"Change in the same way" is only referring to multiplication and division here! If you "add 1 to a" you do NOT "add 1 to b"I have questions in every step. Please help as soon as possible. I am in very much anxiety understanding this. Please explain in simple words. My brain is slow.
I was reading a article where it is said "if a is proportional to b then if any value of a,suppose a0 becomes 2a0 then b0 will change the same way as a0 i.e become 2b0
No. You wrote above, "a ∝ bc", which is correct- a is proportional to bc, not to b and c separately. If a doubles then the product bc doubles. It might be that b doubles while c doesn't change, that c doubles while b doesn't change, or that b is multiplied by 6 while c is divided by 3. Knowing that "a ∝ bc" does not tell you that a is proportional to b and c separately.2a0/2b0=a0/b0=Constant "
if a is 2 and b is 3 then 2a=4 and b will be 2b =6. ok got it .
MY problem starts from here :
after that it is written : "one may have a situation in which more than two variables are proportional to each other. For instance, we might have a situation in which a is proportional both to b and to c . In these situations, the above procedure only works if we keep values other than the two under considerations constant. More generally, using the exact same method as above, we can combine the proportionalities:
a/(b×c)= Constant (Since a is proportional to both b and c so all three variables change “together” ) "
after reading this part my understanding is this :if a is 2 ,b is 3 , c is 4 as they have told a is proportional to both b and c so values of b and c will change “together” in the same way as a changes. I am saying this line because in the direct proportion they have said that "if a is proportional to b then b will change the same way as a changes" . SO in this case of 3 variables i am applying the same logic of a ∝ b just here change is a ∝ bc so both the value of b and c will change depending on how a is changing .
doubt 1: if i double a to 2a =4 ; b will become 2b=6 ; c will become 2c=8 as (a is proportional to both b and c so b and c will change as a changes)
then again if i double 2a to 2 *2a =8, 2b to 2*2b=12, 2*2c=16
but if i do it like this a/(b*c) =k does not hold
they have said all three variables will change together. Am i right? HOW does the changing when there is more than 2 variables. Then arises the next question
doubt 2 : What do they mean by saying "we keep values other than the two under considerations constant" ? and why
I we have an equation like "a= pqbc" and b and c are the "values under consideration" the we keep p and q constant. Of course then their product, pq, is constant so a= constant* bc/
Your problem is that "i know am wrong"! No, you are not wrong. If a is proportional to both b andThen they have said "Note that this equality captures automatically the fact that if a∝b and a∝c then b and c are inversely proportional to each other. The reason is that by “inversely proportional” mean precisely that if b changes a certain way, then c changes in such a way that 1/c (i.e. its inverse) changes in the exact same way. Thus, we can express the inverse proportionality between b and c as b×c= Constant "
NOTE THAT my thinking process is like this : When i read up a∝b and a∝c just like i said before I assured myself thinking of the logic of direct proportionality if a0 becomes 2a0 then b0 will become 2b0 and c will be 2c0 . But i know i am wrong dont know the reason why . I am not understanding actually .
c and changes from a0 to 2a0 then b must change from b0 to 2b0 and c must change from c0 to 2co.[
Doubt 3 :
"k" is a constant. We are saying that, however x changes, y must change so that there product, xy, does not change. Specifically, if x is originally x0 and y is y0, we have (x0)(y0)= k. If x0 changes to 2x0, how must y change? Well we must still have "xy= k" so we must (2x0)(?)= k= x0y0. Divide both sides by 2x0 to get (?)= (x0y0)/(2x0)= (1/2)y0. If x changes from x0 to 2x0 then y must change from y0 to (1/2)y0 so that (2x0)((1/2)y0)= (2)(1/2)x0y0=x0y0= k still.if " b and c are inversely proportional " so i assumed one eg the original value of b is 4 and c is 2 then b *c=8 so if we double the value of b to 2b =8 and new value of c will become half of the previous value 1/2*c i.e 1 so as to maintain constant ratio 8 between them . This is inversely proportional.(product of 2 variables equals to constant)
i know that x*y =k def of inverse but still not getting the right explanation why divide one and multiply other .
Yes, xy= 4*3= 12 but NOT "by definition of inverse"! So far you haven't used an "inverse".I am thinking this way if my first choice of value of x is 4 and y is 3 so in the first stage i know that 4*3=12 by definition of inverse .
So i have to maintain this k (12) throughout my other values of x and y also .So if i try to change my x's value i.e 4 to 8 by multiplying 2 then i know that uptil now x has become 8 so the form is
8 * y=12 so the new value of y has to be 3/2 to maintain constant k so that previous value of y (3) has to be divided by 1/2 . thats why in inverse proportional we multiply and divide two values by the same factor. I hope my intution is correct?
Yes, that is correct.
"a/(bc)= constant" is the same as a= constant* (bc). You can think of "bc" as a single variable, say "d= bc" so that a= constant* d- d is proportional to a.4: i dont understand how this whole thing "a/(b×c)= Constant" is working out? What is the meaning of this whole thing. I understand 2 variable proportionality but more than 3 cannot understand.
in the case of direct proportionality a/b=constant i have made a table and it satisfies the ratio
a | b
2 |4 constant ratio of 2
but when it is more than 2 variables in the form of a/(b*c) = constant how the manupulation of values of variables takes place in compared to 2 variable ?
Did what in what way? And how did it "not work"?i did it in my way in the above in the doubt1 section but it is not working
Suppose we are told that a= pqbc where p, q, b, and c, are some numbers. The "two under consideration", as in the example above, are b and c. So we keep the other values, p and q, constant. That is, since p and q are constant, their product is also: a= constant*b*c as above.i think this is the main key line "we keep values other than the two under considerations constant" that i am not understanding
THIS IS NO DOUBT:
I think you are confusing yourself (you are certainly confusing me!) If a is proportional to b then multiplying a by any number multiplies b by that number and that is what is happening in this case: when a0 is multiplied by 2 so is b0: the ratio, 2a0/2b0, is exactly the same as a0/b0. When a0 is multiplied by 7 so is b: the ratio, 7a0/7b0, is exactly the same as a0/b0. Those are two distinct examples. 2a0/2b0= 7a0/7b0 because both are equal to a0/b0.Another thing i discovered in direct proportion if all the values of a/b gives a constant ratio=2
a | b
4| 2 a0 b0 ---->original value to start with
8 | 4 2a0=a1 2b0=b1
56 |28 7a1=a2 7b1=b2
i discover that the form is coming like this a0/b0= 2a0(i.e a1)/2b0(i.e b1)= 7a1(i.e a2)/7b1(i.e b2)= k(2) we can shorten it like this
a0/b0= 2a0/2b0= 14a0/14b0 =k
at first when i read the article i thought that every term would be like : a0/b0= 2a0(i.e a1)/2b0(i.e b1) =2a1(i.e a2)/2b1(i.e b2)=k
a0/b0= 3a0(i.e a1)/3b0(i.e b1)= 3a1(i.e a2)/3b1(i.e b2) =k
PLEASE READ---> that means at first i thought if i begin the original two values multiplying by a factor "x" then i also need to multiply the next 2 values i.e a1 and b1 to get a2 and b2 (from a1*x ->a2 & b1*x->b2) by "x" only but that is not the case as i have shown in this example
a0/b0= 2a0(i.e a1)/2b0(i.e b1)= 7a1(i.e a2)/7b1(i.e b2)
carefully see --->first two values are multiplied by 2 (a0*2->a1& b0*2->b1) then the next two values a1 and b1 are multiplied by 7 which is a different factor from the previous 2
(a1*7->a2 & b1*7->b2)
it is just that i have to make sure for each transition from a0*x to a1 &b0*x to b1 or from a1*x1 to a2 & b1*x1 to b2 each pair must have its own common factor so that when every pair is divided into simplest terms every pair will give k (constant ratio)