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C4 implicit differentiation question relating to tangent Watch

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Size:  189.9 KB Hi, I don't get the part circled in black and the bit underlined green. For my working out, I wasn't sure what -x-2y=0 was telling me, so I wasn't aware that it is actually an equation. My x & y coordinates are the other way round but I don't really understand why -x-2y=0 is turned into the equation y=-2x...
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    So at the end I deduced that Q is at (-6,3)
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    (Original post by coconut64)
    So at the end I deduced that Q is at (-6,3)
    The gradient of the y-xais is not zero .... the denomnator of dy/dx will be zero.
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    (Original post by Muttley79)
    The gradient of the y-xais is not zero .... the denomnator of dy/dx will be zero.
    But either way, dy/dx will be 0. I know I have expressed it in a wrong way but at the end my numerator still equals to 0... Thanks
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    (Original post by coconut64)
    But either way, dy/dx will be 0. I know I have expressed it in a wrong way but at the end my numerator still equals to 0... Thanks
    No - the x-axis has gradient zero, not the y-axis.

    It's the DENOMINATOR that is zero; please check this.
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    I got -3 = x and 6 = y

    What was the actual answer?

    Also working :

    dy/dx = (-x-2y)/(2x+y) (equal to yours, I think)
    dx/dy is equal to 0 if parallel to Y, hence (2x+y) = 0, y = -2x

    x^2 + 4x(-2x) + (-2x)^2 + 27 = 0
    -8x^2 + 5x^2 + 27 = 0
    3x^2 = 27
    x^2 = 9
    x < 0, hence x = -3 = (9)^(1/2)

    If x = -3 y = -2x = 6.

    (-3,6) ?
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    Sorry my mistake, this is for the gradient of a perpendicular line to the tangent.

    The gradient of a tangent to any curve has the equation -dx/dy

    ie -1/gradient

    So in this case, the tangent to the curve which gradient is -dx/dy which is parallel to the y-axis, a tangent parallel with the y-axis means it is a straight vertical line therefore -1/dy/dx = 0

    Does that help at all?
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    (Original post by itsElliott)
    The gradient of a tangent to any curve has the equation -dx/dy
    That's not even an equation, let alone correct.
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    (Original post by DFranklin)
    That's not even an equation, let alone correct.
    Yes it is, not in the form i've written it because i'm trying to be as general as possible, expressing it out in full in the context of the question, you would produce an equation, don't be so dismissive

    And my mistake reading over, I've stated the gradient of a perpendicular line to the gradient.
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    (Original post by itsElliott)
    Yes it is, not in the form i've written it because i'm trying to be as general as possible, expressing it out in full in the context of the question, you would produce an equation, don't be so dismissive

    And my mistake reading over, I've stated the gradient of a perpendicular line to the gradient.
    Well, my problem was that as it wasn't actually an equation, I couldn't tell if what you'd got wrong was thinking of the normal, or confusing y = kx + C with y+kx = C, or ...

    Which is why words matter. At least if you want anyone to have a hope of working out what you've got wrong when you make a mistake.
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    (Original post by DFranklin)
    Well, my problem was that as it wasn't actually an equation, I couldn't tell if what you'd got wrong was thinking of the normal, or confusing y = kx + C with y+kx = C, or ...

    Which is why words matter. At least if you want anyone to have a hope of working out what you've got wrong when you make a mistake.
    Fair enough I completely bullsed up what I was trying to answer, but the fact I called an expression an equation is hardly something to get confused over when I could literally change "has an equation -dx\dy" to "forms an equation of -dx\dy" .

    But the negative reciprocal of the gradient, is indeed the gradient of a perpendicular line to the tangent, but I can sort of see where you got what I was saying confused.

    my mistake for not thinking about the question correctly
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    (Original post by itsElliott)
    ...I could literally change "has an equation -dx\dy" to "forms an equation of -dx\dy" .
    Neither of these are equations. An equation has an = sign somewhere in it. (The clue's in the name). If you just have something that evaluates to a number, then it's an expression.

    But the negative reciprocal of the gradient, is indeed the gradient of a perpendicular line to the tangent, but I can sort of see where you got what I was saying confused.
    You mean I got confused because you said it was the gradient of the tangent, when you actually meant something completely different? Yeah, that was kind of easy to get confused by...

    I don't think you actually got my point (and I suspect you don't care, which is fair enough), but from my perspective I could see you had an incorrect minus sign in your expression, but I couldn't tell if that was because you were thinking of writing the tangent in the form y + (-\dfrac{dy}{dx}) x = C, or the gradient of the normal -\dfrac{dx}{dy}. So if you'd actually given an equation, it would have helped.

    (In some ways "normal" is the more obvious guess, but you said explicitly it was the gradient, so...)
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    (Original post by DFranklin)
    Neither of these are equations. An equation has an = sign somewhere in it. (The clue's in the name). If you just have something that evaluates to a number, then it's an expression.

    You mean I got confused because you said it was the gradient of the tangent, when you actually meant something completely different? Yeah, that was kind of easy to get confused by...

    I don't think you actually got my point (and I suspect you don't care, which is fair enough), but from my perspective I could see you had an incorrect minus sign in your expression, but I couldn't tell if that was because you were thinking of writing the tangent in the form y + (-\dfrac{dy}{dx}) x = C, or the gradient of the normal -\dfrac{dx}{dy}. So if you'd actually given an equation, it would have helped.

    (In some ways "normal" is the more obvious guess, but you said explicitly it was the gradient, so...)
    Forming the equation would give it a minus sign, saying that it forms an equation of -dx/dy is implying that once you bring it into context you get -dx/dy = ax + b. I explicitly left out an equals sign to not ensue anything - instead giving the general form and stating that "an equation will form in -dx/dy".

    how else would you have expected me to give an equation for the normal i a generalised form without solving it already; like you yourself just expressed the equation of a normal,

    (In some ways "normal" is the more obvious guess, but you said explicitly it was the gradient, so...)
    I did because the gradient of the normal is exactly what I stated.
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    (Original post by itsElliott)
    Forming the equation would give it a minus sign, saying that it forms an equation of -dx/dy is implying that once you bring it into context you get -dx/dy = ax + b. I explicitly left out an equals sign to not ensue anything - instead giving the general form and stating that "an equation will form in -dx/dy".
    No it doesn't. I'm not going to debate what an equation is with you any more, but if you get into the habit of writing things that aren't actually what you mean, with implicit statements you don't give but the reader is supposed to imply, then it will not go well for you in your exams.

    how else would you have expected me to give an equation for the normal
    By saying it was the equation for the normal and not the gradient? Might have helped...

    I did because the gradient of the normal is exactly what I stated.
    (Original post by itsElliot)
    The gradient of a tangent to any curve has the equation -dx/dy
    O RLY?
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    (Original post by DFranklin)
    No it doesn't. I'm not going to debate what an equation is with you any more, but if you get into the habit of writing things that aren't actually what you mean, with implicit statements you don't give but the reader is supposed to imply, then it will not go well for you in your exams.

    By saying it was the equation for the normal and not the gradient? Might have helped...



    O RLY?
    I know exactly what an equation is and setting it up how I stated would give you an equation once you apply it to the question, I don't think you can really debate that. Fair enough some of the wording was rather poor.

    And lastly, I think I've alreay apologised and stated in my OP that it was a mistake and it should have said normal, i never intended to write normal, because when I was writing the post I wrote down the wrong thing for the original question - I am quite aware of that.

    if you took out every time I refereed to anything as the tangent of a curve, and replaced it with "normal" then it would be correct - that is what I am getting at. I wrote down the wrong thing, and I have explicitly said that

    By saying it was the equation for the normal and not the gradient? Might have helped...
    I said "The gradient of a tangent to any curve has the equation -dx/dy "

    Replace tangent to any curve with normal to any curve, that is what I am saying by I made a mistake.
 
 
 
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