Year 13 Maths Help Thread

Original post by AMarques

Did you get that the coordinates of M are

(a\cos^{2}{\phi},a\sin{\phi}\cos{\phi})

? If so, then the next part, you can do by verification.

Edit: this is the Cartesian coordinates.

Alternatively, if you do not seek the method of verification, you can do so by doing this:

As

X = a\cos^{2}{\phi}

we can use the fact that

\cos^{2}{\phi} = \frac{1}{2}(1+\cos{2\phi})

so we get that

\dfrac{2X}{a} = (1 + \cos{2\phi})

so we can obtain that

\cos^{2}{2\phi} = \Big( \dfrac{2X - a}{a}\Big)^{2}

Using the similar approach, notice that

Y = \frac{a}{2}\sin{2\phi}

Hence,

\sin^{2}{2\phi} = \Big( \dfrac{2Y}{a} \Big)^{2}

.Using

\sin^{2}{2\phi} + \cos^{2}{2\phi} = 1 \Longrightarrow \Big( \dfrac{2X-a}{a} \Big)^{2} + \Big( \dfrac{2Y}{a} \Big)^{2} = 1

.

Rearranging,

4X^{2} - 4aX + a^{2} + 4Y^{2} = a^{2} \Longrightarrow X^{2} + Y^{2} = aX

Notice for any varying

\theta

and

r

we can say that

X = r\cos{\theta}, Y = r\sin{\theta}

.

Finally, notice that

X^{2} + Y^{2} = r^2

. Therefore

r^2 = ar\cos{\theta}

.

We can cancel the r, aslong as

C_{2}

covers the point at the pole i.e at

\theta = \pm \frac{\pi}{2}

.

This will give us the required result.

Your working out makes sense and it gets the right answer, however I'm unsure how the get the co-ordinates of M. I'm not used to working with lines in polar form. I can see that the line's equation is

\theta=\phi

but I'm unsure what substitutions go where to get it. I can only see M being

(acos\phi,asin\phi)

(edited 7 years ago)

Reply 203

Original post by RDKGames

\theta=\phi

but I'm unsure what substitutions go where to get it.

I basically turned both equations into Cartesian form, and then subbed one into the other and made x the subject. By doing this, I get three x-coordinates (including the pole which we don't count as either A or B), and hence three y-coordinates.

For the polar equation

\theta = \phi

this is the same as

y = \tan{\phi}x

in Cartesian form. For

C_{1}

we get

x^{2} + y^{2} = a(x \pm \sqrt{x^{2} + y^{2}})

the + sign for a > 0, and - for a < 0.

We then sub

\tan{\phi}x

for

y

in the second equation. By doing this, we can find both A and B, and then M will just be the average of the x/y coordinates.

Out of curiosity how did you find the distance AB without finding A and B explicitly?

(edited 7 years ago)

Reply 204

Original post by AMarques

\theta = \phi

this is the same as

y = \tan{\phi}x

in Cartesian form. For

C_{1}

we get

x^{2} + y^{2} = a(x \pm \sqrt{x^{2} + y^{2}})

the + sign for a > 0, and - for a < 0.

We then sub

\tan{\phi}x

for

y

To find the distance, I firstly knew that

\theta=\phi

and used that substitution into

r=a(1+cos\theta)

and that would give me intersection point A as a function of

\phi

.
For point B (and I'm not entirely sure how to explain this properly), I simply imagined that line at angle

\phi

above the initial line OL and the angle of

\pi-\phi

clockwise, below OL, would give the second point, therefore B would be

r_B=a(1+cos[-(\pi-\phi)])=a(1-cos\phi)

.

Since r represents the length, I simply added

r_A

and

r_B

AB=a(1+cos\phi)+a(1-cos\phi)

thus giving

2a

which is a constant in dependent of

\phi

that I required.

(edited 7 years ago)

Reply 205

B_9710

Original post by RDKGames

\theta=\phi

but I'm unsure what substitutions go where to get it. I can only see M being

(acos\phi,asin\phi)

At point A the coordinates are

(a\cos \phi (1+\cos \phi ), a\sin \phi (1+\cos \phi ))

and the coordinates at B are

(a\cos (\phi - \pi )(1+\cos (\phi - \pi ),a\sin (\phi -\pi )(1+\cos (\phi -\pi )))

. You can simplify for B.
Then the coordinates of the midpoint at point M are

(X, Y) =(\frac{1}{2} (x_A + x_b) , \frac{1}{2} (y_A + y_B ))

.
You can convert from parametric to Cartesian equation in

X

and

Y

and then back to polar form.

(edited 7 years ago)

Reply 206

Original post by RDKGames

To find the distance, I firstly knew that

\theta=\phi

and used that substitution into

r=a(1+cos\theta)

and that would give me intersection point A as a function of

\phi

.
For point B (and I'm not entirely sure how to explain this properly), I simply imagined that line at angle

\phi

above the initial line OL and the angle of

\pi-\phi

clockwise, below OL, would give the second point, therefore B would be

r_B=a(1+cos[-(\pi-\phi)])=a(1-cos\phi)

.

Since r represents the length, I simply added

r_A

and

r_B

AB=a(1+cos\phi)+a(1-cos\phi)

thus giving

2a

which is a constant in dependent of

\phi

that I required.

Oh that is quite a nice way of doing it. However, not sure how using polar coordinates would work to find M. Does what I said previously make sense?

Edit: The post above explains how to do it :smile:

(edited 7 years ago)

Reply 207

Original post by AMarques

Yes it does make sense, thank you. :smile:

I've only started polar stuff yesterday so I do prefer turning everything into Cartesian as of now so I'll keep this method handy, along with my geometric observation as shown. Is there just a method to notice the intersections for A and B straight away or would I necessarily have to do some substitutions before deriving them?

Reply 208

Original post by RDKGames

Yes it does make sense, thank you. :smile:

Using polar equations substitutions is normally the easiest method as demonstrated above, however there is no harm in doing Cartesian. When you say "notice the intersections" what do you mean? I normally sketch the given curves and it becomes apparent where they occur. In this case you could just substitute

\theta = \phi

and

\theta = \phi - \pi

, but in more common polar coordinates questions you'll get, you will normally get equations of the form

r = f(\theta)

(of course they can make it more complicated). In this case, you just substitute for

r

and find the corresponding

\theta

and so on...

Glad I was helpful, let me know how the polar coordinates stuff goes.

Reply 209

[QUOTE=AMarques;66724118]Using polar equations substitutions is normally the easiest method as demonstrated above, however there is no harm in doing Cartesian. When you say "notice the intersections" what do you mean? I normally sketch the given curves and it becomes apparent where they occur. In this case you could just substitute

\theta = \phi

and

\theta = \phi - \pi

, but in more common polar coordinates questions you'll get, you will normally get equations of the form

r = f(\theta)

(of course they can make it more complicated). In this case, you just substitute for

r

and find the corresponding

\theta

and so on...

Glad I was helpful, let me know how the polar coordinates stuff goes.

Ah yeah, thanks. I simply meant whether there is a fast way to get the points of interaction just by inspection. I'll post more polar stuff here which I may not understand so I can give you a tag if you're able to help further on this topic :smile:

(edited 7 years ago)

Reply 210

Original post by RDKGames

Ah yeah, thanks. I simply meant whether there is a fast way to get the points of interaction just by inspection. I'll post more polar stuff here which I may not understand so I can give you a tag if you're able to help further on this topic :smile:

Anytime!

Reply 211

Towards the end when it says "assuming that I has been found, the differential equation becomes..."

Where do they pluck that from? If I is as they define it there as e to the integral of P(x), where do they make that substitution? I am failing to understand how they get that line

ImageUploadedByStudent Room1470220492.982574.jpg

ImageUploadedByStudent Room1470220492.982574.jpg

Posted from TSR Mobile

Reply 212

Original post by RDKGames

Posted from TSR Mobile

The second and third line of working out on the sheet should make you notice how they substituted it. As they have found I to be defined as

e^{\int P.dx}

, then the result in line 3 and 4 of their working holds, so we can just substitute

I\dfrac{dy}{dx} + IPy

for

\dfrac{d}{dx}(Iy)

in line 2 of their working.

(edited 7 years ago)

Reply 213

Zacken

Original post by RDKGames

No, you have

\frac{dy}{dx} + p(x) y =q(x)

multiply this equation through by

I(x)

to get

I(x) \frac{dy}{dx} + I(x)p(x) y = I(x)q(x)

. You would really like this to be an 'exact' differential equation, i.e: of the form

\frac{d}{dx}(R(x)y) = I(x)q(x)

because then you can integrate both sides and be done.

Indeed, we want to choose an

I

such that

\frac{d}{dx}(I(x)y)

is the LHS of our differential equation. Differentiate that using the product rule to get

I(x) \frac{dy}{dx} + I'(x)y

.

Once we have that, we're done since we get

\frac{d}{dx}(Iy) = I(x)q(x) \Rightarrow Iy = \int I(x)q(x) \, \mathrm{d}x

and then divide both sides by

I(x)

.

Notice how this almost looks like our original differential equation multiplied through by

I(x)

? Infact, the only thing that's different is that the coefficient of our

y

term is

I'(x)

here and

I(x) p(x)

there. Since we want to choose

I(x)

such that

\frac{d}{dx}(Iy) = I(x) \frac{dy}{dx} + I(x)p(x)y

and hence allowing us to solve our differential equation, we need to make those two coefficients equal.

i.e:

\displaystyle I(x) = I'(x) p(x) \Rightarrow I(x) = \exp \left(\int p(x) \, \mathrm{d}x\right)

.

That is the motivating thing to do here is to take a DE, multiply it through by a certain function that turns it into an exact differential equation; to find this function, it must satisfy the above.

Reply 214