small changes

Question

It is given that y = 8 - 10/x . Find the small change in x, in terms of p, when the value of y changes from 3 to 3+p.

 

Answer = 2p/5

Answer 1

Hi Joy!

 

The way I looked at this was through Taylor series expansion, which approximate function values at small shifts in input from a certain known value.  If you are not familiar with that, I can suggest how to get to the answer you gave by another approach.

 

The idea is to approximate the value of a function at some x which is very near a certain value a ((x-a) is very small), we can use a Taylor series expansion, written as:

 

f(x) = f(a) + f'(a)(x-a)/1! + f''a(x-a)²/2! + ...

 

The more terms you add, the more accurate this approximation becomes.

 

Usually, we look at small variations in the function input.  But, as presented, the question is asking about variations for the function output.  The way I dealt with this was to invert the function (rewrite it as x in terms of y).  As long as we stay away from any values where the function is undefined, this should be okay.  Once that is done, a change in y represents a change in function input again.

 

After some algebra, I obtained a function for x in terms of y:

 

x = f(y) = 10/(8-y)   (which is only undefined at y = 8, and we are looking at y = 3, so we're good)

 

At y = 3, you can calculate that x = 2.  The problem is interested in the "small change" in x that occurs when y shifts from 3 to 3+p, where p is very small.  I take the "small change" to be the second term in the Taylor series expansion, since the first term represents the value of the function at y = 3, and the second term is the first thing added to adjust that value at y = 3.

 

The first thing I need to do to evaluate that second term in the Taylor series expansion, in this case, is determine an expression for f'(y) (= dx/dy).  So I need to differentiate the function with respect to y.  This requires using the quotient rule for differentiation, which we can talk more about, if you like.  But my result was:

 

f'(y) = 10/(8-y)²

 

The second term in the Taylor series, which represents the "small change" in the value of the function f(y) is:

 

f'(a)(y-a)/1!    (remember that "y" is now our input variable instead of "x")

 

1! is just 1.  a is the "base value" from which we are starting the approximation, which in this case is 3.  And y is 3+p, the small shifted input value.  

 

f'(3) = 10/(8-3)² = 10/5² = 10/25 = 2/5

(y-a) = ((3+p) - 3) = p

 

So the second term in the Taylor series becomes:

 

f'(3)((3+p-3)/1 = (2/5)p = 2p/5, as listed.

 

Hope this helps!  Just let me know if you have any questions, or if the method is unfamiliar.

Answer 2

The question asks you to find (x₂ - x₁), which is often referred to as Δx, given a Δy.

 

3 = 8 - 10/x                           [for y₁=3]

3 + p= 8 - 10/(x+Δx)             [for y₂=3+p; note Δy=p]

 

What is Δx?

  To find x:

    3 = 8 - 10/x  

    3x = 8x - 10        [multiply both sides by x]

    -5x = -10           [subtract 8x from both sides]

      x = 2             [divide both sides by (-5)]

 

Now, substitute for x in second equation to find Δx:

  3 + p = 8 - 10/(2+Δx)

  (3+p)(2+Δx) = 8(2+Δx) - 10                    [multiply both sides by (2+Δx)]

  6 + 3Δx + 2p + pΔx = 16 + 8Δx - 10          [F-O-I-L and distribute]

  3Δx + 2p + pΔx = 8Δx

    (-5+p)Δx = -2p

       Δx = -2p/(-5+p)

       Δx = 2p/(5-p)                [PLZ check my answer vs. yours]