Samacheer Class 12 Maths - Differentials and Partial Derivatives

x = 27
f(x) = $\sqrt[3] { 27 }$ = 3
We need to find the value of $\sqrt[3] { 27.2 }$
We know that
f(x 0 + Δx) = f(x 0 ) + f'(x 0 ) Δx

∴ Approximate value of $\sqrt[3] { 27.2 }$ = 3.0074

Use f(x)=x^{1/n}. (i) For 123^{1/3} take a=125 (5^3). f'(x)=1/(3x^{2/3}), f'(125)=1/75. dx=123-125=-2. f(123)≈5+(1/75)(-2)=5-2/75=4.973333... (ii) For 15^{1/4} take a=16. f(16)=2, f'(x)=1/(4x^{3/4}), f'(16)=1/32. dx=-1 ⇒ f(15)≈2-1/32=1.96875. (iii) For 26^{1/3} take a=27. f(27)=3, f'(27)=1/27. dx=-1 ⇒ f(26)≈3-1/27=2.96296296...

Let f(x)=x^{2/3}. Then f(27)=27^{2/3}=9. f'(x)=(2/3)x^{-1/3} so f'(27)=(2/3)(1/3)=2/9. Linear approximation at a=27: L(x)=f(27)+f'(27)(x-27)=9+(2/9)(x-27). In particular f(27)=9 exactly; for x=26, f(26)≈9+(2/9)(-1)=8.777...

(i) We know that the linear approximation
L(x) = f(x 0 ) + f’(x 0 )(x – x 0 )
f(x) = x³ – 5x + 12
f'(x) = 3x² – 5
f'(x 0 ) = f'(2) = 12 – 5 = 7
f(x 0 ) = f(2) = 8 – 10 + 12 = 10
L(x) = 10 + 7 (x – 2)
= 10 + 7x – 14
= 7x – 4
(ii) g(x) = $\sqrt { x^2+9 }$, x 0 = -4
g(x 0 ) = g(14) = $\sqrt {16+9 }$ = 5

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

Absolute error = measured - actual = 12.65 - 12.5 = 0.15 cm. Relative error = 0.15/12.5 = 0.012. Percentage error = 0.012×100% = 1.2%.

Volume V=(4/3)πr^3 so dV/dr=4πr^2. At r=10 and dr=-0.2: ΔV≈4π(10^2)(-0.2)=-80π ≈ -251.33 cm^3. Surface area S=4πr^2 so dS/dr=8πr. At r=10: ΔS≈8π(10)(-0.2)=-16π ≈ -50.27 cm^2.

T ∝ l^{1/2}. Relative change: ΔT/T ≈ (1/2)(Δl/l). For Δl/l = 2% ⇒ ΔT/T ≈ 1% ⇒ percentage error ≈ 1%.

Let y=x^{1/n}. Then dy = (1/n)x^{1/n-1} dx, so dy/y = (1/n)(dx/x). Converting to percentages: %Δy ≈ (1/n)%Δx.

(ii) y = (3 + sin2x) 2/3

y = f(x) = x 2 + 3x
dy = (2x + 3) dx
(i) dy {when x = 2 and ate = 0.1} = [2(2) + 3] (0.1)
= 7(0.1) = 0.7
(ii) dy {when x = 3 and dx = 0.02} = [2(3) + 3] (0.0.2)
= 9(0.02) = 0.18

y = f(x) = x 2 + 3x
dy = (2x + 3) dx
(i) dy {when x = 2 and ate = 0.1} = [2(2) + 3] (0.1)
= 7(0.1) = 0.7
(ii) dy {when x = 3 and dx = 0.02} = [2(3) + 3] (0.0.2)
= 9(0.02) = 0.18

1003 = 1000(1+0.003). log_{10}1003 = 3 + log_{10}(1+0.003). For small u, log_{10}(1+u) ≈ u log_{10} e = 0.003×0.4343 = 0.0013029. Hence log_{10}1003 ≈ 3.0013029.

Circumference C=πd. ΔC=6 ⇒ Δd=ΔC/π = 6/π ≈ 1.9099 cm. Area A∝d^2 so relative change ≈ 2Δd/d = 2(6/π)/30 = 12/(30π) ≈ 0.12732 ⇒ 12.732% ≈ 12.73%.

Shell volume ΔV = V(R)-V(r) ≈ dV/dr · dr = 4π r^2 dr with r≈5 mm, dr=0.3 mm. So ΔV≈4π(25)(0.3)=30π ≈94.2478 mm^3.

A=πr^2 so dA ≈ 2πr dr. With r=2 mm and dr=0.1 mm: dA≈2π(2)(0.1)=0.4π ≈1.2566 mm^2. Relative change ≈2 dr/r =2(0.1)/2=0.1 ⇒ 10%.

V(t) = 30 + 12t² – t³; dt = 4 $\frac { 1 }{ 6 }$ – 4 = $\frac { 1 }{ 6 }$
V’(t) = (24t – 3t²)dt
= (24(4)-3 (4)²) × $\frac { 1 }{ 6 }$
= (96 – 48) × $\frac { 1 }{ 6 }$
= 48 × $\frac { 1 }{ 6 }$
= 8
Voters in thousands
∴ Approximate change of voters = 8 × 1000 = 8000

y = 52 √x
dy = 52 × $\frac { 1 }{ 2 }$ × x -1/2 dx
x = 1, dx = 0.1
$\frac { 26 }{ √x }$ × 0.1 = 26 × 0.1
= 2.6
≅ 3 words
(ii) y = 52 √y
dy = 52 × $\frac { 1 }{ 2 }$ × x -1/2 dx
x = 4, dx = 0.1
$\frac { 26 }{ √4 }$ × 0.1 = 13 × 0.1
= 1.3
≅ 1 word

Area of the circular plate A = πr²
= π × 10.5 × 105
= 110.25 π
dA = 2πrdr
= 2π × 10.5 × 0.251
= 5.25 π
Approximate percentage change in the area
= $\frac { dA }{ A }$ × 100
= $\frac { 5.25π }{ 110.25π }$ × 100
= 0.04761 × 100
= 4.76%

$\lim _{(x, y) \rightarrow(0,0)}$ cos($\frac { x^3+y^2 }{ x+y+2 }$)
= cos ($\frac { 0+0 }{ 0+0+2 }$)
= cos 0
= $\lim _{(x, y) \rightarrow(0,0)}$ cos($\frac { x^3+y^2 }{ x+y+2 }$) = 1

$\lim _{(x, y) \rightarrow(0,0)}$ cos($\frac { x^3+y^2 }{ x+y+2 }$)
= cos ($\frac { 0+0 }{ 0+0+2 }$)
= cos 0
= $\lim _{(x, y) \rightarrow(0,0)}$ cos($\frac { x^3+y^2 }{ x+y+2 }$) = 1

f(x, y) = $\frac { y^2-xy }{ √x-√y }$

(i) Let g(x, y) = $\frac { x^2y }{ x^4+y^2 }$ for (x, y) ≠ (0, 0) and f(0, 0) = 0
Let y = mx

= 0
Hence proved
(ii) for parabola y = kx²

Hence proved

Let (a, b) ∈ R² be an arbitrary point.
We shall investigate the 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) = $\frac { a^2-b^2 }{ b^2+1 }$ is defined
Next we want t0 find lf $\lim _{(x, y) \rightarrow(a,b)}$ f(x, y) exist or not
so we calculate $\lim _{(x, y) \rightarrow(a,b)}$ x² – y² = a² – b² and
$\lim _{(x, y) \rightarrow(a,b)}$ y² + 1 = b² + 1
By the properties of limit we see that
$\lim _{(x, y) \rightarrow(a,b)}$ f(x, y) = $\frac { x^2-y^2 }{ y^2+1 }$ = $\frac { a^2-b^2 }{ b^2+1 }$ = f(a, b) = L exists
Now, we note that $\lim _{(x, y) \rightarrow(a,b)}$ f(x, y) = L = f(a, b).
Hence f satisfies all the there 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.

(i) f(x, y) = 3x² – 2xy + y² + 5x + 2, (2, -5)

(ii) g(x, y) = 3x² + y² + 5x + 2, (2, -5)

(iii) h(x, y, z) = x sin (xy) + z² x, (2, π/4, 1)

(iv) G(x, y) = e x + 3y log (x² + y²), (-1, 1)

(i) f(x, y) = $\frac { 3x }{ y+sinx }$


(ii) f(x, y) = tan -1 (x/y)

(iii) f(x, y) = cos (x² – 3xy)
$\frac{\partial f}{\partial x}$ = -sin (x² – 3xy) × (2x – 3y)
$\frac{\partial^2 f}{\partial y \partial x}$ = – sin (x² – 3xy) x – 3 + (2x – 3y) × [-cos (x² – 3xy)] × – 3x
= 3 sin (x² – 3xy) + 3x (2x – 3y) cos (x² – 3xy)
$\frac{\partial f}{\partial x}$ = – sin (x² – 3xy) × -3x
= 3x sin (x² – 3xy)
$\frac{\partial^2 f}{\partial x \partial y}$ = 3sin (x² – 3xy) + 3x cos (x² – 3xy) (2x – 3y)
∴ $\frac{\partial^2 f}{\partial y \partial x}$ = $\frac{\partial^2 f}{\partial x \partial y}$

(i) g(x, y) = x e y + 3x²y

(ii) g(x, y) = log (5x + 3y)

(iii) g(x, y) = x² + 3xy – 7y + cos(5x)
g x = 2x+ 3y – sin 5x × 5
g x = 2x+ 3y – 5 sin 5x
g xx = 2 – 25 cos 5x
g yx = 3
g y = 3x – 7
g yy = 0
g xy = 3

First v_y = 3y^2 + xz, so v_{yz} = ∂/∂z(3y^2 + xz) = x. Similarly v_z = 3z^2 + xy, so v_{zy} = ∂/∂y(3z^2 + xy) = x. Hence v_{yz}=v_{zy}=x, proving equality.

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

L (x, y, z) = 2 – (x – 2) – 20 (y + 1)
= 2 – x + 2 – 20y – 20
= -x – 20y – 16
= -(x + 20y + 16)

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

U(x, y, z) = xyz, x = e -t, y = e -t cos t

$\frac { dU }{ dt }$ = -(e -t cos t sin t) e -t + e -t sin t [ e -t (cos t – sin t )] + e -2t cos t (cos t)
= -e -2t cos t sin t – e -2t sin t cos t – e -2t sin²t + e -2t cos²t
= -e -2t (2 sin t cos t + sin²t – cos²t)
= -e -2t [sin 2t – (cos²t – sin²t)]
= -e -2t (sin 2t + cos 2t)

w(x, y) = 6x³ – 3xy + 2y²

z (x, y) = x³ – 3x²y³

(3x² – 6xy³) e t – 9 (xy)² e -t
[3 (s e t )² – 6 (s e t ) (se -t )³]e t
-9 (s e t s e -t )² × e -t
= (3 s² e 2t – 6 s 4 e -2t ) e t – 9 s 4 e -t
= 3s² [(e 2t – 2s² e -2t ) e t -3 e -t s²]
$\frac{\partial z}{\partial s}$ = 3s² e t (e 2t -2s² e -2t – 3 e -2t s²)
$\frac{\partial z}{\partial t}$ = $\frac{\partial z}{\partial x}$ $\frac{\partial x}{\partial t}$ + $\frac{\partial z}{\partial y}$ $\frac{\partial y}{\partial t}$
= (3x² – 6xy³) (s e t )+ (-9x²y²) (- s e -t )
= [3 (s e t )² – 6 (s e t ) (s e -t )³] s e t + 9(se t s e -t )² s e -t
= (3 s² e 2t – 6 s 4 e -2t ) s e t + 9 s 5 e -t
= 3 s³ e 3t – 6 s 5 e -t + 9 s 5 e -t
= 3 s³ e 3t + 3 s 5 e -t
$\frac{\partial y}{\partial t}$ = 3 s³ (e 3t + s 2 e -t )

= (y + z) × 1 + (x + z) × v + (y + x) × 1
= uv + u + v + (u – v + u + v) v+ (uv + u – v)
= uv + u + v + uv + uv + uv + u – v
= 4 uv + 2u
$\frac{\partial w}{\partial u}$ ($\frac{1}{2}$, 1) = 4 × $\frac{1}{2}$ × 1 + 2 × $\frac{1}{2}$ = 2 + 1 = 3
$\frac{\partial w}{\partial v}$ = $\frac{\partial W}{\partial x}$ $\frac{\partial x}{\partial v}$ + $\frac{\partial W}{\partial y}$ $\frac{\partial y}{\partial v}$ + $\frac{\partial W}{\partial z}$ $\frac{\partial z}{\partial v}$
= (y + z) (-1) + (x + z) u + (y + x) × 1
= -y – z + xu + zu + y + x
= -u – v + (u – v) u + (u + v) u + u – v
= -u – v + u² – vu + u² + vu + u – v
= 2u² – 2v
$\frac{\partial W}{\partial v}$ ($\frac{1}{2}$, 1) = 2 × $\frac{1}{4}$ – 2 × 1
= $\frac{1}{2}$ – 2 = –$\frac{3}{2}$

(i) f(x, y) = x²y + 6x³ + 7
f(λx, λy) = λ³x²y + 6λ³x³ + 7
There is no common λ in this equation.
∴ It is not homogeneous

Thus f is homogeneous with degree 3.

Thus g is homogeneous with degree 0.

There is no common λ
∴ It is not homogeneous.

g (x, y) = x log(y/x)
g (tx, ty) = tx log($\frac { ty }{ tx }$)
g is a homogeneous function of degree 1.
∴ By Euler’s Theorem,

Hence verified

u_x = 2x + y, u_y = x + 2y. So x u_x + y u_y = x(2x+y) + y(x+2y) = 2x^2 +2xy +2y^2 +? Combine = 3(x^2 + x y + y^2)=3u.

(b) (1 + xy)e xy
Hint:
f(x, y) = e xy
$\frac{\partial f}{\partial y}$ = x e xy (x) = xe xy
$\frac{\partial^2 f}{\partial x \partial y}$ = x e xy (y) + e xy = e xy (1 + xy)