Mathematics: Post your doubts here!

[whitecorp]

(x-2)^2 + y^2 =r^2 is this a formula used to find the solution for these type of questions?
And it's still not v clear so can u please solve it?
Thnx

If a circle of radius r is centered at (a,b), then its cartesian equation is given by (x-a)^2+ (y-b)^2 =r^2 . In your question, a=2 while b=0.
Hope this helps. Peace.

 

[iKhaled]

The attachment you provided is the pure math portion, not mechanics. Please do check again. Peace.

http://www.xtremepapers.com/papers/CIE/Cambridge International A and AS Level/Mathematics (9709)/9709_w08_qp_4.pdf

pls help me here plsss

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

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

 

[Taiyaba]

Just find the distance between the points. Having the same distance from (2,0) shows its a circle. Just use the hypotunus theorem.

Thnx

 

[Taiyaba]

If a circle of radius r is centered at (a,b), then its cartesian equation is given by (x-a)^2+ (y-b)^2 =r^2 . In your question, a=2 while b=0.
Hope this helps. Peace.

Ok thnx

 

[iKhaled]

The attachment you provided is the pure math portion, not mechanics. Please do check again. Peace.

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

 

[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".