Exercise 8.6 - 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 $8.6$
Question 1.
If $\mathrm{u}(\mathrm{x}, \mathrm{y})=\mathrm{x}^2 \mathrm{y}+3 \mathrm{xy}^4, \mathrm{x}=\mathrm{e}^{\mathrm{t}}$ and $\mathrm{y}=\sin \mathrm{t}$, find $\frac{d u}{d t}$ and evaluate it at $\mathrm{t}=0$.
Solution:
$
\begin{aligned}
& u(x, y)=x^2 y+3 x y^4 ; x=e^t ; y=\sin t \\
& \frac{d u}{d t}=\frac{\partial u}{\partial x} \frac{d x}{d t}+\frac{\partial u}{\partial y} \frac{d y}{d t} \\
& =\left(2 x y+3 y^4\right)\left(e^t\right)+\left(x^2+12 x y^3\right)(\cos t) \\
& =\left(2 e^t \sin t+3 \sin ^4 t\right) e^t+\left[e^{2 t}+12 e^t \sin ^3 t\right] \cos t \\
& =e^t\left[2 e^t \sin t+3 \sin ^4 t+e^t(\cos t)+12 \sin ^3 t \cos t\right] \\
& \text { at } \mathrm{t}=0 \\
& \frac{d u}{d t}=e^0\left[2 e^0 \sin (0)+3 \sin ^4(0)+e^0 \cos (0)+12 \sin ^3(0) \cos (0)\right] \\
& =1[0+0+1+0] \\
& =1 \\
& e^0=1 \\
& \begin{array}{l}
\cos 0=1 \\
\sin 0=0
\end{array} \\
&
\end{aligned}
$

Question 2.
If $u(x, y, z)=x y^2 z^3, x=\sin t, y=\cos t, z=1+e^{2 t}$, find $\frac{d u}{d t}$
Solution:
$
\begin{aligned}
& u=x y^2 z^3 ; x=\sin t ; y=\cos t ; \quad z=1+e^{2 t} \\
& \frac{d u}{d t}=\frac{\partial u}{\partial x} \frac{d x}{d t}+\frac{\partial u}{\partial y} \frac{d y}{d t}+\frac{\partial u}{\partial z} \frac{d z}{d t} \\
& =\left(y^2 z^3\right)(\cos t)+\left(2 x y z^3\right)(-\sin t)+\left(3 x y^2 z^2\right)\left(0+2 e^{2 t}\right) \\
& =y z^2\left[y z \cos t-2 x z \sin t+6 x y e^{2 t}\right] \\
& \text { (Replace } x=\sin t ; y=\cos t \text { and } z=1+e^{2 t} \text { ) } \\
& =\cos t\left(1+e^{2 t}\right)^2\left[\cos ^2 t\left(1+e^{2 t}\right)-2 \sin ^2 t\left(1+e^{2 t}\right)+6 \sin t \cos t e^{2 t}\right] \\
& \frac{d u}{d t}=\left(1+e^{2 t}\right)^2\left[\cos ^3 t\left(1+e^{2 t}\right)-2 \sin t \sin 2 t\left(1+e^{2 t}\right)+6 e^{2 t} \sin t \cos ^2 t\right] \\
&
\end{aligned}
$

Question 3.
If $\mathrm{w}(\mathrm{x}, \mathrm{y}, \mathrm{z})=\mathrm{x}^2+\mathrm{y}^2+\mathrm{z}^2, \mathrm{x}=\mathrm{e}^{\mathrm{t}}, \mathrm{y}=\mathrm{e}^{\mathrm{t}} \sin \mathrm{t}$, and $\mathrm{z}=\mathrm{e}^{\mathrm{t}} \cos \mathrm{t}$, find $\frac{d w}{d t}$
Solution:
$
\begin{aligned}
& \mathrm{w}(\mathrm{x}, \mathrm{y}, \mathrm{z})=\mathrm{x}^2+\mathrm{y}^2+\mathrm{z}^2 ; \mathrm{x}=\mathrm{e}^{\mathrm{t}} ; \mathrm{y}=\mathrm{e}^{\mathrm{t}} \sin \mathrm{t}, \mathrm{z}=\mathrm{e}^{\mathrm{t}} \cos \mathrm{t} \\
& \frac{\partial w}{\partial t}=\frac{\partial w}{\partial x} \frac{d x}{d t}+\frac{\partial w}{\partial y} \frac{d y}{d t}+\frac{\partial w}{\partial z} \frac{d z}{d t} \\
&=(2 x)\left(e^t\right)+(2 y)\left(e^t \cos t+e^t \sin t\right)+(2 z)\left(e^t \cos t-e^t \sin t\right) \\
&=2 e^t[x+y \cos t+y \sin t+z \cos t-z \sin t] \\
& \text { (Replace } x, y, z-\text { values) } \\
&=2 e^t\left[e^t+e^t \sin t \cos t+e^t \sin ^2 t+e^t \cos ^2 t-e^t \cos t \sin t\right] \\
&=2 e^t \cdot e^t\left[1+\sin ^2 t+\cos ^2 t\right] \\
&=2 e^{2 t}[2] \\
&=4 e^{2 t}
\end{aligned}
$

Question 4.
Let $\mathrm{U}(x, y, z)=x y z, x=e^{-t}, y=e^{-t} \cos t, z=\sin t, t \in \mathbf{R}$. Find $\frac{d \mathrm{U}}{d t}$.
Solution:
$
\begin{aligned}
\mathrm{U}(x, y, z) & =x y z ; x=e^{-t} ; y=e^{-t} \cos t ; z=\sin t \\
\frac{d \mathrm{U}}{d t} & =\frac{\partial u}{\partial x} \frac{d x}{d t}+\frac{\partial u}{\partial y} \frac{d y}{d t}+\frac{\partial u}{\partial z} \frac{d z}{d t} \\
& =(y z)\left(-e^{-t}\right)+(x z)\left(-e^{-t} \cos t-e^{-t} \sin t\right)+(x y)(\cos t) \\
& =e^{-t}\left[-y z-x z \cos t-x z \sin t+x y \cos t e^t\right]
\end{aligned}
$
$
\begin{aligned}
& \text { (Replace } \mathrm{x}, \mathrm{y}, \mathrm{z} \text { value) } \\
& =e^{-t}\left[-e^{-t} \cos t \sin \mathrm{t}-e^{-t} \sin t \cos t-e^{-t} \sin ^2 t+e^{-t \cdot} \cdot e^{-t} \cdot \cos ^2 t e^t\right] \\
& =-e^{-t} \cdot e^{-t}\left[\cos t \sin t+\sin t \cos t+\sin ^2 t-\cos ^2 t\right] \\
& =-e^{-2 t}\left[2 \cos t \sin t-\left(\cos ^2 t-\sin ^2 t\right)\right] \\
& =-e^{-2 t}[\sin 2 t-\cos 2 t]
\end{aligned}
$

Question 5.
If $\mathrm{w}(\mathrm{x}, \mathrm{y})=6 \mathrm{x}^3-3 \mathrm{xy}+2 \mathrm{y}^2, \mathrm{x}=\mathrm{e}^{\mathrm{s}}, \mathrm{y}=\cos \mathrm{s}, \mathrm{s} \in \mathrm{R}$, find $\frac{d w}{d s}$, and evaluate $\mathrm{at} \mathrm{s}=0$.
Solution:
$
\begin{aligned}
w(x, y) & =6 x^3-3 x y+2 y^2 ; x=e^s ; y=\cos s \\
\frac{d w}{d s} & =\frac{\partial w}{\partial x} \frac{d x}{d s}+\frac{\partial w}{\partial y} \frac{d y}{d s} \\
& =\left(18 x^2-3 y\right)\left(e^s\right)+(-3 x+4 y)(-\sin s)
\end{aligned}
$
(Replace $x, y$ value)
$
\begin{aligned}
& =18 e^{3 \mathrm{~s}}-3 \cos s e^s+3 e^s \sin s-4 \sin s \cos s \\
& \text { At } s=0 \\
& \frac{d w}{d s}=18-3+0-0 \\
& \frac{d w}{d s}=15 \\
&
\end{aligned}
$
Question 6.
If $\mathrm{z}(\mathrm{x}, \mathrm{y})=\mathrm{x} \tan ^{-1}(\mathrm{x} y), \mathrm{x}=\mathrm{t}^2, \mathrm{y}=\mathrm{se}^{\mathrm{t}}, \mathrm{s}, \mathrm{t} \in \mathrm{R}$, Find $\frac{\partial z}{\partial \mathrm{t}}$ and $\frac{\partial z}{\partial \mathrm{t}}$ at $\mathrm{s}=\mathrm{t}=1$
Solution:
$
z(x, y)=x \tan ^{-1}(x y) ; x=t^2 ; y=s e^t
$

$\begin{aligned}
& \text { (i) } \frac{\partial z}{\partial s}=\frac{\partial z}{\partial x} \frac{\partial x}{\partial s}+\frac{\partial z}{\partial y} \frac{\partial y}{\partial s} \\
& =\left[\frac{x y}{1+(x y)^2}+\tan ^{-1}(x y)\right](0)+\frac{x^2}{1+(x y)^2}\left(e^t\right) \\
& =\frac{x^2 e^t}{1+(x y)^2} \\
& \text { (Replace } x, y \text { values) } \\
& \frac{\partial z}{\partial s}=\frac{e^t t^4}{1+\left(t^2 s e^t\right)^2} \\
& \text { At } s=1 ; t=1 \\
& x=1 ; y=e \\
& \frac{\partial z}{\partial s}=\frac{e}{1+e^2} \\
&
\end{aligned}$

(ii)
$
\begin{aligned}
\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} \\
& =\left[\frac{x y}{1+x^2 y^2}+\tan ^{-1}(x y)\right](2 t)+\left[\frac{x^2}{1+x^2 y^2}\right]\left(s e^t\right)
\end{aligned}
$
At $s=1 ; t=1$
- $x=1 ; y=e$
$
\begin{aligned}
\frac{\partial z}{\partial t} & =2\left[\frac{e}{1+e^2}+\tan ^{-1}(e)\right]+\left[\frac{1}{1+e^2}\right] e \\
& =\frac{2 e}{1+e^2}+2 \tan ^{-1} e+\frac{e}{1+e^2} \\
\frac{\partial z}{\partial t} & =\frac{3 e}{1+e^2}+2 \tan ^{-1} e
\end{aligned}
$

Question 7.
Let $U(x, y)=e^x \sin y$, where $x=s t^2, y=s^2 t, s, t \in R$. Find them at $s=t=1$.

Solution:

(i)
$
\mathrm{U}(x, y)=e^x \sin y ; x=s t^2 ; y=s^2 t
$
$
\begin{aligned}
\frac{\partial \mathrm{U}}{\partial s} & =\frac{\partial \mathrm{U}}{\partial x} \frac{\partial x}{\partial s}+\frac{\partial \mathrm{U}}{\partial y} \frac{\partial y}{\partial s} \\
& =\left(e^x \sin y\right)\left(t^2\right)+\left(e^x \cos y\right)(2 \mathrm{~s} t) \\
& =t e^x[t \sin y+2 s \cos y]
\end{aligned}
$
$
\begin{aligned}
& \text { (Replace } \mathrm{x}, \mathrm{y} \text { values) } \\
& \begin{array}{l}
\frac{\partial \mathrm{U}}{\partial s}=t e^{s t^2}\left[t \sin \left(s^2 t\right)+2 s \cos \left(s^2 t\right)\right] \\
\text { At } s=t=1 \\
\qquad \frac{\partial \mathrm{U}}{\partial s}=e[\sin (1)+2 \cos (1)]
\end{array}
\end{aligned}
$
(ii)
$\frac{\partial \mathrm{U}}{\partial t}=\frac{\partial \mathrm{U}}{\partial x} \cdot \frac{\partial x}{\partial t}+\frac{\partial \mathrm{U}}{\partial y} \cdot \frac{\partial y}{\partial t}$
$=\left(e^x \sin y\right)(2 s t)+\left(e^x \cos y\right)\left(\mathrm{s}^2\right)$
$=s^x[2 t \sin y+s \cos y]$
(Replace $x, y$ values)
$
\frac{\partial \mathrm{U}}{\partial t}=s e^{s t^2}\left[2 t \sin \left(s^2 t\right)+s \cos \left(s^2 t\right)\right]
$
At $s=t=1$
$
\frac{\partial U}{\partial t}=e[2 \sin (1)+\cos (1)]
$

Question 8.
Let $z(x, y)=x^3-3 x^2 y^3$, where $x=s^t, y=s^{-t}, s, t \in R$. Find $\frac{\partial z}{\partial s}$ and $\frac{\partial z}{\partial t}$

Solution:
$(i)$
$
\begin{aligned}
z(x, y) & =x^3-3 x^2 y^3 ; x=s e^t ; y=s e^{-t} \\
\frac{\partial \mathrm{U}}{\partial t} & =\frac{\partial z}{\partial x} \frac{\partial x}{\partial s}+\frac{\partial z}{\partial y} \frac{\partial y}{\partial s} \\
& =\left(3 x^2-6 x y^3\right)\left(e^t\right)+\left(-9 x^2 y^2\right)\left(e^{-t}\right) \\
& =\left(e^t\right)\left[3 x^2-6 x y^3\right]-\left(e^{-t}\right)\left[9 x^2 y^2\right]
\end{aligned}
$
(Replace $x, y$ values)
$
\begin{aligned}
\frac{\partial z}{\partial s} & =e^t\left[3\left(s e^t\right)^2-6\left(s e^t\right)\left(s e^{-t}\right)^3\right]-e^{-t}\left[9\left(s e^t\right)^2\left(s e^{-t}\right)^2\right] \\
& =e^t\left[3 s^2 e^{2 t}-6 s^4 e^t e^{-3 t}\right]-e^{-t}\left[9 s^4\right] \\
& =e^t\left(3 s^2\right)\left[e^{2 t}-2 s^2 e^{-2 t}-3 s^2 e^{-t}\right] \\
& =3 s^2 e^t\left[e^{2 t}-2 s^2 e^{-2 t}-3 s^2 e^{-t}\right]
\end{aligned}
$
(ii)
$
\begin{aligned}
\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} \\
& =\left(3 x^2-6 x y^3\right)\left(s e^t\right)+\left(-9 x^2 y^2\right)\left(-s e^{-t}\right)
\end{aligned}
$
(Replace $x, y$ values)
$
\begin{aligned}
& =\left(s e^t\right)\left[3\left(s e^t\right)^2-6\left(s e^t\right)\left(s e^{-t}\right)^3\right]+\left(s e^{-t}\right)\left[9\left(s e^t\right)^2\left(s e^{-t}\right)^2\right] \\
& =\left(s e^t\right)\left[3 s^2 e^{2 t}-6 s^2 e^t e^{-3 t}\right]+\left(s e^{-t}\right)\left[9 s^4\right] \\
& =3 s^3 e^{3 t}-6 s^3 e^{-t}+9 s^5 e^{-t} \\
& =3 s^3\left[e^{3 t}-2 e^{-t}+3 s^2 e^{-t}\right]
\end{aligned}
$

Question 9.
$\mathrm{W}(\mathrm{x}, \mathrm{y}, \mathrm{z})=\mathrm{xy}+\mathrm{yz}+\mathrm{zx}, \mathrm{x}=\mathrm{u}-\mathrm{v}, \mathrm{y}=\mathrm{uv}, \mathrm{z}=\mathrm{u}+\mathrm{v}, \mathrm{u}, \mathrm{v} \in \mathrm{R}$. Find $\frac{\partial w}{\partial u}, \frac{\partial w}{\partial v}$ them at $\left(\frac{1}{2}, 1\right)$

Solution:

$
\mathrm{W}(x, y, z)=x y+y z+z x ; x=u-v ; y=u v ; z=u+v
$
(i)
$
\begin{aligned}
\frac{\partial w}{\partial u} & =\frac{\partial w}{\partial x} \frac{\partial x}{\partial u}+\frac{\partial w}{\partial y} \frac{\partial y}{\partial u}+\frac{\partial w}{\partial z} \frac{\partial z}{\partial u} \\
& =(y+z)(1)+(x+z)(v)+(x+y)(1) \\
& =y+z+(x+z) v+x+y
\end{aligned}
$
(Replace $x \& y$ values)
$
\begin{aligned}
& =u v+u+v+(u-v+u+v) v+(u-v+u v) \\
& =u v+u+v+2 u v+u-v+u v \\
& =4 u v+2 u \\
\frac{\partial w}{\partial u} & =2 u(1+2 v)
\end{aligned}
$
(ii)
$
\begin{aligned}
\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)+(x+y)(1) \\
& =-y-z+(x+z) u+x+y=-z+(x+z) u+x \\
& =-u-v+(u-v+u+v) u+u-v \\
& =-2 v+2 u^2
\end{aligned}
$
$
\begin{aligned}
\frac{\partial w}{\partial v} & =2\left[u^2-v\right) \\
\frac{\partial w}{\partial u} & =2\left(\frac{1}{2}\right)[1+2(1)]=3 \\
\frac{\partial w}{\partial v} & =2\left[\frac{1}{4}-1\right]=2\left[\frac{-3}{4}\right] \\
& =\frac{-3}{2}
\end{aligned}
$