Question 1

Using differentials, find the approximate value of each of the following up to 3 places of decimals.

(i) $\sqrt{25.3}$

(ii) $\sqrt{49.5}$

(iii) $\sqrt{0.6}$

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

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

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

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

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

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

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

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

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

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

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

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

Accepted Answer

(i) $\sqrt{25.3}$

Consider $y=\sqrt{x}$.

Let $x=25$ and $\Delta x=0.3$

Then,

$$

\begin{aligned}

\Delta y & =\sqrt{x+\Delta x}-\sqrt{x} \

& =\sqrt{25.3}-\sqrt{25} \

& =\sqrt{25.3}-5 \

\Delta y+5 & =\sqrt{25.3}

\end{aligned}

$$

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

$$

\begin{aligned}

d y & =\left(\frac{d y}{d x}ight) \Delta x \

& =\frac{1}{2 \sqrt{x}}(0.3) \quad[\because y=\sqrt{x}] \

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

& =0.03

\end{aligned}

$$

Hence,

$$

\begin{aligned}

\sqrt{25.3} & =0.03+5 \

& =5.03

\end{aligned}

$$

Thus, the approximate value of $\sqrt{25.3}=5.03$.

(ii) $\sqrt{49.5}$

Consider $y=\sqrt{x}$

Let $x=49$ and $\Delta x=0.5$

Then,

$$

\begin{aligned}

\Delta y & =\sqrt{x+\Delta x}-\sqrt{x} \

& =\sqrt{49.5}-\sqrt{49} \

& =\sqrt{49.5}-7 \

\Delta y+7 & =\sqrt{49.5}

\end{aligned}

$$

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

$$

\begin{aligned}

d y & =\left(\frac{d y}{d x}ight) \Delta x \

& =\frac{1}{2 \sqrt{x}}(0.5) \quad[\because y=\sqrt{x}] \

& =\frac{1}{2 \sqrt{49}}(0.5) \

& =\frac{1}{14}(0.5) \

& =0.035

\end{aligned}

$$

Hence,

$$

\begin{aligned}

\sqrt{49.5} & =7+0.035 \

& =7.035

\end{aligned}

$$

Thus, the approximate value of $\sqrt{49.5}=7.035$.

(iii) $\sqrt{0.6}$

Consider $y=\sqrt{x}$.

Let $x=1$ and $\Delta x=-0.4$

Then,

$$

\begin{aligned}

\Delta y & =\sqrt{x+\Delta x}-\sqrt{x} \

& =\sqrt{0.6}-1 \

\Delta y+1 & =\sqrt{0.6}

\end{aligned}

$$

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

$$

\begin{aligned}

d y & =\left(\frac{d y}{d x}ight) \Delta x \

& =\frac{1}{2 \sqrt{x}}(-0.4) \quad[\because y=\sqrt{x}] \

& =\frac{1}{2}(-0.4) \

& =-0.2

\end{aligned}

$$

Hence,

$$

\begin{aligned}

\sqrt{0.6} & =1+(-0.2) \

& =1-0.2 \

& =0.8

\end{aligned}

$$

Thus, the approximate value of $\sqrt{0.6}=0.8$.

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

Consider $y=(x)^{\frac{1}{3}}$

Let $x=0.008$ and $\Delta x=0.001$

Then,

$$

\begin{aligned}

\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 \

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

\end{aligned}

$$

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

$$

\begin{aligned}

d y & =\left(\frac{d y}{d x}ight) \Delta x \

& =\frac{1}{3(x)^{\frac{2}{3}}}(\Delta x) \quad\left[\because y=x^{\frac{1}{3}}ight] \

& =\frac{1}{3 	imes 0.04}(0.001) \

& =\frac{0.001}{0.12} \

& =0.008

\end{aligned}

$$

Hence,

$$

\begin{aligned}

(0.009)^{\frac{1}{3}} & =0.2+0.008 \

& =0.208

\end{aligned}

$$

Thus, the approximate value of $(0.009)^{\frac{1}{3}}=0.208$.

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

Consider $y=(x)^{\frac{1}{10}}$

Let $x=1$ and $\Delta x=-0.001$

Then,

$$

\begin{aligned}

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

& =(0.999)^{\frac{1}{10}}-1 \

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

\end{aligned}

$$

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

$$

\begin{array}{rlr}

d y & =\left(\frac{d y}{d x}ight) \Delta x & \

& =\frac{1}{10(x)^{\frac{9}{10}}}(\Delta x) & {\left[\because y=(x)^{\frac{1}{10}}ight]} \

& =\frac{1}{10}(-0.001) & \

& =-0.0001 &

\end{array}

$$

Hence,

$$

\begin{aligned}

(0.999)^{\frac{1}{10}} & =1+(-0.0001) \

& =0.9999

\end{aligned}

$$

Thus, the approximate value of $(0.999)^{\frac{1}{10}}=0.9999$.

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

Consider $y=(x)^{\frac{1}{4}}$

Let $x=16$ and $\Delta x=-1$

Then,

$$

\begin{aligned}

\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 \

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

\end{aligned}

$$

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

$$

\begin{aligned}

d y & =\left(\frac{d y}{d x}ight) \Delta x \

& =\frac{1}{4(x)^{\frac{3}{4}}}(\Delta x) \quad\left[\because y=(x)^{\frac{1}{4}}ight] \

& =\frac{1}{4(16)^{\frac{3}{4}}}(-1) \

& =\frac{-1}{4 	imes 8} \

& =\frac{-1}{32} \

& =-0.03125

\end{aligned}

$$

Hence,

$$

\begin{aligned}

(15)^{\frac{1}{4}} & =2+(-0.03125) \

& =1.96875

\end{aligned}

$$

Thus, the approximate value of $(15)^{\frac{1}{4}}=1.96875$.

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

Consider $y=(x)^{\frac{1}{3}}$.

Let $x=27$ and $\Delta x=-1$

Then,

$$

\begin{aligned}

\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 \

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

\end{aligned}

$$

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

$$

\begin{aligned}

d y & =\left(\frac{d y}{d x}ight) \Delta x \

& =\frac{1}{3(x)^{\frac{2}{3}}}(\Delta x) \quad\left[\because y=(x)^{\frac{1}{3}}ight] \

& =\frac{1}{3(27)^{\frac{2}{3}}}(-1) \

& =\frac{-1}{27} \

& =-0.0 \overline{370}

\end{aligned}

$$

Hence,

$$

\begin{aligned}

(26)^{\frac{1}{3}} & =3+(-0.0370) \

& =2.9629

\end{aligned}

$$

Thus, the approximate value of $(26)^{\frac{1}{3}}=2.9629$.

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

Consider $y=(x)^{\frac{1}{4}}$

Let $x=256$ and $\Delta x=-1$

Then,

$$

\begin{aligned}

\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 \

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

\end{aligned}

$$

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

$$

\begin{aligned}

d y & =\left(\frac{d y}{d x}ight) \Delta x \

& =\frac{1}{4(x)^{\frac{3}{4}}}(\Delta x) \quad\left[\because y=(x)^{\frac{1}{4}}ight] \

& =\frac{1}{4(256)^{\frac{3}{4}}}(-1) \

& =\frac{-1}{4 	imes 4^{3}} \

& =\frac{-1}{32} \

& =-0.0039

\end{aligned}

$$

Hence,

$$

\begin{aligned}

(255)^{\frac{1}{4}} & =4+(-0.0039) \

& =3.9961

\end{aligned}

$$

Thus, the approximate value of $(255)^{\frac{1}{4}}=3.9961$.

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

Consider $y=(x)^{\frac{1}{4}}$.

Let $x=81$ and $\Delta x=1$

Then,

$$

\begin{aligned}

\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 \

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

\end{aligned}

$$

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

$$

\begin{aligned}

d y & =\left(\frac{d y}{d x}ight) \Delta x \

& =\frac{1}{4(x)^{\frac{3}{4}}}(\Delta x) \quad\left[\because y=(x)^{\frac{1}{4}}ight] \

& =\frac{1}{4(81)^{\frac{3}{4}}}(1) \

& =\frac{1}{4 	imes 3^{3}} \

& =\frac{1}{108} \

& =0.009

\end{aligned}

$$

Hence,

$$

\begin{aligned}

(82)^{\frac{1}{4}} & =3+0.009 \

& =3.009

\end{aligned}

$$

Thus, the approximate value of $(82)^{\frac{1}{4}}=3.009$.

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

Consider $y=(x)^{\frac{1}{2}}$

Let $x=400$ and $\Delta x=1$

Then,

$$

\begin{aligned}

\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 \

\Delta y+20 & =(401)^{\frac{1}{2}}

\end{aligned}

$$

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

$$

\begin{aligned}

d y & =\left(\frac{d y}{d x}ight) \Delta x \

& =\frac{1}{2 \sqrt{x}}(\Delta x) \quad\left[\because y=(x)^{\frac{1}{2}}ight] \

& =\frac{1}{2(20)}(1) \

& =\frac{1}{40} \

& =0.025

\end{aligned}

$$

Hence,

$$

\begin{aligned}

(401)^{\frac{1}{2}} & =20+0.025 \

& =20.025

\end{aligned}

$$

Thus, the approximate value of $(401)^{\frac{1}{2}}=20.025$.

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

Consider $y=(x)^{\frac{1}{2}}$

Let $x=0.0036$ and $\Delta x=0.0001$

Then,

$$

\begin{aligned}

\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 \

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

\end{aligned}

$$

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

$$

\begin{aligned}

d y & =\left(\frac{d y}{d x}ight) \Delta x \

& =\frac{1}{2 \sqrt{x}}(\Delta x) \quad\left[\because y=(x)^{\frac{1}{2}}ight] \

& =\frac{1}{2(0.06)}(0.0001) \

& =\frac{0.0001}{0.12} \

& =0.0008325

\end{aligned}

$$

Hence,

$$

\begin{aligned}

(0.0037)^{\frac{1}{2}} & =0.06+0.00083 \

& =0.06083

\end{aligned}

$$

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

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

Consider $y=(x)^{\frac{1}{3}}$

Let $x=27$ and $\Delta x=-0.43$

Then,

$$

\begin{aligned}

\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 \

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

\end{aligned}

$$

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

$$

\begin{aligned}

d y & =\left(\frac{d y}{d x}ight) \Delta x \

& =\frac{1}{3(x)^{\frac{2}{3}}}(\Delta x) \quad\left[\because y=(x)^{\frac{1}{3}}ight] \

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

& =\frac{-0.43}{27} \

& =-0.015

\end{aligned}

$$

Hence,

$$

\begin{aligned}

(26.57)^{\frac{1}{3}} & =3+(-0.015) \

& =2.984

\end{aligned}

$$

Thus, the approximate value of $(26.57)^{\frac{1}{3}}=2.984$.

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

Consider $y=(x)^{\frac{1}{4}}$

Let $x=81$ and $\Delta x=0.5$

Then,

$$

\begin{aligned}

\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 \

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

\end{aligned}

$$

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

$$

\begin{aligned}

d y & =\left(\frac{d y}{d x}ight) \Delta x \

& =\frac{1}{4(x)^{\frac{3}{4}}}(\Delta x) \quad\left[\because y=(x)^{\frac{1}{4}}ight] \

& =\frac{1}{4(3)^{3}}(0.5) \

& =\frac{0.5}{108} \

& =0.0046

\end{aligned}

$$

Hence,

$$

\begin{aligned}

(81.5)^{\frac{1}{4}} & =3+0.0046 \

& =3.0046

\end{aligned}

$$

Thus, the approximate value of $(81.5)^{\frac{1}{4}}=3.0046$.

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

Consider $y=(x)^{\frac{3}{2}}$

Let $x=4$ and $\Delta x=-0.032$

Then,

$$

\begin{aligned}

\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 \

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

\end{aligned}

$$

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

$$

\begin{aligned}

d y & =\left(\frac{d y}{d x}ight) \Delta x \

& =\frac{3}{2}(x)^{\frac{1}{2}}(\Delta x) \quad\left[\because y=(x)^{\frac{3}{2}}ight] \

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

& -0.096

\end{aligned}

$$

Hence,

$$

\begin{aligned}

(3.968)^{\frac{3}{2}} & =8+(-0.096) \

& =7.904

\end{aligned}

$$

Thus, the approximate value of $(3.968)^{\frac{3}{2}}=7.904$.

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

Consider $y=(x)^{\frac{1}{4}}$

Let $x=32$ and $\Delta x=0.15$

Then,

$$

\begin{aligned}

\Delta y & =(x+\Delta x)^{\frac{1}{4}}-(x)^{\frac{1}{4}} \

& =(32.15)^{\frac{1}{5}}-(32)^{\frac{1}{5}} \

& =(32.15)^{\frac{1}{5}}-2 \

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

\end{aligned}

$$

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

$$

\begin{aligned}

d y & =\left(\frac{d y}{d x}ight) \Delta x \

& =\frac{1}{5(x)^{\frac{4}{5}}}(\Delta x) \quad\left[\because y=(x)^{\frac{1}{5}}ight] \

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

& =\frac{0.15}{80} \

& =0.00187

\end{aligned}

$$

Hence,

$$

\begin{aligned}

(32.15)^{\frac{1}{5}} & =2+0.00187 \

& =2.00187

\end{aligned}

$$

Thus, the approximate value of $(32.15)^{\frac{1}{5}}=2.00187$.

Question 2

Find the approximate value of $f(2.01)$, where $f(x)=4 x^{2}+5 x+2$.

Accepted Answer

Let $x=2$ and $\Delta x=0.01$

Then,

$$

\begin{aligned}

f(2.01) & =f(x+\Delta x) \

& =4(x+\Delta x)^{2}+5(x+\Delta x)+2 \

\Delta y & =f(x+\Delta x)-f(x) \

f(x+\Delta x) & =f(x)+\Delta y \

& \approx f(x)+f^{\prime}(x) \cdot \Delta x \quad(\because d x=\Delta x) \

f(2.01) & \approx\left(4 x^{2}+5 x+2ight)+(8 x-5) \Delta x \

& =\left[4(2)^{2}+5(2)+2ight]+[8(2)+5](0.01) \quad[\because x=2, \Delta x=0.01] \

& =(16+10+2)+(16+5)(0.01) \

& =28+21(0.01) \

& =28+0.21 \

& =28.21

\end{aligned}

$$

Hence, the approximate value of $f(2.01)=28.21$.

Question 3

Find the approximate value of $f(5.001)$, where $f(x)=x^{3}-7 x^{2}+15$.

Accepted Answer

Let $x=5$ and $\Delta x=0.001$

Then,

$$

\begin{aligned}

f(5.001) & =f(x+\Delta x) \

& =(x+\Delta x)^{3}-7(x+\Delta x)^{2}+15 \

\Delta y & =f(x+\Delta x)-f(x) \

f(x+\Delta x) & =f(x)+\Delta y \

& \approx f(x)+f^{\prime}(x) \cdot \Delta x \quad(\because d x=\Delta x) \

f(5.001) & \approx\left(x^{3}-7 x^{2}+15ight)+\left(3 x^{2}-14 xight) \Delta x \

& =\left[(5)^{3}-7(5)^{2}+15ight]+\left[3(5)^{2}-14(5)ight](0.001) \quad \quad[\because x=5, \Delta x=0.001] \

& =(125-175+15)+(75-70)(0.001) \

& =(-35)+(5)(0.001) \

& =-35+0.005 \

& =-34.995

\end{aligned}

$$

Hence, the approximate value of $f(5.001)=-34.995$.

Question 4

Find the approximate change in the volume $V$ of a cube of side $x$ meters caused by increasing the side by $1 \%$.

Accepted Answer

The volume of a cube $V$ of side $x$ is given by $V=x^{3}$

Therefore,

$$

\begin{aligned}

d V & =\left(\frac{d V}{d x}ight) \Delta x \

& =\left(3 x^{2}ight) \Delta x \

& =\left(3 x^{2}ight)(0.01 x) \quad[\because 1 \% 	ext { of } x 	ext { is } 0.01 x] \

& =0.03 x^{3}

\end{aligned}

$$

Hence, the approximate change in the volume of the cube is $0.03 x^{3} m^{3}$.

Question 5

Find the approximate change in the surface area of a cube of side $x$ meters caused by decreasing the side by $1 \%$.

Accepted Answer

The surface area of a cube $(S)$ of side $x$ is given by $S=6 x^{2}$.

Therefore,

$$

\begin{array}{rlr}

d S & =\left(\frac{d S}{d x}ight) \Delta x & \

& =(12 x) \Delta x & \

& =(12 x)(0.01 x) & {[\because 1 \% 	ext { of } x 	ext { is } 0.01 x]} \

& =0.12 x^{2} &

\end{array}

$$

Hence, the approximate change in the surface area of the cube is $0.12 x^{2} m^{2}$.

Question 6

If the radius of a sphere is measured as $7 m$ with an error of $0.02 m$, then find the approximate error in calculating its volume.

Accepted Answer

Let $r$ be the radius of the sphere and $\Delta r$ be the error in measuring the radius.

Then, $r=7 \mathrm{~m}$ and $\Delta r=0.02 \mathrm{~m}$

Now, the volume $V$ of the sphere is given by,

$$

V=\frac{4}{3} \pi r^{3}

$$

Therefore,

$$

\frac{d V}{d r}=4 \pi r^{2}

$$

Hence,

$$

\begin{aligned}

d V & =\left(\frac{d V}{d r}ight) \Delta r \

& =\left(4 \pi r^{2}ight)(0.02) \

& =4 \pi(7)^{2}(0.02) \

& =3.92 \pi

\end{aligned}

$$

Thus, the approximate error in calculating its volume is $3.92 \pi m^{3}$.

Question 7

If the radius of a sphere is measured as $9 m$ with an error of $0.03 m$, then find the approximate error in calculating its surface area.

Accepted Answer

Let $r$ be the radius of the sphere and $\Delta r$ be the error in measuring the radius.

Then, $r=9 m$ and $\Delta r=0.03 m$

Now, the surface area $S$ of the sphere is given by,

$$

S=4 \pi r^{2}

$$

Therefore,

$$

\frac{d S}{d r}=8 \pi r

$$

Hence,

$$

\begin{aligned}

d S & =\left(\frac{d S}{d r}ight) \Delta r \

& =(8 \pi r) \Delta r \

& =8 \pi(9)(0.03) \

& =2.16 \pi

\end{aligned}

$$

Thus, the approximate error in calculating its surface area is $2.16 \pi m^{2}$.

Question 8

If $f(x)=3 x^{2}+15 x+5$, then the approximate value of $f(3.02)$ is

(A) 47.66

(B) 57.66

(C) 67.66

(D) 77.66

Accepted Answer

Let $x=3$ and $\Delta x=0.02$

Then,

$$

\begin{aligned}

f(3.02) & =f(x+\Delta x)=3(x+\Delta x)^{2}+15(x+\Delta x)+5 \

\Delta y & =f(x+\Delta x)-f(x) \

f(x+\Delta x) & =f(x)+\Delta y \

& \approx f(x)+f^{\prime}(x) \cdot \Delta x \quad(\because d x=\Delta x) \

f(3.02) & \approx\left(3 x^{2}+15 x+5ight)+(6 x+15) \Delta x \

& =\left[3(3)^{2}+15(3)+5ight]+[6(3)+15](0.02) \quad[\because x=3, \Delta x=0.02] \

& =(27+45+5)+(18+15)(0.02) \

& =77+(33)(0.02) \

& =77+0.66 \

& =77.66

\end{aligned}

$$

Hence, the approximate value of $f(3.02)=77.66$.

Thus, the correct option is $\mathbf{D}$.

Question 9

The approximate change in the volume of a cube of side $x$ metres caused by increasing the side by is.

(A) $0.06 x^{3} m^{3}$

(B) $0.6 x^{3} m^{3}$

(C) $0.09 x^{3} m^{3}$

(D) $0.9 x^{3} m^{3}$

Accepted Answer

The volume of a cube $(V)$ of side $x$ is given by $V=x^{3}$.

Therefore,

$$

\begin{aligned}

d V & =\left(\frac{d V}{d x}ight) \Delta x \

& =\left(3 x^{2}ight)(0.03 x) \quad[\because 3 \% 	ext { of } x 	ext { is } 0.03 x] \

& =0.09 x^{3}

\end{aligned}

$$

Hence, the approximate change in the volume of the cube $0.09 x^{3} m^{3}$.

Thus, the correct option is $\mathbf{C}$.

\section*{EXERCISE 6.5}