| Increments and Differentials |
Increments and Differentials: Teeny, Tiny, Changes
Say we're dealing with a function that relates the resistance r of a copper wire to the diameter d of the wire. Since such measurements usually include a margin of error, we may have to find the small changes to the function-values that result from teeny, tiny changes or adjustments to the "d" value or independent variable. In other words, say there's a possible 0.5% margin of error when we measure the diameter of the wire. We want to know the resulting teeny, tiny change in the resistance of that wire. These teeny, tiny changes are called increments.
If, in our resistance function f(d) = r, we change d by a small amount, say Êd (read "delta d"), then Êr will be the corresponding change in r. This change in r is f(d + Êd) - f(d); which is a fancy way of saying y2 - y1 or change in the dependent variable -- which is r in this case.
Example 1: Say we have y = 2x 2 - x - 3, and we want to know the change in y (Êy) that results if x changes from 2 to 2.04. We could of course find the y-value at 2.04 and subtract from it the y-value at 2. This would give us the exact change in the y-value or Êy for the specific
change from x = 2 to x = 2.04.
But let's find a formula for the change in the y-value corresponding to any change ( Êx) in the x-value. Then we just substitute the values for x and Êx and we'll know Êy.
f(x + Êx) - f(x) = 2 (x + Êx) 2 - (x + Êx) - 3 - ( 2x 2 - x - 3)
which becomes Êx (2Êx + 4x - 1).
If we set x = 2 and Êx = 0.04, we will find Êy.
Êy = 0.04 [ 2 (0.04) + 4(2) - 1] = 0.2832
To check let's find the exact values.
If x = 2.04, y = 3.2832
If x = 2.00, y = 3 and the difference, Êy = 0.2832
So, to find the increment Êy, the exact change in y related to a change of Êx,
we find a formula for f(x + Êx) - f(x), and substitute values.
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From Increments to Differentials
Let's take another look at our formula for Êy. Since it represents a change in y-values, we need only divide it by the corresponding change in the x-values to make it a slope.
Now we have:
This looks just like Newton's Quotient if we replace "h" with Êx and limit it to 0.
So, now we can say that
This statement says that if we limit the change in x to zero -- if Êx is small,
then
| If y = f(x) is a differentiable function and Êx is an increment or tiny change in x, 1) The differential dx = Êx, 2) The differential dy = f '(x)Êx = f '(x)dx 3) f(x + Êx) = f(x) + f '(x)dx |
The third statement says that the y-value after the change in x is found by taking the original y-value and adding the teeny, tiny change in y which we now call dy rather than Êy.
So, if we need to find only the change in the function value due to a change in the x-value, we'll use statement #2. If we need the new y-value after a change in the x-value, we'll use the third statement.
All of this leads us to the conclusion that we can use the value of dy rather than Êy to approximate the change in the dependent variable due to a small change in the independent variable.
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In this diagram, SR = Êy, the real change in y on the curve y = f(x).
SQ = dy is the change in y on the tangent PQ.
When Êx is kept very small, the value of dy and Êy are approximately the same.
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Example 2: If y = 3x 2 - 5x + 1, use dy to approximate Êy if x changes from 2 to 2.1.
y = f(x) = 3x 2 - 5x + 1, so
dy l f ' (x) = (6x - 5) dx,
since x = 2, and Êx = dx = 0.1,
dy = {(6)(2) - 5} (0.1) = 0.7
When we calculate the actual change in y, we get 0.73
so the approximation is out by 0.03.
Had Êx = dx been smaller, our error would be smaller as well.
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Example 3: Use differentials to approximate the maximum error in the calculated volume of a sphere if the radius measures 12 inches with an error in measurement of ! 0.06 inches.
Solution: The function that relates the volume of a sphere to the measure of its radius is:
We want dV when r = 12 and dr = ! 0.06
dV l 4or 2 dr = 4o (12) 2 (! 0.06) = ! 34.56 o l ! 109 in 3 .
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Example 4: Use differentials to approximate the value of
Solution: Here, we must formulate our own function and determine the values for x and Êx.
Since we want a cube root, we will make f(x) = (x) 1/3 , we'll set x = 27 and Êx = 0.06
We're also looking for the y-value f(x + Êx) instead of the change in y or Êy, so we'll use the 3rd statement from the "info. box" which says f(x + Êx) = f(x) + f '(x)dx.
We have
So, f(x) + f '(x)dx = (27) 1/3 +
When we calculate the cube root of 27.06 we get 3.00220578...
so our approximation is out by 0.00002578 -- which is not very much.
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1) For these functions, find: i) Êy , ii) dy , and iii) dy - Êy
a) y = 4 - 7x - 2x 2 .
b) y = x - 2 .
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2) Use differentials to approximate the change in f(x) if x changes from a to b.
a) f(x) = 4x 5 - 3x 3 + 7x 2 - 3x ; a = 1 , and b = 1.03.
b) f(x) = - 3x 3 + 5x 2 - 9x + 3 ; a = 4 , and b = 3.96.
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3) Use differentials to approximate the increase in volume of a cube when the length of a side increases from 10 cm. to 10.1 cm.
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4) Using differentials , approximate the change in the area of a circle if the radius changes from 2 inches to 1.96 inches. (A = or 2 )
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5) Using differentials , approximate the increase in the surface area of a spherical balloon when the diameter increases from 3 feet to 3.26 feet. (note we're given diameter, not radius!)
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6) Use differentials to approximate the value of
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1) For these functions, find: i) Êy , ii) dy , and iii) dy - Êy
a) y = 4 - 7x - 2x 2 .
i) Êy = f(x + Êx) - f(x) = 4 - 7(x + Êx) - 2(x + Êx) 2 - (4 - 7x - 2x 2)
Êy = - Êx (7 + 4x + 2 Êx )
ii) dy = (-7 - 4x) dx, or ( -7Êx - 4xÊx)
iii) dy - Êy = -7Êx - 4xÊx - ( -7Êx - 4xÊx - 2Êx 2 ) = 2Êx 2
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b) y = x - 2 = (1/ x 2) .
i)
ii)
iii)
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2) Use differentials to approximate the change in f(x) if x changes from a to b.
a) f(x) = 4x 5 - 3x 3 + 7x 2 - 3x ; a = 1 , and b = 1.03.
dy l f '(x) dx ; x = 1 , and dx = 0.03.
f ' (x) = 20x 4 - 9x 2 + 14x - 3 ;
dy = (20 - 9 + 14 - 3)( 0.03) = 0.66
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b) f(x) = - 3x 3 + 5x 2 - 9x + 3 ; a = 4 , and b = 3.96.
dy l f '(x) dx ; x = 4 , and dx = - 0.04.
f ' (x) = - 9x 2 + 10x - 9 ;
dy = {- 9(4) 2 + 10(4) - 9 )}( - 0.04) = 4.52
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3) Use differentials to approximate the increase in volume of a cube when the length of a side increases from 10 cm. to 10.1 cm.
V = x 3 ; x = 10 in and dx = 0.1 in.
dV l (3x 2) dx = 3(100 in 2 )(0.1 in) = 30 in 3
The cube's volume increases by 30 in 3
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4) Using differentials , approximate the change in the area of a circle if the radius changes from 2 inches to 1.96 inches. (A = or 2 )
A = or 2 ; r = 2 in. and dr = - 0.04 in.
dA l (2or) dr = 2o (2 in)(- 0.04 in) = - 0.16o in 2
The circle's area will decrease by 0.16o in 2
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5) Using differentials , approximate the increase in the surface area of a spherical balloon when the diameter increases from 3 feet to 3.26 feet. (note we're given diameter, not radius!)
Since the diameter = 3 ft, r = 1.5 ft and dr = 0.13 ft. Surface area of sphere: A = 4 or 2
dA l (8or) dr = 8o (1.5 ft)( 0.13 ft) = 1.56 ft 2.
The surface area will increase by 1.56 ft 2.
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6) Use differentials to approximate the value of
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We'll let f(x) = x 1 / 5 ; set x = 32 -- (we know its fifth root), and dx = - 0.05
If we calculate the exact value for the 5th root of 31.95 we get 1.9993746
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