Tamilnadu State Board New Syllabus Samacheer Kalvi 12th Maths Guide Pdf Chapter 8 Differentials and Partial Derivatives Ex 8.5 Textbook Questions and Answers, Notes.

## Tamilnadu Samacheer Kalvi 12th Maths Solutions Chapter 8 Differentials and Partial Derivatives Ex 8.5

Question 1.

If w(x, y) = x³ – 3xy + 2y², x, y ∈ R, find the linear approximation for w at (1, -1) linear approximation for w at (1, -1)

Solution:

w(x, y) = x³ – 3xy + 2y², x, y ∈ R, (1, -1)

∴ L(x, y, z) = 6 + 6(x – 1) -7(y + 1)

L (x, y, z) = 6x – 7y – 7

Question 2.

Let z (x, y) = x² y + 3xy^{4}, x, y ∈ R. Find the linear approximation for z at (2, -1).

Solution:

L (x, y, z) = 2 – (x – 2) – 20 (y + 1)

= 2 – x + 2 – 20y – 20

= -x – 20y – 16

= -(x + 20y + 16)

Question 3.

If v (x, y) = x² – xy + \(\frac { 1 }{ 4 }\)y² + 7, x, y ∈ R find the differential dv.

Solution:

First let us find v_{x}, v_{y}

Now, v_{x} = \(\frac{\partial v}{\partial x}\) = 2x – y

v_{y} = \(\frac{\partial v}{\partial y}\) = -x + \(\frac { 1 }{ 2 }\) y

The differential is

dv = v_{x} dx + v_{y} dy

dv = (2x – y) dx + (\(\frac { 1 }{ 2 }\) y – x) dy

Question 4.

Let W (x, y, z) = x² – xy + 3sinz, x, y, z ∈ R. Find the linear approximation at (2, -1, 0).

Solution:

W (x, y, z) = x² – xy + 3sinz, x, y, z ∈ R

L (x, y, z) = 6 + 5 (x – 2)- 2 (y + 1) + 3 (z)

= 6 + 5x – 10 – 2y – 2 + 3z

= 5x – 2y + 3z – 6

Question 5.

Let V (x, y, z) = xy + yz + zx, x, y, z ∈ R. Find the differential dV.

Solution:

V(x, y, z) = xy + yz + zx

V_{x} = y + z

V_{y} = x + z

V_{z} = y + x

The differential is dV = (y + z) dx + (x + z) dy + (y + x) dz