# Given that y - 8 / x^3 , use calculus to determine , in terms of p , where p is small , the approximate change in y as x increases from 4 to 4 + pPlease also explain

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You should evaluate how much y changes if x increases from p to p+4 such that:

`(Delta y)/(Delta x) = (f(x+ Delta x) - f(x))/(Delta x)`

`(Delta y)/(Deltax) = (f(4+p) - f(p))/(p+4-p)`

`(Delta y)/(Delta x) = (8/(4+p)^3 - 8/p^3)/4`

Using the formula `a^3 - b^3 = (a-b)(a^2+ab+b^2)` yields:

`(Delta y)/(Delta x) =(2/(4+p)-2/p)(4/(4+p)^2 + 4(p(p+4)) + 2/p^2)/4`

`(Delta y)/(Delta x) =-2(4/(4+p)^2 + 4(p(p+4)) + 2/p^2)`

**Hence, evaluating the change in y under the given conditions yields `(Delta y)/(Delta x) =-2(4/(4+p)^2 + 4(p(p+4)) + 2/p^2).` **

I suppose you meant y = 8/x^3 . This is a question of Binomial expansion, let me show why it is so !

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*****[I will use the expansion:

(1+x)^n = 1 + nx + n(n-1)(x^2)/2! + ... where x << 1]

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(a) y (x) = 8/x^3

Now

y(4+p) - y(4) = 8 [ 1/(4+p)^3 - 1/4^3]

=> chenge in y = (delta y) = 8 [ (4 + p)^(-3) - 1/4^3]

= 8 * (1/4^3) [{(1+p/4)^(-3)} - 1 ]

= (1/8) [ {1 + (-3) * (p/4) + (-3)(-3-1) (p/4)^2 * (1/2 )

+ negligible terms} - 1 ]

as "p" is very small so p/4 << 1, so we can neglect terms which have higher powers of "p" as it will be smaller and smaller, so to get an aproximate result our above taken terms are sufficient ! Infact simplest result will be to take the least power which gives non zero result, in this sense we can take terms up to (p/4)^1 only but we cannot neglect this term because then the result will be 0 ! So we must take terms up to that which gives nonzero change in y.

So

delta y = (1/8) [ {1 + (-3) * (p/4)} - 1 ]

= (1/8)* [-3 * p /4]

approximate change in y = - 3 * p /32

(b) we repeat the same procedure:

y(1-p) - y(1) = 8 [ 1/(1)^3 - 1/(1-p)^3]

=> chenge in y = (delta y) = 8 [ 1 - {(1-p)^(-3)}]

= 8 * [ 1 - {1 + 3*p + negligible } ]

=> approximate change in y = 8 * (- 3 p) = 24 * p