LINEAR APPROXIMATION WORD PROBLEMS

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 + 12t² − 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  =  52√x, 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  =  52√x

dy  =  52(1/2√x) 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

Apart from the stuff given above, if you need any other stuff in math, please use our google custom search here.

Kindly mail your feedback to v4formath@gmail.com

We always appreciate your feedback.