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Question 1:

1. Using differentials, find the approximate value of each of the following up to 3 places of decimal

(i) (ii) (iii)

(iv) (v) (vi)

(vii) (viii) (ix)

(x) (xi) (xii)

(xiii) (xiv) (xv)

Solution 1:

(i)

Consider. Let x = 25 and Δx = 0.3.

Then,

Now, dy is approximately equal to Δy and is given by,

Hence, the approximate value ofis 0.03 + 5 = 5.03.

(ii)

Consider. Let x = 49 and Δx = 0.5.

Then,

Now, dy is approximately equal to Δy and is given by,

Hence, the approximate value ofis 7 + 0.035 = 7.035.

(iii)

Consider. Let x = 1 and Δx = − 0.4.

Then,

Now, dy is approximately equal to Δy and is given by,

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

(iv)

Consider. Let x = 0.008 and Δx = 0.001.

Then,

Now, dy is approximately equal to Δy and is given by,

Hence, the approximate value ofis 0.2 + 0.008 = 0.208.

(v)

Consider. Let x = 1 and Δx = −0.001.

Then,

Now, dy is approximately equal to Δy and is given by,

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

(vi)

Consider. Let x = 16 and Δx = −1.

Then,

Now, dy is approximately equal to Δy and is given by,

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

(vii)

Consider. Let x = 27 and Δx = −1.

Then,

Now, dy is approximately equal to Δy and is given by,

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

(viii)

Consider. Let x = 256 and Δx = −1.

Then,

Now, dy is approximately equal to Δy and is given by,

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

(ix)

Consider. Let x = 81 and Δx = 1.

Then,

Now, dy is approximately equal to Δy and is given by,

Hence, the approximate value ofis 3 + 0.009 = 3.009.

(x)

Consider. Let x = 400 and Δx = 1.

Then,

Now, dy is approximately equal to Δy and is given by,

Hence, the approximate value ofis 20 + 0.025 = 20.025.

(xi)

Consider. Let x = 0.0036 and Δx = 0.0001.

Then,

Now, dy is approximately equal to Δy and is given by,

Thus, the approximate value ofis 0.06 + 0.00083 = 0.06083.

(xii)

Consider. Let x = 27 and Δx = −0.43.

Then,

Now, dy is approximately equal to Δy and is given by,

Hence, the approximate value ofis 3 + (−0.015) = 2.984.

(xiii)

Consider. Let x = 81 and Δx = 0.5.

Then,

Now, dy is approximately equal to Δy and is given by,

Hence, the approximate value ofis 3 + 0.0046 = 3.0046.

(xiv)

Consider. Let x = 4 and Δx = − 0.032.

Then,

Now, dy is approximately equal to Δy and is given by,

Hence, the approximate value ofis 8 + (−0.096) = 7.904.

(xv)

Consider. Let x = 32 and Δx = 0.15.

Then,

Now, dy is approximately equal to Δy and is given by,

Hence, the approximate value ofis 2 + 0.00187 = 2.00187.

Comments

Harshit Bhardwaj

Question 2:

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

Solution 2:

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

f(2.01) = f(x + 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.

Comments

Harshit Bhardwaj

Question 3:

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

Solution 3:

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

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

Comments

Harshit Bhardwaj

Question 4:

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

Solution 4:

Let $X$ be the side of a cube.

Then volume of the cube is equal to $X^{3}$ .

Now if side of cube is increase by 1% then new side of cube would be 1.1 $X$ .

Similarly volume of new cube is equal to 1.452 $X^{3}$ .

Hence change in volume is equal to 45.2%.

Comments

Rehan Ulla Khan

Question 5:

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

Solution 5:

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

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

Comments

Harshit Bhardwaj

Question 6:

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

Solution 6:

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

Then,

r = 7 m and Δr = 0.02 m

Now, the volume V of the sphere is given by,

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

Comments

Harshit Bhardwaj

Question 7:

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

Solution 7:

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

Then,

r = 9 m and Δr = 0.03 m

Now, the surface area of the sphere (S) is given by,

S = 4πr²

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

Comments

Harshit Bhardwaj

Question 8:

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

A. 47.66 B. 57.66 C. 67.66 D. 77.66

Solution 8:

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

Hence, the approximate value of f(3.02) is 77.66.

The correct answer is D.

Comments

Harshit Bhardwaj

Question 9:

The approximate change in the volume of a cube of side x metres caused by increasing the side by 3% is

A. 0.06 x³ m³B. 0.6 x³ m³C. 0.09 x³ m³D. 0.9 x³ $m^{3}$

Solution 9:

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

Hence, the approximate change in the volume of the cube is 0.09x³ m³.

The correct answer is C.

Comments

Tasneem

All Exercises - Chapter 6 - Application of Derivatives

Exercise 6.1

Exercise 6.2

Exercise 6.3

Exercise 6.4

Exercise 6.5

Miscellaneous Exercise 6

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All Exercises - Chapter 6 - Application of Derivatives

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