Exercise 8.8-Additional Problems - Chapter 8 - Differentials and Partial Derivatives - 12th Maths Guide Samacheer Kalvi Solutions - Tamil Medium

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Additional Problems
Question 1.
If $\mathrm{u}=\mathrm{x}^{\mathrm{y}}$ then is equal to
(a) $y x^{y-1}$
(b) $u \log x$
(c) $u \log y$
(d) $x y^{x-1}$
Solution:
(a) $y x^{y-1}$
Hint:
Given, $u=x^y$
$
\frac{\partial u}{\partial x}=y x^{y-1}
$

Question 2.
If $u=\sin ^{-1}\left(\frac{x^4+y^4}{x^2+y^2}\right)$ and $f=\sin u$ then $f$ is a homogeneous function of degree
(a) 0
(b) 1
(c) 2
(d) 4
Solution:
(c) 2
Hint:
$
\text { Given, } \begin{aligned}
u & =\sin ^{-1}\left(\frac{x^4+y^4}{x^2+y^2}\right) \\
\sin u & =\frac{x^4+y^4}{x^2+y^2}=f \\
f & =\frac{t^4\left(x^4+y^4\right)}{t^2\left(x^2+y^2\right)}=t^2\left(\frac{x^4+y^4}{x^2+y^2}\right)
\end{aligned}
$
$
\text { let } x=t x, y=t y
$
$\therefore f=\sin u$ is a homogeneous function of degree 2 .
Question 3.
If $u=\frac{1}{\sqrt{x^2+y^2}}$, then $x \frac{\partial u}{\partial x}+y \frac{\partial u}{\partial y}$ is equal to
(a) $\frac{1}{2} u$
(b) $u$
(c) $\frac{3}{2} u$
(d) $-\boldsymbol{u}$
Solution:
(d) $-\mathrm{u}$

Hint:
Put $x=t x, y=t y$
$
\text { Given, } u=\frac{1}{\sqrt{x^2+y^2}}
$
$
u=\frac{1}{\sqrt{t^2\left(x^2+y^2\right)}}=\frac{1}{t \sqrt{x^2+y^2}}
$
$
=t^{-1} \frac{1}{\sqrt{x^2+y^2}}
$
$\therefore$ degree of $u=n=-1$
By Euler's theorem $x \frac{\partial u}{\partial x}+y \frac{\partial u}{\partial y}=n u=-u$
Question 4.
The curve $y^2(x-2)=x^2(1+x)$ has
(a) an asymptote parallel to $\mathrm{x}$-axis
(b) an asymptote parallel to y-axis
(c) asymptotes parallel to both axis
(d) no asymptotes
Solution:
(b) an asymptote parallel to y-axis
Hint:
when $x=2, y \rightarrow \infty$
$
\begin{aligned}
y^2(x-2) & =x^2(1+x) \\
y^2 & =\frac{x^2(1+x)}{x-2}
\end{aligned}
$
$\therefore$ The curve has an asymptote at $x=2$
Question 5.
If $\mathrm{x}=\mathrm{r} \cos \theta, \mathrm{y}=\mathrm{r} \sin \theta$, then $\frac{\partial r}{\partial x}$ is equal to $\ldots \ldots$...
(a) $\sec \theta$
(b) $\sin \theta$
(c) $\cos \theta$
(d) $\operatorname{cosec} \theta$
Solution:
(c) $\cos \theta$

Hint:
$
x^2+y^2=r^2 \cos ^2 \theta+r^2 \sin ^2 \theta=r^2
$
Diff w.r.t. ' $x$ ' $\Rightarrow 2 x=2 r \frac{\partial r}{\partial x}$
$
\frac{\partial r}{\partial x}=\frac{x}{r}=\frac{r \cos \theta}{r}=\cos \theta
$
Question 6.
If $u=\log \left(\frac{x^2+y^2}{x y}\right)$ then $x \frac{\partial u}{\partial x}+y \frac{\partial u}{\partial y}$ is
(a) 0
(b) $\mathrm{u}$
(c) $2 \mathrm{u}$
(d) $u-1$
Solution:
(a) 0
Hint:
Given $u=\log _e\left(\frac{x^2+y^2}{x y}\right)$
Changing into exponential form
$\cdot \quad e^u=\frac{x^2+y^2}{x y}=f$
Put $x=t x, y=t y$
$
e^u=\frac{t^2 x^2+t^2 y^2}{t^2 x y}=\frac{t^2\left(x^2+y^2\right)}{t^2(x y)}=t^0\left(\frac{x^2+y^2}{x y}\right)
$
The degree of $f=n=0$
$\therefore$ By Euler's theorem $x \frac{\partial}{\partial x} e^u+y \frac{\partial}{\partial y} e^u=0\left(e^u\right)$
$
\begin{aligned}
e^u\left[x \frac{\partial u}{\partial x}+y \frac{\partial u}{\partial y}\right] & =0 \\
x \frac{\partial u}{\partial x}+y \frac{\partial u}{\partial y} & =0
\end{aligned}
$

Question 7.
The percentage error in the $11^{\text {th }}$ root of the number 28 is approximately times the precentage error is 28.
(a) $\frac{1}{28}$
(b) $\frac{1}{11}$
(c) 11
(d) 28
Solution:
$\frac{1}{11}$
Hint:
Let $y=x^{1 / 11}$
Take log on both sides
$
\begin{aligned}
\log y & =\frac{1}{11} \log x \Rightarrow \frac{d y}{y}=\frac{1}{11} \frac{d x}{x} \\
\frac{d y}{y} \times 100 & =\frac{1}{11} \frac{d x}{x} \times 100
\end{aligned}
$
The percentage error in $x=28 \Rightarrow \frac{d x}{x} \times 100=28$
The percentage error in $y=\frac{d y}{y} \times 100=\frac{1}{11}$ (28)
The percentage error is approximately $\frac{1}{11}$ times the percentage error is 28 .

Question 8.
The curve $a^2 y^2=x^2\left(a^2-x^2\right)$ has ......

(a) only one loop between $x=0$ and $x=a$
(b) two loops between $x=0$ and $x=a$
(c) two loops between $x=-a$ and $x=a$
(d) no loop
Solution:
(c) two loops between $x=-a$ and $x=a$
Hint.
Given, $a^2 y^2=x^2\left(a^2-x^2\right)$
Taking $y=0(1) \Rightarrow 0=x^2\left(a^2-x^2\right)$
$
\begin{aligned}
\therefore x^2 & =0 \text { (or) } a^2-x^2=0 \\
a^2 & =x^2 \Rightarrow x=\pm a
\end{aligned}
$
The points are $(0,0)(-a, 0)(a, 0)$

Question 9.
An asymptote to the curve $y^2(a+2 x)=x^2(3 a-x)$ is
(a) $x=3 a$
(b) $x=-a / 2$
(c) $x=a / 2$
(d) $x=0$
Solution:
(b) $x=-a / 2$
Hint.
$
\begin{aligned}
y^2(a+x) & =x^2(3 a-x) \\
y^2 & =\frac{x^2(3 a-x)}{a+2 x}
\end{aligned}
$
When we put $x=-a / 2, y \rightarrow \infty$
$\therefore x=-a / 2$ is a asymptote.

Question 10.
In which region the curve $y^2(a+x)=x^2(3 a-x)$ does not lie?
(a) $x>0$
(b) $0<x<3 a$ (c) $x \leq-$ a and $x>3 a$
(d) $-a<x<3 a$

Solution:

(c) $x \leq-a$ and $x>3 a$
Hint.
$
y^2(a+x)=x^2(3 a-x)
$
Taking $y=0(1) \Rightarrow 0=x^2(3 a-x)$
$
=\mathrm{x}=0, \mathrm{x}=3 \mathrm{a}
$
$\therefore$ The points are $(0,0)(3 \mathrm{a}, 0)$
There is a loop between $\mathrm{x}=0$ and $\mathrm{x}=3 \mathrm{a}$

when $\mathrm{x}>3$ a, $\mathrm{y}$ imaginary
the curves does not exist beyond $x=3 a$ i.e., $x>3 a$
the curve has asymptote at $x=-a$
the curve does not exist when $\mathrm{x}<-\mathrm{a}$
the curve exists in the region $-a<x<3 a$

Question 11.
If $\mathrm{M}=\mathrm{y} \sin \mathrm{x}$, then $\frac{\partial^2 u}{\partial x \partial y}$ is equal to
(a) $\cos x$
(b) $\cos y$
(c) $\sin x$
(d) 0
Solution:
(a) $\cos x$
Hint:
$
\begin{aligned}
u & =y \sin x \\
\frac{\partial u}{\partial y} & =\sin x \Rightarrow \frac{\partial^2 u}{\partial x \partial y}=\cos x
\end{aligned}
$

Question 12.
If $u=f\left(\frac{y}{x}\right)$ then $x \frac{\partial u}{\partial x}+y \frac{\partial u}{\partial y}$ is equal to
(a) 0
(b) 1
(c) $2 \mathrm{u}$
(d) $\mathrm{u}$
Solution:
(a) 0
Hint:
$\begin{aligned} \text { Given, } u & =f\left(\frac{y}{x}\right)=f\left(\frac{t x}{t y}\right)=f\left[t^0\left(\frac{x}{y}\right)\right]=t^0 f\left(\frac{x}{y}\right) \\ \text { degree of } f & =n\end{aligned}$
By Euler's theorem $x \frac{\partial u}{\partial x}+y \frac{\partial u}{\partial y}=n u=0 u=0$

Question 13.
The curve $9 y^2=x^2\left(4-x^2\right)$ is symmetrical about
(a) $y$ - axis
(b) $\mathrm{x}$ - axis
(c) $y=x$
(d) both the axes
Solution:
(d) both the axes
Hint.
Replace $x$ by $-x$ and $y$ by $-y$
$9\left(-y^2\right)=(-x)^2\left(4-(-x)^2\right)$
The equation is unaltered
$\therefore$ the curve is symmetrical about both the axes.
Question 14.
The curve $a y^2=x^2(3 a-x)$ cuts the $y-a x i s$ at
(a) $x=-3 a, x=0$
(b) $x=0, x=3 a$
(c) $x=0, x=a$
(d) $\mathrm{x}=0$
Solution:
(d) $x=0$

Hint:
Given $a y^2=x^2(3 a-x)$
The point of intersection with $\mathrm{y}$-axis by putting $\mathrm{x}=\mathrm{O}$
$
\text { In (1) }=a y^2=0(3 a-0)
$
ay $=0 ; y=0$
$\therefore$ The curve intersects $\mathrm{y}$-axis at the origin is $\mathrm{x}=0$