SOLUTION: I have a problem with questions relating to variation as the sum of two parts. Here is a worked example. A quantity, p, is the sum of two terms, one of which is constant while th

Question 1203019: I have a problem with questions relating to variation as the sum of two parts.
Here is a worked example.
A quantity, p, is the sum of two terms, one of which is constant while the other varies inversely as the square of q.
When q = 1, p=-1;
when q = 2, p=2;
find the positive value of q when p=2.75
If a and b are constants, p=a+b/q2;
q = 1, p= -1
-1 = a +b/12;
therefore -1 = a+b (equation 1)
when q = 2, p=2
2 = a+b/22
2=a+b/4
8 = 4a+b (equation 2)
Subtracting equation 1 from 2 gives
9=3a
so a =3
Substituting a = 3 into equation 1 we get
-1 = 3+b
b=-4
so p=3-4/q2
so when p = 2.75, 2.75 = 3-4/q2
4/q2 = ¼
q2 = 16
q=+/- 4 and we want the positive value so q = 4
I have tried to apply the same approach to the following problem:
A quantity C is the sum of two parts. The first part varies directly as the cube of t;
the second part varies inversely as the square of t.
C = 74, t = 1
C = 34, t = 2
Find C when t = 3
I wrote C = a3 +b/t2
74 = a3+b/12
74 = a3 +b (equation 1).
34 = a3+b/22
34 = a3+b/4
136 = 4a3+b (equation 2)

Subtract equation 1 from 2
136 = 4a3+b
74 = a3 +b
This gives 62 = 3a3
a3 = 20.67
a = 2.74
Substituting into the first equation
74 = 20.67 + b/12
74 = 20.67+b
b=53.33
To find C when t = 3
C = a3+b/t2
C = 20.67 + 53.33/9
9C = 186.03+53.33
9C = 239.36
C = 25.6 but the answer is 62.
Where have I gone wrong?

Found 2 solutions by Theo, greenestamps:
Answer by Theo(13342) (Show Source): You can put this solution on YOUR website!
your problem seems to be that you assumed that the constant of variation was the same in both parts.
it is not.
this apparently didn't apply to the first problem you solved because one of the parts was a constant and didn't involve a constant of variation.
in the second problem, both parts required a constant of variation.
i initially assumed both constants of variation would be the same but ran into trouble
i then went back and assumed that they were different for each part.
this is how i solved it.
you are given that C = a + b, where a = x * t^3 and b = y / t^2.
both x and y are constants of variations for their part.
you are then given that C = 74 when t = 1 and C = 34 when t = 2
your two equations become:
74 = x * 1^3 + y / 1^2 when t = 1
34 = x * 2^3 + y / 2^2 when t = 2
the first equation becomes 74 = x + y
the second equation bcomes 34 = 8x + y/4
you now want to solve these equations simultaneously.
first equation becomes x + y = 74
second equation becomes 32x + y = 136
subtract the first equation from the second to get 31x = 62
solve for x to get x = 2
since x + y = 74, then y must be equal to 72.
you now have x = 2 and y = 72
x is the constant of variation for the first part and y is the constant of vriation for the second part.
they want you to find C when t = 3.
when t = 3, your equation becomes C = 2 * 3^3 = 72 / 3^2.
this becomes C = 2 * 27 + 72 / 9
this becomes C = 54 + 8 which becomes C = 62.

i'm pretty sure i did it right.
your biggest problem appears that you assumed the constant of variation was the same in both parts.
that's what i think.
i also think think that you assumed b was the constant of variation in the first equation.
it was not.

Answer by greenestamps(13200) (Show Source): You can put this solution on YOUR website!

You are doing most of the work right; but your equation for C in the new problem is not what you intended.

NOTE: For readability, use "^" (shift-6) to denote an exponent -- e.g. "q^2" and "t^3" instead of "q2" and "t3". (It might be that you were using some other special character to represent and exponent, and that special character doesn't show up at all.)

In the new problem, you wrote

But what you meant was

Fix that and solve the problem using the methods you show and you will get the correct answer of 62.

I'll go ahead and finish solving the problem; but you will get more benefit if you finish it yourself and compare your work and final answer to mine.

t=1:
t=2:

The equation is

When t=3...

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