\displaystyle \textbf{Question 1: }\ \ \text{If the radius of a sphere is measured as }9\ \text{cm with an error of }0.03 \text{ cm,} \\ \text{then find the approximating error in calculating its volume.}

\displaystyle \text{Answer:}
\displaystyle  \text{Let } r \text{ be the radius of the sphere and } \Delta r \text{ be the error in measuring the} \\ \text{radius. Then, } r=9\ \text{cm and } \Delta r=0.03\ \text{cm.}
\displaystyle \text{Let } V \text{ be the volume of the sphere. Then,}
\displaystyle V=\frac{4}{3}\pi r^3 \Rightarrow \frac{dV}{dr}=4\pi r^2 \Rightarrow \left(\frac{dV}{dr}\right)_{r=9}=4\pi \times 9^2=324\pi
\displaystyle \text{Let } \Delta V \text{ be the error in } V \text{ due to error } \Delta r \text{ in } r. Then,
\displaystyle \Delta V=\frac{dV}{dr}\Delta r \Rightarrow \Delta V=324\pi \times 0.03=9.72\pi\ \text{cm}^3.
\displaystyle \text{Thus, the approximate error in calculating the volume is }9.72\pi\ \text{cm}^3.

\displaystyle \textbf{Question 2: }\ \ \text{Find the approximate value of } f(3.02),\ \text{where } \\ f(x)=3x^2+5x+3.

\displaystyle \text{Answer:}
\displaystyle  \text{Let } y=f(x),\ x=3 \text{ and } x+\Delta x=3.02.\ \text{Then, } \Delta x=0.02.
\displaystyle \text{For } x=3,\ \text{we get}
\displaystyle y=f(3)=3\times 3^2+5\times 3+3=45
\displaystyle \text{Now,}
\displaystyle y=f(x)\Rightarrow y=3x^2+5x+3 \Rightarrow \frac{dy}{dx}=6x+5 \Rightarrow \left(\frac{dy}{dx}\right)_{x=3}=6\times 3+5=23
\displaystyle \text{Let } \Delta y \text{ be the change in } y \text{ due to change } \Delta x \text{ in } x.\ \text{Then,}
\displaystyle \Delta y=\frac{dy}{dx}\Delta x \Rightarrow \Delta y=23\times 0.02=0.46
\displaystyle \therefore\ f(3.02)=y+\Delta y=45+0.46=45.46.

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