Exercise 8.5 - Chapter 8 Differentials and Partial Derivatives 12th Maths Guide Samacheer Kalvi Solutions - SaraNextGen [2024-2025]

Updated By SaraNextGen

On April 24, 2024, 11:35 AM

Ex $8.5$
Question 1.
If $w(x, y)=x^3-3 x y+2 y^2, x, y \in R$, find the linear approximation for $w$ at $(1,-1)$.

Solution:
$
w(x, y)=x^3-3 x y+2 y^2 ; \text { at }(1,-1)
$
Linear approximation is given by
$
\begin{aligned}
& \mathrm{L}(x, y)=w\left(x_0, y_0\right)+\left(\frac{\partial w}{\partial x}\right)_{\left(x_0, y_0\right)}\left(x-x_0\right)+\left(\frac{\partial w}{\partial y}\right)_{\left(x_0, y_0\right)}\left(y-y_0\right) \\
& \operatorname{Here}\left(x_0, y_0\right)=(1,-1) \\
& \frac{\partial \mathrm{W}}{\partial x}=3 x^2-3 y \\
& \frac{\partial \mathrm{W}}{\partial y}=-3 x+4 y \\
& \text { At }(1,-1) \\
& w=(1)^3-3(1)(-1)+2(-1)^2=6 \\
& \frac{\partial \mathrm{W}}{\partial x}=3(1)^2-3(-1)=6 \\
& \frac{\partial \mathrm{W}}{\partial y}=-3(1)+4(-1)=-7 \\
& \text { (1) } \Rightarrow \operatorname{Let}(x, y)=6+6(x-1)-7(y+1) \\
& =6 x-6-7 y-7 \\
& ==6 \mathrm{x}-7 \mathrm{y}-7 \\
&
\end{aligned}
$
Question 2 .
Let $z(x, y)=x^2 y+3 x y^4, x, y \in R$, Find the linear approximation for $z$ at $(2,-1)$.

Solution:

$
\begin{aligned}
z(x, y) & =x^2 y+3 x y^4 \text { at }(2,-1) \\
\operatorname{Here}\left(x_0, y_0\right) & =(2,-1) \\
\frac{\partial z}{\partial x} & =2 x y+3 y^4 \\
\frac{\partial z}{\partial y} & =x^2+12 x y^3 \\
\text { At }(2,-1) & =(2)^2(-1)+3(2)(-1)^4=-4+6=2 \\
\frac{\partial z}{\partial x} & =2(2)(-1)+3(-1)^4=-4+3=-1 \\
\frac{\partial z}{\partial y} & =(2)^2+12(2)(-1)^3=4-24=-20
\end{aligned}
$
Linear approximation is given by
$
\begin{aligned}
\mathrm{L}(x, y) & =z\left(x_0, y_0\right)+\left(\frac{\partial z}{\partial x}\right)_{\left(x_0, y_0\right)}\left(x-x_0\right)+\left(\frac{\partial z}{\partial y}\right)_{\left(x_0, y_0\right)}\left(y-y_0\right) \\
& =2+(-1)(x-2)-20(y+1) \\
& =2-x+2-20 y-20 \\
& =-x-20 y-16 \\
& =-(x+20 y+16)
\end{aligned}
$

Question 3.
If $v(x, y)=x^2-x y+\frac{1}{4} y^2+7, x, y \in R$, find the differential $d v$.
Solution:
$
\begin{aligned}
v(x, y) & =x^2-x y+\frac{1}{4} y^2+7 \\
v_x & =2 x-y \\
v_y & =-x+\frac{1}{4}(2 y) \\
& =-x+\frac{y}{2}
\end{aligned}
$
The differential is $\mathrm{dv}=(2 \mathrm{x}-\mathrm{y}) \mathrm{dx}+\left(-\mathrm{x}+\frac{y}{2}\right) \mathrm{dy}$

Question 4.
Let $W(x, y, z)=x^2-x y+3 \sin z, x, y, z \in R$. Find the linear approximation at $(2,-1,0)$.
Solution:
$
\begin{aligned}
& \mathrm{w}(\mathrm{x}, \mathrm{y}, \mathrm{z})=\mathrm{x}^2-\mathrm{xy}+3 \sin \mathrm{z} \\
& \operatorname{Here}\left(\mathrm{x}_0, \mathrm{y}_0, \mathrm{y}_0\right)=(2,-1,0) \\
& \frac{\partial \mathrm{W}}{\partial x}=2 x-y \\
& \frac{\partial \mathrm{W}}{\partial y}=-x \\
& \frac{\partial \mathrm{W}}{\partial z}=3 \cos z \\
& \text { At }(2,-1,0) \\
& w=(2)^2-2(-1)+3 \sin (0) \\
& =4+2+3(0)=6 \\
& \frac{\partial \mathrm{W}}{\partial x}=2(2)-(-1)=4+1=5 \\
& \frac{\partial \mathrm{W}}{\partial y}=-(2)=-2 \\
& \frac{\partial \mathrm{W}}{\partial z}=3 \cos (0)=3(1)=3 \\
&
\end{aligned}
$
Linear approximation is given by
$
\begin{aligned}
\mathrm{L}(x, y, z)= & w\left(x_0, y_0, z_0\right)+\left(\frac{\partial w}{\partial x}\right)_{\left(x_0, y_0, z_0\right)}\left(x-x_0\right)+\left(\frac{\partial w}{\partial y}\right)_{\left(x_0, y_0, z_0\right)}\left(y-y_0\right) \\
& +\left(\frac{\partial w}{\partial z}\right)_{\left(x_0, y_0, z_0\right)}\left(z-z_0\right) \\
\mathrm{L}(x, y, z) & =6+5(x-2)-2(y+1)+3(z-0) \\
& =6+5 x-10-2 y-2+3 z \\
& =5 x-2 y+3 z-6
\end{aligned}
$

Question 5.
Let $V(x, y, z)=x y+y z+z x, x, y, z \in R$. Find the differential dV.
Solution:
$
\begin{aligned}
& V(x, y, z)=x y+y z+z x \\
& V_x=y+z
\end{aligned}
$
$
\begin{aligned}
& V_y=x+z \\
& V_z=y+x
\end{aligned}
$
The differential is $d V=(y+z) d x+(x+z) d y+(y+x) d z$