My STEP Question Thread

I did this question a while back. It's best to consider what happens when

\theta = 0, \frac{\pi}{6}, \frac{\pi}{4}, \frac{\pi}{3}, \frac{\pi}{2}, ...

and go about it that way. You'll notice a pattern.

Reply 3

OP

Hmmm, perhaps changing the thread title is a Sub only thing, as I do not seem able to do it. As for the curve, I have got a very (read VERY) rough sketch of it that i found by looking at the values of x when y=0 and the values of y when x=0 and looking at the turning points (which happen to be the same places e_e.) I do still wonder if there is a better way, however.

The problem with the cartesian equation is that you have a rogue

\theta

running about.

Reply 4

OP

Late night STEP ^_^

The gradient

y'

of a curve at a point (x, y) satisfies:

(y')^2 - xy' + y = 0

2y'y'' - y' - xy'' + y' = 0

y''(2y' - x) = 0

y'' = 0

or

2y' = x

y' = a

or

y' = \frac{x}{2}

y = ax + b

or

4y = x^2 + d

That is my working, however, the question says:

Hence show that the curve is either a straight line of the form

y = mx + c

(check), where c = -m^2 =( (This is the bit that I can't do)or the parabola

4y = x^2

(how do I get rid of the d?)

Reply 5

DFranklin

18

To get rid of d, consider what happens when x = 0.

Reply 6

In both case, substitute your sulotions and their derivatives into the original differential equation.

Reply 7

OP

Thanks a lot hash, that worked simply, and thanks for your help again. As for your solution DFranklin I cannot seem to compute what happens when x=0 at this point in time. I've got 6 weeks to build up experience for these exams lol. Thanks again for your help =)

Reply 8

OP

OK, last one, I promise!

The gradient

y'

of a curve at a point (x,y) satisfies:

(x^2 - 1)(y')^2 - 2xyy' + y^2 - 1 = 0

So I used the same method. I need to find that the curve is either a straight line of a form which I should specify, or a circle, the equation of which I should determine, so:

(x^2-1)2y'y'' - 2xyy'' = 0

2y''((x^2-1)y' - xy) = 0

y'' = 0

y = cx + d

By subbing back into the original formula I found that

d^2 - c^2 = 1

that is all I could gather:

y' = -\frac{xy}{x^2-1}

\frac{1}{y} y' = -\frac{x}{x^2-1}

lny = -ln(\sqrt(x^2-1)) + f

y\sqrt(x^2-1) = F

y^2 = \frac{F}{x^2-1}

this is not a circle =( I have tried twice and have got different answers out each time. Both were in the form [latex]y=f(\frac{A}{x^2-1}) so I assume there is something wrong with my method. Where am I going wrong?

Reply 9

You're differentiating the initial equation wrongly - you're treating variables (i.e the

x^2 - 1

as constants.

Reply 10

OP

I thought I applied the chain rule correctly all the way through. There are some terms which cancelled out which I did not include in the equation that I posted here. I will address the question again tomorrow, and if I am still reaching the same conclusions I will post back here, with my full differentiation. Thanks.

Reply 11

Ah, sorry. What you've done is correct, but you've made a sign error - you should be solving

\frac{dy}{dx} = \frac{xy}{x^2 - 1}

.

Reply 12

OP

Oh, haha. That makes thing a load easier, thankyou!

Reply 13

when you solve the above you should not get the equation of a circle.

Reply 14

DeanK22

when you solve the above you should not get the equation of a circle.

I'm guessing the arbitrary constant changes the sign though.

Reply 15

Hashshashin

I'm guessing the arbitrary constant changes the sign though.

Not really. That DE is variable seperable and can be done in your head so to avoid further speculation I am going to write things down [although looking at it you should get something of the form

y^2 - C^2x^2 = -C^2

edit - which will be a hyperbola

Reply 16

DeanK22

Not really. That DE is variable seperable and can be done in your head so to avoid further speculation I am going to write things down [although looking at it you should get something of the form

y^2 - C^2x^2 = -C^2

How about

x^2 - By^2 = 1

?

Reply 17

Hashshashin

How about

x^2 - By^2 = 1

?

erm, yeah but that isn't a circle. (although B > 0 should be specified for obvious reasons - i.e. you want a hyperbola, not an ellipse).

Reply 18

DeanK22

erm, yeah but that isn't a circle. (although B > 0 should be specified for obvious reasons - i.e. you want a hyperbola, not an ellipse).

Yes but I was presuming that substituting into the differential equation gives b = -1. I should probably check though.

Reply 19