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Need help with a differentiation question Watch

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    Find the equation of the normal to y=cosecx at the point (1/2pi,1)

    I differentiated this and got -coesecxcotx so if I want to find the gradient of the tangent, surely I just need to sub in 0.5pi in x. However it doesn't give me a value but it gives me 0 when I used the dy/dx button on my calculator... I think this is to do with the cotx graph not having a gradient at that point...

    Thanks
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    (Original post by coconut64)
    Find the equation of the normal to y=cosecx at the point (1/2pi,1)

    I differentiated this and got -coesecxcotx so if I want to find the gradient of the tangent, surely I just need to sub in 0.5pi in x. However it doesn't give me a value but it gives me 0 when I used the dy/dx button on my calculator... I think this is to do with the cotx graph not having a gradient at that point...

    Thanks
    Then 0 is the gradient of the tangent, the tangent is a line parallel to the x axis.
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    (Original post by NotNotBatman)
    Then 0 is the gradient of the tangent, the tangent is a line parallel to the x axis.
    why does -cot 1/2pi not produce a value then ? Thanks
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    (Original post by coconut64)
    why does -cot 1/2pi not produce a value then ? Thanks
    It does, it is 0. it is -cos(pi/2) / sin(pi/2) = 0/-1 = 0
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    If the tangent is 0, then it is parallel to the x axis, the perpendicular normal is then parallel to the y axis, you need to know where it crosses the x axis to form the equation of the normal.
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    (Original post by NotNotBatman)
    It does, it is 0. it is -cos(pi/2) / sin(pi/2) = 0/-1 = 0
    How come that's the not case with -1/tanx ?
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    (Original post by coconut64)
    How come that's the not case with -1/tanx ?
    because tan(pi/2) is undefined, there is an asymptote on at that point on the y= tanx graph. Also  \frac{sin\frac{\pi}{2}}{cos\frac  {\pi}{2}} is undefined as  cos\frac{\pi}{2} = 0 and division by 0 returns an undefined result, but this isn't the case with cot x.
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    (Original post by coconut64)
    How come that's the not case with -1/tanx ?
    You should think of cotx as cosx/sinx rather than 1/tanx as it causes problems at odd multiple of  \pi /2 .
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    (Original post by B_9710)
    You should think of cotx as cosx/sinx rather than 1/tanx as it causes problems at odd multiple of  \pi /2 .
    Oh okay, so this is the case for pi/2 and what other values ? Thanks
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    (Original post by coconut64)
    Oh okay, so this is the case for pi/2 and what other values ? Thanks
    All the values where tanx=0.
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    (Original post by NotNotBatman)
    because tan(pi/2) is undefined, there is an asymptote on at that point on the y= tanx graph. Also  \frac{sin\frac{\pi}{2}}{cos\frac  {\pi}{2}} is undefined as  cos\frac{\pi}{2} = 0 and division by 0 returns an undefined result, but this isn't the case with cot x.
    Okay, I will bear that in mind thanks. For the equation for the normal, I got y=1 at the end, correct?
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    (Original post by B_9710)
    All the values where tanx=0.
    Is the equation of the normal y=1? thx
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    (Original post by coconut64)
    Is the equation of the normal y=1? thx
    That's the tangent.
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    (Original post by B_9710)
    That's the tangent.
    Why would that be because the gradient of the normal is 0...Thanks
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    (Original post by coconut64)
    Okay, I will bear that in mind thanks. For the equation for the normal, I got y=1 at the end, correct?
    That is a horizontal line; the tangent. You need the normal, which is parallel to the y axis.
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    (Original post by NotNotBatman)
    That is a horizontal line; the tangent. You need the normal, which is parallel to the y axis.
    But the gradient of the normal is also 0 though...
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    (Original post by coconut64)
    But the gradient of the normal is also 0 though...
    No it isn't. It is perpendicular to the tangent. The gradient of the normal is undefined, it has infinite slope.
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    (Original post by NotNotBatman)
    No it isn't. It is perpendicular to the tangent. The gradient of the normal is undefined, it has infinite slope.
    Right... So the normal equation will just be x=0? thanks
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    (Original post by coconut64)
    Right... So the normal equation will just be x=0? thanks
    Almost, it would need to cross the curve at (pi/2, 1).
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    (Original post by NotNotBatman)
    Almost, it would need to cross the curve at (pi/2, 1).
    hence, x=pi/2?
 
 
 
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