LINEAR APPROXIMATION WORD PROBLEMS

Let f : (a,b)→ be a differentiable function, for x ∈ (a,b) and 𐤃x the increment given to x, we define the differential of f by 

df = f '(x) 𐤃x

Problem 1 :

The trunk of a tree has diameter 30 cm. During the following year, the circumference grew 6 cm.

(i) Approximately, how much did the tree’s diameter grow?

(ii) What is the percentage increase in area of the tree’s cross-section?

Solution :

(i)  Circumference of the circle  =  2πr

P  =  2πr

Rate of change in perimeter (dp)  =  6 cm

dp  =  2π(dr)

6  =  2π(dr)

dr  =  6/2π

dr  =  3/π

Change in diameter  =  6/π cm

(ii)  Area of circle (A) =  πr2

radius  =  30/2  ==>  15 cm

dA  =  2πr(dr)

dA  =  2πr(3/π)

dA  =  6r

Percentage increase :

(dA/A)x100%  =  (6r/πr2) x 100%

(dA/A)x100%  =  (6/πr) x 100%

(dA/A)x100%  =  (6/15π) x 100%

(dA/A)x100%  =  (40/π)

Problem 2 :

An egg of a particular bird is very nearly spherical. If the radius to the inside of the shell is 5 mm and radius to the outside of the shell is 5.3 mm, find the volume of the shell approximately.

Solution :

Volume of sphere(V)  =  (4/3)πr3

old radius  =  5 mm and new radius  =  5.3 mm

Difference  =  0.3

dV  =  (4/3)π(3r2)dr

dV  =  4πr2dr

dV  =  4π(5)2(0.3)

dV  =  30π

Problem 3 :

Assume that the cross section of the artery of human is circular. A drug is given to a patient to dilate his arteries. If the radius of an artery is increased from 2 mm to 2.1 mm, how much is cross-sectional area increased approximately?

Solution :

Area of circle (A) =  πr2

dA  =  2πrdr

dr  =  2.1 - 2  ==>  0.1

dA  =  2π(2)(0.1)

dA  =  0.4π mm2

Problem 4 :

In a newly developed city, it is estimated that the voting population (in thousands) will increase according to 

V (t) = 30 + 12t2 − t3

0 ≤ t ≤ 8 where t is the time in years. Find the approximate change in voters for the change from 4 to 4  1/6 years.

Solution :

V (t) = 30 + 12t− t3

V' (t)  =  [24(1) − 3t2] (dt)

dt  =  4 1/6 - 4

dt  =  25/6 - 4  ===> 1/6

V' (t)  =  [24t − 3(4)2] (1/6)

V' (t)  =  [24t − 48] (1/6)

V' (t)  =  [24(4) − 48] (1/6)

V' (t)  =  48/6

V' (t)  =  8

8 thousand

Problem 5 :

The relation between the number of words y a person learns in x hours is given by 

y  =  52x, 0  x  9 

What is the approximate number of words learned when x changes from

(i) 1 to 1.1 hour? (ii) 4 to 4.1 hour?

Solution :

y  =  52x

dy  =  52(1/2x) dx

(i)  x  =  1 and dx  =  1.1-1  ==>  0.1

dy  =  52(1/2√1) (0.1)

dy  =  2.6

So, approximately 3 words.

(ii)  x  =  4 and dx  =  4.1-4  ==>  0.1

dy  =  52(1/2√4) (0.1)

dy  =  52(1/4) (0.1)

dy  =  1.3

So, approximately 1 word.

Problem 6 :

A circular plate expands uniformly under the influence of heat. If it’s radius increases from 10.5 cm to 10.75 cm, then find an approximate change in the area and the approximate percentage change in the area.

Solution :

Area of circle (A)  =  πr2

r  =  10.5 and dr  =  10.75 - 10.5  ==>  0.25

dA  =  2πr (dr)

dA  =  2π(10.5) (0.25)

dA  =  5.25π

(ii)  Percentage change in area  =  (dA/A) x 100%

=  (2πr (dr) / πr2) x 100%

=  (2(0.25) / 10.5) x 100%

=  4.76%

Problem 7 :

A coat of paint of thickness 0 2 cm is applied to the faces of a cube whose edge is 10 cm. Use the differentials to find approximately how many cubic centimeters of paint is used to paint this cube. Also calculate the exact amount of paint used to paint this cube.

Solution :

Volume of cube (V)  =  a3

dv  =  3a2 da

a  =  10 cm and da  =  0.2 cm

dv  =  3(10)2(0.2)

dv  =  60 cm3

Actual quantity of paint used  =  (10.2)3 - 103

=  1061.208 - 1000

=  61.208

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