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## Tamilnadu Samacheer Kalvi 12th Maths Solutions Chapter 8 Differentials and Partial Derivatives Ex 8.1

Question 1.

Let f(x) = \(\sqrt[3]{x}\). Find the linear approximation at x = 27. Use the linear approximation to \(\sqrt[3]{27.2}\).

Solution:

Question 2.

Use the linear approximation to find approximate values of

Solution:

Question 3.

Find a linear approximation for the following functions at the indicated points.

(i) f(x) = x^{3} – 5x + 12, x_{0} = 2

(ii) g(x) = \(\sqrt{x^{2}+9}\) + x_{0} = -4

(iii) h(x) = \(\frac{x}{x+1}\), x_{0} = 1

Solution:

f(x) = x^{3} – 5x + 12

f'(x) = 3x^{2} – 5

f(x_{0}) = f(2) = (2)^{3} – 5(2) + 12 = 8 – 10 + 12 = 10

f'(x_{0}) = f'(2) = 3(2)^{2} – 5 = 12 – 5 = 7

The required linear approximation L(x) = f(x_{0}) + f'(x_{0}) (x – x_{0})

= 10 + 7 (x – 2)

= 10 + 7x – 14

= 7x – 4

Question 4.

The radius of a circular plate is measured as 12.65 cm instead of the actual length 12.5 cm. find the following in calculating the area of the circular plate:

(i) Absolute error

(ii) Relative error

(iii) Percentage error

Solution:

We know that Area of the circular plate A(r) = πr^{2}, A'(r) = 2πr

Change in Area = A’ (12.5) (0.15) = 3.75 π cm^{2}

Exact calculation of the change in Area = A (12.65) – A (12.5)

= 160.0225π – 156.25π

= 3.7725π cm^{2}

(i) Absolute error = Actual value – Approximate value

= 3.7725 π – 3.75 π

= 0.0225 π cm^{2}

Question 5.

A sphere is made of ice having radius 10 cm. Its radius decreases from 10 cm to 9.8 cm. Find approximations for the following:

(i) change in the volume

(ii) change in the surface area

Solution:

(i) We know that Volume of sphere

∴ Volume decreases by 80π cm^{3}

(ii) Surface area of the sphere

S(r) = 4π r^{2}

S'(r) = 8πr

Change in surface area at r = 10 is

= S'(r) [10 – 9.8]

= 8π (10) (0.2) = 16π cm^{2}

∴ Surface Area decreases by 16π cm^{2}

Question 6.

The time T, taken for a complete oscillation of a single pendulum with length l, is given

by the equation T = \(2 \pi \sqrt{\frac{l}{g}}\), where g is a constant. Find the approximate percentage error in the calculated value of T corresponding to an error of 2 percent in the value of 1.

Solution:

Question 7.

Show that the percentage error in the n^{th} root of a number is approximately \(\frac{1}{n}\) times the percentage error in the number.

Solution:

Let x be the number

### Samacheer Kalvi 12th Maths Solutions Chapter 8 Differentials and Partial Derivatives Ex 8.1 Additional Questions Solved

Question 1.

Using differentials, find the approximate value of each of the following upto 3 places of decimal. \((255)^{\frac{1}{4}}\)

Solution:

Question 2.

Using differentials, find the approximate value of each of the following upto 3 places of decimal. \((401)^{\frac{1}{2}}\)

Solution:

Question 3.

Find approximate value of f(5.001) where f(x) = x^{3} – 7x^{2} + 15.

Solution:

Here f(x) = x^{3} – 7x^{2} + 15

f'(x) = 3x^{2} – 14x

Let x = 5 and x + ∆x = 5.001

∴ ∆x (x + ∆x) – x = 5.001 – 5

= 0.001

f(5.001) = f(x + ∆x)

Now ∆y = f(x + ∆x) – f(x)

f(x + ∆x) = f(x) + ∆y

= f(x) + f'(x).∆x [∴ ∆y = f'(x).∆x ]

= (x^{3} – 7x^{2} + 15) + (3x^{2} – 14x)(0.001)

= (5^{3} – 7 × 5^{2} + 15) + (3 × 5^{2} – 14 × 5) [0.001]

= [125 – 175 + 15] + [75 – 70][0.001]

= -35 + 0.005 = – 34.995

Thus approximate value of f(5.001) is – 34.995

Question 4.

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

Solution:

Le r be the radius of the sphere and ∆r be the error in measuring the radius.

Then r = 7 m and ∆r = 0.02 m

Now volume of a sphere is given by