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Need help with a differentiation question

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.
Reply 2
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
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
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.
Reply 5
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 ?
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 sinπ2cosπ2 \frac{sin\frac{\pi}{2}}{cos\frac{\pi}{2}} is undefined as cosπ2=0 cos\frac{\pi}{2} = 0 and division by 0 returns an undefined result, but this isn't the case with cot x.
Reply 7
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 π/2 \pi /2 .
Reply 8
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 π/2 \pi /2 .


Oh okay, so this is the case for pi/2 and what other values ? Thanks
Reply 9
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.
Reply 10
Original post by NotNotBatman
because tan(pi/2) is undefined, there is an asymptote on at that point on the y= tanx graph. Also sinπ2cosπ2 \frac{sin\frac{\pi}{2}}{cos\frac{\pi}{2}} is undefined as cosπ2=0 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?
Reply 11
Original post by B_9710
All the values where tanx=0.


Is the equation of the normal y=1? thx
Reply 12
Original post by coconut64
Is the equation of the normal y=1? thx


That's the tangent.
Reply 13
Original post by B_9710
That's the tangent.


Why would that be because the gradient of the normal is 0...Thanks
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.
Reply 15
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...
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.
Reply 17
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
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).
(edited 7 years ago)
Reply 19
Original post by NotNotBatman
Almost, it would need to cross the curve at (pi/2, 1).


hence, x=pi/2?

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