Posted by Whitecorp on April 2, 2014 at 10:35 AM | comments (0) |
The circle below has a radius of 6 cm and point C is moving clockwise around the circle. Assume that point B is moving away from point A at a rate of 2 cm./sec. i.e. x is increasing at a rate of 2 cm./sec. At this moment , assume θ is pi/3 radians.
a. What is the instantaneous rate of change of θ at this moment. i.e. how fast is θ decreasing at this moment?
b. Also, at this moment, is θ decreasing faster or slower? By how much?
Student X
At any particular instant, let's focus on the right angled triangle ABC where AB= x units and angle CAB = θ, where both x and θ varies with time.
A relationship can be formed between x and theta, which is
x = 6 cos θ
Differentiating both sides wrt y gives
dx/dt = - 6 sin θ * dθ/dt -------(1)
Substituting dx/dt =2 cm/s, θ = π/3,
instantaneous rate of change of θ at this moment
= dθ/dt = 2 ÷ [ -6 * sqrt(3) /2 ] =-0.385 rad/s (shown)
Since x increases at a constant rate of 2cm/s, its second order derivative wrt time is zero, ie d^2 x/dt^2 = 0 cm/s^2
Differentiating (1) on both sides wrt t once again,
d^2 x/dt^2 = -6 [ cosθ *( dθ/dt)^2 + sinθ * (d^2 θ/dt^2 )]
(Be mindful of how the product rule is employed in the RHS)
Substituting d^2 x/dt^2 = 0 cm/s^2, θ = π/3 and dθ/dt = -0.385 rad/s,
0 = -6 [ cos(π/3) *( -0.385)^2 + sin(π/3) * (d^2 θ/dt^2 )]
Solving gives d^2 θ/dt^2 = -0.0856 rad/s^2
Hence, at this moment, θ is decreasing slower (because of the negative sign) by 0.0856 rad/s. (shown)
Hope this helps. Peace.
Best Regards,
Mr Koh
Posted by Whitecorp on March 11, 2014 at 10:35 AM | comments (0) |
Hi,
I don't quite get why:
Var (U-V)= Var(U) + Var (V) - 2Cov( U, V)
I get the covariance bit but why isn't it Var(U) - Var (V) ?
Student X
Understand that every variable introduced involves a certain amount of uncertainty (some may like to call it errors); the more variables, the greater the uncertainty. In this regard, separate variances of variables will reinforce each other as opposed to eliminating each other. Covariance is characterized as a quantity removed as uncertainty overlaps occur when one variable is somewhat dependent on the other.
Hope this clarifies. Peace.
Best Regards,
Mr Koh
Posted by Whitecorp on March 10, 2014 at 9:45 AM | comments (0) |
I have been following the workings in my textbook on autonomous equations and can follow everything except for one step: solving the solution for y.
The general solution is:
To satisfy the initial condition
, ,
Despite having been working on this for a good deal of time, I simply can not see how they have solved for y, giving:
Could you tell me how this is reached?
Student X
Inverting both sides concurrently yields
Now, making use of the substitution
,
we have
Therefore,
Hope this helps. Peace.
Best Regards,
Mr Koh
Posted by Whitecorp on March 9, 2014 at 9:50 PM | comments (0) |
Hello,
I am currently retaking my Maths GCSE atm on a distance learning course. The course lacks a lot of depth in the way they explain things, and doesn't show how they got the solutions, it just solely states the solution. I am stuck on this ratio word problem, would you be willing to help?
Here's the question:
The hotel bill for 4 people staying 7 days comes to £630. How much would 3 people who stayed 10 days have to pay?
I have developed several different answers but none of them are correct. Any help would be much appreciated.
Thanks for reading.:)
Student X
You can break things down by systematic reasoning. Some food for thought:
If the hotel bill for 4 people staying 7 days comes to £630, then 4 people staying for 1 day would need pay £630/7 =£ 90
In consideration of this, what happens if there is only a single person staying at the hotel for 1 day?
Naturally he only pays his share which is equivalent to £ 90/4 =£ 22.5
On the other hand, should this loner choose to stay at the hotel for a stretch of 10 days, the cost would come to £ 22.5*10 =£ 225
I will leave you to figure out the rest, which shouldn't be all that difficult at this point.
Hope my explanation helps. Peace.
Best Regards,
Mr Koh
Posted by Whitecorp on February 28, 2014 at 10:05 AM | comments (0) |
A dog food manufacturer makes 3 types of dog chew, each 10g in weight, which are made from different proportions of 2 basic ingredients. The table below shows this, together with the amount of the two ingredients in stock, and the costs for the three types of chew.
chew type | ingredient 1 | ingredient 2 | cost (p per chew)|
type A 8g 2g 1.8
type B 6g 4g 1.6
type C 5g 5g 1.5
availability 800kg 400kg
The manufacturer wants to make 1600 packets of mixed chews. Each must contain 60 chews, and there must be no more than 30 of each type of chew in a packet.
a) Define appropraite variables and formulate an LP showing that:
8x+6y+5z<=500
2x+4y+5z<=250
b) By eliminating the variable z show that the objective function can be given by c = 90+0.3x+0.1y
and hence define the problem in terms of 2 variables.
c) By using a graphical approach solve the LP stating all the possible solutions.
d) From your graph you should see that 2 of the calculated constraints are unnecessary. Which constraints are these?
ANY HELP IS APPRECIATED!
Student X
Here's getting you started:
Let x, y and z represent the number of each type of chews A, B and C respectively in a packet.
Since 1600 packets are to be produced and a maximum of 800kg of ingredient 1 can be used in total, 800/1600= 0.5 kg or 500 g is the maximum amount of ingredient 1 which can be used in a single packet.
Similarly, since 1600 packets are to be produced and a maximum of 400kg of ingredient 2 can be used in total, 400/1600= 0.25 kg or 250 g is the maximum amount of ingredient 2 which can be used in a single packet.
Hence, since the total amount of ingredients 1 and 2 used in making a single packet of chews are separately represented by 8x+6y+5z and 2x+4y+5z respectively, imposing the above conditions would yield
8x+6y+5z<=500
2x+4y+5z<=250 (shown)
We also know that since each packet must contact exactly 60 chews,
then x+y+z=60 ====> z = 60 -x -y---------(1)
The cost involved in manufacturing one packet is c = 1.8x + 1.6y + 1.5z --------------(2)
Substituting (1) into (2):
c = 1.8x + 1.6y + 1.5(60 -x -y) = 1.8x +1.6y + 90 -1.5x -1.5y
= 90 +0.3x +0.1y (shown)
Hopefully I have given you sufficient reference to continue working on your own through the remaining parts of the question which should be pretty manageable. Peace.
Best Regards,
Mr Koh
Posted by Whitecorp on February 15, 2014 at 9:50 PM | comments (0) |
Hello,
Please help me in this math question-!!
The polynomial f(x) has a factor (x - 1) and leaves a remainder of 3 when divided by (x + 2).
Given that the following relation f(x +1) + f(x +4) = a is true for all real values of x, find the value of a.
I didn't understand the back part of the question and I am lost now so please help me.
Student X
f(1)=0 since x-1 is a factor and f(-2)= 3 since f(x) yields a remainder of 3 when divided by x+2.
f(x +1) + f(x +4) = a is true for all values of x implies we can choose any value of x, substitute into the LHS expression and the RHS will always evaluate to the value of a.
So choosing x=-3,
f(-3+1) + f(-3+4) = f(-2) + f(1) = 3 + 0 = 3
Therefore, the value of a is 3. (shown)
Hope this helps. Peace.
Best Regards,
Mr Koh
Posted by Whitecorp on January 25, 2014 at 10:20 PM | comments (0) |
Express 4/(2+√3+√7) in the form 1 + a√x + b√y where x and y are integers.
Student X
Rationalization is required in this instance, and (a+b)(a-b)= a^2 -b^2 shall be used.
4/(2+√3+√7)= 4/(2+√3+√7) * [(2+√3)-√7]/[(2+√3)-√7]
= 4[(2+√3)-√7]/[ (2+√3)^2 - (√7)^2]
= 4[(2+√3)-√7]/[ 4 + 4√3 + 3 - 7]
= 4[(2+√3)-√7]/4√3
= [(2+√3)-√7]/√3
= [(2+√3)-√7]/√3 * √3/√3
= √3[(2+√3)-√7]/ 3
= 2√3/3 + 1 - √21/3
= √12/3 + 1 - √21/3
= 1 - 1/3*√21 + 1/3*√12 (shown)
Hope this helps. Peace.
Best Regards,
Mr Koh
Posted by Whitecorp on October 7, 2013 at 7:20 AM | comments (0) |
(This comes from a former student of mine now in her second year at Singapore Management University (SMU)-I received this e-mail earlier on 9 September 2013):
Hi Mr Koh!
XXX here. How are you? School has started and stats has come back to haunt. Have a couple of questions, hope you won't mind answering them!
For starters, what is this?
Attached are my lesson slides for context if needed.
Thanks so much and hope to hear from you soon!
Warm regards,
Former Student X
Hi ,
Ah so the new semester has started. Nice ppt slides.
Let me create examples to help you understand better:
(i) E(Y) when Y is a discrete variable
Let's say we have Y=X^2 and the probability distribution function (pdf) of X is as follows:
X 0 1 2 3 4
P(X=x) 0 1/6 1/3 1/12 5/12
Then the probability distributionfunction (pdf) of Y would be:
Y 0 1 4 9 16
P(Y=y) 0 1/6 1/3 1/12 5/12
In that regard, E(Y) = 0(0) +(1)(1/6) + 4(1/3) + 9 (1/12) + 16 (5/12)
= 8.9167 (shown)
(ii) E(Y) when Y is a continuous random variable
Let's say X is defined by the following probability distribution function:
f(x) = 2x + 1 0 ≦ x ≦ 1, 0 otherwise
and Y is related to X such that Y=X^2,
then E(Y) = E(X^2) is obtained by evaluating the integral ∫ (x^2) * (2x+1) dx from x=0 to x=1.
Hope this clarifies. Welcome back to Statistics.
Best Regards,
Mr Koh
Posted by Whitecorp on September 20, 2013 at 8:40 PM | comments (0) |
Hi,
We're currently learning about Taylor series and I'm having a lot of trouble understanding the concept and the math involved in it. I typed into Wolfram Alpha the function e^-(x^2) at point 0 and order 4 and have attached the output:
My understanding is you start by differentiating the function 4 times but I don't understand how to get from there to the equation Wolfram Alpha has given me. I would really appreciate some help with this as the web-sites and books I have looked through so far have only managed to confused me further.
Thanks in advance
Student X
The Taylor series is explained by
f(x) = f(a) + f'(a) * (x-a) + f" (a) *( x-a)^2 /2 ! + f''' (a) *( x-a)^3 / 3! +........
When the function to be represented is about x=0 as mentioned in your problem, the taylor series is reduced to that of a Maclaurin's Series,
ie f(x)= f(0) + f'(0) * x + f" (0) *x^2/ 2! + f'''(0) * x^3/3! +........
Let's work things out to obtain a clearer picture.
Assuming y =f(x)= e^-(x^2) , and therefore implying f'(x)=dy/dx, f"(x)= d^2 y/ dx^2 so on and so forth,
differentiating this once wrt x on both sides gives
dy/dx = -2x * e^-(x^2) = -2xy (Note: The e^-(x^2) component is replaced simply by y)
Differentiating a second time wrt x on both sides
d^2 y/ dx^2 = -2 [ x *dy/dx +y] = -2x*dy/dx -2y
Differentiating a third time wrt x on both sides,
d^3 y/ dx^3 = -2[ x* (d^2 y/ dx^2) +dy/dx] -2 dy/dx
= -2x * (d^2 y/ dx^2) -4 dy/dx
Differentiating one final time wrt x on both sides,
d^4 y/ dx^4 = -2 [ x* (d^3 y/ dx^3) + (d^2 y/ dx^2)] -4 (d^2 y/ dx^2)
Now, we seek to find the values of the various orders of derivatives when x=0, ie
f(0)= value of y when x=0,
f'(0) = value of dy/dx when x=0,
f"(0)= value of (d^2 y/ dx^2) when x=0, so on and so forth
By making a series of substitutions into the above block of differential equations,
When x=0,
y=1, dy/dx= 0, d^2 y/ dx^2 =-2 , d^3 y/ dx^3 = 0 and d^4 y/ dx^4 = 12
Hence,
the expansion of the series is given by
f(x)= 1 + 0 * x + (-2) *x^2 /2! + 0 *x^3/ 3! + (12) *x^4 /4! +.......
= 1 - x^2 + 0.5 * x^4 -.........
This may take a while to internalize, so be patient.
Hope it helps. Peace.
Best Regards,
Mr Koh
Posted by Whitecorp on August 27, 2013 at 11:30 PM | comments (0) |
Hello,
My math teacher gave us a worksheet with six 4-letter words.
Book = 13
Full= 19
Sail= 20
Boil = 11
Silk = 18
Foil = 10
Each letter in the word has a value and they all should equal the number after the word.
also, the B for instance in the word Book has to be the same value as the B in the word Boil, and etc.
I cannot figure this out after hours of trying. How do I do this? Thank you for your time!
Student X
You can formulate a system of linear equations to solve this.
If we let the values of the individual letters be represented by the letter variable themselves,
then
B+ 2O + K = 13
F+ U+ 2L = 19
S + A + I + L = 20
B + O + I + L = 11
S + I + L + K= 18
F + O + I + L =10
The resulting augmented matrix can be written as such:
A B F I K L O S U
0 1 0 0 1 0 2 0 0 13
0 0 1 0 0 2 0 0 1 19
1 0 0 1 0 1 0 1 0 20
0 1 0 1 0 1 1 0 0 11
0 0 0 1 1 1 0 1 0 18
0 0 1 1 0 1 1 0 0 10
A first tier reduction of the matrix gives
A= 12 - 1/2 O - 1/2 S
B= 3 - 3/2 O + 1/2 S
F= 2 - 3/2 O + 1/2 S
I= -1/2 - 1/4 O - 1/4 S + 1/2 U
K= 10 - 1/2 O - 1/2 S
L= 17/2 + 3/4 O - 1/4 S - 1/2 U
O= O
S = S
U= U
Assuming all letters are associated with integer values, then a possible choice would be to let O=2, S=4 and U=6.
Then A=9, B= 2, F=1, I=1, K=7 and L=6.
Naturally there will be many other possible solution sets since this system involves 9 variables yet is only described by 6 equations.
Hope this helps. Peace.
Best Regards,
Mr Koh