Mathematics: Post your doubts here!

[VelaneDeBeaute]

need help with question 7 ii) october/november 2008 mechanics M1

how come the constant is 5 when we integrated the acceleration ?

http://www.xtremepapers.com/papers/...AS Level/Mathematics (9709)/9709_w08_qp_4.pdf

From the glimpse of it (I did not really solve the question but anyway) when you integrate an equation, you always are left with a constant. In this situation, when you integrate the acceleration, you get an equation where you have three unknowns: v, t and c(the constant of integration)! To find c, you substitute the value of v you found in the part (a) as well as the value of corresponding t which leaves you with just an unknown c. Solve the equation to acquire the value of t. Remember! The constant of integration is only ignored or not considered when we're provided with absolute limits.
Hope this clears up!

 

[iKhaled]

someone explain to me question 6 (ii) mechanics may june 2010 p43 pls

http://www.xtremepapers.com/papers/...S Level/Mathematics (9709)/9709_s10_qp_43.pdf

 

[iKhaled]

From the glimpse of it (I did not really solve the question but anyway) when you integrate an equation, you always are left with a constant. In this situation, when you integrate the acceleration, you get an equation where you have three unknowns: v, t and c(the constant of integration)! To find c, you substitute the value of v you found in the part (a) as well as the value of corresponding t which leaves you with just an unknown c. Solve the equation to acquire the value of t. Remember! The constant of integration is only ignored or not considered when we're provided with absolute limits.
Hope this clears up!

btw what do u mean by absolute limits ?

 

[bamteck]

http://www.xtremepapers.com/papers/...S Level/Mathematics (9709)/9709_s10_qp_11.pdf

Please someone help me for no. 8

 

[leosco1995]

Gradient of AB = 2
Since gradient of AB = 2m (given),

2m = 2
m = 1

(ii) C is a point of intersection between AC and BC, so the first thing that should occur to you is obtaining and then solving simultaneous equations.

Gradient of AC = -2m = -2(1) = -2
Now with the gradient of AC and co-ordinates of A, you can form an equation in AC.

Likewise, you can form an equation in BC. C will turn out to be (-1, 6).

(iii) We're talking about the perpendicular bisector of AB. Perpendicular gradient of AB is negative reciprocal of 2, which is -1/2. And the co-ordinates of the point will be the midpoint of AB => (9, 10).

Now make the equation of the perpendicular bisector with (9,10) and the gradient of -1/2. You have the equation of BC as well and since they intersect, you can solve them simultaneously to obtain the coordinates of D.

 

[Ahmedm96]

http://www.xtremepapers.com/papers/...AS Level/Mathematics (9709)/9709_w08_qp_4.pdf no7 why its wrong when i do it a=9.25 given the equation and u=5 t=3-0.5=2.5 and the getting dis=ut+0.5at^2 5(2.5)+4.625(6.25)=41.4 +1.25=42.65

in marking scheme its solved by integrations and i cant understand why

 

[whitecorp]

http://www.xtremepapers.com/papers/CIE/Cambridge International A and AS Level/Mathematics (9709)/9709_w08_qp_4.pdf no7 why its wrong when i do it a=9.25 given the equation and u=5 t=3-0.5=2.5 and the getting dis=ut+0.5at^2 5(2.5)+4.625(6.25)=41.4 +1.25=42.65

in marking scheme its solved by integrations and i cant understand why

Recognize in (ii) that acceleration is no longer a constant( ie not equals to g) and is now a function of time, hence the equations of motions no longer apply; you have to discover the expression for its velocity through one round of integration, and its displacement through a second round of integration.

Hope this helps. Peace.

 

[bamteck]

Gradient of AB = 2
Since gradient of AB = 2m (given),

2m = 2
m = 1

(ii) C is a point of intersection between AC and BC, so the first thing that should occur to you is obtaining and then solving simultaneous equations.

Gradient of AC = -2m = -2(1) = -2
Now with the gradient of AC and co-ordinates of A, you can form an equation in AC.

Likewise, you can form an equation in BC. C will turn out to be (-1, 6).

(iii) We're talking about the perpendicular bisector of AB. Perpendicular gradient of AB is negative reciprocal of 2, which is -1/2. And the co-ordinates of the point will be the midpoint of AB => (9, 10).

Now make the equation of the perpendicular bisector with (9,10) and the gradient of -1/2. You have the equation of BC as well and since they intersect, you can solve them simultaneously to obtain the coordinates of D.

Thank you very much

 

[bamteck]

http://www.xtremepapers.com/papers/...S Level/Mathematics (9709)/9709_s10_qp_12.pdf

Please help me for no. 11 (iii) (iv) (v) only !
Thanks

 

[whitecorp]

http://www.xtremepapers.com/papers/CIE/Cambridge International A and AS Level/Mathematics (9709)/9709_s10_qp_12.pdf

Please help me for no. 11 (iii) (iv) (v) only !
Thanks

(iii) Notice that f(x) lies between 4-3=1 and 4+3=7 inclusive. Solving for f(x)=k can be interpreted as drawing a horizontal like y=k and finding the number of intersections with the y=f(x) curve. Hence, if no solutions arise, there are no intersections, ie k>7 or k<1.

(iv) A function has an inverse if and only if it is either strictly increasing or decreasing. Try sketching out the graph, you should be able to see that maximum value of A= pi radians, because in between pi/2 and pi radians, the function is strictly increasing. Alternatively, you can find dy/dx= -3 cosx, which is > or equals to zero for the domain defined.

(v) Let y=4-3sinx, then making x the subject gives x= arcsin (4-y)/3
Doing the reflection in the y=x line gives inverse of g to be = arcsin (4-x)/3
Hence, when x=3, g^-1 (3) =arcsin (4-3)/3 =arcsin (1/3).

Hope this helps. Peace.

 

[whitecorp]

Please someone help
The question is very simple though bt I m not getting the ryt answer :/
V r suppose to find the gradient
The points given r (p+3,q-5) and (q-5,p+3)
I m getting the gradient as 0 and the answer given in the book is -1 :S
Can someone plz xplain me y?

You probably got zero because you took the y-coordinate of the first point and subtracted it from the x-coordinate of the second point.
For two points (x1, y1) and (x2, y2), the gradient is simply given by m = (y1 - y2)/ (x1 -x2) [ or (y2 - y1)/ (x2 -x1) ]

Hope this helps.

 

[Taiyaba]

You probably got zero because you took the y-coordinate of the first point and subtracted it from the x-coordinate of the second point.
For two points (x1, y1) and (x2, y2), the gradient is simply given by m = (y1 - y2)/ (x1 -x2) [ or (y2 - y1)/ (x2 -x1) ]

Hope this helps.

silly mistake

btw thnx

 

[whitecorp]

silly mistake

btw thnx

No problem, we all make mistakes at times. Peace.

 

[rowanmhayter]

Please Help

 

[VelaneDeBeaute]

btw what do u mean by absolute limits ?

Haha - I meant just "limits".

 

[bamteck]

(iii) Notice that f(x) lies between 4-3=1 and 4+3=7 inclusive. Solving for f(x)=k can be interpreted as drawing a horizontal like y=k and finding the number of intersections with the y=f(x) curve. Hence, if no solutions arise, there are no intersections, ie k>7 or k<1.

(iv) A function has an inverse if and only if it is either strictly increasing or decreasing. Try sketching out the graph, you should be able to see that maximum value of A= pi radians, because in between pi/2 and pi radians, the function is strictly increasing. Alternatively, you can find dy/dx= -3 cosx, which is > or equals to zero for the domain defined.

(v) Let y=4-3sinx, then making x the subject gives x= arcsin (4-y)/3
Doing the reflection in the y=x line gives inverse of g to be = arcsin (4-x)/3
Hence, when x=3, g^-1 (3) =arcsin (4-3)/3 =arcsin (1/3).

Hope this helps. Peace.

OMG ! You are great ! Thanks a lot

 

[iKhaled]

Haha - I meant just "limits".

thanks a lot. this constant thingy annoyed me a lot. sometimes they put it and sometimes they don't and i always used to wonder why sometimes they do and sometimes they dont!

 

[whitecorp]

OMG ! You are great ! Thanks a lot

No problem, am glad it helped. Peace.

 

[VelaneDeBeaute]

thanks a lot. this constant thingy annoyed me a lot. sometimes they put it and sometimes they don't and i always used to wonder why sometimes they do and sometimes they dont!

lol! Its always there but in limits, they just ignore it!

 

[parthrocks]

hey whitecorp
need ur help urgently
shanky631 any idea bro
plzz reply asap!!thanks in advance