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R0L13D

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Hey can anyone solve Q17?

been trying to get it fora while but just can’t get anywhere:
2D601E04-7A90-454A-B936-F196E8A9A980.jpeg828C157F-46F6-47FA-A6E6-42FF1D7970BD.jpeg
 

CM_Tutor

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NOTE: This post has been edited to correct a significant mistake in my earlier reasoning.

The diagonals of a parallelogram bisect the angles only if the sides are equal in length, and thus only if it is a rhombus.

Thus, to bisect the angle we need to adjust the vectors and to be unit vectors, and we can then say that the director of the angle bisector is the same as the direction of , the sum of the unit vectors. Then, since is in this direction, it must be a scalar multiple of this vector and so .

Note: I am using to mean the unit vector in the same direction as , which I would handwrite as a vector with an under tilde and a hat, but the implementation of tex at BoS can't display that (as far as I know) because a package plug-in is required.

--- Part (b) ---
We know that



and



and



We seek to show that



and the RHS of this



So, all you need to do now is to find



and show that it simplifies to

Now, we need to make use of the first part of the question in which we showed that . Recalling that the unit vector of any vector is found by dividing by its magnitude, that is, that



we have that



We can also express as



Equating the components of the two forms of gives



And, equating the components of the two forms of gives



And thus

 
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CM_Tutor

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For question 18
  • Show that
  • Show that
  • Put these into and expand / simplify to get
  • Since this dot product is zero, you have part (a)
  • From this result, you can show that or
  • Since you know the range of possible values for , you can find the range of possible values for
  • Note that one value is excluded as it allows only a single position for
This question appeared on several MX2 trial exams in 2020.
 

CM_Tutor

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Question 19 asks for a proof that the line

where is a unit vector

meets a sphere with centre and radius at two points if


A line will touch a sphere if it is a tangent, and so if its perpendicular distance from the centre of the sphere is equal to the radius. The line will meet the sphere at two distinct points and if its perpendicular distance from the centre is less than the radius. So, the problem here is to find the condition for , the closest point on the line to and which lies somewhere on the interval , to be less than units from . The shortest distance from a point to a line is the perpendicular distance, which in this case means that is perpendicular to the line.

This could also be thought of by taking as any point on the line, letting be the distance between and and seeking to show that the above condition corresponds to the minimum value of , , which occurs when is located at , satisfies . However, I don't think that a calculus-style max/min approach (here in 3 dimensions) is the way to approach the question - I mention it as a way to help visualise the problem.

In either case, I strongly suspect that the problem requires showing that


and I note that, since gives the direction of the line, we know that .

Now, since lies on the line, satisfies the equation and so .


Thus, the position of , the closest point to on the line, corresponds to the point where and so




















And we know that , and so


as required.
 

R0L13D

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For question 18
  • Show that
  • Show that
  • Put these into and expand / simplify to get
  • Since this dot product is zero, you have part (a)
  • From this result, you can show that or
  • Since you know the range of possible values for , you can find the range of possible values for
  • Note that one value is excluded as it allows only a single position for
This question appeared on several MX2 trial exams in 2020.
Do you know which trial exams from what schools? also would u happen to have any resources for mx2 trials?
 

R0L13D

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For question 18
  • Show that
  • Show that
  • Put these into and expand / simplify to get
  • Since this dot product is zero, you have part (a)
  • From this result, you can show that or
  • Since you know the range of possible values for , you can find the range of possible values for
  • Note that one value is excluded as it allows only a single position for
This question appeared on several MX2 trial exams in 2020.
wait I'm confused as to why alpha cant equal pi/2
 

CM_Tutor

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Do you know which trial exams from what schools? also would u happen to have any resources for mx2 trials?
I saw it for the first time on an MX2 trial from a student of mine. I subsequently saw it on the paper from one of the companys that sells papers to schools for use on their trials.

I will post resources / questions here from time to time or when people ask a specific question. There are quite a few 2020 MX2 papers on THSC.
 

CM_Tutor

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wait I'm confused as to why alpha cant equal pi/2
The question says that can take two positions for each value of .

Consider the diagram for the specific case that . Can you see why there can only be one location for ?

It's not unusual for more challenging questions in Maths exams for there to be a difficult-to-get mark that requires some detail to be recognised and addressed.

As a simple example, suppose that solving was a 3 mark question:
  • getting to could be worth 1 mark
  • translating that to or could be the second mark
  • recognising that is excluded (as the original fraction was undefined) and thus the answer is or , could be the final mark
 

R0L13D

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Cm_Tutor , could u help with Q21 (ii), please. I got the answer but was wondering what your method would be:
IMG_9480.jpg
 

CM_Tutor

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We are given that the coordinates of are and so


Now, if the point is represented by the position vector , then we can say that . Since is a point on the surface of the sphere with centre at , the radius must be the vector :


And we know that the length of the radius is 14 units, so the sphere can be expressed in vector form as


Though it was not sought, we can find the Cartesian equation of the sphere by putting :


Now, we know that is the point of contact of the tangent plane with the sphere. The vector for the radius at is


and the tangent plane is perpendicular to this vector.

Let be a point on the tangential plane, then is a vector in the plane given by


As we now have a vector in the plane and a vector perpendicular to it, we know that , which we can use to find the equation of the required plane:

 

R0L13D

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We are given that the coordinates of are and so


Now, if the point is represented by the position vector , then we can say that . Since is a point on the surface of the sphere with centre at , the radius must be the vector :


And we know that the length of the radius is 14 units, so the sphere can be expressed in vector form as


Though it was not sought, we can find the Cartesian equation of the sphere by putting :


Now, we know that is the point of contact of the tangent plane with the sphere. The vector for the radius at is


and the tangent plane is perpendicular to this vector.

Let be a point on the tangential plane, then is a vector in the plane given by


As we now have a vector in the plane and a vector perpendicular to it, we know that , which we can use to find the equation of the required plane:

HeY CM_Tutor. Just a question for Simple HArmonic Motion, how do u know what formula to use? Like for eg with this question:
Capture.PNG

how would you know to use x = c - acosnt

and not x = c + asinnt

or x = c - a sinnt

or x = c + acosnt

or x = c + acos(nt + alpha)

why is it x = c - acosnt?
 

cossine

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HeY CM_Tutor. Just a question for Simple HArmonic Motion, how do u know what formula to use? Like for eg with this question:
View attachment 30605

how would you know to use x = c - acosnt

and not x = c + asinnt

or x = c - a sinnt

or x = c + acosnt

or x = c + acos(nt + alpha)

why is it x = c - acosnt?
Hi ROL13D, the choice does not really matter in this case. The most general approach would be x = acos(nt + alpha) + c.

In this case I would use x = acosnt + c presuming you want time to be zero at 6:13AM. By making this choice a = -0.7, c = 1.2.

In the case of x = c - acosnt, then a = 0.7 and c = 1.2.

Assuming you want the coefficents to be positive for some weird reason then choice of x = c - acosnt will work. But it is not necessary and I personally would just stick with x = acos(nt + alpha) + c or x = c + acosnt (if you working maximum or minimum).
 

R0L13D

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Hi ROL13D, the choice does not really matter in this case. The most general approach would be x = acos(nt + alpha) + c.

In this case I would use x = acosnt + c presuming you want time to be zero at 6:13AM. By making this choice a = -0.7, c = 1.2.

In the case of x = c - acosnt, then a = 0.7 and c = 1.2.

Assuming you want the coefficents to be positive for some weird reason then choice of x = c - acosnt will work. But it is not necessary and I personally would just stick with x = acos(nt + alpha) + c or x = c + acosnt (if you working maximum or minimum).
So the question of x doesn’t matter?
 

CM_Tutor

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Hi @R0L13D,

Any problem in SHM comes back to the second-order DE


which has only one solution, but that can be written in a variety of equivalent forms, including:


The choice of which to use doesn't matter (in the sense that any will work) but does matter in that the choice can make the solutions more or less difficult. It is important to understand what each of the constants do:
  • The in the first two forms gives the amplitude of the motion as the trig function must be in and so . In the latter form, the amplitude is controlled by and .
  • The controls the period of motion, which is
  • The is a phase constant that translates the curve in time and should be zero if the problem is set up wisely.
As you have noted above, additional forms that add a constant are also known, which represent a change in the DE to by the substitution or . Historically, this was not considered sufficient to prove SHM, but I am unsure how it is seen after the syllabus change.

In any case, the key to SHM problems is to set up the problem so that:
  1. The origin of position, , is placed at the centre of the motion.
  2. The origin of time, , is placed sensibly with respect to the motion. This means that it should be defined where the motion is either at its centre or at an extremity (maximum distance from the origin of position).
  3. In solving the DE, if the position at is at an extremity, use a cosine form... if the position at is at the centre of motion, use a sine form.
  4. Following these, you should get or ... in the latter case, the solution can be rewritten to eliminate by expanding the double angle formula.
In the particular case above (and as shown in the attached diagram) I would start by
  1. Drawing a vertical axis, labelled (for height above wharf) and mark on it and .
  2. Draw dashed horizontal lines across from these values, and also a SOLID LINE from the average ().
  3. LEAVING A CENTIMETRE OR TWO SPACE TO THE RIGHT OF THIS AXIS, I would draw a curve between the and .
  4. I would now draw a vertical axis parallel to the axis, but to its right, and passing through the initial minimum at . I would mark where this crosses the solid line at as the origin, and label these axes as and , with the minimum at being at and the position being at .
  5. I can now add (in another colour) the time markings 6:13 am and 12:03 pm, the latter corresponding to . I can also add the time for a complete cycle (at 5:53 pm or ) and the times at the centre of motion (9.08 am and 2:58 pm, half way in time between the extremities), if I so choose. This gives a feeling for when the required time will be, when the height above the wharf is 2.0 m.
  6. I would now write my definitions:
Let be the height of the deck above the wharf at time hours after 6:13 am, so that at .
Let be the difference between and 1.9 m at time , (i.e. ) so that at ,

Now, since the motion is SHM about , we know that , for some constant .

The solution of this DE is , for some constant and where .

At ,

We know the period is

So,

We seek the times when :



So, the height of the deck above the wharf reaches 2.0 m at 3 h 11 min after , that is, at 9:24 am.

Though it is not asked, we see the deck returns to a height of 2.0 m above the wharf after 8 h 29 min, that is, at 2:42 pm.
We can use symmetry to recognise that this answer is correct as we know it took 2 h 39 min to reach its maximum height (at 12: 03 pm) from a height of 2.0 m, and will take the same amount of time to return to this height, and 2:42 pm is indeed 2 h 39 min after 12:03 pm.​
 

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CM_Tutor

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Follow Up:
  • If I chose to use a sine function, I would have or in my equations... though had I chosen at the centre of motion (at 9:08 am) then I would have got and would have found the solution for the same result, a height of 2.0 m above the wharf at 9:24 am (nearest minute).
  • If I chose to work in , rather than in , but still starting from 6:13 am, I would have had .
  • The biggest mistake you could make would be to set at (say) midnight to make the times easier, then would be some irritating decimal.
 

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