, 1). Clearly, all of these curves come into contact at that point, but if they all have different gradients at that point then they cross one another, rather than "touching". By "touching", the question means that the curves are tangential to each other and do not cross, but rather all have the same tangent at that point. For example, consider the family of curves $y = x^n$ where $n \in \mathbb{Z}^+$ . All members of the family pass through the origin. All have a common tangent at the origin of the x-axis. When n is even, the curves all touch at the origin. When n is odd, the tangent is the same but the origin is a point of inflexion and so the curves cross one another, and cross all members of the family where n is even. All members of the family of curves also meet and cross at (1, 1). None touch as they all have different tangents, $y = nx - n + 1$ , and all cross - if $y = x^a$ is above $y = x^b$ for $0 < x < 1$ , with $a,\ b \in \mathbb{Z}^+$ , then $y = x^a$ must be below $y = x^b$ for $x > 1$ .

The proof given uses implicit differentiation to establish that the gradient of the tangent at (1, 1) to the curve is the same regardless of the value of $c$ , and thus concludes that all such curves touch... what it fails to do is establish that there is more than one curve through (1, 1). Indeed, it fails to establish that there is any curve at all through that point.

Take the family of curves that I mentioned above, $y = x^n$ for $n \in \mathbb{Z}^+$ . Every member of the family must have a horizontal tangent at (0, -5), provided they pass through that point. The proof is simple - show that the gradient of the tangent is zero, and thus the curve is horizontal, at the specified point. The problem is that taking the fact that the gradient is zero and using it to conclude that all members of the family touch at this point is flawed when there are no such curves that actually pass through the required point. It is like finding the slope of the unit circle at (10, 500)... I can do the calculation but that doesn't mean that the point is on the circle, and so the result is meaningless. The question posed has only one curve through the given point, so trivially all curves through that point have the same tangent. The flaw is not in the method of proof, it is in the assumption that there are multiple members of the family of curves through the given point. If there were multiple members, the conclusion would be valid.

Whats implicit diff?

Luukas.2 · Dec 24, 2023

eternallyboreduser said:
Whats implicit diff?

Where the solution begins, but adding in every single step involved:

$\begin{align*} x^2 + cxy + y^2 + 1 &= 0 \qquad \text{where $c \neq -2$} \\ \text{Implicitly differentiating with respect to $x$:} \qquad \frac{d}{dx}\left(x^2 + cxy + y^2 + 1\right) &= \frac{d}{dx}\left(0\right) \\ \frac{d}{dx}\left(x^2\right) + \frac{d}{dx}\left(cxy\right) + \frac{d}{dx}\left(y^2\right) + \frac{d}{dx}\left(1\right) &= \frac{d}{dx}\left(0\right) \\ 2x + c\frac{d}{dx}\left(xy\right) + \frac{d}{dx}\left(y^2\right) + 0 &= 0 \\ 2x + c\left[y \times \frac{d}{dx}\left(x\right) + x \times \frac{d}{dx}\left(y\right)\right] + \frac{d}{dx}\left(y^2\right) &= 0 \qquad \text{using the Product Rule} \\ 2x + c\left[y \times 1 + x \times \frac{d}{dy}\left(y\right) \times \frac{dy}{dx}\right] + \frac{d}{dy}\left(y^2\right) \times \frac{dy}{dx} &= 0 \qquad \text{using the Chain Rule} \\ 2x + c\left(y + x \times 1 \times \frac{dy}{dx}\right) + 2y \times \frac{dy}{dx} &= 0 \\ 2x + c\left(y + x\frac{dy}{dx}\right) + 2y\frac{dy}{dx} &= 0 \end{align*}$

This statement is equivalent to the second line of algebra in the solution given:

$2x + c\left(y + x\frac{dy}{dx}\right) + 2y\frac{dy}{dx} = 0 \iff 2x + c\left(y + xy'\right)+ 2yy' = 0 \ \text{where $y' = \frac{dy}{dx}$}$

Continuing the algebra:

$\begin{align*} 2x + c\left(y + x\frac{dy}{dx}\right) + 2y\frac{dy}{dx} &= 0 \\ 2x + cy + cx\frac{dy}{dx} + 2y\frac{dy}{dx} &= 0 \\ \left(cx + 2y\right)\frac{dy}{dx} &= -2x - cy \\ \frac{dy}{dx} &= -\frac{2x + cy}{cx + 2y} \\ \\ \text{So, the tangent at $(1,\ 1)$ has gradient:} \qquad m_{\text{tang}} &= \frac{dy}{dx} = -\frac{2x + cy}{cx + 2y} \qquad \text{where $x = 1$ and $y = 1$} \\ &= -\frac{2(1) + c(1)}{c(1) + 2(1)} \\ &= -\frac{2 + c}{c + 2} \\ &= -1 \qquad \text{provided $c \neq -2$} \end{align*}$

Since the gradient is independent of $c$ at the pointe (1, 1), all members of the family of curves through this point have the same gradient, and hence the same tangent, and thus much touch.

As I've said above, the only problem is that this conclusion is true whether "all members of the family of curves" is no curves, one curve, or many curves. The wording of the question, with a focus on "various values of $c$ ," renders the result only meaningful if there are multiple curves (multiple $c$ values) that pass through (1, 1)... unfortunately, there is only one member of the family through that point.

eternallyboreduser · Dec 26, 2023

Luukas.2 said:
Where the solution begins, but adding in every single step involved:

$\begin{align*} x^2 + cxy + y^2 + 1 &= 0 \qquad \text{where $c \neq -2$} \\ \text{Implicitly differentiating with respect to $x$:} \qquad \frac{d}{dx}\left(x^2 + cxy + y^2 + 1\right) &= \frac{d}{dx}\left(0\right) \\ \frac{d}{dx}\left(x^2\right) + \frac{d}{dx}\left(cxy\right) + \frac{d}{dx}\left(y^2\right) + \frac{d}{dx}\left(1\right) &= \frac{d}{dx}\left(0\right) \\ 2x + c\frac{d}{dx}\left(xy\right) + \frac{d}{dx}\left(y^2\right) + 0 &= 0 \\ 2x + c\left[y \times \frac{d}{dx}\left(x\right) + x \times \frac{d}{dx}\left(y\right)\right] + \frac{d}{dx}\left(y^2\right) &= 0 \qquad \text{using the Product Rule} \\ 2x + c\left[y \times 1 + x \times \frac{d}{dy}\left(y\right) \times \frac{dy}{dx}\right] + \frac{d}{dy}\left(y^2\right) \times \frac{dy}{dx} &= 0 \qquad \text{using the Chain Rule} \\ 2x + c\left(y + x \times 1 \times \frac{dy}{dx}\right) + 2y \times \frac{dy}{dx} &= 0 \\ 2x + c\left(y + x\frac{dy}{dx}\right) + 2y\frac{dy}{dx} &= 0 \end{align*}$

This statement is equivalent to the second line of algebra in the solution given:

$2x + c\left(y + x\frac{dy}{dx}\right) + 2y\frac{dy}{dx} = 0 \iff 2x + c\left(y + xy'\right)+ 2yy' = 0 \ \text{where $y' = \frac{dy}{dx}$}$

Continuing the algebra:

$\begin{align*} 2x + c\left(y + x\frac{dy}{dx}\right) + 2y\frac{dy}{dx} &= 0 \\ 2x + cy + cx\frac{dy}{dx} + 2y\frac{dy}{dx} &= 0 \\ \left(cx + 2y\right)\frac{dy}{dx} &= -2x - cy \\ \frac{dy}{dx} &= -\frac{2x + cy}{cx + 2y} \\ \\ \text{So, the tangent at $(1,\ 1)$ has gradient:} \qquad m_{\text{tang}} &= \frac{dy}{dx} = -\frac{2x + cy}{cx + 2y} \qquad \text{where $x = 1$ and $y = 1$} \\ &= -\frac{2(1) + c(1)}{c(1) + 2(1)} \\ &= -\frac{2 + c}{c + 2} \\ &= -1 \qquad \text{provided $c \neq -2$} \end{align*}$

Since the gradient is independent of $c$ at the pointe (1, 1), all members of the family of curves through this point have the same gradient, and hence the same tangent, and thus much touch.

As I've said above, the only problem is that this conclusion is true whether "all members of the family of curves" is no curves, one curve, or many curves. The wording of the question, with a focus on "various values of $c$ ," renders the result only meaningful if there are multiple curves (multiple $c$ values) that pass through (1, 1)... unfortunately, there is only one member of the family through that point.

ahhh icic thank you

)

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