Class CBSE Class 12 Applications of Derivatives Question

Question

A spherical ball has a variable diameter $\frac{5}{2}(3x+1).$ The rate of change of its volume w.r.t. x, when $x=1$, is :

Accepted Answer

Correct Answer: . Explanation: The diameter of the spherical ball is given by $D = \frac{5}{2}(3x+1)$.
The radius of the ball is $r = \frac{D}{2} = \frac{1}{2} \cdot \frac{5}{2}(3x+1) = \frac{5}{4}(3x+1)$.

The volume of a sphere is $V = \frac{4}{3}\pi r^3$.
Substitute the expression for $r$:
$V = \frac{4}{3}\pi \left(\frac{5}{4}(3x+1)ight)^3$
$V = \frac{4}{3}\pi \left(\frac{125}{64}(3x+1)^3ight)$
$V = \frac{125\pi}{48}(3x+1)^3$

To find the rate of change of its volume with respect to $x$, we differentiate $V$ with respect to $x$:
$\frac{dV}{dx} = \frac{d}{dx}\left(\frac{125\pi}{48}(3x+1)^3ight)$
$\frac{dV}{dx} = \frac{125\pi}{48} \cdot 3(3x+1)^{3-1} \cdot \frac{d}{dx}(3x+1)$
$\frac{dV}{dx} = \frac{125\pi}{48} \cdot 3(3x+1)^2 \cdot 3$
$\frac{dV}{dx} = \frac{125\pi \cdot 9}{48}(3x+1)^2$
$\frac{dV}{dx} = \frac{125\pi \cdot 3}{16}(3x+1)^2$
$\frac{dV}{dx} = \frac{375\pi}{16}(3x+1)^2$

Now, we evaluate this rate of change when $x=1$:
$\left.\frac{dV}{dx}ight|_{x=1} = \frac{375\pi}{16}(3(1)+1)^2$
$\left.\frac{dV}{dx}ight|_{x=1} = \frac{375\pi}{16}(4)^2$
$\left.\frac{dV}{dx}ight|_{x=1} = \frac{375\pi}{16} \cdot 16$
$\left.\frac{dV}{dx}ight|_{x=1} = 375\pi$

The final answer is $\boxed{	ext{375\pi}}$.

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