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

2 Answers | Add Yours

sciencesolve's profile pic

sciencesolve | Teacher | (Level 3) Educator Emeritus

Posted on

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).`

quantatanu's profile pic

quantatanu | Student, Undergraduate | (Level 1) Valedictorian

Posted on

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

 

 __________________________________________

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

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

__________________________________________

 

(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

                                       

We’ve answered 318,916 questions. We can answer yours, too.

Ask a question