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    Hey.

    Today in a C4 paper I came across the parametric equations of a curve:
    x=a(t^3), y=a/(1+(t^2)).

    I found the gradient function dy/dx to be -2/(3t((1+(t^2))^2)).

    However, finding this function forced me to assume that t was unequal to 0 although t=0 does create a point on the curve. (Because a=0 would only create a point and not a curve I was fairly confident a could not be 0 (please challenge this if you think it is wrong )).

    Now, dy/dx=dy/dt x dt/dx. I can see that if dy/dx is undefined, then dy/dt or dt/dx must also be undefined. However, if dy/dt or dt/dx is undefined then how can I be sure that dy/dx is undefined? I cannot see clearly how the reverse logic works. If this second statement is indeed true then I can be confident in saying dy/dx is undefined when t=0 (which it appears to be when I consider the graph of the curve).

    The reason why I have given this thread the above title is because this situation appears to be a similar situation to that when one considers the gradient of y=|x| at x=0. Why is the gradient undefined there? (This may help me to understand the above problem).

    Please add anything you can.
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    Look here for a plot of the graph defined by your parametric equations (with a=1). You'll notice it has a sharp point at (0,1) where t=0 -- this is similar to what happens to the graph y=|x| at x=0. It's no coincidence that \dfrac{dy}{dx} is undefined at sharp points, because the notion of a "gradient" doesn't make sense there.

    Going back to first principles, the gradient of a function f at a point x is given by the limit of \dfrac{f(x+h)-f(x)}{h} as h \to 0 (if this limit exists). In the example y=|x| we have:

    \dfrac{dy}{dx} = \displaystyle \lim_{h \to 0} \dfrac{|x+h|-|x|}{h} = \lim_{h \to 0} \dfrac{(|x+h|-|x|)(|x+h|+|x|)}{h(|x+h|+|x|)}

    = \displaystyle \lim_{h \to 0} \dfrac{(x+h)^2-x^2}{h(|x+h|+|x|)} = \lim_{h \to 0} \dfrac{2x-h}{|x+h|+|x|}

    This is all well and good when x \ne 0; taking the limit simply gives \dfrac{x}{|x|} which is as we expect: 1 when x is positive and -1 when it's negative. But when x=0, we have to take the limit of \dfrac{-h}{|h|} as h \to 0, which changes value depending on whether h is coming from above or below, meaning the gradient is undefined.

    Another way of thinking about it is deciding where the tangent would go. If you think of the tangent to a curve at a point as being the limit of a cord on the curve as the two points of intersection of the cord and the curve tend to the point in question, you could essentially generate a "tangent" of any gradient between -1 and 1, simply by choosing cords which intersect the curve at different points and slide towards the origin at different speeds.

    If any of the above makes no sense to you, don't worry too much about it :p:
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    (Original post by nuodai)
    Look here for a plot of the graph defined by your parametric equations (with a=1). You'll notice it has a sharp point at (0,1) where t=0 -- this is similar to what happens to the graph y=|x| at x=0. It's no coincidence that \dfrac{dy}{dx} is undefined at sharp points, because the notion of a "gradient" doesn't make sense there.

    Going back to first principles, the gradient of a function f at a point x is given by the limit of \dfrac{f(x+h)-f(x)}{h} as h \to 0 (if this limit exists). In the example y=|x| we have:

    \dfrac{dy}{dx} = \displaystyle \lim_{h \to 0} \dfrac{|x+h|-|x|}{h} = \lim_{h \to 0} \dfrac{(|x+h|-|x|)(|x+h|+|x|)}{h(|x+h|+|x|)}

    = \displaystyle \lim_{h \to 0} \dfrac{(x+h)^2-x^2}{h(|x+h|-|x|)} = \lim_{h \to 0} \dfrac{2x-h}{|x+h|+|x|}
    There should be h(|x+h|+|x|) at the denominator before the last limit.
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    (Original post by nuodai)
    Look here for a plot of the graph defined by your parametric equations (with a=1). You'll notice it has a sharp point at (0,1) where t=0 -- this is similar to what happens to the graph y=|x| at x=0. It's no coincidence that \dfrac{dy}{dx} is undefined at sharp points, because the notion of a "gradient" doesn't make sense there.

    Going back to first principles, the gradient of a function f at a point x is given by the limit of \dfrac{f(x+h)-f(x)}{h} as h \to 0 (if this limit exists). In the example y=|x| we have:

    \dfrac{dy}{dx} = \displaystyle \lim_{h \to 0} \dfrac{|x+h|-|x|}{h} = \lim_{h \to 0} \dfrac{(|x+h|-|x|)(|x+h|+|x|)}{h(|x+h|+|x|)}

    = \displaystyle \lim_{h \to 0} \dfrac{(x+h)^2-x^2}{h(|x+h|+|x|)} = \lim_{h \to 0} \dfrac{2x-h}{|x+h|+|x|}

    This is all well and good when x \ne 0; taking the limit simply gives \dfrac{x}{|x|} which is as we expect: 1 when x is positive and -1 when it's negative. But when x=0, we have to take the limit of \dfrac{-h}{|h|} as h \to 0, which changes value depending on whether h is coming from above or below, meaning the gradient is undefined.

    Another way of thinking about it is deciding where the tangent would go. If you think of the tangent to a curve at a point as being the limit of a cord on the curve as the two points of intersection of the cord and the curve tend to the point in question, you could essentially generate a "tangent" of any gradient between -1 and 1, simply by choosing cords which intersect the curve at different points and slide towards the origin at different speeds.

    If any of the above makes no sense to you, don't worry too much about it :p:
    Almost all of it makes sense to me and thank-you for the detailed response. So your point is that, for the gradient of a curve to be defined at that point, the forward difference approximation limit for positive h and negative h must converge to the same number?

    This is not in the case of y=|x| at the origin. How would I show it is also not the case in terms of my parametric equations? And do you think my reasoning is correct for saying a is unequal to 0?

    Furthermore, is it right to conclude that, because dt/dx is undefined for t=0, dy/dx is undefined for t=0, using that reasoning?
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    (Original post by Magu1re)
    So your point is that, for the gradient of a curve to be defined at that point, the forward difference approximation limit for positive h and negative h must converge to the same number?
    Yes; the idea is that a limit only exists if you get the same result regardless of how you approach it. It's for this very reason that we can't define \dfrac{0}{0}. To illustrate, consider the fraction \dfrac{ax}{x} as x \to 0. Both the numerator and denominator tend to zero, but the fraction itself is equal to a when x \ne 0 (and so must tend to a as x tends to 0); but we can choose a to be whatever we like, so we can't define \dfrac{0}{0} to have any one value.

    (Original post by Magu1re)
    How would I show it is also not the case in terms of my parametric equations? And do you think my reasoning is correct for saying a is unequal to 0?
    In order to find \dfrac{dy}{dx} you have to divide by \dfrac{dx}{dt}. Since \dfrac{dx}{dt} = 0 when t \to 0, the graph of y against x must become vertical when x=0, and so its gradient isn't finite. The graph (see link in my previous post) shows that the gradient is very large and positive before x=0 and very large and negative after x=0.

    As for saying that a \ne 0, well a is just a fixed constant. I think given the context you can assume that a \ne 0, since if it were equal to zero then the graph would, very boringly, just be a dot at the origin (and we obviously can't define the gradient of a point).

    (Original post by Magu1re)
    Furthermore, is it right to conclude that, because dt/dx is undefined for t=0, dy/dx is undefined for t=0, using that reasoning?
    Sort of. I was maybe a bit misleading with the explanation in terms of the graph having a point, I think in this case it's more to do with the fact that the graph becomes vertical when x=0. It's similar to what happens with the graph of y=\dfrac{1}{x} as x \to 0, with the main difference being that your function is defined at x=0 whereas 1/x is not.
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    (Original post by nuodai)
    Yes; the idea is that a limit only exists if you get the same result regardless of how you approach it. It's for this very reason that we can't define \dfrac{0}{0}. To illustrate, consider the fraction \dfrac{ax}{x} as x \to 0. Both the numerator and denominator tend to zero, but the fraction itself is equal to a when x \ne 0 (and so must tend to a as x tends to 0); but we can choose a to be whatever we like, so we can't define \dfrac{0}{0} to have any one value.

    In order to find \dfrac{dy}{dx} you have to divide by \dfrac{dx}{dt}. Since \dfrac{dx}{dt} = 0 when t \to 0, the graph of y against x must become vertical when x=0, and so its gradient isn't finite. The graph (see link in my previous post) shows that the gradient is very large and positive before x=0 and very large and negative after x=0.

    As for saying that a \ne 0, well a is just a fixed constant. I think given the context you can assume that a \ne 0, since if it were equal to zero then the graph would, very boringly, just be a dot at the origin (and we obviously can't define the gradient of a point).


    Sort of. I was maybe a bit misleading with the explanation in terms of the graph having a point, I think in this case it's more to do with the fact that the graph becomes vertical when x=0. It's similar to what happens with the graph of y=\dfrac{1}{x} as x \to 0, with the main difference being that your function is defined at x=0 whereas 1/x is not.
    Okay. The first, third and fourth paragraphs of your above post made sense to me; the other is not as clear in my head. lol.

    Knowing, dy/dt and dx/dt can I be sure that the gradient is undefined when t=0? The reason why I am asking this is, although I can see that if I could write y as a function of x then I could satisfy myself the gradient tended to +- infinity from different sides and so would be underfined at the point in question, but this is just a little C4 question with an annoying property (which I hate we're supposed to ignore). Is it possible to correctly deduce this from my workings so far?

    i.e. dy/dx = dy/dt x dt/dx

    If the right-hand side is undefined does the left-hand side necessarily have to be undefined?

    Also, can a point be described as a curve (because that would also categorically tell me a is unequal to 0 due to the use of the word "curve" in the question)?
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    (Original post by Magu1re)
    Hey.

    Today in a C4 paper I came across the parametric equations of a curve:
    x=a(t^3), y=a/(1+(t^2)).

    I found the gradient function dy/dx to be -2/(3t((1+(t^2))^2)).

    However, finding this function forced me to assume that t was unequal to 0 although t=0 does create a point on the curve. (Because a=0 would only create a point and not a curve I was fairly confident a could not be 0 (please challenge this if you think it is wrong )).

    Now, dy/dx=dy/dt x dt/dx. I can see that if dy/dx is undefined, then dy/dt or dt/dx must also be undefined. However, if dy/dt or dt/dx is undefined then how can I be sure that dy/dx is undefined? I cannot see clearly how the reverse logic works. If this second statement is indeed true then I can be confident in saying dy/dx is undefined when t=0 (which it appears to be when I consider the graph of the curve).
    I think your logic so does not work.
    Start with that if dy/dx is defined then dy/dt and dx/dt must be defined
    because \frac{dy}{dx}=\frac{\frac{dy}{dt  }}{\frac{dx}{dt}},
    so this fraction is defined.
    If You state the negation of above that is if dy/dx is undefined then dy/dt or dx/dt
    must also be undefined is not works, because both maybe defined but when dx/dt is zero then dy/dx is undefined.
    So you can state that
    dy/dx exists imply that (dy/dt)/(dx/dt) exists , and if (dx/dt)/(dx/dt) does not exists
    imply that dy/dx does not exists
    (modus tollens in the logic)
    Similarly if dx/dt or dy/dt does not exist then (dx/dt)/(dx/dt) does not exists
    but in your example they exist for all t.
    But there is another case here: when dx/dt=0. This means that the fraction of
    (dx/dt)/(dx/dt) does not exists and this imply that dy/dx does not exist.
    dx/dt =0 when t=0
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    (Original post by ztibor)
    I think your logic so does not work.
    Start with that if dy/dx is defined then dy/dt and dx/dt must be defined
    because \frac{dy}{dx}=\frac{\frac{dy}{dt  }}{\frac{dx}{dt}},
    so this fraction is defined.
    If You state the negation of above that is if dy/dx is undefined then dy/dt or dx/dt
    must also be undefined is not works, because both maybe defined but when dx/dt is zero then dy/dx is undefined.
    So you can state that
    dy/dx exists imply that (dy/dt)/(dx/dt) exists , and if (dx/dt)/(dx/dt) does not exists
    imply that dy/dx does not exists
    (modus tollens in the logic)
    Similarly if dx/dt or dy/dt does not exist then (dx/dt)/(dx/dt) does not exists
    but in your example they exist for all t.
    But there is another case here: when dx/dt=0. This means that the fraction of
    (dx/dt)/(dx/dt) does not exists and this imply that dy/dx does not exist.
    dx/dt =0 when t=0
    Okay. I get some of that having studied a bit of logic.

    If ((dy/dt)/(dx/dt)) is undefined, does that necessarily mean dy/dx is undefined? (I am sorry if this is a stupid/easy question but I truly do not know the answer).
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    (Original post by Magu1re)
    Okay. I get some of that having studied a bit of logic.

    If ((dy/dt)/(dx/dt)) is undefined, does that necessarily mean dy/dx is undefined? (I am sorry if this is a stupid/easy question but I truly do not know the answer).
    Yes.
    When true that if dy/dx defined means that (dy/dt) and (dx/dt) defined and dx/dt is not zero(this is true by the equation)
    you should be more strict. If dx/dt or dx/dt is undefined or dx/dt is zero, that means dy/dx is undefined
    When A \rightarrow B then notB \rightarrow notA
    A: dy/dx defined
    Bdy/dt) and (dx/dt) defined and dx/dt is not 0
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    (Original post by Magu1re)
    Knowing, dy/dt and dx/dt can I be sure that the gradient is undefined when t=0?
    Yes; if \dfrac{dx}{dt} = 0 at any point then the gradient is not defined at that point. There are a few ways of thinking about this, and the best is probably to think about the "speed" of a curve, but to think about it this way you need to have an idea of what all this parametric stuff really means.

    Essentially, what you get when you write \begin{cases} x = f(t) \\ y = g(t) \end{cases} is a way of describing the path of a dot as it travels along the curve. At some time t=a the dot is in the position \begin{pmatrix} x \\ y \end{pmatrix} = \begin{pmatrix} f(a) \\ g(a) \end{pmatrix}. If we think as time passing at some constant rate, we can consider the velocity of the dot as it moves around, and we look at its horizontal and vertical velocities separately (in the same way that we look at its horizontal position (x) and vertical position (y) separately). The horizontal velocity is given by f'(t) and the vertical velocity is given by g'(t) (where the ' represents differentiation w.r.t. t; that is, f'(t)=dx/dt and so on).

    Now, the vector given by \begin{pmatrix} f'(a) \\ g'(a) \end{pmatrix} at some point a tells you what direction the particle is moving at that point. In other words, this vector is precisely the tangent vector to the curve at the point a. Now, when f'(a)=0, this tangent vector points upwards, because the dot is not moving in the x-direction but is moving in the y-direction. However, we limit ourselves by not working with vectors to our definition of the "direction" of a curve to be its gradient. Whereas a line with slope +\infty and a line with slope -\infty look the same (i.e. vertical lines), unfortunately the difference between the two, and even merely dealing with infinity, means that we can't define the gradient to be any real number. This is why whenever \dfrac{dx}{dt}=0 we cannot define \dfrac{dy}{dx}.

    This covers the case when the graph goes vertical. You'll find in cases where there is a sharp point that one of \dfrac{dy}{dt} or \dfrac{dx}{dt} is also undefined. For example you can parametrise the graph y=|x| simply by \begin{cases} x=|t| \\ y=t \end{cases}, and then clearly you bump into the same problem when trying to find \dfrac{dx}{dt}.


    (Original post by Magu1re)
    (which I hate we're supposed to ignore)
    Indeed, but it's good that you're asking questions!

    (Original post by Magu1re)
    i.e. dy/dx = dy/dt x dt/dx

    If the right-hand side is undefined does the left-hand side necessarily have to be undefined?
    I think you're thinking about this the wrong way. If one side of the equation is undefined then by definition it cannot be an equation. An equation is something which says that the two values on either side represent the same number (okay not always numbers, but here they're numbers). If one side does not represent a number, then it's not an equation.

    However, we can consider limits (like before). If one side of an equation tends to a value at some point then so does the other side. Similarly if one side has no limit at some point then neither can the other side -- by nature of equality, whenever one side is defined, the other side is defined and has the same value (or at least tends to that value, as in the case of 1 = \dfrac{x}{x} as x \to 0). [As such, if the LHS were undefined and the RHS were defined, then by equality the LHS would be defined. But we just said it's undefined... so it can't happen.]

    (Original post by Magu1re)
    Also, can a point be described as a curve (because that would also categorically tell me a is unequal to 0 due to the use of the word "curve" in the question)?
    I speak loosely when I say "curve" (so do many mathematicians; it can mean many things). What I really mean is "the image of the function which maps t \mapsto \begin{pmatrix}f(t) \\ g(t) \end{pmatrix}", but to be honest I think it's implied that a \ne 0 even without mention of the word "curve".
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    (Original post by ztibor)
    Yes, when true that if dy/dx defined means that (dy/dt)/(dx/dt) is defined (but this is true by the equation)
    When A \rightarrow B then notB \rightarrow notA
    A: dy/dx defined
    Bdy/dt)/(dx/dt) defined
    I accept this is true if A implies B. Why does dy/dx= dy/dt x dt/dx always? I can see why it is true for a point where the gradient is defined, but why must it hold true if the gradient is not defined?
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    (Original post by Magu1re)
    I accept this is true if A implies B. Why does dy/dx= dy/dt x dt/dx always? I can see why it is true for a point where the gradient is defined, but why must it hold true if the gradient is not defined?
    It neither does nor doesn't hold true when the gradient is not defined, because when the gradient isn't defined, neither side of the equation makes sense.
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    (Original post by nuodai)
    It neither does nor doesn't hold true when the gradient is not defined, because when the gradient isn't defined, neither side of the equation makes sense.
    Thanks for the long post above this and for this comment too.

    I can see why the graph would be vertical and thus dy/dx would be undefined if dx/dt=0 (only because I know a vertical slope has an undefined gradient.

    Why is it that the difference between a curve with a slope of - infinity or infinity mean we cannot define it?

    P.S. I think the labels for x and y have been mixed up for the parametric equations.
 
 
 
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