Additional Maths (p2)

[Padapop]

Coming up tmmr.

It's almost the 24 hours since i've done it. Thought paper 1 was relatively hard, i got all of the questions except for question 9 and 12. 12 was hell for me

What do u think would come up for the next paper? (ps. Please tell me when the 24 hour mark is )

[vanibharutham]

erm guys! :S :S my teacher told us that approximate change and increments are not coming in the exam, i looked at the syllabus properly, and they are there... i have no clue how to solve approximate change and small/large increments

Can anyone help?

i know that it is an application of differentials...

and i looked up on the net

the

the change in y = dy/dx * change in x

is that right?
anything i need to know?

[Padapop]

erm guys! :S :S my teacher told us that approximate change and increments are not coming in the exam, i looked at the syllabus properly, and they are there... i have no clue how to solve approximate change and small/large increments

Can anyone help?

i know that it is an application of differentials...

and i looked up on the net

the

the change in y = dy/dx * change in x

is that right?
anything i need to know?

Yep thats right/ Really nothing else u need to know. Change in y = dy/dx * change in x

[smashbros]

hey, i am not sure but i think dat is not it!!
a change/increase in y=dy/dt
as it is over time.
then they ask u to find out the change in y

but they give u the change in x!!!
so next u equate

as change in x=dx/dt

dy/dt=dy/dx x dx/dt

we differentiate 2 find dy/dx and then put it in the equation
^^^^^^^^
and we get the change in y over time!!!!

tada!!

[vanibharutham]

smashbros, what your talking about it CONNECTED RATE OF CHANGE

we were finding the approximate change

[divineobsidian]

i think the person above was correct. its

change in x/change in y = dx/dy and you rearrange

[Ghost Of Highbury]

rate of change
the rate of change of a variable x with respect to time t is dx/dt

if x and y are related by the equaton y=f(x). then the rate of change dx/dt and dy/dt are related by
dy/dt = dy/dx * dx/dt

if its small change
then
&y/&x = dy/dx

the percentage change in x is &x/x *100

--
hope this helped
 

[Ghost Of Highbury]

for e.g  (m = pi)
the area of a circle increases at a rate of  2m cm^2/s. Calculate the rate of increase of the radius at radius = 6cm

Ans: dA/dr = dt/dr  * dA/dt
 A= mr^2 therefore dA/dr = 2mr

at r = 6cm 2mr = 12m
so dA/dr = 12m
dA/dt = 2m cm^2/s

so the equation is

12m = dt/dr  *  2m
therefore dt/dr = 12m/2m = 6

dt/dr = 6
therefore dr/dt = 1/6  (inverse)

easy..
hope it helped
thank u

note: m = pi (i cud not put the symbol of pi in the question)
 

[vanibharutham]

Thanks + rep