trig question (1 Viewer)

[grimreaper]

Hey I got this question from a past paper I cant get out:

a)draw the graph of y=5sin2x, and then draw a line through the origin with positive gradient that touches the graph in exactly 5 places

b) hence find the positive value of m for which the equation 5sin2x = mx has exactly 5 solutions

I can do a) but not b), so any help would be appreciated

 

[nike33]

not exactly sure as we havent done this in class.. is the answer

9pi/20???

 

[grimreaper]

Originally posted by nike33
not exactly sure as we havent done this in class.. is the answer

9pi/20???

I have no idea I havent got the answers, and I dont know how to do it

 

[George W. Bush]

Well by inspection of the graph, you have four solutions from cutting the arcs above the x-axis (I hope you can tell what I'm talking about). Now, for exactly 5 solutions, we have to just touch the top of the 5th arc, so the line doesn't cut twice, which would be six solutions. You know the point at the top of that arc, and that gives you the gradient of the line with just 5 solutions i.e. m.

 

[grimreaper]

Originally posted by George W. Bush
Well by inspection of the graph, you have four solutions from cutting the arcs above the x-axis (I hope you can tell what I'm talking about). Now, for exactly 5 solutions, we have to just touch the top of the 5th arc, so the line doesn't cut twice, which would be six solutions. You know the point at the top of that arc, and that gives you the gradient of the line with just 5 solutions i.e. m.

hrm I'm not sure if I'm not understanding the question or youre not. The way I see it , the straight line should cut once at the origin, and then twice on either side ie it will cut the curve once on either side and touch it once on either side in order to have 5 solutions. However I'm really not sure how to find out the gradient. By the way, I'm pretty sure the straight line doesnt touch at the top of an arc - only a straight line parallel to the x axis would I think

 

[KeypadSDM]

Easy enough to solve, I suppose...

 

[grimreaper]

well yeah thats a graph of it keypad, but I can do that. It's a question from a past mid course paper so it cant be too hard to solve it...

 

[mushroom_head]

umm.... 4/pi ??
m= rise/run
=5/ (5pi/4)
=5 x 4/5pi
=4/pi

i dunno..... i have a feeling this is wrong

 

[grimreaper]

Originally posted by mushroom_head
umm.... 4/pi ??
m= rise/run
=5/ (5pi/4)
=5 x 4/5pi
=4/pi

i dunno..... i have a feeling this is wrong

Yes I think this is wrong - as far as I can tell the line doesnt touch the curve at 5pi/4, only a line parallel to the x axis would

 

[Xayma]

so it would just be a tangent to the curve at the last point

 

[mushroom_head]

eek! how do u do this Q??

 

[googleplex]

Hmm I think i remember a question like this and the teacher said all u can do is approximate the x values.

However I have seen questions like this in 4unit, except they are just normal graphs.

 

[grimreaper]

Originally posted by Xayma
so it would just be a tangent to the curve at the last point

But what is the last point?
Anyway, I think the only way is to do it by inspection

 

[CM_Tutor]

Originally posted by grimreaper
But what is the last point?
Anyway, I think the only way is to do it by inspection

By Extension 1 methods, all you can do is estimate the value - you could narrow it down by Newton's method if you were really keen, or elseapproximate it (by eye, by halving the interval, etc) - but you can't solve the equation.

 

[AGB]

Originally posted by CM_Tutor
By Extension 1 methods, all you can do is estimate the value - you could narrow it down by Newton's method if you were really keen, or elseapproximate it (by eye, by halving the interval, etc) - but you can't solve the equation.

when you say it cant be done by extention 1 methods....duz that mean there is a way to do it using extention 2 methods???

 

[KeypadSDM]

It means there is a way of doing it mathematically.

Not necessarily extension II methods either.

 

[CM_Tutor]

Originally posted by KeypadSDM
It means there is a way of doing it mathematically.

Not necessarily extension II methods either.

I concur - and AGB: off hand, I can't think of an Extn II method either.