Exercise 8.1-Additional Problems - Chapter 8 - Differentials and Partial Derivatives - 12th Maths Guide Samacheer Kalvi Solutions - Tamil Medium

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Additional Questions Solved
Question 1.
Using differentials, find the approximate value of each of the following upto 3 places of decimal. $(255)^{\frac{1}{4}}$ Solution:
$
\text { Let } \begin{aligned}
y & =x^{\frac{1}{4}} \\
\frac{d y}{d x} & =\frac{1}{4 x^{\frac{3}{4}}}
\end{aligned}
$
Let $x=256$ and $x+\Delta x=255$
$
\begin{aligned}
\Delta x & =(x+\Delta x)-x=255-256=-1 \\
\Delta y & =(x+\Delta x)^{\frac{1}{4}}-(x)^{\frac{1}{4}} \\
& =(255)^{\frac{1}{4}}-(256)^{\frac{1}{4}}=(255)^{\frac{1}{4}}-4
\end{aligned}
$
$
\therefore(255)^{\frac{1}{4}}=4+\Delta y
$
Since $d y$ is approximately equal to $\Delta y$ and is given by
$
\begin{aligned}
d y & =\frac{d y}{d x} \cdot \Delta x=\frac{1}{4 x^{3 / 4}} \times(-1) \\
& =\frac{-1}{4 \times(256)^{\frac{3}{4}}}=\frac{-1}{4 \times 64}=-0.004 \\
\therefore(255)^{\frac{1}{4}} & =4-0.004=3.996
\end{aligned}
$
Thus approximately value of $(255)^{\frac{1}{4}}$ is $3.996$.

Question 2.
Using differentials, find the approximate value of each of the following upto 3 places of decimal. $(401)^{\frac{1}{2}}$
Solution:
Let $\begin{aligned} y & =x^{\frac{1}{2}} \\ \frac{d y}{d x} & =\frac{1}{2 \sqrt{x}}\end{aligned}$
Let $x=400$ and $x+\Delta x=401$
$
\begin{aligned}
\Delta x & =(x+\Delta x)-x=401-400=1 \\
\Delta y & =(x+\Delta x)^{\frac{1}{2}}-(x)^{\frac{1}{2}} \\
\Delta y & =(x+\Delta x)^{\frac{1}{2}}-(x)^{\frac{1}{2}} \\
& =(401)^{\frac{1}{2}}-(400)^{\frac{1}{2}} \\
& =(401)^{\frac{1}{2}}-20
\end{aligned}
$

$
\therefore(401)^{\frac{1}{2}}=20+\Delta y
$
Since $d y$ is approximately equal to $\Delta y$ and is given by
$
\begin{aligned}
d y & =\frac{d y}{d x} \cdot \Delta x=\frac{1}{2 \sqrt{x}} \cdot 1 \\
& =\frac{1}{2 \sqrt{400}}=\frac{1}{2 \times 20}=0.025 \\
\therefore \quad(401)^{\frac{1}{2}} & =20+0.025=20.025
\end{aligned}
$
Thus approximately value of $(401)^{\frac{1}{2}}$ is $20.025$.

Question 3.
Find approximate value of $f(5.001)$ where $f(x)=x^3-7 x^2+15$.
Solution:
Here $f(x)=x^3-7 x^2+15$
$f^{\prime}(x)=3 x^2-14 x$
Let $\mathrm{x}=5$ and $\mathrm{x}+\Delta \mathrm{x}=5.001$
$\therefore \Delta \mathrm{x}(\mathrm{x}+\Delta \mathrm{x})-\mathrm{x}=5.001-5$
$=0.001$
$\mathrm{f}(5.001)=\mathrm{f}(\mathrm{x}+\Delta \mathrm{x})$
Now $\Delta \mathrm{y}=\mathrm{f}(\mathrm{x}+\Delta \mathrm{x})-\mathrm{f}(\mathrm{x})$
$\mathrm{f}(\mathrm{x}+\Delta \mathrm{x})=\mathrm{f}(\mathrm{x})+\Delta \mathrm{y}$
$=\mathrm{f}(\mathrm{x})+\mathrm{f}(\mathrm{x}) \cdot \Delta \mathrm{x}\left[\therefore \Delta \mathrm{y}=\mathrm{f}^{\prime}(\mathrm{x}) \cdot \Delta \mathrm{x}\right]$
$=\left(\mathrm{x}^3-7 \mathrm{x}^2+15\right)+\left(3 \mathrm{x}^2-14 \mathrm{x}\right)(0.001)$
$
\begin{aligned}
& =\left(5^3-7 \times 5^2+15\right)+\left(3 \times 5^2-14 \times 5\right)[0.001] \\
& =[125-175+15]+[75-70][0.001] \\
& =-35+0.005=-34.995
\end{aligned}
$
Thus approximate value of $\mathrm{f}(5.001)$ is $-34.995$

Question 4.
If the radius of a sphere is measured as $7 \mathrm{~m}$ with an error of $0.02 \mathrm{~m}$ then find the approximate error in calculating its volume.
Solution:
Le $\mathrm{r}$ be the radius of the sphere and $\Delta \mathrm{r}$ be the error in measuring the radius.
Then $\mathrm{r}=7 \mathrm{~m}$ and $\Delta \mathrm{r}=0.02 \mathrm{~m}$
Now volume of a sphere is given by
$
\begin{aligned}
v=\frac{4}{3} & \pi r^3 \\
\frac{d v}{d r} & =\frac{4}{3} \pi \times 3 r^2=4 \pi r^2 \\
\therefore \quad d v & =\frac{d v}{d r} \cdot \Delta r=\left(4 \pi r^2\right) \cdot \Delta r \\
& =4 \pi \times 7^2 \times 0.02=3.92 \pi \mathrm{m}^3
\end{aligned}
$