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Mathematics: Post your doubts here!

lxelle

lxelle

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sagar65265 said:
(This solution is for 7(ii), not for 7(i), 7(iii) or 7(iv), so in case you'd like a discussion on those too, do post here)

From solving part (i), we know that α satisfies the equation (99/100)x = sin(x). We can rearrange this equation by subtracting sin(x) from both sides to give us a zero equality, namely

(99/100)x - sin(x) = 0

Therefore, the root of this equation will satisfy this equality, and any values just above or below it shouldn't (unless they themselves are roots).

Also, note that this function is continuous - there are no points where this function is invalid, so at all points the equation should be valid, and since the graph drawn from this would be a smooth curve, it is continuous at all points - you can confirm this by checking that the gradient of this curve is a valid function - differentiating this equation, we get

f'(x) = (99/100) - cos(x)

(Contrast this with the f(x) = |x| graph - the graph takes a sudden turn at x=0, and so the gradient at x=0 cannot be verified by looking at the graph. However, with the above curve, the gradient at all points is verifiable, so it is smooth).

This smoothness is an important point - for there to be a root, the curve must touch or cross the x-axis at some point on the other, at which point the value of f(x) will be equal to zero (what we wanted from the solution). Since the curve is continuous, it has to go from above the x-axis to below the x-axis (or vice-versa, from below the x-axis to above the x-axis), cross the x-axis in the process, giving us a root.

Therefore, the one-line solution for this is to prove that the curve has opposite signs at each end of the range in which there is a root. In other words, we have to prove that the sign of f(0.1) is the opposite of the sign of f(0.5) - this means that the curve will cross the x-axis between these values of x, and thus give us a root somewhere in between.

(Working in radians,) we substitute the value x = 0.1 radians in f(x) and find that

f(0.1) = (99/100) * 0.1 - sin(0.1 radians) = -8.3 * 10^-4

which is clearly a negative value - when x = 0.1 radians, the function holds a negative value, i.e. (99/100)x is slightly less than sin(x). Now let's try x = 0.5 radians:

f(0.5) (99/100) * 0.5 - sin(0.5 radians) = + 0.479

which has the opposite sign, i.e. f(0.5) is positive, and (99/100)x is slightly greater than sin(x). Therefore, since we are looking for the value of x for which
(99/100)x = sin(x) (and for which (99/100)x - sin(x) = 0), we must find it somewhere between 0.1 and 0.5 - therefore, a root of this equation exists between x = 0.1 and x = 0.5.

There is an interesting case that i've not mentioned here, and that is if the curve just touches the x-axis and continues with the same sign after doing so - take the equation of (x-2)2 = 0 , which only has one root at x = 2, and is positive on both sides of the root (and always even otherwise, since it is a square). In this case, this root check will not be possible, but I doubt such situations will ever come up in your exams.

One thing in these situations is that as long as the equation is continuous, the gradient of the function at the root point will be equal to zero. In our above example, if we differentiate (x-2)2=0 , we get f'(x) = 2(x-2) , which is indeed zero at x=2, the point at which we find our root.

Hope this helped!
Good Luck for all your exams!
Click to expand...
THANKSSS! Could you please help me with 4ii and iii http://www.xtremepapers.com/papers/...AS Level/Mathematics (9709)/9709_s06_qp_7.pdf
 
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sagar65265

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iqbal said:
Please somebody help me out with this question....Q4
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4) i)

Since we know that the speed of each particle is the same as that of the other when Q has reached point B and P has reached point A, our plan of action can be equating the speeds of the particles at those points, and using that equation to find the initial velocity of Q.

Firstly, we know that Q starts out with an unknown initial velocity and accelerates with a non-uniform acceleration magnitude toward point B, while P starts out with a known initial velocity and accelerates with uniform acceleration magnitude toward point A. We also know that the time taken for Q to reach B is the same as the time taken for P to reach A.

Armed with these facts, let's begin.

Since P moves with a constant acceleration, we can say that the velocity of P at point A will be given by

v = u + at

and putting the values we know (a = 0.1 ms^-2 , u = 1.3 ms^-1 and t = 20 seconds), we get

v = 1.3 + 0.1 * 20 = 3.3 ms^-1

As for Q, we can write the acceleration equation as

a = dv/dt = (0.016t) ms^-2

which has the variable "t" on the right hand side, and a derivative with respect to this very variable on the left hand side - a differential equation. If we multiply both sides by "dt", we get

dv = (0.016t) * dt

and this we can integrate - on the right hand side, we integrate with respect to time, on the left hand side, with respect to velocity, giving us

v = 0.016 * (t2/2) + C

Where C is a constant representing initial conditions (if you put t = 0, you get v = C). Substituting t = 20 seconds in this equation, we get

v = 0.016 * (400/2) + C

We know that the velocity of particle Q at point B is the same as the velocity of particle P at point A, and since we calculated the velocity of particle P at point A to be 3.3 ms^-1, we can insert this value of v into the left hand side, giving us

3.3 = 3.2 + C
Therefore, C = 3.3-3.2 0.1 ms^-1. Therefore, our final equation is

v(t) = 0.008t2 + 0.1

and if we put t = 0 seconds , we get v(0) = 0.1 ms^-1, our answer to part (i).

4) ii)

Here, since P travels from O to A and Q travels from O to B, if we find the distance P travels in 20 seconds (when it reaches A) and add it to the distance Q travels in 20 seconds (when it reaches B) we have found the total distance from A to B, which is the length AB we are looking for.

Since P travels with a constant acceleration, we can find the distance it travels with the equation

s = ut + (1/2)at2

which gives us

s = (1.3)(20) + 0.5*0.1*400 = 46 meters

Which is the length OA. Now for length OB, we need to find the distance traveled by the particle Q, which we can get from the equation v(t) = 0.008t2 + 0.1, by writing it as

v(t) = dx/dt = 0.008t2 + 0.1

and if we multiply both sides by dt, we get

dx = (0.008t2 + 0.1) * dt

And on integrating, we get

x = 0.008 * (t3/3) + 0.1t + C

and if we put t=0 seconds and x = 0 meters (at t = 0), we see that C = 0. Therefore,

x(t) = 0.008 * (t3/3) + 0.1t

and if we put t = 20 seconds into this equation (which is when Q reaches B) we get

x(20) =23.33 meters.

So adding these two values of x, we get 23.33 + 46 = our answer = 69.33 m.

Hope this helped!

Good Luck for all your exams!
 
L

librement

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can someone help me out with this question, part (ii) only? Thanks!
 

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I

iqbal

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sagar65265 said:
4) i)

Since we know that the speed of each particle is the same as that of the other when Q has reached point B and P has reached point A, our plan of action can be equating the speeds of the particles at those points, and using that equation to find the initial velocity of Q.

Firstly, we know that Q starts out with an unknown initial velocity and accelerates with a non-uniform acceleration magnitude toward point B, while P starts out with a known initial velocity and accelerates with uniform acceleration magnitude toward point A. We also know that the time taken for Q to reach B is the same as the time taken for P to reach A.

Armed with these facts, let's begin.

Since P moves with a constant acceleration, we can say that the velocity of P at point A will be given by

v = u + at

and putting the values we know (a = 0.1 ms^-2 , u = 1.3 ms^-1 and t = 20 seconds), we get

v = 1.3 + 0.1 * 20 = 3.3 ms^-1

As for Q, we can write the acceleration equation as

a = dv/dt = (0.016t) ms^-2

which has the variable "t" on the right hand side, and a derivative with respect to this very variable on the left hand side - a differential equation. If we multiply both sides by "dt", we get

dv = (0.016t) * dt

and this we can integrate - on the right hand side, we integrate with respect to time, on the left hand side, with respect to velocity, giving us

v = 0.016 * (t2/2) + C

Where C is a constant representing initial conditions (if you put t = 0, you get v = C). Substituting t = 20 seconds in this equation, we get

v = 0.016 * (400/2) + C

We know that the velocity of particle Q at point B is the same as the velocity of particle P at point A, and since we calculated the velocity of particle P at point A to be 3.3 ms^-1, we can insert this value of v into the left hand side, giving us

3.3 = 3.2 + C
Therefore, C = 3.3-3.2 0.1 ms^-1. Therefore, our final equation is

v(t) = 0.008t2 + 0.1

and if we put t = 0 seconds , we get v(0) = 0.1 ms^-1, our answer to part (i).

4) ii)

Here, since P travels from O to A and Q travels from O to B, if we find the distance P travels in 20 seconds (when it reaches A) and add it to the distance Q travels in 20 seconds (when it reaches B) we have found the total distance from A to B, which is the length AB we are looking for.

Since P travels with a constant acceleration, we can find the distance it travels with the equation

s = ut + (1/2)at2

which gives us

s = (1.3)(20) + 0.5*0.1*400 = 46 meters

Which is the length OA. Now for length OB, we need to find the distance traveled by the particle Q, which we can get from the equation v(t) = 0.008t2 + 0.1, by writing it as

v(t) = dx/dt = 0.008t2 + 0.1

and if we multiply both sides by dt, we get

dx = (0.008t2 + 0.1) * dt

And on integrating, we get

x = 0.008 * (t3/3) + 0.1t + C

and if we put t=0 seconds and x = 0 meters (at t = 0), we see that C = 0. Therefore,

x(t) = 0.008 * (t3/3) + 0.1t

and if we put t = 20 seconds into this equation (which is when Q reaches B) we get

x(20) =23.33 meters.

So adding these two values of x, we get 23.33 + 46 = our answer = 69.33 m.

Hope this helped!

Good Luck for all your exams!
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Thanks
 
I

iYuuki

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Hey guys, could someone help me out with question 6 (ii)?? Really appreciate it! >.<
 

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AnujaK

AnujaK

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SitiPutri said:
Hi. Would've been better if you could compile all the images in one pdf file for each exam. It'd be much easier, instead of downloading the images one by one. Thanks for making the site, btw. :)
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Thanks for the tip! I'll try that :)
 
Haya Ahmed

Haya Ahmed

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Zain Salman Dar said:
Draw a circle of radius 2 cm at (0,2)
Draw an angle of 45 degrees (-2,0)
Calculate distance between (-2,0) and the circle+diameter
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how and why did you calculate the distance for modZ ... isn't it supposed to be from the (0,0) to the farthest point on the shaded region ?!
 
Haider Ejaz

Haider Ejaz

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MATH HELP NEEDED...PLEASE HELP

How to solve this inequality question?
(3x+1)(x+1)>0

Truly its an easy question but i am confused that if we take (3x+1)>0 and (x+1)>0 then it shows answer wrong.... why is this so??????? please help
 
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