Secondary Maths questions, give me your maths questions

[mdc]

I draw the graph, work out which bit is above and below the x axis; in this case between pi/6 and pi is below, between pi and 7pi/6 is above.

right.

So i do -S (-2cos(x/2)dx (from pi/6 to pi) + S-2cos(pi/2)dx from pi to 7pi/6

I think somewhere in there i'm making some errors, sorry about the confusing working out (the s is meant to be the anti derivitive sign)

I assume the bold bit is supposed to be x, not pi. Otherwise it looks ok.

Without your working I can't really tell you where you could be going wrong. Best I can do is remind you that Scos(x/a)dx = asin(x/a). I.e. no - sign should appear during the integration bit.

 

[Subaru Impreza]

With these integration of trig functions I like to draw the graph first.

I also did the manual working out, the negative area, treat it as a modulus (or absolute value)

 

[magic_johnson!]

With these integration of trig functions I like to draw the graph first.

I also did the manual working out, the negative area, treat it as a modulus (or absolute value)

Thank you!!!!!!

 

[Speck]

How do you find the anti derivative of:

x^3+1 dx
x^2+1

I can't for the life of me figure out what to do.

 

[jinny1]

hmmm. Im on my iphone.. But have u tried using partial fractions to spilt it up. And u will end up with 3 terms. 2 of which will be easy to antidiff. The third one would have "x" as the numerator. So u use linear substitution to antidiff thatterm.( im probly wrong lol got no paper to try wprking out)

 

[mdc]

How do you find the anti derivative of:

x^3+1 dx
x^2+1

I can't for the life of me figure out what to do.

Do polynomial division, and you should get something like x + (1-x)/(x^2+1). Split the latter term to get:

x + 1/(x^2+1) - x/(x^2+1), and you should be able to do those three integrals.

If you don't know polynomial division, just note that x^3+1 = x(x^2+1) - x + 1 and go from there.

 

[Speck]

hmmm. Im on my iphone.. But have u tried using partial fractions to spilt it up. And u will end up with 3 terms. 2 of which will be easy to antidiff. The third one would have "x" as the numerator. So u use linear substitution to antidiff thatterm.( im probly wrong lol got no paper to try wprking out)

Thanks mate but I'm pretty sure you can't split that into partial fractions (not to my knowledge anyway).

Do polynomial division, and you should get something like x + (1-x)/(x^2+1). Split the latter term to get:

x + 1/(x^2+1) - x/(x^2+1), and you should be able to do those three integrals.

If you don't know polynomial division, just note that x^3+1 = x(x^2+1) - x + 1 and go from there.

Yeah that it, seems easy now once I know where to start. Thanks a heap.

 

[SJ]

Heres one for you guys:

$10,000

needs to be repaid in 5 periods

interest rate is 15%

how much would be paid back if, payment was made at the start of the period? how much would be paid in payment was made at the end of the period?

Compound interest of 15% p.a.?

If payments are at the start of the period, is the first payment made at t = 0?

 

[Dylz48]

Experiments are being carried out on a new ‘high-tech’ swing in a playground, the motion of which follows the model of

y=(e^(-ct))*cos(at)

Where y is the distance in meters from the equilibrium point of the swing, t is the time in minutes from midday on Sunday, and a, c are real constants

a=pi/6, c=0.04 and e=eulers number

so: y=(e^(0.04t))*cos((pi/6)t)

You are then asked to graph this on a CAS calculator in the domain of [0,20] of t.
Which i had no problem with.

The question then asks:

1. Find the greatest distance the swing travels in one direction, to the nearest centimeter.

I assumed this was the greatest y value in the given domain as y is the distance from the equilibrium point. So I moved onto Q2:

2. Find the greatest length for the pathway of the swing, travelling in one direction, to the nearest centimeter (note: the pathway of the swing is the Arc of the swing and can be found using CAS)

I am unsure of what the second question is asking

3. Hence, find the time, to the nearest second, that this swing is furthest from the equilibrium point after initially being let go?
I'm assuming this is the x value of first stationary point however I'm guessing I need my answer from question 2 to show some form of working out due to the word 'hence'

The length of 351 seconds that a swing is in the air, from first letting go, to the point of first returning to equilibrium point, is quite unreasonable. This being so, the City Council has decided to employ an engineer, and to give up their pondering of a design for this new ‘high-tech’ swing.

The firm of engineers of plans to have this distance, from the letting go to the first return to the equilibrium point, be covered in a much shorter time. For ease, they plan to distance to still be 1 metre to the positive side of the equilibrium point, and 0.8 metre to the negative side. They plan to period of the cosine part of the function y=e^(-ct) cos(at) to be 1 minute instead of 12 minutes in the previous function.
For the function y=e^(-ct) cos(at), where y is the distance in metres from the equilibrium point of the swing, and t is the time in minutes, find approximate values for a and c that allow for this design.

I could work out a=2*pi however have no idea how to work out c

Can someone show me how to get the answer and explain the mathmatics behind it as I struggle to understand the way my teacher explains and have no idea where to go with these questions.

 

[mdc]

1. This won't be the greatest y-value, but rather the greatest difference between y-values. In this case, picture your swing being released at y = 1m, and swinging all the way to slightly less than 1m on the other side (where the first [negative] turning point occurs).

2. Maybe someone familiar with CAS can help.

3. This should be your t-value for the turning point you found in question 1.

4. To solve for c, use the t-value from Part 3 and the new y-value of 0.8 from the instructions in the question.

 

[Dylz48]

1. This won't be the greatest y-value, but rather the greatest difference between y-values. In this case, picture your swing being released at y = 1m, and swinging all the way to slightly less than 1m on the other side (where the first [negative] turning point occurs).

2. Maybe someone familiar with CAS can help.

3. This should be your t-value for the turning point you found in question 1.

4. To solve for c, use the t-value from Part 3 and the new y-value of 0.8 from the instructions in the question.

thanks your info for question 1 and 3 makes sense however with question 4part 4 doesn't the t value of the first turning point change as the period is now 2*pi instead of pi/6

 

[mdc]

thanks your info for question 1 and 3 makes sense however with question 4part 4 doesn't the t value of the first turning point change as the period is now 2*pi instead of pi/6

yep, you're right, just divide the original t-value by 12.

 

[SchumiUCD]

For part 4, you're right that a = 2*pi.

To get c, use the fact that the first stationary point "on the negative side" is 0.8m. At that point, the Cos part must equal -1. That gives:
Cos(2*pi*t) = -1
2*pi*t = pi
t = 0.5

-0.8 = e^(-0.5*c)*(-1)
c = -2*ln(0.8)

Incidentally, my best guess is that Part 2 wants the (curved) length the swing travels between the first positive and negative stationary parts. It's not a well worded question to say the least.

 

[DREAM TEAM BUFF]

I got this in a year 11 test and was a long wayout

1).Find the co-ordinates of the point A on the line x= -3 such that the line joining A to B(5,3) is perpendicular to the line 2x+5y =12.

2). Using the vertices A(2a,2b), B(-2c, 0), C(2c,0).
(i) prove that the medians of a triangle are concurrent. (a median is a line joining a vertex to a midpoint of the opposite sides; the lines are concurrent if they have a common point of intersection);
(ii) show that the centroid is 2a/3, 2b/3. (a centroid is the point of intersection of the medians).

 

[Subaru Impreza]

I got this in a year 11 test and was a long wayout

1).Find the co-ordinates of the point A on the line x= -3 such that the line joining A to B(5,3) is perpendicular to the line 2x+5y =12.

2). Using the vertices A(2a,2b), B(-2c, 0), C(2c,0).
(i) prove that the medians of a triangle are concurrent. (a median is a line joining a vertex to a midpoint of the opposite sides; the lines are concurrent if they have a common point of intersection);
(ii) show that the centroid is 2a/3, 2b/3. (a centroid is the point of intersection of the medians).

I did Question 1 here. m1 x m 2 = -1 , for the 2 gradients to be perpendicular

 

[Subaru Impreza]

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[Subaru Impreza]

The manufacturer of a certain fabris estimates that if she charges $10 per metre she can sell 500 m of the fabric each week. Her experience tells her that each 20 cents per metre price reduction she makes from this $10 will increase her weekly sales of the fabric by 25 metres.


Find the amount she should charge per metre to maximise her revenue.


(Revenue = Number of metres sold x price per metre)


Answer is $7. Any working out would be much appreciated.

 

[mdc]

The manufacturer of a certain fabris estimates that if she charges $10 per metre she can sell 500 m of the fabric each week. Her experience tells her that each 20 cents per metre price reduction she makes from this $10 will increase her weekly sales of the fabric by 25 metres.


Find the amount she should charge per metre to maximise her revenue.


(Revenue = Number of metres sold x price per metre)


Answer is $7. Any working out would be much appreciated.

Let R = revenue, P = price per metre, L = metres sold

R = P * L, so;

R = (10-0.2D) * (500+25D), where D is just the variable representing the amount of discounting, or the number of reductions of 20c/metre that she makes.

From there, it's just a matter of maximising R.

 

[magic_johnson!]

Just a few questions, left study to the last minute and have no time to ask the teacher before sac...

1) Given 3(x+1)y^2 = 4(2x-y-1), find dy/dx in terms of x and y.
Just struggling to group the two y's together

2) f'(x) = x(1-x^2)^1/2
f(x) = ??

3) y=xsinx. Find the volume of the 3D shape when rotate around the x axis between x=0 and x=2.

Have no idea how to anti differentiate y=xsinx, or basically any trig function with an x before it I can do the rest, just can't find that bit.

 

[jinny1]

Just a few questions, left study to the last minute and have no time to ask the teacher before sac...

1) Given 3(x+1)y^2 = 4(2x-y-1), find dy/dx in terms of x and y.
Just struggling to group the two y's together

2) f'(x) = x(1-x^2)^1/2
f(x) = ??

3) y=xsinx. Find the volume of the 3D shape when rotate around the x axis between x=0 and x=2.

Have no idea how to anti differentiate y=xsinx, or basically any trig function with an x before it I can do the rest, just can't find that bit.

1. Put the "y"s in the same expression, complete the square and then square root both sides to get y alone.

2. Oops my bad. Isnt this just a straight forward question requiring product rule?? I just did it on paper and didnt have any probs

3. To solve the question you would actually need to integrate (xsinx)^2 dx (defined for 0~2, and times by pi) . This question is definitley a calculator question.

Oh btw are you doing specialist?? Gl for the sac

 

[SchumiUCD]

Just a few questions, left study to the last minute and have no time to ask the teacher before sac...

1) Given 3(x+1)y^2 = 4(2x-y-1), find dy/dx in terms of x and y.
Just struggling to group the two y's together

2) f'(x) = x(1-x^2)^1/2
f(x) = ??

3) y=xsinx. Find the volume of the 3D shape when rotate around the x axis between x=0 and x=2.

Have no idea how to anti differentiate y=xsinx, or basically any trig function with an x before it I can do the rest, just can't find that bit.

You don't need to group the ys together in #1. The question asks for dy/dx in terms of x and y so just differentiate each bit as is it appears in the question. When you differentiate y, you get dy/dx; when you differentiate y^2, you get 2y(dy/dx), etc. That'll give you an equation in dy/dx.

#2 is an integration problem jinny1. Let u = (1 - x^2) and it should work out.

For #3: is integration by parts on your course? That's the only way I could get an answer but it's very messy.

 

[jinny1]

#2 is an integration problem jinny1. Let u = (1 - x^2) and it should work out.

For #3: is integration by parts on your course? That's the only way I could get an answer but it's very messy.

How embarrassing...

Yes for #2, As Schumi said, use substitution where the derivative is present in the integrand.

and if you are doing VCE specialist, Integration by parts isn't on the syllabus so use your calculator

 

[magic_johnson!]

You don't need to group the ys together in #1. The question asks for dy/dx in terms of x and y so just differentiate each bit as is it appears in the question. When you differentiate y, you get dy/dx; when you differentiate y^2, you get 2y(dy/dx), etc. That'll give you an equation in dy/dx.

#2 is an integration problem jinny1. Let u = (1 - x^2) and it should work out.

For #3: is integration by parts on your course? That's the only way I could get an answer but it's very messy.

How embarrassing...

Yes for #2, As Schumi said, use substitution where the derivative is present in the integrand.

and if you are doing VCE specialist, Integration by parts isn't on the syllabus so use your calculator

Thanks for the help! Good about question three, was a bit worried about that... Tell you how i go by tomorrow/end of the week (it's the 40 mark sac...)

 

[SchumiUCD]

Good about question three, was a bit worried about that...

Ha ha, sorry about that! I'm not too familiar with the Australian syllabus, there isn't much on numerical methods in secondary school here.

 

[magic_johnson!]

OK, last one, hopefully... I think i got about 13 or 14 out of 19 for part one of three, but that's meant to be the easiest part so who knows. Not too disappointed so far...

Euler.
h (step size) = 0.5
interval = [0,2]
dy/dx = 2x^2 + 3x

y(0)=1