## Rajasthan Board RBSE Class 12 Maths Chapter 8 Application of Derivatives Ex 8.4

Use differential to approximate the following :

[Q. 1 to Q. 11]

Question 1.

(0.009)^{1/3}

Solution:

Let y = x^{1/3} and x = 0.008

∆x = 0.009 – 0.008 = 0.001

∵ y = x^{1/3}

Diff. w.r.t. x,

Question 2.

(0.999)^{1/10}

Solution:

Let y = x^{1/10}, x = 1, y = 1

∆x = 0.999 – 1 = -0.001

∵ y = x^{1/10}

Diff. w.r.t. x

Question 3.

Solution:

Let y = , x = 0.0036, y = 0.06

∆x = 0.0037 – 0.0036 = -0.001

∵ y =

Diff. w.r.t. x

Question 4.

Solution:

Question 5.

(15)^{1/4}

Solution:

Let y = x^{1/4}

x = 16, y = (16)^{1/4} = 2

∆x= 15 – 16 = – 1

∵ y = x^{1/14}

Diff. w.r.t. x

Question 6.

Solution:

Let

y = x^{1/2}, x = 400, y = 20,

∆x = 401 – 400 = 1

∵ y = x^{1/2}

Diff. w.r.t. x

Question 7.

(3.968)^{3/2}

Solution:

Let y = x^{3/2}, x = 4

y = (4)^{3/2} = (22)^{3/2} = 23 = 8

∆x = 3.968 – 4 = -0.032

∵ y = x^{3/2}

Diff. w.r.t. x

Question 8.

(32.15)^{1/5}

Solution:

Let y = x^{1/5}, x = 32, y = (32)^{1/5} = 2,

∆x = 32.15 – 32 = 0.15

∵ y = x^{1/5}

Diff. w.r.t. x

Question 9.

Solution:

Let y = , x = 0.64, y = 0.8

∆x = 0.6 – 0.64 = -0.04

∵ y =

Diff. w.r.t. x

Question 10.

log_{10} (10.1), when log_{10} e = 0.4343

Solution:

Let y = log_{10} x

where x = 10, ∆x = 0.1

⇒ x + ∆x = 10.1

y = log_{10 }x

= log10^{e}. log_{e }x

Diff. w.r.t. x,

Now, from equation (i)

y + ∆y = log_{10} (x + ∆x)

⇒ log_{10} x + ∆y = log_{10} (x + ∆x)

⇒ log_{10} 10 + 0.004343 = log_{10} (10.1)

⇒ log_{10} (10.1)= 1 + 0.004343

log_{10} (10.1)= 1.004343

Hence, approximate value of log_{10} (10.1) is 1.004343.

Question 11.

loge (10.02), when log_{e}10 = 2.3026

Solution:

Let y = log_{e} x

Where x = 10, ∆x = 0.02

and x + ∆x = 10.02

∵ y = log_{e} x

Diff. w.r.t. x,

From equation (i),

y + ∆y = log_{e} (x + ∆x)

log_{e} x + ∆y = log_{e} (x + ∆x)

⇒ log_{e} 10 + 0.002 = log_{e} (10.02)

⇒ log_{e} (10.02) = 2.3026 + 0.002

⇒ log_{e} (10.02)= 2.3046

Hence, approximate value of loge (10.02) is 2.3046.

Question 12.

If y = x^{2} + 4 and x changes from 3 to 3.1, then use differential to approximate change in y.

Solution:

Given, y = x^{2} + 4

At x = 3

y = (3)^{2} + 4 = 9 + 4 = 13

∆x = 3.1 – 3 = 0.1

∵ y = x^{2} + 4

Diff. w.r.t. x,

Hence, approximate value of change in y is 0.6.

Question 13.

Show that the relative error in computing the volume of a cubical box, due to an error in measuring the edge, is approximately equal to the three times the relative error in the edge.

Solution:

Let length of edge of cubical box is x and volume is V, then

V = x^{3}

Diff. w.r.t. x,

Question 14.

The radius of a sphere shrinks from 10 cm to 9.8 cm. Find approximate decrease in its volume.

Solution:

Given

Radius of sphere = 10 cm

∆r = radius shrinks

= 9.8 – 10

= – 0.2 cm

Volume of sphere, V = πr^{3}

Diff. w.r.t. r,

= π 3r^{2} = 4 πr^{2}

∵ Approximation error in calculation of volume of sphere,

dV = × (∆r)

dV= 4πr^{2} × (∆r)

⇒ dV= 4π × (10)^{2} × (- 0.2)

⇒ dV = – 400 × 0.2 π cm^{3}

= -80 π cm^{3}

Hence approximation error in volume is 80 π cm^{3}

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