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maths help pls

y=sqrt(x) makes an angle of a with the line y=0
find a
Reply 1
Original post by gomc
y=sqrt(x) makes an angle of a with the line y=0
find a


Why don't you try plotting the graph and seeing what arises.

Hint:

Spoiler

Reply 2
Original post by Zacken
Why don't you try plotting the graph and seeing what arises.

Hint:

Spoiler



so differentiate?
Reply 3
Original post by gomc
y=sqrt(x) makes an angle of a with the line y=0
find a


Have you written the question properly, because, upon reflection we have that y=12xy' = \frac{1}{2\sqrt{x}} which is surely undefined at (0,0)(0,0) - unless, of course the angle is the trivial a=π2=90a = \frac{\pi}{2} = 90^{\circ} because the tangent is essentially the y-axis?
Original post by Zacken
Why don't you try plotting the graph and seeing what arises.

Hint:

Spoiler




Whenever there's a thread to do with maths, I could almost guess you'd be the first one to post on it lool :tongue:
Reply 5
Original post by ravioliyears
Whenever there's a thread to do with maths, I could almost guess you'd be the first one to post on it lool :tongue:


Haha, I've got some free time, so I've just come to the maths forum and blaze out a few helping hands. :rofl:
Reply 6
Original post by Zacken
Have you written the question properly, because, upon reflection we have that y=12xy' = \frac{1}{2\sqrt{x}} which is surely undefined at (0,0)(0,0) - unless, of course the angle is the trivial a=π2=90a = \frac{\pi}{2} = 90^{\circ} because the tangent is essentially the y-axis?


yes thats all thats given
and its a non-calc question if that helps
Original post by ravioliyears
Whenever there's a thread to do with maths, I could almost guess you'd be the first one to post on it lool :tongue:

Or TeeEm!!
Reply 8
Original post by gomc
yes thats all thats given
and its a non-calc question if that helps


I actually do think the angle is 9090^{\circ} using my calculus method.

The only other method I can think of is using the tangent formula for angles between lines but x\sqrt{x} certainly isn't a line... what chapter/topic is this question from?

Edit: I'm sorry to have given you the full solution, but I only did so because I was uncertain about the question itself, many apologies.
Reply 9
Original post by Zacken
I actually do think the angle is 9090^{\circ} using my calculus method.

The only other method I can think of is using the tangent formula for angles between lines but x\sqrt{x} certainly isn't a line... what chapter/topic is this question from?

Edit: I'm sorry to have given you the full solution, but I only did so because I was uncertain about the question itself, many apologies.


i mean non calculator
its a core 1 extension question
no worries haha
Reply 10
Original post by gomc
i mean non calculator
its a core 1 extension question
no worries haha


In that case, I think the intended method was y=12xy' = \frac{1}{2\sqrt{x}}, so tangent at (0,0) is the y-axis, angle between x-axis and y-axis is 90 degrees. Boom. Done.
Reply 11
Original post by ghostwalker
Since you want the angle with the y-axis, turn your graph on its side and express x as a function of y. Answer becomes obvious.

Note: You want the angle with the y-axis, and it's not 90.


I'm probably being thick but isn't y=0y=0 the xx-axis?
Original post by Zacken
I'm probably being thick but isn't y=0y=0 the xx-axis?


Yep - ignore me.

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