Mathematics: Post your doubts here!

[Anum96]

Thats part iii where it asks to use iterative formula to find value of a
How to prove that it's converging to a ?

Oh thaT.
this is how u do it.

 

[***amd***]

View attachment 58958
Parts ii and iii please

I cant recall any other method. Maybe ***amd*** can help.

I guess you just need to rearrange
a sin(a) + cos(a) = 1.5
and prove by induction that when a = 1, left-hand-side expression equals 1.38, and when a = 1/2 pi, left-hand-side expression equals 1.57.
As the value of LHS is increasing from a=1 to a=0.5pi, the value of 'a' for which LHS = 1.5 must lie b/w 1 and 0.5pi. 'a' is then greater than 1.
since the graph of 'a sina + cos a' against 'a' is going to be a wavy kinda one, you jest prove that the part of it which lies between a=1 and a=0.5pi is increasing with value of LHS = 1.5 inside it.
http://fooplot.com/#W3sidHlwZSI6MCw...3MDQ5OTk0NiIsIjE0Ljk4NzMyNjk1NDQ5OTk1Il19XQ-- just zoom out and check out the shape of graph *_*

P.S. you should consult examiner report of mark scheme, I am not sure if my method is right or not. This should be acceptable though.

 

[Anum96]

Can you please do it ... ms says separate variables and integrate to obtain ln(x+2) of one side... I'm not getting that

There

 

[nehaoscar]

how to solve??

 

[nehaoscar]

Can anyone please solve to find values of A,B and C in these partial fractions ??

[A/(x+1)] + [B/(x+2)] + [C/(4x+3)] = the part ii shown above
So how to find A,B and C?? I can do it with 2 denominators... with the 3 I'm getting it wrong :/
w15-33

 

[Anum96]

View attachment 58988
Can anyone please solve to find values of A,B and C in these partial fractions ??

[A/(x+1)] + [B/(x+2)] + [C/(4x+3)] = the part ii shown above
So how to find A,B and C?? I can do it with 2 denominators... with the 3 I'm getting it wrong :/
w15-33

If your factorization is correct then you should get the correct answers. Take x as -1 then -3/4 and then -2

 

[nehaoscar]

how to solve??

 

[nehaoscar]

If your factorization is correct then you should get the correct answers. Take x as -1 then -3/4 and then -2

So do you take x=-1 on both sides and then solve like so: A(x+2)(4x+3) + B(x+1)(4x+3) + C(x+1)(x+2) so that B and C terms become 0 when x=-1 ??

 

[Anum96]

So do you take x=-1 on both sides and then solve like so: A(x+2)(4x+3) + B(x+1)(4x+3) + C(x+1)(x+2) so that B and C terms become 0 when x=-1 ??

Yup. You will get A. Next value will give u B and third C

 

[The Sarcastic Retard]

Integrate : sqrt(25 - x^2)
lower bound 0 and uper bound is 5

 

[Anum96]

Integrate : sqrt(25 - x^2)
lower bound 0 and uper bound is 5

(25 -x^2)^(1/2)
((25 - x^2)^(1/2 + 1))/(1/2+1 * -2x)
-((25-x^2)^(3/2))/3x

put the limits in and calculate.

 

[The Sarcastic Retard]

(25 -x^2)^(1/2)
((25 - x^2)^(1/2 + 1))/(1/2+1 * -2x)
-((25-x^2)^(3/2))/3x

put the limits in and calculate.

Hey thanks, I tried the same thing, but not getting the answer.
Lower bound is zero, hence the denominator will be 0, hence infinity.

 

[holoholo]

Can you please show me the steps for question (iii) ?

 

[Eugene99]

Can you please show me the steps for question (iii) ? View attachment 58996

which paper?

 

[sj0007]

Can anyone tell me as to why the speed in part (iii) would not exceed 20? (The method used to calculate this 20 is assuming a=0............. which is when there is max speed.......... so how is it the least value?)

Thnx

 

[qwertypoiu]

Can anyone tell me as to why the speed in part (iii) would not exceed 20? (The method used to calculate this 20 is assuming a=0............. which is when there is max speed.......... so how is it the least value?)
View attachment 59004
Thnx

I haven't read the question in detail, but one thing you say has caught my attention.

Acceleration being zero doesn't mean maximum speed, it could mean either max OR minimum speed.

It's just like the derivative of a function. When it's zero, it represents a total maxima OR minima of the original function. To find out which one it is (Max or Min) you have to do the sign changing test or evaluate the Second Derivative and see it it's greater than or less than zero.

Hope that makes sense.

 

[qwertypoiu]

Can you please show me the steps for question (iii) ? View attachment 58996

Take both the roots you obtained in part (i), and cube it, if it evaluates to -1, that's proof that it satisfies the given equation in the question. Hope that makes sense.

 

[Rizwan Javed]

Can anyone tell me as to why the speed in part (iii) would not exceed 20? (The method used to calculate this 20 is assuming a=0............. which is when there is max speed.......... so how is it the least value?)
View attachment 59004
Thnx

(ii) F = P/V

F ~ 1/V

Since the power is constant, the force is inversely proportional to V. In this situation, the acceleration will only be zero when the forward force will be equal to the resistive force. So in order to achieve this, the forward force must increase. As this will happen, the velocity of the car will decrease; therefore the velocity won't be a maximum when a=0 in this situation. Thus, this value of v is a minimum value here.

 

[sj0007]

I haven't read the question in detail, but one thing you say has caught my attention.

Acceleration being zero doesn't mean maximum speed, it could mean either max OR minimum speed.

It's just like the derivative of a function. When it's zero, it represents a total maxima OR minima of the original function. To find out which one it is (Max or Min) you have to do the sign changing test or evaluate the Second Derivative and see it it's greater than or less than zero.

Hope that makes sense.

Ohhhhhh, Ohkayyyyyy!
Yup that makes sense, though I haven't studied the last part

 

[sj0007]

(ii) F = P/V

F ~ 1/V

Since the power is constant, the force is inversely proportional to V. In this situation, the acceleration will only be zero when the forward force will be equal to the resistive force. So in order to achieve this, the forward force must increase. As this will happen, the velocity of the car will decrease; therefore the velocity won't be a maximum when a=0 in this situation. Thus, this value of v is a minimum value here.

Oooooh okay, ryt!
Thnx!!!!