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Republic of the Philippines
ISABELA STATE UNIVERSITY
Echague, Isabela
COLLEGE OF ENGINEERING
CE 111 – CALCULUS 1
- Chapter 6: THE DIFFERENTIAL
- Overview/Introduction
- Differentials
- Approximate Formulas
- Discontinuous Derivatives.Learning Outcome/Objective
At the end of the discussion, the student should be able to:
- Understand the concept of derivatives and differentials;
- Find the differential of a function; and;
- Solve approximation problems using differentials.
- Learning Content/Topic
I. DIFFERENTIALS
Consider an interval in which a curve relating 𝑥 and 𝑦 has a slope 𝑦’. Let 𝑃: (𝑥, 𝑦) be a
point on the curve, as shown in Figures 1 and 2. A change ∆𝑥 in the value of 𝑥 changes
y by some amount ∆𝑦. In the Figures 𝑃’ is the point (𝑥 + ∆𝑥, 𝑦 + ∆𝑦); ∆𝑦 is the distance
𝑄𝑃’. Unless the equation of the curve is particularly simple, it may be difficult to compute
∆𝑦. We seek for ∆𝑦 an approximation which must satisfy two requirements: First it must
be possible for us to prove that the difference between the approximation and ∆𝑦 can be
made arbitrarily small by taking ∆𝑥 sufficiently small; second, the approximation must be
easy to compute.
In Figure 1 and Figure 2, the tangent line at 𝑃 intersects the ordinate through 𝑃’ at the
point 𝑅. Examination of the figures shows that it is plausible that the length 𝑄𝑅 is an
Figure 1 Figure 2
approximation to 𝑄𝑃’ = 𝛥𝑦 for small 𝛥𝑥. Let us see whether 𝑄𝑅 satisfies our two
requirements.
At 𝑃, the slope of the curve is
ொோ
ொ
. Now 𝑃𝑄 = ∆𝑥, so that we obtain
e
We already know how easy it is to compute the slope y’. Hence our second requirement
is satisfied by 𝑄𝑅.
The difference between 𝑄𝑅 and 𝑄𝑃’ is given by
ᇱ
ᇱ
ᇱ
Our first requirement demands that we show that 𝑃’𝑅 → 0 as ∆𝑥 → 0.
We shall do even better by showing that
ᇲ
ோ
∆௫
→ 0 as ∆𝑥 → 0. Indeed,
e
Lim
∆௫→
ᇱ
= Lim
∆௫→
ᇱ
ᇱ
ᇱ
because
Lim
∆௫→
since we are working in an interval where the slope exists. In a sense (2) shows that
𝑃’𝑅 → 0 more rapidly than ∆𝑥 → 0.
The quantity 𝑄𝑅 is called the differential of 𝑦 and is denoted by 𝑑𝑦.
By equation (1)
Theoretically, we are still at liberty to define dx – i.e., the differential of the independent
variable – in any way we please. But if in (3) we put
ᇱ
the result is
Thus, in order to avoid conflict when (3) is applied to the function y=x, we adopt (4) as
our definition. That is, the differential of the independent variable is equal to the increment
of that variable.
We may therefore write
ᇱ
and state the definition as follows:
ଶ
When the radius increases by an amount ∆𝑟, the area increases by an amount ∆𝐴 which
is approximated by
(Since 𝑟 is the independent variable, 𝑑𝑟 = ∆𝑟.) Hence the area of a narrow circular ring
is approximately the product of the circumference by the width.
Example (). Find an approximate value for √8.
Put 𝑦 = √
𝑥, from which
For 𝑥 we choose a number which is close to 8.73 and for which we know the square root.
Choose
So that 𝑥 + 𝑑𝑥 = 8.73. From (1) it follows that
Then
approximately. To five decimal places the correct value is 2.95466.
- Teaching and Learning Activities
EXERCISES
ସ
ଷ
ସ
ି
య
మ
ସ
ହ
ଶ
ଷ
√ଵିଶఈ
ଶ
ଵ
ଷ
ଷ
ଵ
௧
మ
ସ
ଶ
ଶ
భ
మ
௩
య
௩ ା ଶ
௧
మ
√
ଵ ି ௧
ଶ
ଶ
, 𝑎 held constant.
ଶ
ଶ
, 𝑥 held constant.
ଶ
ଶ
య
మ
ଶ
ଶ
భ
మ
௦
√ଵ ି ௦
మ
√ଵି௦
మ
௦
ସ
௧
(ଶିଷ௧ )
ర
(ଵି௫ )
మ
(ଵିଶ௫ )
మ
ଵ – ௦
ଵ ା ௦
൫௫
య
ି ௫൯
మ
௫ ା ଵ
ଶ
௫
మ
మ
௬
మ
మ
ଶ
ଶ
ଶ
ଶ
ଶ
௫
మ
௫
మ
ି
మ
ଶ
௫
య
ଶ ି ௫
ଷ
௫
మ
ି ௫
ଷ
ଷ
- Find approximately the volume of a thin spherical shell.
- Find an approximate formula for the volume of a thin cylindrical shell of given height.
- Find approximately the volume of wood required to make a cubical box, of edge
length 6ft., using boards ½ in. thick.
- The base of a right triangle is fixed at 3 ft., the hypothenuse is 5 ft. long and subject
to change. Find the approximate change in altitude when the hypothenuse is changed
by a small amount ∆ℎ.
- The diameter of a circle is measured and found to be 6 ft. with a maximum error of
0.1 in. Find the approximate maximum error in the computed area.
- The diameter of a sphere is measured and found to be 3 ft. with a maximum error of
0.1 in. Find the approximate maximum error in the computed volume.
- The diameter of a circle is to be measured, and its volume computed. If the diameter
can be measured with a maximum error of 0.001 in., and the area must be accurate
to within 0.1 sq. in., find the largest diameter for which the process can be used.
- The diameter of a sphere is to be measured, and its volume computed. If the diameter
can be measured with a maximum error of 0.001 in., and the volume must be accurate
to within 0.1 cu. in., find the largest diameter for which the process can be used.
- Find approximately the change in the reciprocal of a number 𝑥 produced by a small
change in the number. Investigate also the case when the number itself is small.
- The volume of a body of gas is measured; the pressure is then computed from the
formula
If the allowable error in 𝑝 is 0.001𝑘, and the maximum error in measuring 𝑣 is 0.6 cu.
ft., what is the volume of the smallest container to which the process can be applied?
In the following exercises, use differentials to approximate to the desired number.
- The square root of 627.
- The square root of 398.
- The square root of 193.
- The square root of 287.
- The square root of 0.253.
- The square root of 98.8.
- The cube root of 26.
- The cube root of 0.009.
- The cube root of 1.35.
- The cube root of 3.3.
- The fourth root of 17.
- The fourth root of 255.
- Flexible Teaching Learning Modality (FTLM) adapted
Module, Messenger Rooms/ Google Classroom / Google docs
Exercises and Problem Sets.
- Assessment Task
Assessment Task will be a 30-point quiz covering the topics discussed in this module.
In addition, a long exam will be given at the end of the term
