Exercise 8.2-Additional Problems - Chapter 8 - Differentials and Partial Derivatives - 12th Maths Guide Samacheer Kalvi Solutions - Tamil Medium

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Additional Questions
Question 1.
Find the differential $d y$ and evaluate $d y$ for the given values of $x$ and $d x \cdot y=x^4-3 x^2+x-1, x=$ $2, \mathrm{dx}=0.1$
Solution:
$
\begin{aligned}
& \begin{array}{l}
y=x^4-3 x^2+x-1 \\
\qquad \frac{d y}{d x}
\end{array}=4 x^3-6 x+1 ; d y=\left(4 x^3-6 x+1\right) d x \\
& \text { At } x=2, d x=0.1 \\
& d y=[4(8)-6(2)+1](0.1)=(32-12+1)(0.1) \\
& =(21)(0.1)=2.1 \\
& d y=\left(4 x^3-6 x+1\right) d x ; d y=2.1
\end{aligned}
$
Question 2.
Find the differential dy and evaluate dy for the given values of $\mathrm{x}$ and $\mathrm{dx} . \mathrm{y}=\sqrt{1-x}, \mathrm{x}=0, \mathrm{dx}=$ $0.02$

Solution:
$
\begin{aligned}
y & =\sqrt{1-x} \\
\frac{d y}{d x} & =\frac{1}{2 \sqrt{1-x}}(-1)=\frac{-1}{2 \sqrt{1-x}} ; \quad \therefore d y=\frac{-1}{2 \sqrt{1-x}} d x \\
\text { At } x & =0, d x=0.02 \\
d y & =\frac{-1}{2 \sqrt{1-0}}(0.02)=\frac{-0.02}{2}=-0.01 \\
d y & =\frac{-1}{2 \sqrt{1-x}} d x ;-0.01
\end{aligned}
$

Question 3.
The edge of a cube was found to be $30 \mathrm{~cm}$ with a possible error in measurement of $0.1 \mathrm{~cm}$. Use differentials to estimate the maximum possible error in computing
(i) the volume of the cube and
(ii) the surface area of cube.
Solution:
The side of the cube $=\mathrm{a}=30 \mathrm{~cm}$
Error in $\mathrm{a}=\mathrm{da}=0.1 \mathrm{~cm}$
(i)
$
\begin{aligned}
\text { Volume } & =\mathrm{V}=a^3 ; \therefore \frac{d \mathrm{~V}}{d a}=3 a^2 \text { (or) } d \mathrm{~V}=3 a^2 d a \\
\text { At } a & =30 \mathrm{~cm}, d a=0.1 \mathrm{~cm} \\
d \mathrm{~V} & =3(30)^2(0.1)=3(900)(0.1)=270 \mathrm{~cm}^3
\end{aligned}
$

(ii)
$
\begin{aligned}
\mathrm{A} & =6 a^2 \\
\frac{d \mathrm{~A}}{d r} & =6(2 a)=12 a \\
d \mathrm{~A} & =12 a d a \\
\text { At } a & =30 \mathrm{~cm}, d a=0.1 \mathrm{~cm}, \\
d \mathrm{~A} & =12(30)(0.1)=36 \mathrm{~cm}^2
\end{aligned}
$

Question 4.
The radius of a circular disc is given as $24 \mathrm{~cm}$ with a maximum error in measurement of $0.02 \mathrm{~cm}$ (i) Use differentials to estimate the maximum error in the calculated area of the disc, (ii) Compute the relative error.
Solution:
$
\mathrm{r}=24 \mathrm{~cm} \text {, }
$
$
\mathrm{dr}=0.02 \mathrm{~cm}
$
(i) Area $=\mathrm{A}=\pi r^2 ; \frac{d \mathrm{~A}}{d r}=\pi(2 r) ; \quad \therefore d \mathrm{~A}=2 \pi r d r$
$
\text { At } \begin{aligned}
r & =24 \mathrm{~cm} \text { and } d r=0.02 \mathrm{~cm} \\
d \mathrm{~A} & =2 \pi(24)(0.02)=0.96 \pi \mathrm{cm}^2
\end{aligned}
$
(ii)
$
\frac{d \mathrm{~A}}{\mathrm{~A}}=\frac{0.96 \pi}{\pi r^2}=\frac{0.96 \pi}{\pi \times 24 \times 24}=0.0016666=0.001667
$