SELECT * FROM question_mgmt as q WHERE id=1180 AND status=1 SELECT id,question_no,question,chapter FROM question_mgmt as q WHERE courseId=3 AND subId=6 AND chapterId=93 and ex_no='1' AND status=1 ORDER BY CAST(question_no AS UNSIGNED)

Question:

A balloon, which always remains spherical, has a variable diameter

\begin{align} \frac{3}{2}(2x+1)\end{align}

Find the rate of change of its volume with respect to x.

Answer:

The volume of a sphere (V) with radius (r) is given by,

\begin{align} V=\frac{4}{3}\pi r^3 \end{align}

It is given that:

\begin{align} Diameter =\frac{3}{2}(2x+1) \end{align}

\begin{align} \Rightarrow r =\frac{3}{4}(2x+1) \end{align}

\begin{align} \therefore V =\frac{4}{3}\pi(\frac{3}{4})^3(2x+1)^3=\frac{9}{16}\pi\times(2x+1)^3 \end{align}

Hence, the rate of change of volume with respect to x is as

\begin{align} \frac{dV}{dx}=\frac{9}{16}\pi\frac{d}{dx}(2x+1)^3=\frac{9}{16}\pi\times3(2x+1)^2 \times2=\frac{27}{8}\pi(2x+1)^2\end{align}


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