Using differentials, find the approximate value of each

Solution:

(i) $\sqrt{25.3}$

Consider $y=\sqrt{x}$. Let $x=25$ and $\Delta x=0.3$.

Then,

$\Delta y=\sqrt{x+\Delta x}-\sqrt{x}=\sqrt{25.3}-\sqrt{25}=\sqrt{25.3}-5$

 

$\Rightarrow \sqrt{25.3}=\Delta y+5$

Now, dy is approximately equal to Δy and is given by,

$d y=\left(\frac{d y}{d x}\right) \Delta x=\frac{1}{2 \sqrt{x}}(0.3) \quad[$ as $y=\sqrt{x}]$

$=\frac{1}{2 \sqrt{25}}(0.3)=0.03$

Hence, the approximate value of $\sqrt{25.3}$ is $0.03+5=5.03$.

(ii) $\sqrt{49.5}$

Consider $y=\sqrt{x}$. Let $x=49$ and $\Delta x=0.5$.

Then,

$\Delta y=\sqrt{x+\Delta x}-\sqrt{x}=\sqrt{49.5}-\sqrt{49}=\sqrt{49.5}-7$

$\Rightarrow \sqrt{49.5}=7+\Delta y$

Now, dy is approximately equal to Δy and is given by,

$d y=\left(\frac{d y}{d x}\right) \Delta x=\frac{1}{2 \sqrt{x}}(0.5) \quad[$ as $y=\sqrt{x}]$

$=\frac{1}{2 \sqrt{49}}(0.5)=\frac{1}{14}(0.5)=0.035$

Hence, the approximate value of $\sqrt{49.5}$ is $7+0.035=7.035$.

(iii) $\sqrt{0.6}$

Consider $y=\sqrt{x}$. Let $x=1$ and $\Delta x=-0.4$.

Then,

$\Delta y=\sqrt{x+\Delta x}-\sqrt{x}=\sqrt{0.6}-1$

 

$\Rightarrow \sqrt{0.6}=1+\Delta y$

Now, dy is approximately equal to $\Delta y$ and is given by,

$d y=\left(\frac{d y}{d x}\right) \Delta x=\frac{1}{2 \sqrt{x}}(\Delta x) \quad[$ as $y=\sqrt{x}]$

$=\frac{1}{2}(-0.4)=-0.2$

Hence, the approximate value of $\sqrt{0.6}$ is $1+(-0.2)=1-0.2=0.8$.

(iv) $(0.009)^{\frac{1}{3}}$

Consider $y=x^{\frac{1}{3}}$. Let $x=0.008$ and $\Delta x=0.001$.

Then,

$\Delta y=(x+\Delta x)^{\frac{1}{3}}-(x)^{\frac{1}{3}}=(0.009)^{\frac{1}{3}}-(0.008)^{\frac{1}{3}}=(0.009)^{\frac{1}{3}}-0.2$

$\Rightarrow(0.009)^{\frac{1}{3}}=0.2+\Delta y$

Now, dy is approximately equal to Δy and is given by,

$d y=\left(\frac{d y}{d x}\right) \Delta x=\frac{1}{3(x)^{\frac{2}{3}}}(\Delta x) \quad\left[\right.$ as $\left.y=x^{\frac{1}{3}}\right]$

$=\frac{1}{3 \times 0.04}(0.001)=\frac{0.001}{0.12}=0.008$

Hence, the approximate value of $(0.009)^{\frac{1}{3}}$ is $0.2+0.008=0.208$.

(v) $(0.999)^{\frac{1}{10}}$

Consider $y=(x)^{\frac{1}{10}} .$ Let $x=1$ and $\Delta x=-0.001$

Then,

$\Delta y=(x+\Delta x)^{\frac{1}{10}}-(x)^{\frac{1}{10}}=(0.999)^{\frac{1}{10}}-1$

$\Rightarrow(0.999)^{\frac{1}{10}}=1+\Delta y$

Now, dy is approximately equal to Δy and is given by,

$d y=\left(\frac{d y}{d x}\right) \Delta x=\frac{1}{10(x)^{\frac{9}{10}}}(\Delta x) \quad\left[\right.$ as $\left.y=(x)^{\frac{1}{10}}\right]$

$=\frac{1}{10}(-0.001)=-0.0001$

Hence, the approximate value of $(0.999)^{\frac{1}{10}}$ is $1+(-0.0001)=0.9999$.

(vi) $(15)^{\frac{1}{4}}$

Consider $y=x^{\frac{1}{4}}$. Let $x=16$ and $\Delta x=-1$.

Then,

$\Delta y=(x+\Delta x)^{\frac{1}{4}}-x^{\frac{1}{4}}=(15)^{\frac{1}{4}}-(16)^{\frac{1}{4}}=(15)^{\frac{1}{4}}-2$

 

$\Rightarrow(15)^{\frac{1}{4}}=2+\Delta y$

Now, dy is approximately equal to Δy and is given by,

$d y=\left(\frac{d y}{d x}\right) \Delta x=\frac{1}{4(x)^{\frac{3}{4}}}(\Delta x) \quad\left[\right.$ as $\left.y=x^{\frac{1}{4}}\right]$

$=\frac{1}{4(16)^{\frac{3}{4}}}(-1)=\frac{-1}{4 \times 8}=\frac{-1}{32}=-0.03125$

Hence, the approximate value of $(15)^{\frac{1}{4}}$ is $2+(-0.03125)=1.96875$.

(vii) $(26)^{\frac{1}{3}}$

Consider $y=(x)^{\frac{1}{3}}$. Let $x=27$ and $\Delta x=-1$.

Then,

$\Delta y=(x+\Delta x)^{\frac{1}{3}}-(x)^{\frac{1}{3}}=(26)^{\frac{1}{3}}-(27)^{\frac{1}{3}}=(26)^{\frac{1}{3}}-3$

$\Rightarrow(26)^{\frac{1}{3}}=3+\Delta y$

Now, dy is approximately equal to Δy and is given by,

$d y=\left(\frac{d y}{d x}\right) \Delta x=\frac{1}{3(x)^{\frac{2}{3}}}(\Delta x) \quad\left[\right.$ as $\left.y=(x)^{\frac{1}{3}}\right]$

$=\frac{1}{3(27)^{\frac{2}{3}}}(-1)=\frac{-1}{27}=-0.0 \overline{370}$

Hence, the approximate value of $(26)^{\frac{1}{3}}$ is $3+(-0.0370)=2.9629$.

(viii) $(255)^{\frac{1}{4}}$

Consider $y=(x)^{\frac{1}{4}}$. Let $x=256$ and $\Delta x=-1$.

Then,

$\Delta y=(x+\Delta x)^{\frac{1}{4}}-(x)^{\frac{1}{4}}=(255)^{\frac{1}{4}}-(256)^{\frac{1}{4}}=(255)^{\frac{1}{4}}-4$

$\Rightarrow(255)^{\frac{1}{4}}=4+\Delta y$

Now, dy is approximately equal to Δy and is given by,

$d y=\left(\frac{d y}{d x}\right) \Delta x=\frac{1}{4(x)^{\frac{3}{4}}}(\Delta x) \quad\left[\right.$ as $\left.y=x^{\frac{1}{4}}\right]$

$=\frac{1}{4(256)^{\frac{3}{4}}}(-1)=\frac{-1}{4 \times 4^{3}}=-0.0039$

Hence, the approximate value of $(255)^{\frac{1}{4}}$ is $4+(-0.0039)=3.9961$.

(ix) $(82)^{\frac{1}{4}}$

Consider $y=x^{\frac{1}{4}}$. Let $x=81$ and $\Delta x=1$.

Then,

$\Delta y=(x+\Delta x)^{\frac{1}{4}}-(x)^{\frac{1}{4}}=(82)^{\frac{1}{4}}-(81)^{\frac{1}{4}}=(82)^{\frac{1}{4}}-3$

$\Rightarrow(82)^{\frac{1}{4}}=\Delta y+3$

Now, dy is approximately equal to Δy and is given by,

$d y=\left(\frac{d y}{d x}\right) \Delta x=\frac{1}{4(x)^{\frac{3}{4}}}(\Delta x) \quad\left[\right.$ as $\left.y=x^{\frac{1}{4}}\right]$

$=\frac{1}{4(81)^{\frac{3}{4}}}(1)=\frac{1}{4(3)^{3}}=\frac{1}{108}=0.009$

Hence, the approximate value of $(82)^{\frac{1}{4}}$ is $3+0.009=3.009$.

(x) $(401)^{\frac{1}{2}}$

Consider $y=x^{\frac{1}{2}}$. Let $x=400$ and $\Delta x=1$.

Then,

$\Delta y=\sqrt{x+\Delta x}-\sqrt{x}=\sqrt{401}-\sqrt{400}=\sqrt{401}-20$

 

$\Rightarrow \sqrt{401}=20+\Delta y$

Now, dy is approximately equal to Δy and is given by,

$d y=\left(\frac{d y}{d x}\right) \Delta x=\frac{1}{2 \sqrt{x}}(\Delta x) \quad\left[\right.$ as $\left.y=x^{\frac{1}{2}}\right]$

$=\frac{1}{2 \times 20}(1)=\frac{1}{40}=0.025$

Hence, the approximate value of $\sqrt{401}$ is $20+0.025=20.025$.

(xi) $(0.0037)^{\frac{1}{2}}$

Consider $y=x^{\frac{1}{2}}$. Let $x=0.0036$ and $\Delta x=0.0001$.

Then,

$\Delta y=(x+\Delta x)^{\frac{1}{2}}-(x)^{\frac{1}{2}}=(0.0037)^{\frac{1}{2}}-(0.0036)^{\frac{1}{2}}=(0.0037)^{\frac{1}{2}}-0.06$

 

$\Rightarrow(0.0037)^{\frac{1}{2}}=0.06+\Delta y$

Now, dy is approximately equal to Δy and is given by,

$d y=\left(\frac{d y}{d x}\right) \Delta x=\frac{1}{2 \sqrt{x}}(\Delta x) \quad\left[\right.$ as $\left.y=x^{\frac{1}{2}}\right]$

$=\frac{1}{2 \times 0.06}(0.0001)$

$=\frac{0.0001}{0.12}=0.00083$

Thus, the approximate value of $(0.0037)^{\frac{1}{2}}$ is $0.06+0.00083=0.06083$.

(xii) $(26.57)^{\frac{1}{3}}$

Consider $y=x^{\frac{1}{3}}$. Let $x=27$ and $\Delta x=-0.43$.

Then,

$\Delta y=(x+\Delta x)^{\frac{1}{3}}-x^{\frac{1}{3}}=(26.57)^{\frac{1}{3}}-(27)^{\frac{1}{3}}=(26.57)^{\frac{1}{3}}-3$

$\Rightarrow(26.57)^{\frac{1}{3}}=3+\Delta y$

Now, dy is approximately equal to Δy and is given by,

$d y=\left(\frac{d y}{d x}\right) \Delta x=\frac{1}{3(x)^{\frac{2}{3}}}(\Delta x) \quad\left[\right.$ as $\left.y=x^{\frac{1}{3}}\right]$

$=\frac{1}{3(9)}(-0.43)$

 

$=\frac{-0.43}{27}=-0.015$

Hence, the approximate value of $(26.57)^{\frac{1}{3}}$ is $3+(-0.015)=2.984$.

(xiii) $(81.5)^{\frac{1}{4}}$

Consider $y=x^{\frac{1}{4}}$. Let $x=81$ and $\Delta x=0.5$.

Then,

$\Delta y=(x+\Delta x)^{\frac{1}{4}}-(x)^{\frac{1}{4}}=(81.5)^{\frac{1}{4}}-(81)^{\frac{1}{4}}=(81.5)^{\frac{1}{4}}-3$

 

$\Rightarrow(81.5)^{\frac{1}{4}}=3+\Delta y$

Now, dy is approximately equal to Δy and is given by,

$d y=\left(\frac{d y}{d x}\right) \Delta x=\frac{1}{4(x)^{\frac{3}{4}}}(\Delta x) \quad\left[\right.$ as $\left.y=x^{\frac{1}{4}}\right]$

$=\frac{1}{4(3)^{3}}(0.5)=\frac{0.5}{108}=0.0046$

Hence, the approximate value of $(81.5)^{\frac{1}{4}}$ is $3+0.0046=3.0046$.

(xiv) $(3.968)^{\frac{3}{2}}$

Consider $y=x^{\frac{3}{2}}$. Let $x=4$ and $\Delta x=-0.032$

Then,

$\Delta y=(x+\Delta x)^{\frac{3}{2}}-x^{\frac{3}{2}}=(3.968)^{\frac{3}{2}}-(4)^{\frac{3}{2}}=(3.968)^{\frac{3}{2}}-8$

$\Rightarrow(3.968)^{\frac{3}{2}}=8+\Delta y$

Now, dy is approximately equal to Δy and is given by,

$d y=\left(\frac{d y}{d x}\right) \Delta x=\frac{3}{2}(x)^{\frac{1}{2}}(\Delta x) \quad\left[\right.$ as $\left.y=x^{\frac{3}{2}}\right]$

$=\frac{3}{2}(2)(-0.032)$

 

$=-0.096$

Hence, the approximate value of $(3.968)^{\frac{3}{2}}$ is $8+(-0.096)=7.904$.

$(\mathrm{xv})(32.15)^{\frac{1}{5}}$

Consider $y=x^{\frac{1}{5}}$. Let $x=32$ and $\Delta x=0.15$

Then,

$\Delta y=(x+\Delta x)^{\frac{1}{5}}-x^{\frac{1}{5}}=(32.15)^{\frac{1}{5}}-(32)^{\frac{1}{5}}=(32.15)^{\frac{1}{5}}-2$

 

$\Rightarrow(32.15)^{\frac{1}{5}}=2+\Delta y$

Now, dy is approximately equal to Δy and is given by,

$d y=\left(\frac{d y}{d x}\right) \Delta x=\frac{1}{5(x)^{\frac{4}{5}}} \cdot(\Delta x) \quad\left[\right.$ as $\left.y=x^{\frac{1}{5}}\right]$

$=\frac{1}{5 \times(2)^{4}}(0.15)$

$=\frac{0.15}{80}=0.00187$

Hence, the approximate value of $(32.15)^{\frac{1}{5}}$ is $2+0.00187=2.00187$.