By (14), Linear approximation is given by
L (x, y, z) =  U(x₀, y₀, z₀) + \({ \frac { { \partial }U }{ { \partial x } } | }_{ { x }_{ 0 },{ y }_{ 0 },{ z }_{ 0 } }\) (x-x₀)+\({ \frac { { \partial }U }{ { \partial y } } | }_{ { x }_{ 0 },{ y }_{ 0 },{ z }_{ 0 } }\) (y-y₀)+\({ \frac { { \partial }U }{ { \partial z } } | }_{ { x }_{ 0 },{ y }_{ 0 },{ z }_{ 0 } }\) (z-z₀)
Now U_x = 2x -y, Uy = -xand U_z = 3cos z.
Here (x₀, y₀, z₀) = (2,−1,0 )
hence U_x (2, −1,0) = 5, U_y (2, −1,0) = −2 and U_z (2,-1,0) = 3.
Thus L(x, y, z) = 6 + 5(x − 2) − 2( y +1) + 3(z − 0) = 5x − 2y + 3z − 6 is the required linear approximation for U at (2,−1,0).

First let us find w_x, w_y, and w_z
Now w_x = 2xy + z², w_y = 2yz +x² and w_z = 2zx + y².
Thus,by (15), the differential is
dw = (2xy + z² )dx + (2yz + x² )dy+ (2zx + y² )dz.

Observe that the function g is defined for all (x, y)∈R² It is easy to check, as in the above examples, that g is continuous at all point (x, y) ≠ (0, 0). Next, we shall check the continuity of g at (0, 0). For that we see if g has a limit L at (0, 0) and if L = g(0, 0) = 0. So we consider
\(\left| g\left( x,y \right) -g\left( 0,0 \right) \right| =\left| \frac { 2{ x }^{ 2 }y }{ { x }^{ 2 }+{ y }^{ 2 } } -0 \right| =\frac { 2\left| { x }^{ 2 }y \right| }{ \left| { x }^{ 2 }+{ y }^{ 2 } \right| } =\frac { 2\left| xy \right| \left| x \right| }{ { x }^{ 2 }+{ y }^{ 2 } } \le \frac { \left( { x }^{ 2 }+{ y }^{ 2 } \right) \left| x \right| }{ { x }^{ 2 }+{ y }^{ 2 } } \le \left| x \right| \)   ...(9)
Note that in the final step above we have used 2 \(\left| xy \right| \) \(\le \) x² + y² (which follows by considering 0\(\le \) (x - y)² for all  x, y∈ R . Note that (x, y)→(0, 0) implies |x| → 0. Then from (9) it follows that \(\underset { (x,y)\longrightarrow (0,0) }{ lim } \) \(\frac { 2{ x }^{ 2 }y }{ { x }^{ 2 }+{ y }^{ 2 } } \) = 0 = g (0, 0) which proves that g is continuous at (0, 0). So g is continuous at every point of R²

Note that f is defined for every (x, y)∈R². First let us check the continuity at (a, b) ≠ (0, 0).
Let us say, just for instance, (a, b) = (2, 5). Then f(2, 5) = \(\frac{10}{29}\). Then, as in the above example, we callculate \(\underset { (x,y)\longrightarrow (2,5) }{ lim } \) xy = 2(5) and \(\underset { (x,y)\longrightarrow (2,5) }{ lim } \)  x²+y² = 2²+5² = 29 ≠ 0.
Hence,  \(\underset { (x,y)\longrightarrow (2,5) }{ lim } \) \(\frac { xy }{ { x }^{ 2 }+{ y }^{ 2 } } \) = \(\frac { 10 }{ 29 } \).
Since f(2,5) = \(\frac { 10 }{ 29 } \) \(\underset { (x,y)\longrightarrow (2,5) }{ lim } \) \(\frac { xy }{ { x }^{ 2 }+{ y }^{ 2 } } \) it follows that f is continuous at (2, 5)
Exactly by similar arguments we can show that f is continuous at every point (a, b) ≠ (0, 0). Now let us check the continuity at (0, 0). Note that f (0, 0) = 0 by definition. Next we want to find if \(\underset { (x,y)\longrightarrow (0,0) }{ lim } \) \(\frac { xy }{ { x }^{ 2 }+{ y }^{ 2 } } \) exists or not.
First let us check the limit along the straight lines y = mx , passing through (0,0) .
 \(\underset { (x,y)\longrightarrow (0,0) }{ lim } \) \(\frac { xy }{ { x }^{ 2 }+{ y }^{ 2 } } \)  = \(\underset { x\longrightarrow 0 }{ lim } \) \(\frac { m{ x }^{ 2 } }{ \left( 1+{ m }^{ 2 } \right) { x }^{ 2 } } =\frac { m }{ 1+{ m }^{ 2 } } \neq \) f (0, 0), if m ≠0.
So for different values of m, we get different values \(\frac { m }{ 1+{ m }^{ 2 } } \) and hence we conclude that   \(\underset { (x,y)\longrightarrow (0,0) }{ lim } \) \(\frac { xy }{ { x }^{ 2 }+{ y }^{ 2 } } \) does not exist. Hence f cannot be continuous at (0, 0).

Let (a,b)∈R² be an arbitrary point. We shall investigate continuity of f at (a,b).
That is, we shall check if all the three conditions for continuity hold for f at (a,b).
To check first condition, note that f (a, b) = \(\frac { 3x-5y+8 }{ { x }^{ 2 }+{ y }^{ 2 }+1 } \) is defined.
Next we want to find if \(\underset { (x,y)\longrightarrow (a,b) }{ lim } \) exists or not.
So we calculate (a,b) \(\underset { (x,y)\longrightarrow (a,b) }{ lim } \) (3x -5y +8) = 3a-5b+8 and \(\underset { (x,y)\longrightarrow (a,b) }{ lim } \) (x² +y²+1) = a²+b²+1 ≠ 0
Thus, by the properties of limits, we see that
\(\underset { (x,y)\longrightarrow (a,b) }{ lim } \) f(x, y) = \(\frac { \underset { (x,y)\longrightarrow (a,b) }{ lim } \left( 3x-5y+8 \right) }{ \underset { (x,y)\longrightarrow (a,b) }{ lim } \left( { x }^{ 2 }+{ y }^{ 2 }+1 \right) } \) = \(\frac { 3a-5b+8 }{ { a }^{ 2 }+{ b }^{ 2 }+1 } \) = f(a, b) = L exists.
Now we note that \(\underset { (x,y)\longrightarrow (a,b) }{ lim } \) f(x, y) = L = f(a, b). Hence f satisfies all the three conditions for continuity of f at (a, b). Since (a, b) is an arbitrary point in R², we conclude that f is continuous at every point of R².

Recall that surface area of a sphere with radius r is given by S(r) = 4\(\pi \)r³. Note that even though we can calculate the exact change using this formula, we shall try to approximate the change using the linear approximation. So, using (4), we have
Change in the surface area = S(5.2) - S(5) ≈ S'(5)(0.2)
= 8\(\pi \)(5)(0.2)
= 8\(\pi \) cm²
Exact calculation of the change in the surface gives
S(5.2) − S(5) = 108.16\(\pi \)-100\(\pi \) = cm².
Percentage error = relative error \(\times\)100 = \(\frac { 8.16\pi -8\pi }{ 8.16\pi } \)\(\times\)100 = 1.9607%

We need to find an approximate value of \(\sqrt { 9.2 } \) using linear approximation. Now by (3), we have f(x₀+Δx) ≈ f(x₀)+f'(x₀)Δx. To do this, we have to identify an appropriate function f, a point x₀ and Δx. Our choice should be such that the right side of the above approximate equality, should be computable without the help of a calculator. So, we choose  
f(x) = \(\sqrt { x,{ x }_{ 0 } } \) = 9  and Δx = 0.2. Then f'(x₀) = \(\frac { 1 }{ 2\sqrt { 9 } } \) and hence.
 \(\sqrt { 9.2 } \) ≈ f(9) + f'(9)(0.2) = 3+\(\frac { 0.2 }{ 6 } \) = 3.03333
Now if we use a calculator, just to compare, we find \(\sqrt { 9.2 } \) = 3.03315. We see that our approximation is accurate to three decimal places and the error is 3.03315 - 3.03333 = 0.00018. [Also note that one could choose f (x) = \(\sqrt { 1+x,{ x }_{ 0 } } =8\) and Δx = 0.2. So the choice of f and x₀ - are not necessarily unique].
 So in the above example, the absolute error is 3.03315-3.03333 = -0.00018. Note that the absolute error says how much the error; but it does not say how good the approximation is. For instance, let us consider two simple cases
Case 1 : Suppose that the actual value of something is 5 and its approximated value is 4, then the absolute error is 5 − 4 = 1.
Case 2 : Suppose that the actual value of something is 100 and its approximated value is 95. In this case, the absolute error is 100 − 95 = 5. So the absolute error in the first case is smaller when compared to the second case.
Among these two approximations, which is a better approximation; and why? The absolute error does not give a clear picture about whether an approximation is a good one or not. On the other hand, if we calculate relative error or percentage of error (defined below), it will be easy to see how good an approximation is. If the actual value is zero, then we do know how close our approximate answer is to the actual value. So if the actual value is not zero.

We know from (4), that L(x) = f(x₀) +f'(x₀)(x-x₀) We have x₀ = 3, \(\Delta \)x = 0.2 and hence f(3) = \(\sqrt { 1+3 } \) = 2. Also,
f′(x) = \(\frac { 1 }{ 2\sqrt { 1+x } } \) and hence f'(3) = \(\frac { 1 }{ 2\sqrt { 1+3 } } \) = \(\frac { 1 }{ 4 } \)
Thus, L(x) = 2 +\(\frac { 1 }{ 4 } \)(x-3) = \(\frac { x }{ 4 } \)+\(\frac { 5 }{ 4 } \) gives the required linear approximation.
Now, f(3.2) = \(\sqrt { 4.2 } \) ≈ L(3.2) = \(\frac { 3.2 }{ 4 } \)+\(\frac { 5 }{ 4 } \) = 2.050
Actually, if we use a calculator to calculate we get \(\sqrt { 4.2 } \) = 2.04939

Note that the function u is not homogeneous. So we cannot apply Euler’s Theorem for u.
However, note that f(x,y) = \(\frac { x+y }{ \sqrt { x } +\sqrt { y } }\) = sin u is homogeneous; because
f(tx,ty) = \(\frac { tx+ty }{ \sqrt { tx } +\sqrt { ty } } \) = t^1/2 f(x, y), \(\forall \) x, y, t\(\ge \)0
Thus f is homogeneous with degree \(\frac { 1 }{ 2 } \) and so by Euler’s Theorem we have
\(x\frac { \partial f }{ \partial x } +y\frac { \partial f }{ \partial y } =\frac { 1 }{ 2 } f(x,y)\).
Now substituting f = sin u in the above equation, we obtain
\(x\frac { \partial (sinu) }{ \partial x } +y\frac { \partial (sinu) }{ \partial y } =\frac { 1 }{ 2 } sin \ u\)
\(x\quad cosu\frac { \partial u }{ \partial x } +y\quad cosu\frac { \partial u }{ \partial x } =\frac { 1 }{ 2 } sin \ u\)  ...(19)
Dividing both sides by cosu we obtain
\(x\frac { \partial u }{ \partial x } +y\frac { \partial u }{ \partial y } =\frac { 1 }{ 2 } tan \ u\)
Note:
Solving this problem by direct calculation will be possible; but will involve lengthy calculations.

Recall that volume of a right circular cylinder is given by V = \(\pi \)r²h where r is the radius and h is the height. So we have V (r) = \(\pi \)r²h = 20\(\pi \)r²
V (10.1) −V (10)≈ \(\frac { dV }{ dr } { { | }_{ r=10 } }\) (10.1 10) = 20\(\pi \)²(10(0.1))
Thus the estimate for the change in the volume is 40 \(\pi \) cm³
Exact calculation of the volume change gives
V (10.1) −V (10) = 2040.2\(\pi \) -2000\(\pi \) = 40.2\(\pi \) cm³.
So relative error = \(\frac { 40.2\pi -40\pi }{ 40.2\pi } \) = \(\frac { 1 }{ 201 } \) = 0.00497 and hence
the percentage error = relative error x 100 = \(\frac { 1 }{ 201 } \)x100 = 0.497% 

We compute
F(\(\lambda \)x, \(\lambda \)y) = \(\frac { { \left( { \lambda x } \right) }^{ 2 }+5\left( \lambda x \right) \left( \lambda y \right) -10(\lambda y)^2 }{ 3\lambda x+7\lambda y } =\frac { { \lambda }^{ 2 } }{ \lambda } \left( \frac { { x }^{ 2 }+5xy-10{ y }^{ 2 } }{ 3x+7y } \right) =\lambda F(x,y)\)
for all \(\lambda \) ∈ R. So F is a homogeneous function of degree 1.