1. Given that $ \displaystyle y = 3x^3 - 7x^2 + 8$, find the value of $ \displaystyle \frac{dy}{dx}$ at the point $ \displaystyle (2, 4)$. Hence, find the approximate increase in $ \displaystyle x$ which will cause $ \displaystyle y$ to increase from $ \displaystyle 4$ to $ \displaystyle 4.04$.
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$ \displaystyle \begin{array}{l}\ \ \ y=3{{x}^{3}}-7{{x}^{2}}+8\\\\\ \ \ \text{Let}\ ({{x}_{0}},{{y}_{0}})=(2,4)\ \text{and}\ {{y}_{1}}=4.04\\\\\therefore \ \delta y={{y}_{1}}-{{y}_{0}}\\\\\ \ \ \ \ \ \ \ =4.04-4\\\\\ \ \ \ \ \ \ \ =0.04\end{array}$
$ \displaystyle \ \ \ \frac{{dy}}{{dx}}=9{{x}^{2}}-14x$
$ \displaystyle \ \ \ {{\left. {\frac{{dy}}{{dx}}} \right|}_{{({{x}_{0}},{{y}_{0}})}}}={{\left. {\frac{{dy}}{{dx}}} \right|}_{{(2,4)}}}=9{{(2)}^{2}}-14(2)=8$
$ \displaystyle \ \ \ \text{By linear approximation,}$
$ \displaystyle \ \ \ \delta y\simeq {{\left. {\frac{{dy}}{{dx}}} \right|}_{{({{x}_{0}},{{y}_{0}})}}}\cdot \delta x$
$ \displaystyle \begin{array}{l}\therefore \ 0.04\simeq 8\cdot \delta x\\\\\therefore \delta x\simeq 0.005\\\\\therefore \ x\ \text{is increased by 0}\text{.005}\text{.}\end{array}$
$ \displaystyle \ \ \ \frac{{dy}}{{dx}}=9{{x}^{2}}-14x$
$ \displaystyle \ \ \ {{\left. {\frac{{dy}}{{dx}}} \right|}_{{({{x}_{0}},{{y}_{0}})}}}={{\left. {\frac{{dy}}{{dx}}} \right|}_{{(2,4)}}}=9{{(2)}^{2}}-14(2)=8$
$ \displaystyle \ \ \ \text{By linear approximation,}$
$ \displaystyle \ \ \ \delta y\simeq {{\left. {\frac{{dy}}{{dx}}} \right|}_{{({{x}_{0}},{{y}_{0}})}}}\cdot \delta x$
$ \displaystyle \begin{array}{l}\therefore \ 0.04\simeq 8\cdot \delta x\\\\\therefore \delta x\simeq 0.005\\\\\therefore \ x\ \text{is increased by 0}\text{.005}\text{.}\end{array}$
2. Given that $ \displaystyle y = 2x^2 + 3x$, find the approximate percentage change in $ \displaystyle y$ when $ \displaystyle x$ decreases from $ \displaystyle 2$ to $ \displaystyle 1.97$.
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$ \displaystyle \begin{array}{l}\ \ \ \ y=2{{x}^{2}}+3x\\\\\ \ \ \ \text{Let}\ {{x}_{0}}=2\ \text{and}\ {{x}_{1}}=1.97\\\\\therefore \ \ \delta x=1.97-2=-0.03\\\\\ \ \ \ \text{When}\ {{x}_{0}}=2,\ {{y}_{0}}=14\end{array}$
$ \displaystyle \ \ \ \ \frac{{dy}}{{dx}}=4x+3$
$ \displaystyle \ \ \ \ {{\left. {\frac{{dy}}{{dx}}} \right|}_{{{{x}_{0}}}}}={{\left. {\frac{{dy}}{{dx}}} \right|}_{2}}=4(2)+3=11$
$ \displaystyle \ \ \ \ \delta y\simeq {{\left. {\frac{{dy}}{{dx}}} \right|}_{{{{x}_{0}}}}}\cdot \delta x$
$ \displaystyle \begin{array}{l}\ \ \ \ \ \ \ \ \ =11(-0.03)\\\\\ \ \ \ \ \ \ \ \ =-0.33\\\\\ \ \ \ \text{Approximate percentage change in }y\end{array}$
$ \displaystyle \ \ \ \ =\frac{{\delta y}}{{{{y}_{0}}}}\times 100 $ %
$ \displaystyle \ \ \ \ =\frac{{-0.33}}{{14}}\times 100$ %
$ \displaystyle \ \ \ \ =-2.36$ %
$ \displaystyle \therefore \ \ \text{ }y\ \text{is decreased by}\ 2.36$ %.
$ \displaystyle \ \ \ \ \frac{{dy}}{{dx}}=4x+3$
$ \displaystyle \ \ \ \ {{\left. {\frac{{dy}}{{dx}}} \right|}_{{{{x}_{0}}}}}={{\left. {\frac{{dy}}{{dx}}} \right|}_{2}}=4(2)+3=11$
$ \displaystyle \ \ \ \ \delta y\simeq {{\left. {\frac{{dy}}{{dx}}} \right|}_{{{{x}_{0}}}}}\cdot \delta x$
$ \displaystyle \begin{array}{l}\ \ \ \ \ \ \ \ \ =11(-0.03)\\\\\ \ \ \ \ \ \ \ \ =-0.33\\\\\ \ \ \ \text{Approximate percentage change in }y\end{array}$
$ \displaystyle \ \ \ \ =\frac{{\delta y}}{{{{y}_{0}}}}\times 100 $ %
$ \displaystyle \ \ \ \ =\frac{{-0.33}}{{14}}\times 100$ %
$ \displaystyle \ \ \ \ =-2.36$ %
$ \displaystyle \therefore \ \ \text{ }y\ \text{is decreased by}\ 2.36$ %.
3. Use approximation to approximate the following values $ \displaystyle \sqrt{80}$.
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$ \displaystyle \begin{array}{l}\ \ \ \ \text{Let}\ y=\sqrt{x}\ \text{and}\ {{x}_{0}}=81\\\\\therefore \ \ {{y}_{0}}=9\ \ \\\\\ \ \ \ \text{Let}\ {{x}_{1}}=\ 80.\\\\\therefore \ \ {{y}_{1}}=\sqrt{{80}}\\\\\therefore \ \ \delta x=80-81=-1\end{array}$
$ \displaystyle \ \ \ \ \frac{{dy}}{{dx}}=\frac{1}{{2\sqrt{x}}}$
$ \displaystyle \ \ \ \ {{\left. {\frac{{dy}}{{dx}}} \right|}_{{{{x}_{0}}}}}={{\left. {\frac{{dy}}{{dx}}} \right|}_{{81}}}=\frac{1}{{2\sqrt{{81}}}}=\frac{1}{{18}}$
$ \displaystyle \ \ \ \ \delta y\simeq {{\left. {\frac{{dy}}{{dx}}} \right|}_{{{{x}_{0}}}}}\cdot \delta x$
$ \displaystyle \ \ \ \ \ \ \ \ \ =\frac{1}{{18}}(-1)$
$ \displaystyle \begin{array}{l}\ \ \ \ \ \ \ \ \ =-0.056\\\\\therefore \ \ {{y}_{1}}={{y}_{0}}+\delta y=9-0.056=8.944\\\\\therefore \ \ \sqrt{{80}}\simeq 8.944\end{array}$
$ \displaystyle \ \ \ \ \frac{{dy}}{{dx}}=\frac{1}{{2\sqrt{x}}}$
$ \displaystyle \ \ \ \ {{\left. {\frac{{dy}}{{dx}}} \right|}_{{{{x}_{0}}}}}={{\left. {\frac{{dy}}{{dx}}} \right|}_{{81}}}=\frac{1}{{2\sqrt{{81}}}}=\frac{1}{{18}}$
$ \displaystyle \ \ \ \ \delta y\simeq {{\left. {\frac{{dy}}{{dx}}} \right|}_{{{{x}_{0}}}}}\cdot \delta x$
$ \displaystyle \ \ \ \ \ \ \ \ \ =\frac{1}{{18}}(-1)$
$ \displaystyle \begin{array}{l}\ \ \ \ \ \ \ \ \ =-0.056\\\\\therefore \ \ {{y}_{1}}={{y}_{0}}+\delta y=9-0.056=8.944\\\\\therefore \ \ \sqrt{{80}}\simeq 8.944\end{array}$
4. The side of a square is $ \displaystyle 5\ \text{cm}$ . How much will the area of the square increase when the side expands by $ \displaystyle 0.01\ \text{cm}$?
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$ \displaystyle \begin{array}{l}\ \ \ \ \text{Let the length of the side of a square be }x\\\\\ \ \ \ \operatorname{and}\ \text{the area of the square be }A.\\\\\therefore \ \ A={{x}^{2}}\\\\\ \ \ \ \text{Let }{{x}_{0}}=5\ \text{then}\ {{A}_{0}}=\ 25\\\\\ \ \ \ \text{When the sides expand 0}\text{.01 cm, }\\\\\ \ \ \ \delta x=0.01\ \text{cm}\text{.}\end{array}$
$ \displaystyle \ \ \ \ \frac{{dA}}{{dx}}=2x$
$ \displaystyle \ \ \ \ {{\left. {\frac{{dA}}{{dx}}} \right|}_{{{{x}_{0}}}}}={{\left. {\frac{{dA}}{{dx}}} \right|}_{5}}=2(5)=10$
$ \displaystyle \ \ \ \ \delta A\simeq {{\left. {\frac{{dA}}{{dx}}} \right|}_{{{{x}_{0}}}}}\cdot \delta x$
$ \displaystyle \begin{array}{l}\ \ \ \ \ \ \ \ \ =10(0.01)\\\\\ \ \ \ \ \ \ \ \ =0.1\ \text{c}{{\text{m}}^{2}}\\\\\therefore \ \ \text{The area of the square is increased by }0.1\ \text{c}{{\text{m}}^{2}}.\end{array}$
$ \displaystyle \ \ \ \ \frac{{dA}}{{dx}}=2x$
$ \displaystyle \ \ \ \ {{\left. {\frac{{dA}}{{dx}}} \right|}_{{{{x}_{0}}}}}={{\left. {\frac{{dA}}{{dx}}} \right|}_{5}}=2(5)=10$
$ \displaystyle \ \ \ \ \delta A\simeq {{\left. {\frac{{dA}}{{dx}}} \right|}_{{{{x}_{0}}}}}\cdot \delta x$
$ \displaystyle \begin{array}{l}\ \ \ \ \ \ \ \ \ =10(0.01)\\\\\ \ \ \ \ \ \ \ \ =0.1\ \text{c}{{\text{m}}^{2}}\\\\\therefore \ \ \text{The area of the square is increased by }0.1\ \text{c}{{\text{m}}^{2}}.\end{array}$
5. If the sides of a square are increased by $ \displaystyle 20$ %, by what percentage does the area of the square increase?
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$ \displaystyle \ \ \ \ \text{Solution (1) - Using Approximation}$
$ \displaystyle \begin{array}{l}\ \ \ \ \text{Let the length each side of a square be}\ x\\\ \ \ \ \text{and}\ \text{the area be }A.\text{ }\end{array}$
$ \displaystyle \therefore \ \ A={{x}^{2}}.$
$ \displaystyle \therefore \ \ \frac{{dA}}{{dx}}=2x$
$ \displaystyle \ \ \ \ \text{When the lengths of the sides are incresed by 20}\ \text{%,}$
$ \displaystyle \ \ \ \ \frac{{\delta x}}{x}=\text{20 } \text{% }=0.2\Rightarrow \delta x=0.2x$
$ \displaystyle \ \ \ \ \delta A\simeq \frac{{dA}}{{dx}}\times \delta x$
$ \displaystyle \begin{array}{l}\ \ \ \ \delta A\simeq 2x\times 0.2x\\\\\therefore \ \ \delta A\simeq 0.4{{x}^{2}}\\\\\ \ \ \ \delta A\simeq 0.4A\end{array}$
$ \displaystyle \therefore \ \ \frac{{\delta A}}{A}\simeq \text{0}\text{.4}=\text{40}\ \text{%}$
$ \displaystyle \therefore \ \ \text{The area of the rectangle is increased by 40}\ \text{%.}$
$ \displaystyle \ \ \ \ \text{Solution (2) - Without Calculus}$
$ \displaystyle \begin{array}{l}\ \ \ \ \text{Let the length each side of a square be}\ x\\\\\ \ \ \ \text{and}\ \text{the area be }A.\\\\\therefore \ \ {{A}_{0}}={{x}^{2}}\\\\\ \ \ \ \text{When the lengths of the sides are incresed by 20}\ \text{%,}\\\\\ \ \ \ {{x}_{1}}=x+0.2x=1.2x\\\\\therefore \ \ {{A}_{1}}={{(1.2x)}^{2}}=1.44{{x}^{2}}\\\\\ \ \ \ \delta A\simeq {{A}_{1}}-{{A}_{0}}=0.44{{x}^{2}}\\\\\therefore \ \ \ \delta A=0.44A\end{array}$
$ \displaystyle \therefore \ \ \frac{{\delta A}}{A}=\text{0}\text{.44}=\text{44}\ \text{% }$
$ \displaystyle \therefore \ \ \text{The area of the rectangle is increased by 44}\ \text{%.}$
$ \displaystyle \begin{array}{l}\ \ \ \ \text{Let the length each side of a square be}\ x\\\ \ \ \ \text{and}\ \text{the area be }A.\text{ }\end{array}$
$ \displaystyle \therefore \ \ A={{x}^{2}}.$
$ \displaystyle \therefore \ \ \frac{{dA}}{{dx}}=2x$
$ \displaystyle \ \ \ \ \text{When the lengths of the sides are incresed by 20}\ \text{%,}$
$ \displaystyle \ \ \ \ \frac{{\delta x}}{x}=\text{20 } \text{% }=0.2\Rightarrow \delta x=0.2x$
$ \displaystyle \ \ \ \ \delta A\simeq \frac{{dA}}{{dx}}\times \delta x$
$ \displaystyle \begin{array}{l}\ \ \ \ \delta A\simeq 2x\times 0.2x\\\\\therefore \ \ \delta A\simeq 0.4{{x}^{2}}\\\\\ \ \ \ \delta A\simeq 0.4A\end{array}$
$ \displaystyle \therefore \ \ \frac{{\delta A}}{A}\simeq \text{0}\text{.4}=\text{40}\ \text{%}$
$ \displaystyle \therefore \ \ \text{The area of the rectangle is increased by 40}\ \text{%.}$
$ \displaystyle \ \ \ \ \text{Solution (2) - Without Calculus}$
$ \displaystyle \begin{array}{l}\ \ \ \ \text{Let the length each side of a square be}\ x\\\\\ \ \ \ \text{and}\ \text{the area be }A.\\\\\therefore \ \ {{A}_{0}}={{x}^{2}}\\\\\ \ \ \ \text{When the lengths of the sides are incresed by 20}\ \text{%,}\\\\\ \ \ \ {{x}_{1}}=x+0.2x=1.2x\\\\\therefore \ \ {{A}_{1}}={{(1.2x)}^{2}}=1.44{{x}^{2}}\\\\\ \ \ \ \delta A\simeq {{A}_{1}}-{{A}_{0}}=0.44{{x}^{2}}\\\\\therefore \ \ \ \delta A=0.44A\end{array}$
$ \displaystyle \therefore \ \ \frac{{\delta A}}{A}=\text{0}\text{.44}=\text{44}\ \text{% }$
$ \displaystyle \therefore \ \ \text{The area of the rectangle is increased by 44}\ \text{%.}$
6. One side of a rectangle is three times the other. If the perimeter increases by $ \displaystyle 2$ % what is the percentage increase in area?
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$ \displaystyle \begin{array}{l}\ \ \ \ \text{As one side of the rectangle is three times the other},\\\ \ \ \ \text{let the length and the breadth be 3}x\ \text{and}\ x.\end{array}$
$ \displaystyle \begin{array}{l}\ \ \ \ \text{Let}\ \text{the perimetre and area of the rectangle be }\\\ \ \ \ p\ \text{and }A\ \text{respectively}\text{.}\end{array}$
$ \displaystyle \therefore \ \ p=2(3x+x)=8x\ \text{and}\ A=3{{x}^{2}}.$
$ \displaystyle \therefore \ \ x=\frac{p}{8}\Rightarrow A=3{{\left( {\frac{p}{8}} \right)}^{2}}=\frac{{3{{p}^{2}}}}{{64}}$
$ \displaystyle \therefore \ \ \frac{{dA}}{{dp}}=\frac{{3p}}{{32}}$
$ \displaystyle \ \ \ \ \text{When the perimeter is incresed by 2}$ %,
$ \displaystyle \ \ \ \ \frac{{\delta p}}{p}=\text{2 %}=0.02\Rightarrow \delta p=0.02p$
$ \displaystyle \ \ \ \ \delta A\simeq \frac{{dA}}{{dp}}\times \delta p$
$ \displaystyle \ \ \ \ \delta A\simeq \frac{{3p}}{{32}}\times 0.02p$
$ \displaystyle \therefore \ \ \delta A\simeq \frac{{3{{p}^{2}}}}{{32\times 2}}\times 0.02\times 2$
$ \displaystyle \therefore \ \ \delta A\simeq \frac{{3{{p}^{2}}}}{{64}}\times 0.04$
$ \displaystyle \ \ \ \ \delta A\simeq 0.04A$
$ \displaystyle \therefore \ \ \frac{{\delta A}}{A}\simeq \text{0}\text{.04}=\text{4}$ %
$ \displaystyle \therefore \ \ \text{The area of the rectangle is increased by 4}$ %.
$ \displaystyle \begin{array}{l}\ \ \ \ \text{Let}\ \text{the perimetre and area of the rectangle be }\\\ \ \ \ p\ \text{and }A\ \text{respectively}\text{.}\end{array}$
$ \displaystyle \therefore \ \ p=2(3x+x)=8x\ \text{and}\ A=3{{x}^{2}}.$
$ \displaystyle \therefore \ \ x=\frac{p}{8}\Rightarrow A=3{{\left( {\frac{p}{8}} \right)}^{2}}=\frac{{3{{p}^{2}}}}{{64}}$
$ \displaystyle \therefore \ \ \frac{{dA}}{{dp}}=\frac{{3p}}{{32}}$
$ \displaystyle \ \ \ \ \text{When the perimeter is incresed by 2}$ %,
$ \displaystyle \ \ \ \ \frac{{\delta p}}{p}=\text{2 %}=0.02\Rightarrow \delta p=0.02p$
$ \displaystyle \ \ \ \ \delta A\simeq \frac{{dA}}{{dp}}\times \delta p$
$ \displaystyle \ \ \ \ \delta A\simeq \frac{{3p}}{{32}}\times 0.02p$
$ \displaystyle \therefore \ \ \delta A\simeq \frac{{3{{p}^{2}}}}{{32\times 2}}\times 0.02\times 2$
$ \displaystyle \therefore \ \ \delta A\simeq \frac{{3{{p}^{2}}}}{{64}}\times 0.04$
$ \displaystyle \ \ \ \ \delta A\simeq 0.04A$
$ \displaystyle \therefore \ \ \frac{{\delta A}}{A}\simeq \text{0}\text{.04}=\text{4}$ %
$ \displaystyle \therefore \ \ \text{The area of the rectangle is increased by 4}$ %.
7. The kinetic energy $ \displaystyle K$ of a body of mass $ \displaystyle m$ moving with speed $ \displaystyle v$ is given by $ \displaystyle K =\frac{1}{2}mv^2$ . If a body’s speed is increased by $ \displaystyle 1.5$ % what is the approximate percentage change in the kinetic energy?
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$ \displaystyle \ \ \ \ \text{Kinetic Energy}=\ K,\text{ mass}=m,\text{ speed}=v$
$ \displaystyle \ \ \ \ K=\frac{1}{2}m{{v}^{2}}$
$ \displaystyle \ \ \ \ \text{If a body's speed is incresed by 1.5}$ %,
$ \displaystyle \ \ \ \ \frac{{\delta v}}{v}=\text{1}\text{.5}\ \text{%}=0.015\Rightarrow \delta v=0.015v$
$ \displaystyle \ \ \ \ \frac{{dK}}{{dv}}=mv$
$ \displaystyle \ \ \ \ \delta K\simeq \frac{{dK}}{{dv}}\times \delta v$
$ \displaystyle \begin{array}{l}\ \ \ \ \delta K\simeq mv\times \delta v\\\\\therefore \ \ \delta K\simeq mv\times 0.015v\end{array}$
$ \displaystyle \therefore \ \ \delta K\simeq \frac{1}{2}m{{v}^{2}}\times 0.03$
$ \displaystyle \therefore \ \ \delta K\simeq K\times 0.03$
$ \displaystyle \therefore \ \ \frac{{\delta K}}{K}\simeq 0.03=3$ %
$ \displaystyle \therefore \ \ \text{The kinetic energy is increased by}$ 3%.
$ \displaystyle \ \ \ \ K=\frac{1}{2}m{{v}^{2}}$
$ \displaystyle \ \ \ \ \text{If a body's speed is incresed by 1.5}$ %,
$ \displaystyle \ \ \ \ \frac{{\delta v}}{v}=\text{1}\text{.5}\ \text{%}=0.015\Rightarrow \delta v=0.015v$
$ \displaystyle \ \ \ \ \frac{{dK}}{{dv}}=mv$
$ \displaystyle \ \ \ \ \delta K\simeq \frac{{dK}}{{dv}}\times \delta v$
$ \displaystyle \begin{array}{l}\ \ \ \ \delta K\simeq mv\times \delta v\\\\\therefore \ \ \delta K\simeq mv\times 0.015v\end{array}$
$ \displaystyle \therefore \ \ \delta K\simeq \frac{1}{2}m{{v}^{2}}\times 0.03$
$ \displaystyle \therefore \ \ \delta K\simeq K\times 0.03$
$ \displaystyle \therefore \ \ \frac{{\delta K}}{K}\simeq 0.03=3$ %
$ \displaystyle \therefore \ \ \text{The kinetic energy is increased by}$ 3%.
8. The two equal sides of an isosceles triangle with fixed base $ \displaystyle b$ are decreasing at the rate of $ \displaystyle 3\ \text{cm}$ per second. How fast is the area decreasing when the two equal sides are equal to the base ?
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$ \displaystyle \therefore \ \ \text{By Pythagoras' } \text{Theorem,}$
$ \displaystyle \ \ \ \ h=\sqrt{{{{x}^{2}}-{{{\left( {\frac{b}{2}} \right)}}^{2}}}}=\sqrt{{{{x}^{2}}-\frac{{{{b}^{2}}}}{4}}}$
$ \displaystyle \ \ \ \ \text{By the problem, }\frac{{dx}}{{dt}}=3\ \text{cm/s.}$
$ \displaystyle \ \ \ \ \text{Let the area of the triangle be }A.$
$ \displaystyle \therefore \ \ A=\frac{1}{2}bh=\frac{1}{2}b\sqrt{{{{x}^{2}}-\frac{{{{b}^{2}}}}{4}}}$
$ \displaystyle \therefore \ \ \frac{{dA}}{{dx}}=\frac{1}{2}b\times \frac{{2x}}{{2\sqrt{{{{x}^{2}}-\frac{{{{b}^{2}}}}{4}}}}}=\frac{{bx}}{{2\sqrt{{{{x}^{2}}-\frac{{{{b}^{2}}}}{4}}}}}$
$ \displaystyle \ \ \ \ \text{When }x=b,\frac{{dA}}{{dx}}=\frac{{{{b}^{2}}}}{{2\sqrt{{{{b}^{2}}-\frac{{{{b}^{2}}}}{4}}}}}=\frac{{{{b}^{2}}}}{{2b\sqrt{3}}}=\frac{b}{{\sqrt{3}}}$
$ \displaystyle \ \ \ \ \text{By Chain Rule,}\frac{{dA}}{{dt}}=\frac{b}{{\sqrt{3}}}\times 3=\sqrt{3}b\ \text{c}{{\text{m}}^{2}}\text{/s.}$
$ \displaystyle \therefore \ \ \text{The area of the triangle is decreasing at a rate of }\sqrt{3}b\ \text{c}{{\text{m}}^{2}} \text{/s.}$
9. The volume of the cube is increasing at a rate of $ \displaystyle 9\ \text{cm}^3/ \text{s}$. How fast is the surface area increasing when the edge is $ \displaystyle 10\ \text{cm}$ long ?
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$ \displaystyle \begin{array}{l}\ \ \ \ \text{Let the length of each sides be }x,\ \\\ \ \ \ \text{the volume be }V\text{and the surface area be }A.\end{array}$
$ \displaystyle \therefore \ \ V={{x}^{3}}\ \text{and}\ A=6{{x}^{2}}$
$ \displaystyle \therefore \ \ \frac{{dV}}{{dx}}=3{{x}^{2}}\ \text{and}\ \frac{{dA}}{{dx}}=12x.$
$ \displaystyle \ \ \ \ \text{By the problem, }\frac{{dV}}{{dt}}=9\ \text{c}{{\text{m}}^{3}}\text{/s}\text{.}$
$ \displaystyle \ \ \ \ \text{Since }\frac{{dV}}{{dt}}=\frac{{dV}}{{dx}}\times \frac{{dx}}{{dt}},\ \ \ \ \ \left[ {\text{Chain Rule}} \right]$
$ \displaystyle \ \ \ \ 3{{x}^{2}}\times \frac{{dx}}{{dt}}=9\Rightarrow \frac{{dx}}{{dt}}=\frac{3}{{{{x}^{2}}}}$
$ \displaystyle \ \ \ \ \text{Similarly }\frac{{dA}}{{dt}}=\frac{{dA}}{{dx}}\times \frac{{dx}}{{dt}}$
$ \displaystyle \therefore \ \ \frac{{dA}}{{dt}}=12x\left( {\frac{3}{{{{x}^{2}}}}} \right)=\frac{{36}}{x}$
$ \displaystyle \ \ \ \ \text{When }x=10\ \text{cm},\ \frac{{dA}}{{dt}}=\frac{{36}}{{10}}=3.6\ \text{c}{{\text{m}}^{2}}\text{/s.}$
$ \displaystyle \therefore \ \ \text{The surface area of the cube is increasing at a rate of }3.6\ \text{c}{{\text{m}}^{2}}\text{/s.}$
$ \displaystyle \therefore \ \ V={{x}^{3}}\ \text{and}\ A=6{{x}^{2}}$
$ \displaystyle \therefore \ \ \frac{{dV}}{{dx}}=3{{x}^{2}}\ \text{and}\ \frac{{dA}}{{dx}}=12x.$
$ \displaystyle \ \ \ \ \text{By the problem, }\frac{{dV}}{{dt}}=9\ \text{c}{{\text{m}}^{3}}\text{/s}\text{.}$
$ \displaystyle \ \ \ \ \text{Since }\frac{{dV}}{{dt}}=\frac{{dV}}{{dx}}\times \frac{{dx}}{{dt}},\ \ \ \ \ \left[ {\text{Chain Rule}} \right]$
$ \displaystyle \ \ \ \ 3{{x}^{2}}\times \frac{{dx}}{{dt}}=9\Rightarrow \frac{{dx}}{{dt}}=\frac{3}{{{{x}^{2}}}}$
$ \displaystyle \ \ \ \ \text{Similarly }\frac{{dA}}{{dt}}=\frac{{dA}}{{dx}}\times \frac{{dx}}{{dt}}$
$ \displaystyle \therefore \ \ \frac{{dA}}{{dt}}=12x\left( {\frac{3}{{{{x}^{2}}}}} \right)=\frac{{36}}{x}$
$ \displaystyle \ \ \ \ \text{When }x=10\ \text{cm},\ \frac{{dA}}{{dt}}=\frac{{36}}{{10}}=3.6\ \text{c}{{\text{m}}^{2}}\text{/s.}$
$ \displaystyle \therefore \ \ \text{The surface area of the cube is increasing at a rate of }3.6\ \text{c}{{\text{m}}^{2}}\text{/s.}$
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