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# A balloon, which always remains spherical, has a variable diameter $\large\frac{3}{2}$$(2x+1). Find the rate of change of its volume with respect to x. \begin{array}{1 1} (A)\;\frac{27}{8}\pi(2x+1)^2 \\ (B)\;\frac{-27}{8}\pi(2x+1)^2 \\ (C)\;\frac{-27}{4}\pi(2x+1)^2 \\ (D)\;\frac{27}{4}\pi(2x+1)^2 \end{array} Can you answer this question? ## 1 Answer 0 votes Toolbox: • If y=f(x),then \large\frac{dy}{dx} measures the rate of change of y w.r.t x. • \big(\large\frac{dy}{dx}\big)_{x=x_0} represents the rate of change of y w.r.t x at x=x_0 Step 1: Diameter of the sphere =\large\frac{3}{2}$$(2x+1)$
Therefore radius $r=\large\frac{3}{4}$$(2x+1) Volume of the sphere v=\large\frac{4}{3}$$\pi r^3$
Substituting for $r$ we get,
$v=\large\frac{4}{3}$$\pi\big[\large\frac{3}{4}$$(2x+1)\big]^3$
Step 2:
Differentiating w.r.t $x$ on both sides,
$\large\frac{dv}{dx}=\frac{4}{3}$$x\times \big(\large\frac{3}{4}\big)^3$$\times 3(2x+1)^2(2)$
$\quad\;\;=\large\frac{27}{8}$$\pi(2x+1)^2$