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

## 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) .
Solution:
w(x, y) = x3 – 3xy + 2y2 ; at (1, -1)
Linear approximation is given by

(1) ⇒ Let(x, y) = 6 + 6(x – 1) – 7(y + 1)
= 6x – 6 – 7y – 7
= = 6x – 7y – 7

Question 2.
Let z(x, y) = x2y + 3xy4, x, y ∈ R, Find the linear approximation for z at (2, -1).
Solution:

Linear approximation is given by

Question 3.
If v(x, y) = x2 – xy + $$\frac{1}{4}$$y2 + 7, x, y ∈ R, find the differential dv.
Solution:

The differential is dv = (2x -y) dx + (- x + $$\frac{y}{2}$$)dy

Question 4.
Let W(x, y, z) = x2 – xy + 3 sin z, x, y, z ∈ R. Find the linear approximation at (2, -1, 0).
Solution:
w (x, y, z) = x2 – xy + 3 sin z
Here(x0, y0, y0) = (2, -1, 0)

Linear approximation is given by

Question 5.
Let V (x, y, z) = xy + yz + zx, x, y, z ∈ R. Find the differential dV.
Solution:
First let us find Vx, Vy, Vz
Now Vx = $$\frac{\partial v}{\partial x}$$ = y + z
Vy = $$\frac{\partial v}{\partial y}$$ = x + z
Vz = $$\frac{\partial v}{\partial z}$$ = y + x
The differential is
dv = vx dx + vy dy + vz dz = (y + z) dx + (x + z) dy +(y + x) dz