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Application of Derivatives Ex-6.4 Questions Answers

Subjects

Question 1 :

Using differentials, find the approximatevalue of each of the following up to 3 places of decimal

Answer 1 : (i) 

Consider. Let x = 25 and Δx =0.3.

Then,

Now, dy is approximatelyequal to Δy and is given by,

Hence, the approximatevalue ofis 0.03 + 5 = 5.03.

(ii) 
Consider. Let x = 49 and Δx =0.5.

Then,

Now, dy is approximatelyequal to Δy and is given by,

Hence, the approximatevalue ofis 7 + 0.035 = 7.035.

(iii) 
Consider. Let = 1 and Δx =− 0.4.

Then,

Now, dy is approximatelyequal to Δy and is given by,

Hence, the approximatevalue ofis 1 + (−0.2) = 1 − 0.2 = 0.8.

(iv) 
Consider. Let x = 0.008 and Δx =0.001.

Then,

Now, dy is approximatelyequal to Δy and is given by,

Hence, the approximatevalue ofis 0.2 + 0.008 = 0.208.

(v) 
Consider. Let x = 1 and Δx =−0.001.

Then,

Now, dy is approximatelyequal to Δy and is given by,

Hence, the approximatevalue ofis 1 + (−0.0001) = 0.9999.

(vi) 
Consider. Let x = 16 and Δx =−1.

Then,

Now, dy is approximatelyequal to Δy and is given by,

Hence, the approximatevalue ofis 2 + (−0.03125) = 1.96875.

(vii) 
Consider. Let x = 27 and Δx =−1.

Then,

Now, dy is approximatelyequal to Δy and is given by,

Hence, the approximatevalue ofis 3 + (−0.0370) = 2.9629.

(viii) 
Consider. Let = 256 and Δx =−1.

Then,

Now, dy is approximatelyequal to Δy and is given by,

Hence, the approximatevalue ofis 4 + (−0.0039) = 3.9961.

(ix) 
Consider. Let x = 81 and Δx =1.

Then,

Now, dy is approximatelyequal to Δy and is given by,

Hence, the approximatevalue ofis 3 + 0.009 = 3.009.

(x) 
Consider. Let x = 400 and Δx =1.

Then,

Now, dy is approximatelyequal to Δy and is given by,

Hence, the approximatevalue ofis 20 + 0.025 = 20.025.


Question 2 :

Find the approximate value of f (2.01),where f (x) = 4x2 +5x + 2

Answer 2 :

Let x = 2 and Δx =0.01. Then, we have:

f(2.01)= f(+ Δx) = 4(x + Δx)2 +5(x + Δx) + 2

Now, Δy = f(x +Δx) − f(x)

 f(x + Δx) = f(x)+ Δy

Hence, the approximate value of f (2.01)is 28.21.

Question 3 :

Find the approximate value of f (5.001),where f (x) = x3 −7x2 + 15.

Answer 3 :

Let x = 5 and Δx =0.001. Then, we have:

Hence, the approximate value of f (5.001)is −34.995.

Question 4 :

Find the approximate change in thevolume V of a cube of side x metres caused byincreasing side by 1%.

Answer 4 :

The volume of a cube (V) ofside x is given by V = x3.

Hence, the approximate change in the volumeof the cube is 0.03x3 m3.

Question 5 :

Find the approximate change in the surfacearea of a cube of side x metres caused by decreasing the sideby 1%

Answer 5 :

The surface area of a cube (S) ofside x is given by S = 6x2.

Hence, the approximate change in the surfacearea of the cube is 0.12x2 m2.

Question 6 :

If the radius of a sphere is measured as 7 m with an errorof 0.02m, then find the approximate error in calculating its volume.

Answer 6 :

Let r be the radius of thesphere and Δr be the error in measuring the radius.

Then,

r = 7 mand Δr = 0.02 m

Now, the volume V of thesphere is given by,

Hence, the approximate error in calculatingthe volume is 3.92 π m3.

Question 7 :

If the radius of a sphere is measured as 9 m with an errorof 0.03 m, then find the approximate error in calculating in surface area.

Answer 7 :

Let be the radius of thesphere and Δr be the error in measuring the radius.

Then,

r = 9 mand Δr = 0.03 m

Now, the surface area of the sphere (S) isgiven by,

S = 4πr2

Hence, the approximate error in calculatingthe surface area is 2.16π m2.

Question 8 :

If f (x) = 3x2 +15x + 5, then the approximate value of (3.02) is

A. 47.66 B. 57.66 C. 67.66 D. 77.66

Answer 8 :

Let x = 3 and Δx =0.02. Then, we have: