\displaystyle \textbf{Question 1: }~\text{If }y=\sin x\text{ and }x\text{ changes from }\frac{\pi}{2}\text{ to }\frac{22}{14}, \\ \text{ what is the approximate change in }y?
\displaystyle \text{Answer:}

\displaystyle \text{Let } x = \frac{\pi}{2}.

\displaystyle x + \Delta x = \frac{22}{14}.

\displaystyle \Rightarrow \Delta x = \frac{22}{14} - \frac{\pi}{2} = 0.

\displaystyle \text{Now, } y = \sin x.

\displaystyle \Rightarrow \frac{dy}{dx} = \cos x.

\displaystyle \Rightarrow \left(\frac{dy}{dx}\right)_{x=\frac{\pi}{2}} = \cos\left(\frac{\pi}{2}\right) = 0.

\displaystyle \therefore \Delta y = \frac{dy}{dx}\,\Delta x = 0 \times 0 = 0.

\displaystyle \Rightarrow \Delta y = 0.

\displaystyle \text{Hence, there is no change in the value of } y.

\displaystyle \textbf{Question 2: }~\text{The radius of a sphere shrinks from }10\text{ cm to } 9.8\text{ cm. Find approximately} \\ \text{the decrease in its volume.}
\displaystyle \text{Answer:}

\displaystyle \text{Let } r \text{ be the radius of the sphere.}

\displaystyle r = 10\,\text{cm}.

\displaystyle r + \Delta r = 9.8\,\text{cm}.

\displaystyle \Rightarrow \Delta r = 9.8 - 10 = -0.2\,\text{cm}.

\displaystyle \text{Volume of the sphere, } V = \frac{4}{3}\pi r^3.

\displaystyle \Rightarrow \frac{dV}{dr} = 4\pi r^2.

\displaystyle \Rightarrow \left(\frac{dV}{dr}\right)_{r=10} = 4\pi (10)^2 = 400\pi\,\text{cm}^3/\text{cm}.

\displaystyle \text{Change in the volume of the sphere, } \Delta V = \frac{dV}{dr}\,\Delta r.

\displaystyle \Rightarrow \Delta V = 400\pi \times (-0.2).

\displaystyle \Rightarrow \Delta V = -80\pi\,\text{cm}^3.

\displaystyle \textbf{Question 3: }~\text{A circular metal plate expands under heating so that its radius } \\ \text{increases by }k\%. \text{ Find the approximate increase in the area of the plate,} \\ \text{if the radius of the plate before heating is }10\text{ cm.}
\displaystyle \text{Answer:}

\displaystyle \text{Let at any time, } x \text{ be the radius and } y \text{ be the area of the plate.}

\displaystyle \text{Then, } y = \pi x^2.

\displaystyle \text{Let } \Delta x \text{ be the change in the radius and } \Delta y \text{ be the change in the area of the plate.}

\displaystyle \text{We have } \frac{\Delta x}{x} \times 100 = k.

\displaystyle \text{When } x = 10, \text{ we get}

\displaystyle \Delta x = \frac{10k}{100} = \frac{k}{10}.

\displaystyle \Rightarrow \frac{dy}{dx} = 2\pi x.

\displaystyle \Rightarrow \left(\frac{dy}{dx}\right)_{x=10} = 20\pi\,\text{cm}^2/\text{cm}.

\displaystyle \therefore \Delta y \approx dy = \frac{dy}{dx}\,\Delta x.

\displaystyle \Rightarrow \Delta y = 20\pi \times \frac{k}{10}.

\displaystyle \Rightarrow \Delta y = 2k\pi\,\text{cm}^2.

\displaystyle \text{Hence, the approximate change in the area of the plate is } 2k\pi\,\text{cm}^2.

\displaystyle \textbf{Question 4: }~\text{Find the percentage error in calculating the surface area of a cubical} \\ \text{box if an error of }1\% \text{ is made in measuring the lengths of edges of the cube.}
\displaystyle \text{Answer:}

\displaystyle \text{Let } x \text{ be the edge of the cube and } y \text{ be the surface area.}

\displaystyle \text{Then, } y = 6x^2.

\displaystyle \text{Let } \Delta x \text{ be the error in } x \text{ and } \Delta y \text{ be the corresponding error in } y.

\displaystyle \text{We have } \frac{\Delta x}{x} \times 100 = 1.

\displaystyle \Rightarrow \Delta x = \frac{x}{100}.

\displaystyle \text{Now, } y = 6x^2.

\displaystyle \Rightarrow \frac{dy}{dx} = 12x.

\displaystyle \therefore \Delta y \approx dy = \frac{dy}{dx}\,\Delta x.

\displaystyle \Rightarrow \Delta y = 12x \times \frac{x}{100}.

\displaystyle \Rightarrow \Delta y = \frac{12x^2}{100}.

\displaystyle \Rightarrow \Delta y = \frac{2y}{100}.

\displaystyle \Rightarrow \frac{\Delta y}{y} = \frac{2}{100}.

\displaystyle \Rightarrow \frac{\Delta y}{y} \times 100 = 2.

\displaystyle \text{Hence, the percentage error in calculating the surface area is } 2.

\displaystyle \textbf{Question 5: }~\text{If there is an error of }0.1\%\text{ in the measurement of the radius of a sphere, } \\ \text{find approximately the percentage error in the calculation of the volume of the sphere.}
\displaystyle \text{Answer:}

\displaystyle \text{Let } x \text{ be the radius and } y \text{ be the volume of the sphere.}

\displaystyle y = \frac{4}{3}\pi x^3.

\displaystyle \text{Let } \Delta x \text{ be the error in the radius and } \Delta y \text{ be the error in the volume.}

\displaystyle \text{Then, } \frac{\Delta x}{x} \times 100 = 0.1.

\displaystyle \Rightarrow \frac{dx}{x} = \frac{1}{1000}.

\displaystyle \text{Now, } y = \frac{4}{3}\pi x^3.

\displaystyle \Rightarrow \frac{dy}{dx} = 4\pi x^2.

\displaystyle \Rightarrow dy = 4\pi x^2\,dx.

\displaystyle \Rightarrow \frac{dy}{y} = \frac{4\pi x^2\,dx}{\frac{4}{3}\pi x^3} = \frac{3}{x}\,dx.

\displaystyle \Rightarrow \frac{dy}{y} = \frac{3}{1000}.

\displaystyle \Rightarrow \frac{\Delta y}{y} \times 100 = 0.3.

\displaystyle \text{Hence, the percentage error in the calculation of the volume of the sphere is } 0.3.

\displaystyle \textbf{Question 6: }~\text{The pressure }p\text{ and the volume }v\text{ of a gas are connected by the relation } \\ pv^{1.4}=\text{constant. Find the percentage error in }p\text{ corresponding to a decrease of }\frac{1}{2}\%\text{ in }v.
\displaystyle \text{Answer:}

\displaystyle \text{We have } pv^{1.4} = \text{constant} = k \text{ (say)}.

\displaystyle \text{Taking logarithm on both sides, we get}

\displaystyle \log(pv^{1.4}) = \log k.

\displaystyle \Rightarrow \log p + 1.4\log v = \log k.

\displaystyle \text{Differentiating both sides with respect to } v, \text{ we get}

\displaystyle \frac{1}{p}\frac{dp}{dv} + \frac{1.4}{v} = 0.

\displaystyle \Rightarrow \frac{dp}{p} = -1.4\,\frac{dv}{v}.

\displaystyle \text{Now, percentage error in } p = \frac{dp}{p} \times 100.

\displaystyle \Rightarrow \frac{dp}{p} \times 100 = -1.4\left(\frac{dv}{v} \times 100\right).

\displaystyle \Rightarrow \frac{dp}{p} \times 100 = -1.4 \times \left(-\frac{1}{2}\right) \quad \left[\because \text{there is a } \frac{1}{2}\% \text{ decrease in } v\right].

\displaystyle \Rightarrow \frac{dp}{p} \times 100 = 0.7.

\displaystyle \text{Hence, the percentage error in } p \text{ is } 0.7\%.

\displaystyle \textbf{Question 7: }~\text{The height of a cone increases by }k\%,\text{ its semi-vertical angle } \\ \text{remaining the same. What is the approximate percentage increase (i) in total } \\ \text{surface area, and (ii) in the volume, assuming that } k\text{ is small?}
\displaystyle \text{Answer:}

\displaystyle \text{Let } h \text{ be the height, } r \text{ be the radius, } l \text{ be the slant height, } \\ T \text{ be the total surface area and } V \text{ be the volume of the cone.}

\displaystyle \text{Let } \Delta h, \Delta r \text{ and } \Delta l \text{ be the corresponding changes.}

\displaystyle \text{Since the semi-vertical angle remains constant,}

\displaystyle \frac{\Delta h}{h} = \frac{\Delta r}{r} = \frac{\Delta l}{l}.

\displaystyle \text{Also, } \frac{\Delta h}{h} \times 100 = k.

\displaystyle \Rightarrow \frac{\Delta r}{r} \times 100 = \frac{\Delta l}{l} \times 100 = k. \quad (1)

\displaystyle \text{(i) Total surface area of the cone: } T = \pi r l + \pi r^2.

\displaystyle \Rightarrow \frac{dT}{dr} = \pi l + \pi r\frac{dl}{dr} + 2\pi r.

\displaystyle \text{From (1), } \frac{dl}{dr} = \frac{l}{r}.

\displaystyle \Rightarrow \frac{dT}{dr} = \pi l + \pi l + 2\pi r.

\displaystyle \Rightarrow \frac{dT}{dr} = 2\pi(l + r).

\displaystyle \therefore \Delta T \approx \frac{dT}{dr}\Delta r = 2\pi(l + r)\frac{kr}{100}.

\displaystyle \Rightarrow \frac{\Delta T}{T} \times 100 = \frac{2\pi r(l + r)}{\pi r(l + r)} \times k = 2k.

\displaystyle \text{Hence, the percentage increase in the total surface area of the cone is } 2k\%.

\displaystyle \text{(ii) Volume of the cone: } V = \frac{1}{3}\pi r^2 h.

\displaystyle \Rightarrow \frac{dV}{dh} = \frac{1}{3}\pi r^2 + \frac{1}{3}\pi h \cdot 2r\frac{dr}{dh}.

\displaystyle \text{From (1), } \frac{dr}{dh} = \frac{r}{h}.

\displaystyle \Rightarrow \frac{dV}{dh} = \frac{1}{3}\pi r^2 + \frac{2}{3}\pi r^2.

\displaystyle \Rightarrow \frac{dV}{dh} = \pi r^2.

\displaystyle \therefore \Delta V \approx \frac{dV}{dh}\Delta h = \pi r^2 \cdot \frac{kh}{100}.

\displaystyle \Rightarrow \frac{\Delta V}{V} \times 100 = \frac{\pi r^2 h}{\frac{1}{3}\pi r^2 h} \times k = 3k.

\displaystyle \text{Hence, the percentage increase in the volume of the cone is } 3k\%.

\displaystyle \textbf{Question 8: }~\text{Show that the relative error in computing the volume of a sphere,} \\ \text{due to an error in measuring the radius, is approximately equalto three times} \\ \text{ the relative error in the radius.}
\displaystyle \text{Answer:}

\displaystyle \text{Let } x \text{ be the radius of the sphere and } y \text{ be its volume.}

\displaystyle \text{Let } \Delta x \text{ be the error in the radius and } \Delta y \text{ be the approximate error in the volume.}

\displaystyle y = \frac{4}{3}\pi x^3.

\displaystyle \Rightarrow \frac{dy}{dx} = 4\pi x^2.

\displaystyle \Rightarrow \Delta y \approx dy = \frac{dy}{dx}\,\Delta x = 4\pi x^2\,\Delta x.

\displaystyle \Rightarrow \Delta y = 3 \times \frac{4}{3}\pi x^3 \cdot \frac{\Delta x}{x}.

\displaystyle \Rightarrow \Delta y = 3y \cdot \frac{\Delta x}{x}.

\displaystyle \Rightarrow \frac{\Delta y}{y} = 3\frac{\Delta x}{x}.

\displaystyle \text{Hence proved.}

\displaystyle \textbf{Question 9: }~\text{Using differentials, find the approximate values of the following:}
\displaystyle \text{(i) }\sqrt{25.02}

\displaystyle \text{Answer:}

\displaystyle \text{Consider the function } y = f(x) = \sqrt{x}.

\displaystyle \text{Let } x = 25 \text{ and } x + \Delta x = 25.02.

\displaystyle \Rightarrow \Delta x = 0.02.

\displaystyle \text{For } x = 25, \quad y = \sqrt{25} = 5.

\displaystyle \text{Let } dx = \Delta x = 0.02.

\displaystyle \text{Now, } y = \sqrt{x}.

\displaystyle \Rightarrow \frac{dy}{dx} = \frac{1}{2\sqrt{x}}.

\displaystyle \Rightarrow \left(\frac{dy}{dx}\right)_{x=25} = \frac{1}{10}.

\displaystyle \therefore \Delta y \approx dy = \frac{dy}{dx}\,dx = \frac{1}{10} \times 0.02 = 0.002.

\displaystyle \Rightarrow \Delta y = 0.002.

\displaystyle \therefore \sqrt{25.02} \approx y + \Delta y = 5 + 0.002 = 5.002.

\displaystyle \text{(ii) } (0.009)^{1/3}

\displaystyle \text{Answer:}

\displaystyle \text{Consider the function } y = f(x) = \sqrt[3]{x}.

\displaystyle \text{Let } x = 0.008 \text{ and } x + \Delta x = 0.009.

\displaystyle \Rightarrow \Delta x = 0.001.

\displaystyle \text{For } x = 0.008, \quad y = \sqrt[3]{0.008} = 0.2.

\displaystyle \text{Let } dx = \Delta x = 0.001.

\displaystyle \text{Now, } y = \sqrt[3]{x}.

\displaystyle \Rightarrow \frac{dy}{dx} = \frac{1}{3x^{2/3}}.

\displaystyle \Rightarrow \left(\frac{dy}{dx}\right)_{x=0.008} = \frac{1}{3(0.008)^{2/3}} = \frac{1}{3(0.04)} = \frac{1}{0.12}.

\displaystyle \therefore \Delta y \approx dy = \frac{dy}{dx}\,dx = \frac{1}{0.12} \times 0.001 = \frac{1}{120}.

\displaystyle \Rightarrow \Delta y = 0.008333.

\displaystyle \therefore \sqrt[3]{0.009} \approx y + \Delta y = 0.2 + 0.008333 = 0.208333.

\displaystyle \text{(iii) } (0.007)^{1/3}

\displaystyle \text{Answer:}

\displaystyle \text{Consider the function } y = f(x) = \sqrt[3]{x}.

\displaystyle \text{Let } x = 0.008 \text{ and } x + \Delta x = 0.007.

\displaystyle \Rightarrow \Delta x = -0.001.

\displaystyle \text{For } x = 0.008, \quad y = \sqrt[3]{0.008} = 0.2.

\displaystyle \text{Let } dx = \Delta x = -0.001.

\displaystyle \text{Now, } y = \sqrt[3]{x}.

\displaystyle \Rightarrow \frac{dy}{dx} = \frac{1}{3x^{2/3}}.

\displaystyle \Rightarrow \left(\frac{dy}{dx}\right)_{x=0.008} = \frac{1}{3(0.008)^{2/3}} = \frac{1}{3(0.04)} = \frac{1}{0.12}.

\displaystyle \therefore \Delta y \approx dy = \frac{dy}{dx}\,dx = \frac{1}{0.12} \times (-0.001) = -\frac{1}{120}.

\displaystyle \Rightarrow \Delta y = -0.008333.

\displaystyle \therefore \sqrt[3]{0.007} \approx y + \Delta y = 0.2 - 0.008333 = 0.191667.

\displaystyle \text{(iv) } \sqrt{401}
\displaystyle \text{Answer:}

\displaystyle \text{Consider the function } y = f(x) = \sqrt{x}.

\displaystyle \text{Let } x = 400 \text{ and } x + \Delta x = 401.

\displaystyle \Rightarrow \Delta x = 1.

\displaystyle \text{For } x = 400, \quad y = \sqrt{400} = 20.

\displaystyle \text{Let } dx = \Delta x = 1.

\displaystyle \text{Now, } y = \sqrt{x}.

\displaystyle \Rightarrow \frac{dy}{dx} = \frac{1}{2\sqrt{x}}.

\displaystyle \Rightarrow \left(\frac{dy}{dx}\right)_{x=400} = \frac{1}{40}.

\displaystyle \therefore \Delta y \approx dy = \frac{dy}{dx}\,dx = \frac{1}{40} \times 1 = \frac{1}{40}.

\displaystyle \Rightarrow \Delta y = 0.025.

\displaystyle \therefore \sqrt{401} \approx y + \Delta y = 20 + 0.025 = 20.025.

\displaystyle \text{(v) } (15)^{1/4}
\displaystyle \text{Answer:}

\displaystyle \text{Consider the function } y = f(x) = x^{\frac{1}{4}}.

\displaystyle \text{Let } x = 16 \text{ and } x + \Delta x = 15.

\displaystyle \Rightarrow \Delta x = -1.

\displaystyle \text{For } x = 16, \quad y = 16^{\frac{1}{4}} = 2.

\displaystyle \text{Let } dx = \Delta x = -1.

\displaystyle \text{Now, } y = x^{\frac{1}{4}}.

\displaystyle \Rightarrow \frac{dy}{dx} = \frac{1}{4x^{\frac{3}{4}}}.

\displaystyle \Rightarrow \left(\frac{dy}{dx}\right)_{x=16} = \frac{1}{4(16)^{\frac{3}{4}}} = \frac{1}{32}.

\displaystyle \therefore \Delta y \approx dy = \frac{dy}{dx}\,dx = \frac{1}{32} \times (-1) = -\frac{1}{32}.

\displaystyle \Rightarrow \Delta y = -0.03125.

\displaystyle \therefore 15^{\frac{1}{4}} \approx y + \Delta y = 2 - 0.03125 = 1.96875.

\displaystyle \text{(vi) } (255)^{1/4}
\displaystyle \text{Answer:}

\displaystyle \text{Consider the function } y = f(x) = x^{\frac{1}{4}}.

\displaystyle \text{Let } x = 256 \text{ and } x + \Delta x = 255.

\displaystyle \Rightarrow \Delta x = -1.

\displaystyle \text{For } x = 256, \quad y = 256^{\frac{1}{4}} = 4.

\displaystyle \text{Let } dx = \Delta x = -1.

\displaystyle \text{Now, } y = x^{\frac{1}{4}}.

\displaystyle \Rightarrow \frac{dy}{dx} = \frac{1}{4x^{\frac{3}{4}}}.

\displaystyle \Rightarrow \left(\frac{dy}{dx}\right)_{x=256} = \frac{1}{4(256)^{\frac{3}{4}}} = \frac{1}{256}.

\displaystyle \therefore \Delta y \approx dy = \frac{dy}{dx}\,dx = \frac{1}{256} \times (-1) = -\frac{1}{256}.

\displaystyle \Rightarrow \Delta y = -0.003906.

\displaystyle \therefore 255^{\frac{1}{4}} \approx y + \Delta y = 4 - 0.003906 = 3.996094.

\displaystyle \text{(vii) } \frac{1}{(2.002)^2}
\displaystyle \text{Answer:}

\displaystyle \text{Consider the function } y = f(x) = \frac{1}{x^2}.

\displaystyle \text{Let } x = 2 \text{ and } x + \Delta x = 2.002.

\displaystyle \Rightarrow \Delta x = 0.002.

\displaystyle \text{For } x = 2, \quad y = \frac{1}{2^2} = \frac{1}{4}.

\displaystyle \text{Let } dx = \Delta x = 0.002.

\displaystyle \text{Now, } y = \frac{1}{x^2}.

\displaystyle \Rightarrow \frac{dy}{dx} = -\frac{2}{x^3}.

\displaystyle \Rightarrow \left(\frac{dy}{dx}\right)_{x=2} = -\frac{2}{(2)^3} = -\frac{1}{4}.

\displaystyle \therefore \Delta y \approx dy = \frac{dy}{dx}\,dx = -\frac{1}{4} \times 0.002 = -0.0005.

\displaystyle \Rightarrow \Delta y = -0.0005.

\displaystyle \therefore \frac{1}{(2.002)^2} \approx y + \Delta y = 0.25 - 0.0005 = 0.2495.

\displaystyle \text{(viii) }\log_e 4.04,\text{ it being given that }\log_{10}4=0.6021\text{ and }\log_{10}e=0.4343,
\displaystyle \text{Answer:}

\displaystyle \text{Consider the function } y = f(x) = \log_e x.

\displaystyle \text{Let } x = 4 \text{ and } x + \Delta x = 4.04.

\displaystyle \Rightarrow \Delta x = 0.04.

\displaystyle \text{For } x = 4,

\displaystyle y = \log_e 4 = \frac{\log_{10} 4}{\log_{10} e} = \frac{0.6021}{0.4343} = 1.386368.

\displaystyle \text{Let } dx = \Delta x = 0.04.

\displaystyle \text{Now, } y = \log_e x.

\displaystyle \Rightarrow \frac{dy}{dx} = \frac{1}{x}.

\displaystyle \Rightarrow \left(\frac{dy}{dx}\right)_{x=4} = \frac{1}{4}.

\displaystyle \therefore \Delta y \approx dy = \frac{dy}{dx}\,dx = \frac{1}{4} \times 0.04 = 0.01.

\displaystyle \Rightarrow \Delta y = 0.01.

\displaystyle \therefore \log_e 4.04 \approx y + \Delta y = 1.386368 + 0.01 = 1.396368.

\displaystyle \text{(ix) } \log_e 10.02,\text{ it being given that }\log_e 10=2.3026,
\displaystyle \text{Answer:}

\displaystyle \text{Consider the function } y = f(x) = \log_e x.

\displaystyle \text{Let } x = 10 \text{ and } x + \Delta x = 10.02.

\displaystyle \Rightarrow \Delta x = 0.02.

\displaystyle \text{For } x = 10, \quad y = \log_e 10 = 2.3026.

\displaystyle \text{Let } dx = \Delta x = 0.02.

\displaystyle \text{Now, } y = \log_e x.

\displaystyle \Rightarrow \frac{dy}{dx} = \frac{1}{x}.

\displaystyle \Rightarrow \left(\frac{dy}{dx}\right)_{x=10} = \frac{1}{10}.

\displaystyle \therefore \Delta y \approx dy = \frac{dy}{dx}\,dx = \frac{1}{10} \times 0.02 = 0.002.

\displaystyle \Rightarrow \Delta y = 0.002.

\displaystyle \therefore \log_e 10.02 \approx y + \Delta y = 2.3026 + 0.002 = 2.3046.

\displaystyle \text{(x) } \log_{10}10.1,\text{ it being given that }\log_{10}e=0.4343,
\displaystyle \text{Answer:}

\displaystyle \text{Consider the function } y = f(x) = \log_{10} x.

\displaystyle \text{Let } x = 10 \text{ and } x + \Delta x = 10.1.

\displaystyle \Rightarrow \Delta x = 0.1.

\displaystyle \text{For } x = 10, \quad y = \log_{10} 10 = 1.

\displaystyle \text{Let } dx = \Delta x = 0.1.

\displaystyle \text{Now, } y = \log_{10} x = \frac{\log_e x}{\log_e 10}.

\displaystyle \Rightarrow \frac{dy}{dx} = \frac{1}{x\log_e 10}.

\displaystyle \Rightarrow \left(\frac{dy}{dx}\right)_{x=10} = \frac{1}{10\log_e 10} = \frac{1}{23.025} = 0.04343.

\displaystyle \therefore \Delta y \approx dy = \frac{dy}{dx}\,dx = 0.04343 \times 0.1 = 0.004343.

\displaystyle \Rightarrow \Delta y = 0.004343.

\displaystyle \therefore \log_{10} 10.1 \approx y + \Delta y = 1 + 0.004343 = 1.004343.

\displaystyle \text{(xi) } \cos 61^\circ,\text{ it being given that }\sin 60^\circ=0.86603\text{ and }1^\circ=0.01745\text{ radian},
\displaystyle \text{Answer:}

\displaystyle \text{Consider the function } y = f(x) = \cos x.

\displaystyle \text{Let } x = 60^\circ \text{ and } x + \Delta x = 61^\circ.

\displaystyle \Rightarrow \Delta x = 1^\circ = \frac{\pi}{180} = 0.01745 \text{ rad}.

\displaystyle \text{For } x = 60^\circ, \quad y = \cos 60^\circ = 0.5.

\displaystyle \text{Let } dx = \Delta x = 0.01745.

\displaystyle \text{Now, } y = \cos x.

\displaystyle \Rightarrow \frac{dy}{dx} = -\sin x.

\displaystyle \Rightarrow \left(\frac{dy}{dx}\right)_{x=60^\circ} = -\sin 60^\circ = -\frac{\sqrt{3}}{2} = -0.86603.

\displaystyle \therefore \Delta y \approx dy = \frac{dy}{dx}\,dx = -0.86603 \times 0.01745 = -0.01511.

\displaystyle \Rightarrow \Delta y = -0.01511.

\displaystyle \therefore \cos 61^\circ \approx y + \Delta y = 0.5 - 0.01511 = 0.48489.

\displaystyle \text{(xii) } \frac{1}{\sqrt{25.1}}
\displaystyle \text{Answer:}

\displaystyle \text{Consider the function } y = f(x) = \frac{1}{\sqrt{x}}.

\displaystyle \text{Let } x = 25 \text{ and } x + \Delta x = 25.1.

\displaystyle \Rightarrow \Delta x = 0.1.

\displaystyle \text{For } x = 25, \quad y = \frac{1}{\sqrt{25}} = 0.2.

\displaystyle \text{Let } dx = \Delta x = 0.1.

\displaystyle \text{Now, } y = \frac{1}{\sqrt{x}}.

\displaystyle \Rightarrow \frac{dy}{dx} = -\frac{1}{2x^{3/2}}.

\displaystyle \Rightarrow \left(\frac{dy}{dx}\right)_{x=25} = -\frac{1}{2(25)^{3/2}} = -\frac{1}{250} = -0.004.

\displaystyle \therefore \Delta y \approx dy = \frac{dy}{dx}\,dx = (-0.004)(0.1) = -0.0004.

\displaystyle \Rightarrow \Delta y = -0.0004.

\displaystyle \therefore \frac{1}{\sqrt{25.1}} \approx y + \Delta y = 0.2 - 0.0004 = 0.1996.

\displaystyle \text{(xiii) } \sin\!\left(\frac{22}{14}\right)
\displaystyle \text{Answer:}

\displaystyle \text{Consider the function } y = f(x) = \sin x.

\displaystyle \text{Let } x = \frac{22}{7} \text{ and } x + \Delta x = \frac{22}{14}.

\displaystyle \Rightarrow \Delta x = \frac{22}{14} - \frac{22}{7} = -\frac{22}{14}.

\displaystyle \text{For } x = \frac{22}{7} \approx \pi, \quad y = \sin\!\left(\frac{22}{7}\right) \approx 0.

\displaystyle \text{Let } dx = \Delta x = -\frac{22}{14}.

\displaystyle \text{Now, } y = \sin x.

\displaystyle \Rightarrow \frac{dy}{dx} = \cos x.

\displaystyle \Rightarrow \left(\frac{dy}{dx}\right)_{x=\frac{22}{7}} = \cos\!\left(\frac{22}{7}\right) \approx -1.

\displaystyle \therefore \Delta y \approx dy = \frac{dy}{dx}\,dx = (-1)\left(-\frac{22}{14}\right) = \frac{22}{14}.

\displaystyle \Rightarrow \Delta y \approx 1.57.

\displaystyle \therefore \sin\!\left(\frac{22}{14}\right) \approx y + \Delta y \approx 1.57.

\displaystyle \text{(xiv) } \cos\!\left(\frac{11\pi}{36}\right)
\displaystyle \text{Answer:}

\displaystyle \text{Consider the function } y = f(x) = \cos x.

\displaystyle \text{Let } x = \frac{\pi}{3} \text{ and } x + \Delta x = \frac{11\pi}{36}.

\displaystyle \Rightarrow \Delta x = \frac{11\pi}{36} - \frac{\pi}{3} = \frac{11\pi - 12\pi}{36} = -\frac{\pi}{36}.

\displaystyle \text{For } x = \frac{\pi}{3}, \quad y = \cos\!\left(\frac{\pi}{3}\right) = \frac{1}{2}.

\displaystyle \text{Let } dx = \Delta x = -\frac{\pi}{36}.

\displaystyle \text{Now, } y = \cos x.

\displaystyle \Rightarrow \frac{dy}{dx} = -\sin x.

\displaystyle \Rightarrow \left(\frac{dy}{dx}\right)_{x=\frac{\pi}{3}} = -\sin\!\left(\frac{\pi}{3}\right) = -\frac{\sqrt{3}}{2} = -0.86603.

\displaystyle \therefore \Delta y \approx dy = \frac{dy}{dx}\,dx = (-0.86603)\left(-\frac{\pi}{36}\right).

\displaystyle \Rightarrow \Delta y \approx 0.0756.

\displaystyle \therefore \cos\!\left(\frac{11\pi}{36}\right) \approx y + \Delta y = 0.5 + 0.0756 = 0.5756.

\displaystyle \text{(xv) } (80)^{1/4}

\displaystyle \text{Answer:}

\displaystyle \text{Consider the function } y = f(x) = x^{\tfrac14}.

\displaystyle \text{Let } x = 81 \text{ and } x + \Delta x = 80.

\displaystyle \Rightarrow \Delta x = 80 - 81 = -1.

\displaystyle \text{For } x = 81, \quad y = 81^{\tfrac14} = 3.

\displaystyle \text{Let } dx = \Delta x = -1.

\displaystyle \text{Now, } y = x^{\tfrac14}.

\displaystyle \Rightarrow \frac{dy}{dx} = \frac{1}{4x^{\tfrac34}}.

\displaystyle \Rightarrow \left(\frac{dy}{dx}\right)_{x=81} = \frac{1}{4(81)^{\tfrac34}} = \frac{1}{4(27)} = \frac{1}{108}.

\displaystyle \therefore \Delta y \approx dy = \frac{dy}{dx}\,dx = \frac{1}{108}(-1).

\displaystyle \Rightarrow \Delta y \approx -0.00926.

\displaystyle \therefore 80^{\tfrac14} \approx y + \Delta y = 3 - 0.00926 = 2.99074.

\displaystyle \text{(xvi) } (29)^{1/3}
\displaystyle \text{Answer:}

\displaystyle \text{Consider the function } y = f(x) = x^{\tfrac13}.

\displaystyle \text{Let } x = 27 \text{ and } x + \Delta x = 29.

\displaystyle \Rightarrow \Delta x = 2.

\displaystyle \text{For } x = 27, \quad y = 27^{\tfrac13} = 3.

\displaystyle \text{Let } dx = \Delta x = 2.

\displaystyle \text{Now, } y = x^{\tfrac13}.

\displaystyle \Rightarrow \frac{dy}{dx} = \frac{1}{3x^{\tfrac23}}.

\displaystyle \Rightarrow \left(\frac{dy}{dx}\right)_{x=27} = \frac{1}{3(27)^{\tfrac23}} = \frac{1}{3(9)} = \frac{1}{27}.

\displaystyle \therefore \Delta y \approx dy = \frac{dy}{dx}\,dx = \frac{1}{27} \times 2.

\displaystyle \Rightarrow \Delta y \approx 0.074.

\displaystyle \therefore 29^{\tfrac13} \approx y + \Delta y = 3 + 0.074 = 3.074.

\displaystyle \text{(xvii) } (66)^{1/3}
\displaystyle \text{Answer:}

\displaystyle \text{Consider the function } y = f(x) = x^{\tfrac13}.

\displaystyle \text{Let } x = 64 \text{ and } x + \Delta x = 66.

\displaystyle \Rightarrow \Delta x = 2.

\displaystyle \text{For } x = 64, \quad y = 64^{\tfrac13} = 4.

\displaystyle \text{Let } dx = \Delta x = 2.

\displaystyle \text{Now, } y = x^{\tfrac13}.

\displaystyle \Rightarrow \frac{dy}{dx} = \frac{1}{3x^{\tfrac23}}.

\displaystyle \Rightarrow \left(\frac{dy}{dx}\right)_{x=64} = \frac{1}{3(64)^{\tfrac23}} = \frac{1}{3(16)} = \frac{1}{48}.

\displaystyle \therefore \Delta y \approx dy = \frac{dy}{dx}\,dx = \frac{1}{48} \times 2.

\displaystyle \Rightarrow \Delta y \approx 0.042.

\displaystyle \therefore 66^{\tfrac13} \approx y + \Delta y = 4 + 0.042 = 4.042.

\displaystyle \text{(xviii) } \sqrt{26} \hspace{10.0cm} [\text{CBSE 2000}]
\displaystyle \text{Answer:}

\displaystyle \text{Consider the function } y = f(x) = \sqrt{x}.

\displaystyle \text{Let } x = 25 \text{ and } x + \Delta x = 26.

\displaystyle \Rightarrow \Delta x = 1.

\displaystyle \text{For } x = 25, \quad y = \sqrt{25} = 5.

\displaystyle \text{Let } dx = \Delta x = 1.

\displaystyle \text{Now, } y = x^{\tfrac12}.

\displaystyle \Rightarrow \frac{dy}{dx} = \frac{1}{2\sqrt{x}}.

\displaystyle \Rightarrow \left(\frac{dy}{dx}\right)_{x=25} = \frac{1}{10}.

\displaystyle \therefore \Delta y \approx dy = \frac{dy}{dx}\,dx = \frac{1}{10} \times 1 = 0.1.

\displaystyle \Rightarrow \Delta y = 0.1.

\displaystyle \therefore \sqrt{26} \approx y + \Delta y = 5 + 0.1 = 5.1.

\displaystyle \text{(xix) } \sqrt{37} \hspace{10.0cm} [\text{CBSE 2005}]
\displaystyle \text{Answer:}

\displaystyle \text{Consider the function } y = f(x) = \sqrt{x}.

\displaystyle \text{Let } x = 36 \text{ and } x + \Delta x = 37.

\displaystyle \Rightarrow \Delta x = 1.

\displaystyle \text{For } x = 36, \quad y = \sqrt{36} = 6.

\displaystyle \text{Let } dx = \Delta x = 1.

\displaystyle \text{Now, } y = x^{\tfrac12}.

\displaystyle \Rightarrow \frac{dy}{dx} = \frac{1}{2\sqrt{x}}.

\displaystyle \Rightarrow \left(\frac{dy}{dx}\right)_{x=36} = \frac{1}{12}.

\displaystyle \therefore \Delta y \approx dy = \frac{dy}{dx}\,dx = \frac{1}{12} \times 1 = 0.0833.

\displaystyle \Rightarrow \Delta y = 0.0833.

\displaystyle \therefore \sqrt{37} \approx y + \Delta y = 6 + 0.0833 = 6.0833.

\displaystyle \text{(xx) } \sqrt{0.48} \hspace{10.0cm} [\text{CBSE 2002C}]
\displaystyle \text{Answer:}

\displaystyle \text{Consider the function } y = f(x) = \sqrt{x}.

\displaystyle \text{Let } x = 0.49 \text{ and } x + \Delta x = 0.48.

\displaystyle \Rightarrow \Delta x = -0.01.

\displaystyle \text{For } x = 0.49, \quad y = \sqrt{0.49} = 0.7.

\displaystyle \text{Let } dx = \Delta x = -0.01.

\displaystyle \text{Now, } y = x^{\tfrac12}.

\displaystyle \Rightarrow \frac{dy}{dx} = \frac{1}{2\sqrt{x}}.

\displaystyle \Rightarrow \left(\frac{dy}{dx}\right)_{x=0.49} = \frac{1}{2(0.7)} = \frac{1}{1.4}.

\displaystyle \therefore \Delta y \approx dy = \frac{dy}{dx}\,dx = \frac{1}{1.4} \times (-0.01) = -0.007143.

\displaystyle \Rightarrow \Delta y = -0.007143.

\displaystyle \therefore \sqrt{0.48} \approx y + \Delta y = 0.7 - 0.007143 = 0.692857 \approx 0.693.

\displaystyle \text{(xxi) } (82)^{1/4}
\displaystyle \text{Answer:}

\displaystyle \text{Consider the function } y = f(x) = x^{\tfrac14}.

\displaystyle \text{Let } x = 81 \text{ and } x + \Delta x = 82.

\displaystyle \Rightarrow \Delta x = 1.

\displaystyle \text{For } x = 81, \quad y = 81^{\tfrac14} = 3.

\displaystyle \text{Let } dx = \Delta x = 1.

\displaystyle \text{Now, } y = x^{\tfrac14}.

\displaystyle \Rightarrow \frac{dy}{dx} = \frac{1}{4x^{\tfrac34}}.

\displaystyle \Rightarrow \left(\frac{dy}{dx}\right)_{x=81} = \frac{1}{4(81)^{\tfrac34}} = \frac{1}{4(27)} = \frac{1}{108}.

\displaystyle \therefore \Delta y \approx dy = \frac{dy}{dx}\,dx = \frac{1}{108} \times 1 = 0.009259.

\displaystyle \Rightarrow \Delta y = 0.009259.

\displaystyle \therefore 82^{\tfrac14} \approx y + \Delta y = 3 + 0.009259 = 3.009259.

\displaystyle \text{(xxii) } \left(\frac{17}{81}\right)^{1/4}
\displaystyle \text{Answer:}

\displaystyle \text{Consider the function } y = f(x) = x^{\tfrac14}.

\displaystyle \text{Let } x = \frac{16}{81} \text{ and } x + \Delta x = \frac{17}{81}.

\displaystyle \Rightarrow \Delta x = \frac{1}{81}.

\displaystyle \text{For } x = \frac{16}{81}, \quad y = \left(\frac{16}{81}\right)^{\tfrac14} = \frac{2}{3}.

\displaystyle \text{Let } dx = \Delta x = \frac{1}{81}.

\displaystyle \text{Now, } y = x^{\tfrac14}.

\displaystyle \Rightarrow \frac{dy}{dx} = \frac{1}{4x^{\tfrac34}}.

\displaystyle \Rightarrow \left(\frac{dy}{dx}\right)_{x=\tfrac{16}{81}} = \frac{1}{4\left(\tfrac{16}{81}\right)^{\tfrac34}} = \frac{1}{4\left(\tfrac{8}{27}\right)} = \frac{27}{32}.

\displaystyle \therefore \Delta y \approx dy = \frac{dy}{dx}\,dx = \frac{27}{32} \times \frac{1}{81} = \frac{1}{96}.

\displaystyle \Rightarrow \Delta y \approx 0.01042.

\displaystyle \therefore \left(\frac{17}{81}\right)^{\tfrac14} \approx y + \Delta y = \frac{2}{3} + \frac{1}{96} = 0.6771.

\displaystyle \text{(xxiii) } (33)^{1/5}
\displaystyle \text{Answer:}

\displaystyle \text{Consider the function } y = f(x) = x^{\tfrac15}.

\displaystyle \text{Let } x = 32 \text{ and } x + \Delta x = 33.

\displaystyle \Rightarrow \Delta x = 1.

\displaystyle \text{For } x = 32, \quad y = 32^{\tfrac15} = 2.

\displaystyle \text{Let } dx = \Delta x = 1.

\displaystyle \text{Now, } y = x^{\tfrac15}.

\displaystyle \Rightarrow \frac{dy}{dx} = \frac{1}{5x^{\tfrac45}}.

\displaystyle \Rightarrow \left(\frac{dy}{dx}\right)_{x=32} = \frac{1}{5(32)^{\tfrac45}} = \frac{1}{5(16)} = \frac{1}{80}.

\displaystyle \therefore \Delta y \approx dy = \frac{dy}{dx}\,dx = \frac{1}{80} \times 1 = 0.0125.

\displaystyle \Rightarrow \Delta y = 0.0125.

\displaystyle \therefore 33^{\tfrac15} \approx y + \Delta y = 2 + 0.0125 = 2.0125.

\displaystyle \text{(xxiv) } \sqrt{36.6}
\displaystyle \text{Answer:}

\displaystyle \text{Consider the function } y = f(x) = \sqrt{x}.

\displaystyle \text{Let } x = 36 \text{ and } x + \Delta x = 36.6.

\displaystyle \Rightarrow \Delta x = 0.6.

\displaystyle \text{For } x = 36, \quad y = \sqrt{36} = 6.

\displaystyle \text{Let } dx = \Delta x = 0.6.

\displaystyle \text{Now, } y = x^{\tfrac12}.

\displaystyle \Rightarrow \frac{dy}{dx} = \frac{1}{2\sqrt{x}}.

\displaystyle \Rightarrow \left(\frac{dy}{dx}\right)_{x=36} = \frac{1}{12}.

\displaystyle \therefore \Delta y \approx dy = \frac{dy}{dx}\,dx = \frac{1}{12} \times 0.6 = 0.05.

\displaystyle \Rightarrow \Delta y = 0.05.

\displaystyle \therefore \sqrt{36.6} \approx y + \Delta y = 6 + 0.05 = 6.05.

\displaystyle \text{(xxv) } 25^{1/3}
\displaystyle \text{Answer:}

\displaystyle \text{Consider the function } y = f(x) = x^{\tfrac13}.

\displaystyle \text{Let } x = 27 \text{ and } x + \Delta x = 25.

\displaystyle \Rightarrow \Delta x = -2.

\displaystyle \text{For } x = 27, \quad y = 27^{\tfrac13} = 3.

\displaystyle \text{Let } dx = \Delta x = -2.

\displaystyle \text{Now, } y = x^{\tfrac13}.

\displaystyle \Rightarrow \frac{dy}{dx} = \frac{1}{3x^{\tfrac23}}.

\displaystyle \Rightarrow \left(\frac{dy}{dx}\right)_{x=27} = \frac{1}{3(27)^{\tfrac23}} = \frac{1}{3(9)} = \frac{1}{27}.

\displaystyle \therefore \Delta y \approx dy = \frac{dy}{dx}\,dx = \frac{1}{27} \times (-2) = -\frac{2}{27}.

\displaystyle \Rightarrow \Delta y \approx -0.07407.

\displaystyle \therefore 25^{\tfrac13} \approx y + \Delta y = 3 - 0.07407 = 2.92593 \approx 2.9259.

\displaystyle \text{(xxvi) } \sqrt{495} \hspace{10.0cm} [\text{CBSE 2012}]
\displaystyle \text{Answer:}

\displaystyle \text{Consider the function } y = f(x) = \sqrt{x}.

\displaystyle \text{Let } x = 49 \text{ and } x + \Delta x = 49.5.

\displaystyle \Rightarrow \Delta x = 0.5.

\displaystyle \text{For } x = 49, \quad y = \sqrt{49} = 7.

\displaystyle \text{Let } dx = \Delta x = 0.5.

\displaystyle \text{Now, } y = x^{\tfrac12}.

\displaystyle \Rightarrow \frac{dy}{dx} = \frac{1}{2\sqrt{x}}.

\displaystyle \Rightarrow \left(\frac{dy}{dx}\right)_{x=49} = \frac{1}{14}.

\displaystyle \therefore \Delta y \approx dy = \frac{dy}{dx}\,dx = \frac{1}{14} \times 0.5 = 0.0357.

\displaystyle \Rightarrow \Delta y = 0.0357.

\displaystyle \therefore \sqrt{49.5} \approx y + \Delta y = 7 + 0.0357 = 7.0357.

\displaystyle \text{(xxvii) } (3.968)^{3/2}
\displaystyle \text{Answer:}

\displaystyle \text{Consider the function } y = f(x) = x^{\tfrac32}.

\displaystyle \text{Let } x = 4 \text{ and } x + \Delta x = 3.968.

\displaystyle \Rightarrow \Delta x = -0.032.

\displaystyle \text{For } x = 4, \quad y = 4^{\tfrac32} = 8.

\displaystyle \text{Let } dx = \Delta x = -0.032.

\displaystyle \text{Now, } y = x^{\tfrac32}.

\displaystyle \Rightarrow \frac{dy}{dx} = \frac{3}{2}\sqrt{x}.

\displaystyle \Rightarrow \left(\frac{dy}{dx}\right)_{x=4} = \frac{3}{2}\times 2 = 3.

\displaystyle \therefore \Delta y \approx dy = \frac{dy}{dx}\,dx = 3 \times (-0.032) = -0.096.

\displaystyle \Rightarrow \Delta y = -0.096.

\displaystyle \therefore (3.968)^{\tfrac32} \approx y + \Delta y = 8 - 0.096 = 7.904.

\displaystyle \text{(xxviii) } (1.999)^5
\displaystyle \text{Answer:}

\displaystyle \text{Consider the function } y = f(x) = x^5.

\displaystyle \text{Let } x = 2 \text{ and } x + \Delta x = 1.999.

\displaystyle \Rightarrow \Delta x = -0.001.

\displaystyle \text{For } x = 2, \quad y = 2^5 = 32.

\displaystyle \text{Let } dx = \Delta x = -0.001.

\displaystyle \text{Now, } y = x^5.

\displaystyle \Rightarrow \frac{dy}{dx} = 5x^4.

\displaystyle \Rightarrow \left(\frac{dy}{dx}\right)_{x=2} = 5(2)^4 = 80.

\displaystyle \therefore \Delta y \approx dy = \frac{dy}{dx}\,dx = 80 \times (-0.001) = -0.08.

\displaystyle \Rightarrow \Delta y = -0.08.

\displaystyle \therefore (1.999)^5 \approx y + \Delta y = 32 - 0.08 = 31.92.

\displaystyle \text{(xxix) } \sqrt{0.082}
\displaystyle \text{Answer:}

\displaystyle \text{Consider the function } y = f(x) = \sqrt{x}.

\displaystyle \text{Let } x = 0.0841 \text{ and } x + \Delta x = 0.082.

\displaystyle \Rightarrow \Delta x = 0.082 - 0.0841 = -0.0021.

\displaystyle \text{For } x = 0.0841, \quad y = \sqrt{0.0841} = 0.29.

\displaystyle \text{Let } dx = \Delta x = -0.0021.

\displaystyle \text{Now, } y = x^{\tfrac12}.

\displaystyle \Rightarrow \frac{dy}{dx} = \frac{1}{2\sqrt{x}}.

\displaystyle \Rightarrow \left(\frac{dy}{dx}\right)_{x=0.0841} = \frac{1}{2(0.29)} = \frac{1}{0.58} = \frac{50}{29}.

\displaystyle \therefore \Delta y \approx dy = \frac{dy}{dx}\,dx = \frac{50}{29} \times (-0.0021) = -0.00362.

\displaystyle \Rightarrow \Delta y \approx -0.00362.

\displaystyle \therefore \sqrt{0.082} \approx y + \Delta y = 0.29 - 0.00362 = 0.28638 \approx 0.2864.

\displaystyle \textbf{Question 10: }~\text{Find the approximate value of }f(2.01),\text{ where } \\ f(x)=4x^2+5x+2.
\displaystyle \text{Answer:}

\displaystyle \text{Let } x = 2 \text{ and } x + \Delta x = 2.01.

\displaystyle \Rightarrow \Delta x = 0.01.

\displaystyle \text{Given } f(x) = 4x^2 + 5x + 2.

\displaystyle \text{For } x = 2, \quad f(2) = 4(2)^2 + 5(2) + 2 = 16 + 10 + 2 = 28.

\displaystyle \text{Now, } y = f(x).

\displaystyle \Rightarrow \frac{dy}{dx} = 8x + 5.

\displaystyle \therefore dy \approx \Delta y = \frac{dy}{dx}\,dx = (8x + 5)(0.01).

\displaystyle \Rightarrow \Delta y = (16 + 5)\times 0.01 = 0.21.

\displaystyle \therefore f(2.01) \approx y + \Delta y = 28 + 0.21 = 28.21.

\displaystyle \textbf{Question 11: }~\text{Find the approximate value of }f(5.001),\text{ where } \\ f(x)=x^3-7x^2+15.
\displaystyle \text{Answer:}

\displaystyle \text{Let } x = 5 \text{ and } x + \Delta x = 5.001.

\displaystyle \Rightarrow \Delta x = 0.001.

\displaystyle \text{Given } f(x) = x^3 - 7x^2 + 15.

\displaystyle \text{For } x = 5, \quad y = f(5) = 5^3 - 7(5)^2 + 15 = 125 - 175 + 15 = -35.

\displaystyle \text{Now, } y = f(x).

\displaystyle \Rightarrow \frac{dy}{dx} = 3x^2 - 14x.

\displaystyle \therefore dy \approx \Delta y = \frac{dy}{dx}\,dx = (3x^2 - 14x)\times 0.001.

\displaystyle \Rightarrow \Delta y = (75 - 70)\times 0.001 = 0.005.

\displaystyle \therefore f(5.001) \approx y + \Delta y = -35 + 0.005 = -34.995.

\displaystyle \textbf{Question 12: }~\text{Find the approximate value of }\log_{10}1005,\text{ given that } \\ \log_{10}e=0.4343.
\displaystyle \text{Answer:}

\displaystyle \text{Let } y = f(x) = \log_{10} x.

\displaystyle \text{Here, let } x = 1000 \text{ and } x + \Delta x = 1005.

\displaystyle \Rightarrow \Delta x = 5.

\displaystyle \text{Let } dx = \Delta x = 5.

\displaystyle \text{For } x = 1000, \quad y = \log_{10} 1000 = \log_{10}(10^3) = 3.

\displaystyle \text{Now, } y = \log_{10} x = \frac{\log_e x}{\log_e 10}.

\displaystyle \Rightarrow \frac{dy}{dx} = \frac{1}{x \log_e 10} = \frac{0.4343}{x}.

\displaystyle \Rightarrow \left(\frac{dy}{dx}\right)_{x=1000} = \frac{0.4343}{1000} = 0.0004343.

\displaystyle \therefore \Delta y \approx dy = \frac{dy}{dx}\,dx = 0.0004343 \times 5 = 0.0021715.

\displaystyle \therefore \log_{10} 1005 \approx y + \Delta y = 3 + 0.0021715 = 3.0021715.

\displaystyle \textbf{Question 13: }~\text{If the radius of a sphere is measured as }\\ 9\text{ cm with an error of }0.03\text{ cm, find the approximate error in calculating its surface area.}
\displaystyle \text{Answer:}

\displaystyle \text{Let } x \text{ be the radius and } y \text{ be the surface area of the sphere}.

\displaystyle \text{Then, let } x = 9\text{ cm}.

\displaystyle \Delta x = 0.03\text{ m} = 3\text{ cm}.

\displaystyle \Rightarrow x + \Delta x = 9 + 3 = 12\text{ cm}.

\displaystyle \text{Now, } y = 4\pi x^2.

\displaystyle \text{For } x = 9, \quad y = 4\pi(9)^2 = 324\pi.

\displaystyle \Rightarrow \frac{dy}{dx} = 8\pi x.

\displaystyle \Rightarrow \left(\frac{dy}{dx}\right)_{x=9} = 72\pi.

\displaystyle \therefore \Delta y \approx dy = \frac{dy}{dx}\,dx = 72\pi \times 3 = 216\pi\ \text{cm}^2.

\displaystyle \therefore \text{The approximate error in the surface area is } 216\pi\ \text{cm}^2.

\displaystyle \textbf{Question 14: }~\text{Find the approximate change in the surface area of a cube of side }\\ x\text{ metres caused by decreasing the side by }1\%.
\displaystyle \text{Answer:}

\displaystyle \text{Let } x \text{ be the edge of the cube and } y \text{ be its surface area}.

\displaystyle \text{Then, } y = 6x^2.

\displaystyle \text{We are given } \frac{\Delta x}{x} \times 100 = 1.

\displaystyle \Rightarrow \Delta x = \frac{x}{100}.

\displaystyle \text{Now, } \frac{dy}{dx} = 12x.

\displaystyle \therefore \Delta y \approx dy = \frac{dy}{dx}\,dx = 12x \times \frac{x}{100} = 0.12x^2.

\displaystyle \therefore \text{The approximate change in the surface area of the cube is } 0.12x^2.

\displaystyle \textbf{Question 15: }~\text{If the radius of a sphere is measured as }\\ 7\text{ m with an error of }0.02\text{ m, find the approximate error in calculating its volume.}
\displaystyle \text{Answer:}

\displaystyle \text{Let } x \text{ be the radius of the sphere and } y \text{ be its volume}.

\displaystyle y = \frac{4}{3}\pi x^3.

\displaystyle \text{Let } \Delta x \text{ be the error in the radius}.

\displaystyle x = 7,\quad \Delta x = 0.02.

\displaystyle \frac{dy}{dx} = 4\pi x^2.

\displaystyle \left( \frac{dy}{dx} \right)_{x=7} = 196\pi.

\displaystyle \therefore \Delta y \approx dy = \frac{dy}{dx}\,dx = 196\pi \times 0.02 = 3.92\pi.

\displaystyle \therefore \text{The approximate error in calculating the volume of the sphere is } 3.92\pi \text{ cubic units}.

\displaystyle \textbf{Question 16: }~\text{Find the approximate change in the volume of a cube of side }\\ x\text{ metres caused by increasing the side by }1\%.
\displaystyle \text{Answer:}

\displaystyle \text{Volume of the cube } V = x^3.

\displaystyle \text{We have } \Delta x = 0.01x.

\displaystyle \frac{dV}{dx} = 3x^2.

\displaystyle \therefore \Delta V \approx dV = \frac{dV}{dx}\,dx = 3x^2 \times 0.01x = 0.03x^3.

\displaystyle \therefore \text{The approximate change in the volume of the cube is } 0.03x^3 \text{ cubic units}.

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